tag:blogger.com,1999:blog-38321360178937494972017-08-20T20:31:42.669-04:00Elliptic ComposabilityFlorin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.comBlogger224125tag:blogger.com,1999:blog-3832136017893749497.post-19506398177029549802017-08-20T20:31:00.000-04:002017-08-20T20:31:42.704-04:00<h2 style="text-align: center;">Impressions from Yellowstone</h2><div><br /></div><div>I was on vacation for a week in Yellowstone and I will put the physics post on hold want to share what I saw. First, the park is simply amazing and I highly recommend to visit if you have the chance. You need at least 3 days as a bare minimum. The main road is like the number 8 and on the west (left) side you get to see lots of fuming hot spots ejecting steam and sulfur.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-cG4nVoFoS1c/WZoeZL3xCjI/AAAAAAAABPg/jIGbPHT1Qrg88j4s-vcrURlBY2HfbDqIQCLcBGAs/s1600/DSCN1412.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://1.bp.blogspot.com/-cG4nVoFoS1c/WZoeZL3xCjI/AAAAAAAABPg/jIGbPHT1Qrg88j4s-vcrURlBY2HfbDqIQCLcBGAs/s320/DSCN1412.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-jMPAbCknCfk/WZojgRn20vI/AAAAAAAABQQ/6MlzDeC5yAQNlkBoopa1hAYn3ML22zPEQCLcBGAs/s1600/DSCN1393.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://2.bp.blogspot.com/-jMPAbCknCfk/WZojgRn20vI/AAAAAAAABQQ/6MlzDeC5yAQNlkBoopa1hAYn3ML22zPEQCLcBGAs/s320/DSCN1393.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">The colors are due to bacteria and different bacteria live at different temperatures giving the hot spots rings of color.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">On the south side you get the geysers and Old Faithful which erupts every 90 minutes.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-ZQQwo8QkOU4/WZoe5Gqs_3I/AAAAAAAABPo/fKaxLBzy7B4HPyExgJsYtudquZ_RNzmfQCLcBGAs/s1600/DSCN1388.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="320" src="https://4.bp.blogspot.com/-ZQQwo8QkOU4/WZoe5Gqs_3I/AAAAAAAABPo/fKaxLBzy7B4HPyExgJsYtudquZ_RNzmfQCLcBGAs/s320/DSCN1388.JPG" width="240" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: left;">You need to be there approximately 1 hour before the eruption to get a sit on the benches which surround Old Faithful. There are other geysers but you don't know when they erupt.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">On the east side at the bottom of the 8 there is Yellowstone lake which gives rise to Yellowstone river and the Yellowstone canyon. Not much to do at the lake, the water is very cold. The river forms two large waterfalls and you can visit them on both sides.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-hf-X6rRLEIc/WZogJLcl2sI/AAAAAAAABP0/Strexi5ZwhsjyZFYxo2anDj9uhItfZ0-QCLcBGAs/s1600/DSCN1498.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="320" src="https://1.bp.blogspot.com/-hf-X6rRLEIc/WZogJLcl2sI/AAAAAAAABP0/Strexi5ZwhsjyZFYxo2anDj9uhItfZ0-QCLcBGAs/s320/DSCN1498.JPG" width="240" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-K7QxESPaWiE/WZogUmbdBTI/AAAAAAAABP4/u-J2XC8ymLgjKS6VjZX4UzSLfzgAnEtZwCLcBGAs/s1600/DSCN1504.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://3.bp.blogspot.com/-K7QxESPaWiE/WZogUmbdBTI/AAAAAAAABP4/u-J2XC8ymLgjKS6VjZX4UzSLfzgAnEtZwCLcBGAs/s320/DSCN1504.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Coming north on the east side, you encounter more waterfalls and a bit of bisons. If you are lucky you get to see in the distance bears usually eating a dead moose. By the way, there is a big business ripoff in terms of bear sprays. You can buy one for $50, but you should rent one for $10/day when you hike in the forest. Even better just buy a $1 bell to wear to let the wildlife you are there (bears avoid people if they can hear them coming).</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">You can hike Mt. Washburn (4 hour round trip hike) to get a panoramic view of the park 50 miles in any direction.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-BY6MjBuX3AQ/WZoj2eZyNzI/AAAAAAAABQY/hMRnVI0vECEnY-jWBoMSxDR0hUCi6ZWnACLcBGAs/s1600/DSCN1576.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://2.bp.blogspot.com/-BY6MjBuX3AQ/WZoj2eZyNzI/AAAAAAAABQY/hMRnVI0vECEnY-jWBoMSxDR0hUCi6ZWnACLcBGAs/s320/DSCN1576.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">There is nothing to see in the east-west part of the road at the middle of the 8, and on the the north of the east road there is another road leading east in Lamar's valley. Here is where you see a ton of wildlife: bisons, moose, wolves. Literally there are thousands of bisons in big herds which often cross the road.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-cKu7aOHjiZE/WZoi1ZGO46I/AAAAAAAABQI/hYfHPNz-c_QHmoB-FyylPSv0oRfgHAA6QCLcBGAs/s1600/DSCN1659.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://4.bp.blogspot.com/-cKu7aOHjiZE/WZoi1ZGO46I/AAAAAAAABQI/hYfHPNz-c_QHmoB-FyylPSv0oRfgHAA6QCLcBGAs/s320/DSCN1659.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-iMGboZXpMMM/WZojBmTVSEI/AAAAAAAABQM/xEJay1yRZSEtqW7cmisAvO06BERGPsS2wCLcBGAs/s1600/DSCN1649.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://2.bp.blogspot.com/-iMGboZXpMMM/WZojBmTVSEI/AAAAAAAABQM/xEJay1yRZSEtqW7cmisAvO06BERGPsS2wCLcBGAs/s320/DSCN1649.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-dr2OlNiwGGE/WZokL04OHQI/AAAAAAAABQc/dED9cjQfYOIjcqV3J9Xe2kQyTHmh65ecwCLcBGAs/s1600/DSCN1439.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1200" data-original-width="1600" height="240" src="https://1.bp.blogspot.com/-dr2OlNiwGGE/WZokL04OHQI/AAAAAAAABQc/dED9cjQfYOIjcqV3J9Xe2kQyTHmh65ecwCLcBGAs/s320/DSCN1439.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">Driving in the park is slow (25 mph) due to many attractions on the side and the traffic jams caused by animals. You need one day for north part, one day for the south loop, and one day for Lamar valley.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Yellowstone is at the spot of a supervolcano which erupted 7 times in the past: when it erupts it covers half of US with volcanic ash. There is a stationary hot spot of magma and because the tectonic plate moves different eruptions occur in different places. The past eruption locations trace a clear path on the map.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-vD96n-tw31s/WZonZq_EO5I/AAAAAAAABQo/cnV5k-VoGd0jOW339lu0knAmLAWdsQCiwCLcBGAs/s1600/HotspotsSRP_update2013.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="730" data-original-width="1063" height="219" src="https://3.bp.blogspot.com/-vD96n-tw31s/WZonZq_EO5I/AAAAAAAABQo/cnV5k-VoGd0jOW339lu0knAmLAWdsQCiwCLcBGAs/s320/HotspotsSRP_update2013.JPG" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">Yellowstone park is located in the caldera (the volcano crater) of the last eruption.</div><div class="separator" style="clear: both; text-align: left;"><br /></div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-27456631207218784912017-08-06T21:07:00.000-04:002017-08-06T21:07:58.925-04:00<h2 style="text-align: center;">The origins of gauge theory</h2><br />After a bit of absence I am back resuming my usual blog activity. However I am extremely busy and I will create new posts every two weeks from now on. I am starting now a series explaining gauge theory and today I will start at the beginning with Hermann Weyl's proposal.<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-2hpX48sDNZQ/WYSVe7rANVI/AAAAAAAABNg/lETh39eQAboVSAxsV9MuxB2WV487XWsZgCLcBGAs/s1600/Hermann_Weyl.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="479" data-original-width="462" height="320" src="https://1.bp.blogspot.com/-2hpX48sDNZQ/WYSVe7rANVI/AAAAAAAABNg/lETh39eQAboVSAxsV9MuxB2WV487XWsZgCLcBGAs/s320/Hermann_Weyl.jpg" width="308" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">In 1918 Hermann Weyl attempted to unify gravity with electromagnetism (the only two forces known at the time) and in the process he introduce the idea of gauge theory. He espouse his ideas in his book "Space Time Matter" and this is a book which I personally find hard to read. Usually the leading physics people have crystal clear original papers: von Neumann, Born, Schrodinger, but Weyl's book combines mathematical musings with metaphysical ideas in an unclear direction. The impression I got was of a mathematical, physical and philosophical random walk testing in all possible ways and directions and see where he could make progress. He got lucky and his lack of cohesion saved the day because he could not spot simple counter arguments against his proposal which could have stopped him cold in his tracks. But what was his motivation and what was his approach?</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Weyl like the local character of general relativity and proposed (from pure philosophical reasons) the idea that all physical measurements are relative. I particular, the norm of a vector should not be thought as an absolute value, but as a value that can change at various point of spacetime. To compare at different points, you need a "gauge", like a device used in train tracks to make sure the train tracks remained at a fixed distance from each other. Another word he used was "calibration", but the name "gauge" stuck.</div><br />So now suppose we have a norm \(N(x)\) of a vector and we do a shift to \(x + dx\). Then:<br /><br />\(N(x+dx) = N(x) + \partial_{\mu}N dx^{\mu}\)<br /><br />Also suppose that there is a scaling factor \(S(x)\):<br /><br />\(S(x+dx) = S(x) + \partial_{\mu}S dx^{\mu}\)<br /><br />and so to first order we get that N changes by:<br /><br />\(( \partial_{\mu} + \partial_{\mu} S) N dx^{\mu} \)<br />Since for a second gauge \(\Lambda\), \(S\) transforms like:<br /><br />\(\partial_{\mu} S \rightarrow \partial_{\mu} S +\partial_{\mu} \Lambda \)<br /><br />and since in electromagnetism the potential changes like:<br /><br />\(A_{\mu} \rightarrow A_{\mu} S +\partial_{\mu} \Lambda \)<br /><br /><div>Weyl conjectured that \(\partial_{\mu} S = A_{\mu}\).</div><div><br /></div><div>However this is disastrous because (as pointed by Einstein to Weyl on a postcard) it implies that the clocks would change their frequencies based on the paths they travel (and since you can make atomic clocks it implies that the atomic spectra is not stable).</div><div><br /></div><div>Later on with the advent of quantum mechanics Weyl changed his idea of scale change into that of a phase change for the wavefunction and the original objections became mute. Still more needed to be done for gauge theory to become useful.</div><div><br /></div><div>Next time I will talk about Bohm-Aharonov and the importance of potentials in physics as a segway into the proper math for gauge theory. </div><div><br /></div><div>Please stay tuned.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-8975140737566540782017-07-10T00:04:00.000-04:002017-07-10T00:04:32.066-04:00<h2 style="text-align: center;">The main problem of MWI is the concept of probability</h2><div style="text-align: center;"><br /></div><div style="text-align: left;">Now it is my turn to present the counter arguments against many worlds. All known derivations of Born rule in MWI have (documented) issues of circularity: in the derivation the Born rule is injected in some form or another. However the problem is deeper: <b>there is no good way to define probability in MWI</b>.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">Probability can be defined either in the <a href="https://en.wikipedia.org/wiki/Frequentist_probability" target="_blank">frequentist approach</a> as limit of frequency for large trial numbers, or subjectively as information update in the <a href="https://en.wikipedia.org/wiki/Bayesian_probability" target="_blank">Bayesian approach</a>. Both those approaches are making the same predictions. </div><div style="text-align: left;"><br /></div><div style="text-align: left;">It is generally assumed by all MWI supporters that branch counting leads to incorrect predictions and because of this the focused is changed on subjective probabilities and the "apparent emergence" of Born rule. However this implicitly breaks the frequentist-subjective probability relationship. The only way one can use the frequentist approach is by using branch counting. Let's have a simple example.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">Suppose you work at a factory which makes fair (quantum) coins which land 50% up and 50% down. Your job is quality assurance and you are tasked with finding the defective coins. Can you do your job in a MWI quantum universe? The only thing you can do is to flip the coin many times and see if it lands about 50% up and 50% down. For a fair coin there is no issue. However<b> for a biased coin (say 80%-20%) you get the very same outcomes as in the case of the fair coins and you cannot do your job</b>.</div><div style="text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-v3a8EYXG0W8/WWL2JRYuN6I/AAAAAAAABMs/GBhaM1Hm4Acud3qnEzz3TglGGtE4kdfOACLcBGAs/s1600/dice.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="490" data-original-width="700" height="140" src="https://3.bp.blogspot.com/-v3a8EYXG0W8/WWL2JRYuN6I/AAAAAAAABMs/GBhaM1Hm4Acud3qnEzz3TglGGtE4kdfOACLcBGAs/s200/dice.jpg" width="200" /></a></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div><div style="text-align: left;">There is only one way to fix the problem: consider that the world does not split in 2 up and down branches, but say in 1 million up and 1 million down branches. In this case you can think that in the unfair case the world splits in 1.6 million up worlds, and 400 thousand down worlds. <b>This would fix the concept of probability in MWI restoring the link between frequentist and subjective probabilities, but <u>this is not what MWI supporters claim</u>. </b>Plus, this has problems of its own with irrational numbers and the solution is only approximate to some limit of precision which can be refuted by any experiment run long enough.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">So to boil the problem down, in MWI there is no outcome difference in case of a fair coin versus an unfair coin toss: in both cases you get an "up world" and a "down world". Repeating the coin toss any number of times does not change the nature of the problem in any way. Physics is an experimental science and we test the validity of the theories against experiments. <b>Discarding branch counting in MWI is simply unscientific</b>. </div><div style="text-align: left;"><br /></div><div style="text-align: left;">Now in the last post Per argued for MWI. I asked him to show what would happen if we flip a fair and an unfair coin three times to simply run through his argument on an elementary example and not hid behind general equations. After some back and forth, Per computed the distribution \(\rho\) in the fair and unfair case (to match quantum mechanics predictions) but <b>the point is that</b> \(\rho\) <b>must arise out of the relative frequencies and not be computed by hand</b>. <b>Because the relative frequencies are identical in the two cases</b> \(\rho\) must be injected by a different mechanism. His computation of \(\rho\) is the point where circularity is introduced in the explanation. If you look back in his post, this comes from his equation 5 which is derived from equation 3. <b>Equation 3 assumes Born rule and is the root cause of circularity in his argument. <u>Per's equation 7 recovers the Born rule in the limit case after assuming Born rule in equation 3 -</u></b><u> <b>q.e.d.</b></u></div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com30tag:blogger.com,1999:blog-3832136017893749497.post-81022975360025661612017-06-25T15:58:00.000-04:002017-07-02T22:31:26.672-04:00<h2 style="text-align: center;">Guest Post defending MWI</h2><div><i><br /></i></div><div><i>As promised, here is a guest post from Per Arve. I am not interjecting my opinion in the main text but I will ask questions in the comments section.</i><br /><i><br /></i><i>Due to the popularity of this post I am delaying the next post for a week.</i></div><div><br /></div>The reason to abandon the orthodox interpretation of quantum mechanics is its incompleteness. Bohr and Heisenberg refused the possibility to describe the measurement process as a physical process. This is encoded in Bohr's claim that the quantum world cannot be understood. Such an attitude served to avoid endless discussions about the weirdness of quantum mechanics and divert attention to the description of microscopic physics with quantum mechanics. Well done! A limited theory is better than no theory. <br /><br />But, we should always try to find theories that in a unified way describes the larger set of processes. The work by Everett and the later development of decoherence theory by Zeh, Zurek and others have given us elements to describe also the measurement process as a quantum mechanical process. Their analysis of the measurement process implies that the unitary quantum evolution leads to the emergence of separate new "worlds". The appearance of separate "worlds" can only be avoided if there is some mechanism that breaks unitarity.<br /><br />The most well-known problem of Everett's interpretation is that of the derivation of the Born rule. I describe the solution of that problem here. (You can also check my article on the arxiv <a href="http://arxiv.org/abs/1603.01625" target="_blank">[1603.01625] Postulates for and measurements in Everett's quantum mechanics</a>)<br /><br />The main point is to prove that physicists experience the Born rule. That is by taking an outside view of the parallel worlds created in a measurement situation. The question, what probability is from the perspective of an observer inside a particular branch, is more a matter of philosophy than of science.<br /><br />The natural way to find out where something is located is to test with some force and find out where we find resistance. The force should not be so strong that it modifies the system we want to probe. This corresponds to the first order perturbation of the energy due to the external potential U(x),<br /><div><br /></div><div>\(\Delta E =\int d^3 x {|\psi (x)|}^2 U(x)\) (1)<br /><br />This shows that \({|\psi(x)|}^2\) gives where the system is located. (Here, spin and similar indexes are omitted.)<br /><br />The argumentation for the Born rule relies on that one may ignore the presence of the system in regions, where integrated value of the wave function absolute square is very small. <br /><br />In order to have a well defined starting point I have formulated two postulates for Everett's quantum mechanics.<br /><br /><b>EQM1 </b>The state is a complex function of positions and a discrete index j for spin etc,<br /><br />\(\Psi = \psi_j (t, x_1, x_2, ...) \) (2)<br /><br />Its basic interpretation is given by that the density </div><div><br /></div><div>\(\rho_j (t, x_1, x_2,...) = {|\psi_j (t, x_1, x_2, ...)|}^2 \) (3) <br /><br />answers where the system is in position, spin, etc.</div><div><br />It is absolute square integrable normalized to one </div><div><br />\( \int \int···dx_1dx_2 ··· \sum_j {|\psi_j (t, x_1, x_2, ...)|}^2 = 1\) (4)</div><div><br />This requirement signifies that the system has to be somewhere, not everywhere. If the value of the integral is zero, the system doesn’t exist anywhere. <br /><br /><b>EQM2 </b>There is a unitary time development of the state, e.g.,</div><div><br /></div><div>\(i \partial_t \Psi = H\Psi \), <br /><br />where H is the hermitian Hamiltonian. The term unitary signifies that the value of the left hand side in (4) is constant for any state (2). <br /><br />Consider the typical measurement where something happens in a reaction and what comes out is collected in an array of detectors, for instance the Stern-Gerlach experiment. Each detector will catch particles that have a certain value of the quantity B we want measure.<br /><br />Write the state that enter the array of detectors as sum of components that enter the individual detectors, \(|\psi \rangle = \sum c_b |b\rangle\), where b is one of the possible values of B. When that state has entered the detectors we can ask, where is it? The answer is that it is distributed over the individual detectors. The distribution is </div><div><br /></div><div>\(\rho_b = {|c_b|}^2 \) (5)</div><div><br />This derived by integrate the density (3) over the detector using that the states \(|b\rangle\) have support only inside its own detector. </div><div><br />The interaction between \(|\psi \rangle\) and the detector array will cause decoherence. The total system of detector array and \(|\psi \rangle\) splits into separate "worlds" such that the different values b of the quantity B will belong to separate "worlds".<br /><br />After repeating the measurement N times, the distribution that answer how many times have the value \(b=u\) been measured is</div><div><br />\(\rho(m:N | u)= b(N,m) {(\rho_u)}^m{(\rho_{¬u})}^{N−m} \) (6)</div><div><br />where \(b(N,m)\) is the binomial coefficient \(N\) over \(m\) and \(\rho_{¬u}\) is the sum over all \(ρ_b\) except \(b=u\).</div><div><br />The relative frequency \(z=m/N\) is then given by<br /><br />\(\rho(z|u) \approx \sqrt{(N/(2\pi \rho_u \rho_{¬u}))} exp( −N{(z−\rho_u)}^2/(2\rho_u \rho_{¬u}) ) \) (7) <br /><br />This approaches a Dirac delta \(\delta(z − \rho_u)\). If the tails of (7) with low integrated value are ignored, we are left with a distribution with \(z \approx u\). This shows that the observer experiences a relative frequency close to the Born value. Reasonably, the observer will therefore believe in the Born rule.<br /><br />The palpability of the densities (6) and (7) may be seen by replacing the detectors by a mechanism that captures and holds the system at the different locations. Then, we can measure to what extent the system is at the different locations (4) using an external perturbation (1). In principle, also the distribution from N measurements is directly measurable if we consider N parallel experiments. The relative frequency distribution (7) is then also in principle a directly measurable quantity.<br /><br />A physicist that believes in the Born rule will use that for statistical inference in quantum experiments. According to the analysis above, it will work just as well as we expect it to do using the Born rule in a single world theory. <br /><br />A physicist who believes in a single world will view the Born rule as a law about probabilities. A many-worlder may view it as a rule that can be used for inference about quantum states as if the Born rule is about probabilities.<br /><br />With my postulates, Everett's quantum mechanics describe the world as we see it. That is what should be discussed. Not whether it pleases anybody or not.<br /><br />If the reader is interested what to do in a quantum russian roulette situation, I have not much to offer. How to decide your future seems to be a philosophical and psychological question. As a physicist, I don't feel obliged to help you with that. <br /><br />Per Arve, Stockholm June 24, 2017 </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com34tag:blogger.com,1999:blog-3832136017893749497.post-638933484799339292017-06-18T18:22:00.000-04:002017-06-18T18:22:16.086-04:00<h2 style="text-align: center;">Impressions from FQMT 2017</h2><br /><img alt="Poster, FQMT" src="https://lnu.se/ImageVault/publishedmedia/sct81szxsxvuiy402jjl/FQMT_affisch_170612.jpg" /><br /><br />I just came back from Vaxjo where I had a marvelous time. It does sounds cliche, but this year was the best conference organized by Professor Khrennikov and I got many pleasant and unexpected surprises.<br /><br />The conference did had one drawback: everyday after the official talks we continue the discussions about quantum mechanics well past midnight at "The Bishops Arms" where we drank too many beers causing me to gained a few pounds :)<br /><br />At the conference I had a chance to meet and talk with Phillipe Grangier (he worked with Aspect on the famous Bell experiment) and I witness him giving the best cogent comments on all talks: from experimental to theoretical. He even surprised me when he asked at the end of my presentation why I am using time to derive Leibniz identity, where any other symmetry will do? Indeed this is true, but the drawback is that any other symmetry lacks generality later on during composition arguments. Suppose we compose two physical systems: one with with a continuous symmetry and another without, then the composed system will lack that symmetry. The advantage of using time is that it works for all cases where energy is conserved.<br /><br />Grangier presented his approach on quantum mechanics reconstruction using contextuality and continuity (like in Hardy's 5 reasonable axioms paper). The problem of continuity is that it lacks physical intuition/motivation. Why not impose right away the C* condition: \(||a^* a|| = {||a||}^2\) and recover everything from it?<br /><br />Bob Coecke and Aleks Kissinger book on the pictorial formalism: "Picturing Quantum Processes" was finally ready and was advertised at the conference. If you go to <a href="http://www.cambridge.org/pqp" target="_blank">www.cambridge.org/pqp</a> you can get with a 20% discount when you enter the code COECKE2017 at the checkout.<br /><br />Coecke 's talk was about causal theories and his main idea was: "time reversal of any causal theory = eternal noise". This looks deep, but it is really a trivial observation: you can't get anything meaningful and you can't control signals which have an information starting point because the starting point corresponds to the notion of false and anything is derivable from false.<br /><br />Robert Raussendorf from University of Vancouver had a nice talk about measurement based quantum computations where measurements are used to control the computation and he identified a cohomological framework.<br /><br />One surprise talk for me was the one given by Marcus Appleby from University of Sydney who presented a framework of equivalence for Quantum Mechanics between finite and infinite dimensional cases. This is of particular importance to me as I recovered quantum mechanics in the finite dimensional case only and I am searching for an approach to handle the infinite dimensional case.<br /><br />I made new friends there and I got very good advice and ideas - a big thank you. I also got to give many in person presentations of my quantum reconstruction program.<br /><br />There was one person claiming he solved the puzzles of the many worlds interpretation. I sat next to him at the conference dinner and I invited him to have a guest post at this blog to present his solution. As a disclaimer, I think MWI lacks the proper notion of probability and I am yet to see a solution but I am open to listen to new arguments. What I would like to see is an explanation of how to reconcile the world split of 50-50% when the quantum probabilities are 80-20%? I did not see this explained in his presentation to my satisfaction, but maybe I was not understating the argument properly.Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-65337870302605349522017-06-10T18:25:00.000-04:002017-06-10T18:25:27.893-04:00<h2 style="text-align: center;">Jordan-Banach, Jordan-Lie-Banach, C* algebras, and quantum mechanics reconstruction</h2><div><br /></div><div>This a short post written as waiting for my flight at Dulles Airport on my way to Vaxjo Sweden for a physics conference. </div><div><br /></div><div>First some definitions. a Jordan-Banach algebra is a Jordan algebra with the usual norm properties of a Banach algebra. A Jordan-Lie-Banach algebra is a Jordan-Banach algebra which is a Lie algebra at the same time. A Jordan-Lie algebra is the composability two-product algebra which we obtained using category theory arguments.</div><div><br /></div><div>Last time I hinted about this week's topic which is the final step in reconstructing quantum using category theory arguments. What we obtain from category theory is a Jordan-Lie algebra which in the finite dimensional case has the spectral properties for free because the spectrum in uniquely defined in an algebraic fashion (things gets very tricky in the infinite dimensional case). So in the finite dimensional case JL=JLB.</div><div><br /></div><div>But how can we go from Jordan-Banach algebra to C*? In general it cannot be done. C* algebras correspond to quantum mechanics and on the Jordan side we have the octonionic algebra which is exceptional. Thus cannot be related to quantum mechanics because octonions are not associative. However we can define state spaces for both Jordan-Banach and C* algebras and we can investigate their geometry. The geometry is definable in terms if projector elements which obey: \(a*a = a\). In turn this defines the pure states as the boundary of the state spaces. If the two geometries are identical, we are in luck. </div><div><br /></div><div>Now the key question is: <b>under what circumstances can we complexify a Jordan-Banach algebra to get a C* algebra?</b></div><div><br /></div><div>In nature, observables play a dual role as both observables and generators. In literature this is called <b>dynamic correspondence</b>. Dynamic correspondence is the essential ingredient: any <b>JB algebra with dynamic correspondence is the self-adjoint part of a C* algebra. This result holds in general and can be established by comparing the geometry of the state spaces for JB and C* algebras.</b></div><div><b><br /></b></div><div>Now for the punch line: a JL algebra comes with dynamic correspondence and I showed that in prior posts. The conclusion is therefore:</div><div><b><br /></b></div><div><b>in the finite dimensional case: JL is a JLB algebra which gives rise to a C* algebra by complexification and by GNS construction we obtain the standard formulation of quantum mechanics. </b></div><div><b><br /></b></div><div><b><u>Quantum mechanics is fully reconstructed in the finite dimensional case from physical principles using category theory arguments! </u></b></div><div><b><u><br /></u></b></div><div>By the way this is what I'll present at the conference (the entire series on QM reconstruction).</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-59230500583289665442017-06-04T22:06:00.000-04:002017-06-04T22:06:58.044-04:00<h2 style="text-align: center;">From composability two-product algebra to quantum mechanics</h2><div>Last time we introduced the composability two-product algebra consisting of the Lie algebra \(\alpha\) and the Jordan algebra \(\sigma\) along with their compatibility relationship. This structure was obtained by categorical arguments using two natural principles of nature:</div><div><br /></div><div>- laws of nature are invariant under time evolution</div><div>- laws of nature are invariant under system composition</div><div><br /></div><div>What we did not obtain were spectral properties. <b>However, in the finite dimensional case, we do not need spectral properties and we can fully recover quantum mechanics <u>in this particular case</u>. </b>The trick is to classify all possible two-product algebras because there are only a handful of them. This is achieved with the help of the <a href="https://en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem" target="_blank">Artin-Weddenburn theorem</a>. </div><div><br /></div><div>First some preliminary. We need to introduce a Lie-Jordan-Banach (JLB) algebra by augmenting the composability two-product algebra with spectral properties:</div><div>-a JLB-algebra is a composability two-product algebra with the following two additional properties:</div><div><ul><li>\(||x\sigma x|| = {||x||}^{2}\)</li><li>\(||x\sigma x||\leq ||x\sigma x + y\sigma y||\)</li></ul><div>Then we can define a C* algebra by compexification of a JLB algebra where the C* norm is:</div></div><div><br /></div><div>\(||a+ib|| = \sqrt{{||a||}^{2}+{||b||}^{2}}\)</div><div><br /></div><div>Conversely from a C* algebra we define a JLB algebra as the self-adjoint part and where the Jordan part is:</div><div><br /></div><div>\(a\sigma b = \frac{1}{2}(ab+ba)\)</div><div><br /></div><div>and the Lie part is:</div><div><br /></div><div>\(a\alpha b = \frac{i}{\hbar}(ab-ba)\)</div><div><br /></div><div>From C* algebra we recover the usual quantum mechanics formulation by <a href="https://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction" target="_blank">GNS construction</a> which gets for us:</div><div><br /></div><div>- a Hilbert space H</div><div>- a distinguished vector \(\Omega\) on H arising out of the identity of the C* algebra</div><div>- a representation \(\pi\) of the algebra as linear operators on H</div><div>- a state \(\omega\) on C* represented as \(\omega (A) = \langle \Omega, \pi (A)\Omega\rangle_{H}\)</div><div><br /></div><div>Conversely, from quantum mechanics a C* algebra arises as bounded operators on the Hilbert space.</div><div><br /></div><div>The infinite dimensional case is a much harder <b>open</b> problem. Jumping from the Jordan-Banach operator algebra side to the C* and von Neuman algebras is very tricky and this involves characterizing the state spaces of operator algebras. Fortunately all this is already settled by the works of Alfsen, Shultz, Stormer, Topping, Hanche-Olsen, Kadison, Connes. </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-41965847399232794912017-05-21T23:32:00.000-04:002017-05-29T21:38:58.652-04:00<h2 style="text-align: center;">The algebraic structure of quantum and classical mechanics</h2><div style="text-align: center;"><br /></div><div style="text-align: left;">Let's recap on what we derived so far. We started by considering <a href="http://fmoldove.blogspot.com/2017/03/time-as-continous-functor-to-recall.html" target="_blank">time as a continous functor</a> and we derived Leibniz identity from it. Then for a particular kind of time evolution which allows a representation as a product we were able to derive two products \(\alpha\) and \(\sigma\) for which we derived the <a href="http://fmoldove.blogspot.com/2017/04/the-fundamental-bipartite-relations.html" target="_blank">fundamental bipartite relations</a>.<br /><br />Repeated applications of Leibniz identity resulted in proving \(\alpha\) as a Lie algebra, and \(\sigma\) as a Jordan algebra and an associator identity between them:<br /><br />\([A,B,C]_{\sigma} + \frac{J^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0\)<br /><br />where \(J\) is a map between generators and observables encoding Noether's theorem.<br /><br />Now we can combine the Jordan and Lie algebra as:<br /><br />\(\star = \sigma\pm \frac{J \hbar}{2}\alpha\)<br /><br />and it is not hard to show that this product is associative (pick \(\hbar = 2\) for convenience):<br /><br />\([f,g,h]_{\star} = (f\sigma g \pm J f\alpha g)\star h - f\star(g\sigma h \pm J g\alpha h)=\)<br />\((f\sigma g)\sigma h \pm J(f\sigma g)\alpha h \pm J(f\alpha g)\sigma h + J^2 (f\alpha g)\alpha h \)<br />\(−f\sigma (g\sigma h) \mp J f\sigma (g\alpha h) \mp J f\alpha (g\sigma h) − J^2 f\alpha (g\alpha h) =\)<br />\([f, g, h]_{\sigma} + J^2 [f, g, h]_{\alpha} ±J\{(f\sigma g)\alpha h + (f\alpha g)\sigma h − f\sigma (g\alpha h) − f\alpha (g\sigma h)\} = 0\)<br /><br />because the first part is zero by associator identity and the second part is zero by applying Leibniz identity. In Hilbert space representation the star product is nothing but the complex number multiplication in ordinary quantum mechanics<br /><br />Now we can introduce the algebraic structure of quantum (and classical) mechanics:<br /><br /><b>A composability two-product algebra is a real vector space equipped with two bilinear maps</b> \(\sigma \) <b>and </b>\(\alpha \) <b>such that the following conditions apply:</b><br /><br />- \(\alpha \) is a Lie algebra,<br />- \(\sigma\) is a Jordan algebra,<br />- \(\alpha\) is a derivation for \(\sigma\) and \(\alpha\),<br />- \([A, B, C]_{\sigma} + \frac{J^2 \hbar^2}{4} [A, B, C]_{\alpha} = 0\),<br />where \(J \rightarrow (−J)\) is an involution mapping generators and observables, \(1\alpha A = A\alpha 1 = 0\), \(1\sigma A = A\sigma 1 = A\)<br /><br /><b>For quantum mechanics </b>\(J^2 = -1\). <b>In the finite dimensional case the composability two-product algebra is enough to fully recover the full formalism of quantum mechanics</b> by using the <a href="https://en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem" target="_blank">Artin-Wedderburn theorem</a>.<br /><br /><b>The same structure applies to classical mechanics with only one change:</b> \(J^2 = 0\).<br /><br />In classical mechanics case, in phase space, the usual Poisson bracket representation for product \(\alpha\) can be constructively derived from above:<br />\(f\alpha g = \{f,g\} = f \overset{\leftrightarrow}{\nabla} g = \sum_{i=1}^{n} \frac{\partial f}{\partial q^i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q^i}\)<br /><br />and the product \(\sigma\) is then the regular function multiplication.<br /><br />In quantum mechanics case in the Hilbert space representation we have the commutator and the Jordan product:<br /><br />\(A\alpha B = \frac{i}{\hbar} (AB − BA)\)<br />\(A\sigma B = \frac{1}{2} (AB + BA)\)<br /><br />or in the phase space representation the Moyal and cosine brackets:<br /><br />\(\alpha = \frac{2}{\hbar}\sin (\frac{\hbar}{2} \overset{\leftrightarrow}{\nabla})\)<br />\(\sigma = \cos (\frac{\hbar}{2} \overset{\leftrightarrow}{\nabla})\)<br /><br />where the associative product is the <a href="https://en.wikipedia.org/wiki/Moyal_product" target="_blank">star product</a>.<br /><br /><b><u>Update:</u></b> Memorial Day holiday interfered with this week's post. I was hoping to make it back home on time to write it today, but I got stuck on horrible traffic for many hours. I'll postpone the next post for a week.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-75836958396042359732017-05-15T00:46:00.000-04:002017-05-15T07:27:59.121-04:00<h2 style="text-align: center;">The Jordan algebra of observables</h2><div><br /></div><div>Last time, from concrete representations of the products \(\alpha\) and \(\sigma\) we derived this identity:</div><div><br /></div><div>\([A,B,C]_{\sigma} + \frac{i^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0\)</div><div><br /></div><div>Let's use this in a particular case when \(C = A\sigma A\). What does the left hand side say?</div><div><br /></div><div>\([A,B,C]_{\sigma} = (A\sigma B) \sigma (A\sigma A)) - A\sigma (B \sigma (A \sigma A))\) </div><div><br /></div><div>which if we drop \(\sigma\) for convenience sake reads:</div><div><br /></div><div>\((AB)(AA) - A(B(AA))\)</div><div><br /></div><div>If the right hand side is zero then we get the <a href="https://en.wikipedia.org/wiki/Jordan_algebra" target="_blank">Jordan identity</a>:</div><div><br /></div><div>\((xy)(xx) = x(y(xx))\) where \(xy = yx\)</div><div><br /></div><div>Now let's compute the right hand side and show it is indeed zero:</div><div><br /></div><div>\([A,B,A\sigma A]_{\alpha} = (A\alpha B) \alpha (A\sigma A)) - A\alpha (B \alpha (A \sigma A))\)</div><div><br /></div><div>Using Leibniz identity in the second term we get:</div><div><br /></div><div>\((A\alpha B) \alpha (A\sigma A)) - (A\alpha B) \alpha (A\sigma A) - B \alpha (A\alpha (A\sigma A))) = - B \alpha (A\alpha (A\sigma A))\)</div><div><br /></div><div>But \(A\alpha (A\sigma A) = 0 \) because</div><div><br /></div><div>\(A\alpha (A\sigma A) = (A\alpha A) \sigma A + A\sigma (A\alpha A) \)</div><div><br /></div><div>and \(A\alpha A = -A\alpha A = 0\) by skew symmetry.</div><div><br /></div><div>Therefore due to the associator identity, the product \(\sigma\) is a Jordan algebra. Now we need to arrive at the associator identity using only the ingredients derived so far. This is tedious but it can be done using only Jacobi and Leibniz identity. Grgin and Petersen derived it in 1976 and you can see the proof <a href="https://projecteuclid.org/download/pdf_1/euclid.cmp/1103900192" target="_blank">here</a>. </div><div><br /></div><div>The associator identity is better written as:</div><div><br /></div><div>\([A,B,C]_{\sigma} + \frac{J^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0\)</div><div><br /></div><div>where \(J\) is a map from the the product \(\alpha\) to the product \(\sigma\). <b>The existence of this map is equivalent with Noether's theorem. </b>It just happens that in quantum mechanics case \(J^2 = -1\) and the imaginary unit maps anti-Hermitean generators to Hermitean observables. </div><div><br /></div><div>In classical physics case, \(J^2 = 0\) and this means that the product \(\sigma\) is associative (in fact it is the ordinary function multiplication) and the product \(\alpha\) can be proven to be the Poisson bracket, but that is a topic for another day as we will continue to derive the mathematical structure of quantum mechanics. Please stay tuned. </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-38246176897686707642017-05-07T19:31:00.000-04:002017-05-08T00:44:08.433-04:00<h2 style="text-align: center;">Lie, Jordan algebras and the associator identity</h2><div><br /></div><div>Before I continue the quantum mechanics algebraic series, I want to first state my happiness for the defeat of the far (alt)-right candidate in France despite Putin's financial and hacking support. Europe has much better antibodies against the scums like Trump than US. In US the diseases caused by the inoculation of hate perpetuated over many years by Fox News has to run its course before things will get better.</div><div><br /></div><div>Back to physics, first I will show that the product \(\alpha\) is indeed a <a href="https://en.wikipedia.org/wiki/Lie_algebra" target="_blank">Lie algebra</a>. This is utterly trivial because we need to show antisymmetry and the Jacobi identity:</div><div><br /></div><div>\(a\alpha b = -b\alpha a\)</div><div>\(a\alpha (b\alpha c) + c\alpha (a\alpha b) + b\alpha (c\alpha a) = 0\)</div><div><br /></div><div>We already know that the product \(\alpha\) is antisymmetric and we know that the it obeys Leibniz identity:</div><div><br /></div><div>\(a\alpha (b\circ c) = (a\alpha b) \circ c + b\circ (a\alpha c) \)</div><div><br /></div><div>where \(\circ\) can stand for either \(\alpha\) or \(\sigma\). When \(\circ = \alpha\) we get:</div><div><br /></div><div>\(a\alpha (b\alpha c) = (a\alpha b) \alpha c + b\alpha (a\alpha c) \)</div><div><br /></div><div>which by antisymmetry becomes</div><div><br /></div><div><div>\(a\alpha (b\alpha c) = - c \alpha (a\alpha b) - b\alpha (c\alpha a) \)</div></div><div><br /></div><div>In other words, the Jacobi identity.</div><div><br /></div><div>Therefore the product \(\alpha\) is in fact a Lie algebra. Now we want to prove that the product \(\sigma\) is a <a href="https://en.wikipedia.org/wiki/Jordan_algebra" target="_blank">Jordan algebra</a>.</div><div><br /></div><div>This is not as simple as proving the Lie algebra, and we will do it with the help of a new concept: the <a href="https://en.wikipedia.org/wiki/Associator" target="_blank">associator</a>. Let us first define it. The associator of an arbitrary product \(\circ\) is defined as follows:</div><div><br /></div><div>\([a,b,c]_{\circ} = (a\circ b)\circ c - a\circ (b\circ c)\)</div><div><br /></div><div>as such it measures the lack of associativity. </div><div><br /></div><div>It is helpful now to look at the concrete realizations of the products \(\alpha\) and \(\sigma\) in quantum mechanics to know where we want to arrive. In quantum mechanics the product alpha is the commutator, and the product sigma is the anticommutator:</div><div><br /></div><div>\(A \alpha B = \frac{i}{\hbar}[A,B] = \frac{i}{\hbar}(AB - BA)\)</div><div>\(A\sigma B = \frac{1}{2}\{A, B\} = \frac{1}{2}(AB+BA)\)</div><div><br /></div><div>Let's compute alpha and sigma associators:</div><div><br /></div><div>\([A,B,C]_{\alpha} = \frac{-1}{\hbar^2}([AB-BA, C] - [A, BC-CB]) = \)</div><div>\(=\frac{-1}{\hbar^2}(ABC-BAC-CAB+CBA - ABC+ACB+BCA-CBA)\)</div><div>\(= \frac{-1}{\hbar^2}(-BAC-CAB +ACB+BCA)\)</div><div><br /></div><div><br /></div><div>\([A,B,C]_{\sigma} = \frac{1}{4}(\{AB+BA, C\} - \{A, BC+CB\}) = \)</div><div>\(=\frac{1}{4}(ABC+BAC+CAB+CBA - ABC-ACB-BCA-CBA) = \)</div><div>\(=\frac{1}{4}(BAC+CAB -ACB-BCA) \)</div><div><br /></div><div>and so we have the remarkable relationship:</div><div><br /></div><div>\([A,B,C]_{\sigma} + \frac{i^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0\)</div><div><br /></div><div><b>What is remarkable about this is that the Jordan and Lie algebras lack associativity in precisely the same way and because of this they can be later combined into a single operation. The identity above also holds the key for proving the Jordan identity.</b></div><div><b><br /></b></div><div>Next time I'll show how to derive the identity above using only the ingredients we proved so far and then I'll show how Jordan identity arises out of it. Please stay tuned.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-24171504867372669852017-04-30T00:55:00.000-04:002017-04-30T00:55:56.784-04:00<h2 style="text-align: center;">The origin of the symmetries of the quantum products</h2><div><br /></div><div>Quantum mechanics has three quantum products: </div><div><ul><li>the Jordan product of observables</li><li>the commutator product used for time evolution</li><li>the complex number multiplication of operators </li></ul><div>The last product is a composite construction of the first two and it is enough to study the Jordan product and the commutator. In the prior posts notation, the Jordan product is called \(\sigma\), and the commutator is called \(\alpha\). We will derive their full properties using category theory arguments and the Leibniz identity. Bur before doing this, I want to review a bit the two products. The commutator is well known and I will not spend time on it. Instead I will give the motovation for the Jordan product. </div></div><div><br /></div><div>In quantum mechanics the observables are represented as self-adjoint operators: \(O = O^{\dagger}\) If we want to create another self-adjoint operator out of two self-adjoint operators A and B, the simple multiplication won't work because \((AB)^{\dagger} = B^{\dagger} A^{\dagger} = BA \ne AB\). The solution is to have a symmetrized product: \(A\sigma B = (AB+BA)/2\). A lot of the quantum mechanics formalism transfers to the Jordan algebra of observables, but this is a relatively forgotten approach because it is rather cumbersome (the Jordan product is not associative but power associative) and (as it is expected) it does not produce any different predictions than the standard formalism based on complex numbers.<br /><br />Now back to obtaining the symmetry properties of the Jordan product \(\sigma\) and commutator \(\alpha\), at first we cannot say anything about the symmetry of the product \(\sigma\). However we do know that the product \(\alpha\) obeys the Leibniz identity. We have already use it to derive the fundamental composition relationships, so what else can we do? <b>We can apply it to a bipartite system:</b><br /><br />\(f_{12}\alpha_{12}(g_{12}\alpha_{12}h_{12}) = g_{12}\alpha_{12}(f_{12}\alpha_{12}h_{12}) + (f_{12}\alpha_{12}g_{12})\alpha_{12} h_{12}\)<br /><br />where<br /><br />\(\alpha_{12} = \alpha\otimes \sigma + \sigma\otimes\alpha\)<br /><br />Now <b>the key observation is that in the right hand side, </b>\(f\) <b>and</b> \(g\) <b>appear in reverse order. </b>Remember that the functions involved in the relationship above are free of constraints, by judicious picks of their value can lead to great simplifications because \(1 \alpha f = f\alpha 1 = 0\). The computation is tedious and I will skip it, but what you get in the end is this:<br /><br />\(f_1\alpha h_1 \otimes [f_2 \alpha g_2 + g_2 \alpha f_2 ] = 0\)<br /><br />which means that the product alpha is anti-symmetric \(f\alpha g = -g\alpha f\)<br /><br />If we use this property in the fundamental bypartite relationship we obtain in turn that the product sigma is symmetric: \(f\sigma g = g\sigma f\)<br /><br />Next time we will prove that \(\alpha\) is a Lie algebra and that \(\sigma\) is a Jordan algebra. Please stay tuned.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-63379044842014083972017-04-23T20:52:00.000-04:002017-04-23T20:52:04.238-04:00<h2 style="text-align: center;">Politics and a bit on the symmetry properties of the commutator and the Jordan products</h2><div><br /></div><div>This week I am thorn between a physics and politics. On one hand I have the scheduled physics topics to talk about, and on the other hand there are very juicy political topics. So let me start with some political commentary which I will attempt to keep at the minimum.</div><div><br /></div><div>If you do not live in US, it is hard to understand the amount of political pressure on science which comes from the right. GOP is at war with science because of three factors: the political elites are corrupt and depend on lobby money from corporations who make more money when they wreck the environment, the religious right is at war with evolution, and lastly the inbred rednecks swallow hook, line, and sinker the toxic sludge of propaganda of Fox News in the name of "freedom". </div><div><br /></div><div>So it was a breath of fresh air the recent march for science in which people sick and tired of the GOP war on science took a stand for the facts that 2+2 is still 4, humans cause global warming, and Earth is older than 5000 years. And then I opened my email and I see an alert about a new post by Lubos Motl defending Bill O'Reilly. I normally delete those notifications and I don't really know how I am subscribed to them because I only get about one a week - it is strangely inconsistent. So I said to myself: how fitting. The (naked) emperor of physics who wanted once to reclassify an archive paper to the general physics section <b>after it was published in PRL, </b>the climate change denier and the open apologist of the murderer Putin, con-man Trump, and white trash Sarah Palin throws his support behind the another toxic sorry propagandist like himself. Would have been too much to expect him to defend science instead? Out of curiosity I followed the link to see the pro O'Reilly rant, and I was not disappointed: it was choke full of imbecilic nonsense as I expected. But then I saw the icing on the cake: I see in the history list that Lubos did write a rant against the march for science too calling it misguided and unethical. Wow! Now in France (like in US or UK) there is no shortage of stupidity which just propelled Marine LePen into the final for presidency. The global village idiots will flock to her side and I have no doubt Lubos will support her too.</div><div><br /></div><div>OK, the political topics took too much and I want to continue with the series topic on quantum mechanics reconstruction. Let me just say what the products \(\alpha\) and \(\sigma\) will turn out to be. In the classical mechanics case it can be constructively proven that \(\alpha\) is the Poisson bracket while in the quantum case, \(\alpha\) is the commutator. The other product \(\sigma\) is the regular function multiplication in classical mechanics and the Jordan product (the anti-commutator) in quantum mechanics. </div><div><br /></div><div>Now the Poisson bracket and the commutators are anti-symmetric: \(f\alpha g = - g\alpha f\), and the regular function multiplication and the Jordan product are symmetric products: \(f\sigma g = g\sigma f\). The symmetry properties are preserved under system composition as we can see from the fundamental relationships:<br /><br />\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha \)<br />\(\Delta (\sigma) = \sigma \otimes \sigma - \alpha \otimes \alpha\)<br /><br />because S*S = S, S*A = A, A*A = S<br /><br />Incidentally, this observation opens up another way into quantum mechanics reconstruction (from the operational point of view) but I will not talk about it in this series. Instead next time I will show how to prove the fact that the product \(\alpha\) is anti-symmetric. Again Leibniz identity will come to the rescue. Then using the fundamental relationship we must have that the product \(\sigma\) is symmetric. Eventually all their mathematical properties will be obtained. Please stay tuned. </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-28911436502100318092017-04-16T23:52:00.001-04:002017-04-16T23:52:31.915-04:00<h2 style="text-align: center;">The fundamental bipartite relations</h2><div><br /></div><div>Continuing from where we left off last time, we introduced the most general composite products for a bipartite system:</div><div><br /></div><div><div>\(\alpha_{12} = a_{11}\alpha \otimes \alpha + a_{12} \alpha\otimes\sigma + a_{21} \sigma\otimes \alpha + a_{22} \sigma\otimes\sigma\)</div><div>\(\sigma_{12} = b_{11}\alpha \otimes \alpha + b_{12} \alpha\otimes\sigma + b_{21} \sigma\otimes \alpha + b_{22} \sigma\otimes\sigma\)<br /><br />The question now becomes: are the \(a\)'s and \(b\)'s parameters free, or can we say something abut them? To start let's normalize the products \(\sigma\) like this:<br /><br />\(f\sigma I = I\sigma f = f\)<br /><br />which can always be done. Now in:<br /><br /><div>\((f_1 \otimes f_2)\alpha_{12}(g_1\otimes g_2) = \)</div><div>\(=a_{11}(f_1 \alpha g_1)\otimes (f_2 \alpha g_2) + a_{12}(f_1 \alpha g_1) \otimes (f_2 \sigma g_2 ) +\)</div><div>\(+a_{21}(f_1 \sigma g_1)\otimes (f_2 \alpha g_2) + a_{22}(f_1 \sigma g_1) \otimes (f_2 \sigma g_2 )\)<br /><br />if we pick \(f_1 = g_1 = I\) :<br /><br /><div>\((I \otimes f_2)\alpha_{12}(I\otimes g_2) = \)</div><div>\(=a_{11}(I \alpha I)\otimes (f_2 \alpha g_2) + a_{12}(I \alpha I) \otimes (f_2 \sigma g_2 ) +\)</div><div>\(+a_{21}(I \sigma I)\otimes (f_2 \alpha g_2) + a_{22}(I \sigma I) \otimes (f_2 \sigma g_2 )\)<br /><br />and recalling from last time that \(I\alpha I = 0\) from Leibniz identity we get:<br /><br />\(f_2 \alpha g_2 = a_{21} (f_2 \alpha g_2 ) + a_{22} (f_2 \sigma g_2)\)<br /><br />which demands \(a_{21} = 1\) and \(a_{22} = 0\).<br /><br />If we make the same substitution into:<br /><br /> \((f_1 \otimes f_2)\sigma_{12}(g_1\otimes g_2) = \)<br /><div>\(=b_{11}(f_1 \alpha g_1)\otimes (f_2 \alpha g_2) + b_{12}(f_1 \alpha g_1) \otimes (f_2 \sigma g_2 ) +\)</div><div>\(+b_{21}(f_1 \sigma g_1)\otimes (f_2 \alpha g_2) + b_{22}(f_1 \sigma g_1) \otimes (f_2 \sigma g_2 )\)<br /><br />we get:<br /><br />\(f_2 \sigma g_2 = b_{21} (f_2 \alpha g_2 ) + b_{22} (f_2 \sigma g_2)\)<br /><br />which demands \(b_{21} = 0\) and \(b_{22} = 1\)<br /><br />We can play the same game with \(f_2 = g_2 = I\) and (skipping the trivial details) we get two additional conditions: \(a_{12} = 1\) and \(b_{12} = 0\).<br /><br />In coproduct notation what we get so far is:<br /><br />\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha + a_{11} \alpha \otimes \alpha\)<br />\(\Delta (\sigma) = \sigma \otimes \sigma + b_{11} \alpha \otimes \alpha\)<br /><br />By applying Leibniz identity on a bipartite system, one can show after some tedious computations that \(a_{11} = 0\). The only remaining free parameters is \(b_{11}\) which can be normalized to be ether -1, 0, or 1 (or elliptic, parabolic, and hyperbolic). <b>Each choice corresponds to a potential theory of nature</b>. For example 0 corresponds to classical mechanics, and -1 to quantum mechanics.<br /><br /><b>Elliptic composability is quantum mechanics! The bipartite products obey:</b><br /><b><br /></b><b><br /></b>\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha \)<br />\(\Delta (\sigma) = \sigma \otimes \sigma - \alpha \otimes \alpha\)<br /><br /><b>Please notice the similarity with complex number multiplication. This is why complex numbers play a central role in quantum mechanics.</b><br /><b><br /></b>Now at the moment the two products do not respect any other properties. But we can continue this line of argument and prove their symmetry/anti-symmetry. And from there we can derive their complete properties arriving constructively at the standard formulation of quantum mechanics. Please stay tuned.</div></div></div></div></div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-75737353711244180922017-04-09T20:49:00.001-04:002017-04-09T20:49:47.734-04:00<h2 style="text-align: center;">Time evolution for a composite system</h2><div><br /></div><div>Continuing where we left off last time, let me first point out one thing which I glossed over too fast: the representation of \(D\) as a product \(\alpha\): \(Dg = f\alpha g\). This is highly nontrivial and <b>not all time evolutions respect it</b>. In fact, the statement above is nothing but a reformulation of <a href="https://en.wikipedia.org/wiki/Noether%27s_theorem" target="_blank">Noether's theorem</a> in the Hamiltonian formalism. I did not build up the proper mathematical machinery to easily show this, so take my word on it for now. I might revisit this at a later time.</div><div><br /></div><div>Now what I want to do is explore what happens to the product \(\alpha\) when we consider two physical systems 1 and 2. First, let's introduce the unit element of our category, and let's call it "I":</div><div><br /></div><div>\(f\otimes I = I\otimes f = f\)</div><div><br /></div><div>for all \(f \in C\)</div><div><br /></div><div>Then we have \((f_1\otimes I) \alpha_{12} (g_1\otimes I) = f \alpha g\)</div><div><br /></div><div>On the other hand suppose in nature there exists only the product \(\alpha\). Then the only way we can construct a composite product \(\alpha_{12}\) out of \(\alpha_1\) and \(\alpha_2\) is:</div><div><br /></div><div>\((f_1\otimes f_2) \alpha_{12} (g_1 \otimes g_2) = a(f_1 \alpha_1 g_1)\otimes (f_2\alpha_2 g_2)\)</div><div><br /></div><div>where \(a\) is a constant. </div><div><br /></div><div>Now if we pick \(f_2 = g_2 = I\) we get:</div><div><br /></div><div>\((f_1\otimes I) \alpha_{12} (g_1 \otimes I) = a(f_1 \alpha_1 g_1)\otimes (I \alpha_2 I) \)</div><div>which is the same as \(f \alpha g\) by above. </div><div><br /></div><div>But what is \(I\alpha I\)? Here we use the Leibniz identity and prove it is equal with zero:</div><div><br /></div><div>\(I \alpha (I\alpha A) = (I \alpha I) \alpha A + I \alpha (I \alpha A)\)</div><div><br /></div><div>for all \(A\) and hence \(I\alpha I = 0\)</div><div><br /></div><div>But this means that <b>a single product alpha by itself is not enough! Therefore we need a second product </b>\(\sigma\)! Alpha will turn out to be the commutator, and sigma the Jordan product of observables, but we will derive this in a constructive fashion.</div><div><br /></div><div>Now that we have two products in our theory of nature, let's see how can we build the composite products out of individual systems. Basically we try all possible combinations:</div><div><br /></div><div>\(\alpha_{12} = a_{11}\alpha \otimes \alpha + a_{12} \alpha\otimes\sigma + a_{21} \sigma\otimes \alpha + a_{22} \sigma\otimes\sigma\)</div><div>\(\sigma_{12} = b_{11}\alpha \otimes \alpha + b_{12} \alpha\otimes\sigma + b_{21} \sigma\otimes \alpha + b_{22} \sigma\otimes\sigma\)</div><div><br /></div><div>which is shorthand for (I am spelling out only the first case):</div><div><br /></div><div>\((f_1 \otimes f_2)\alpha_{12}(g_1\otimes g_2) = \)</div><div>\(=a_{11}(f_1 \alpha g_1)\otimes (f_2 \alpha g_2) + a_{12}(f_1 \alpha g_1) \otimes (f_2 \sigma g_2 ) +\)</div><div>\(+a_{21}(f_1 \sigma g_1)\otimes (f_2 \alpha g_2) + a_{22}(f_1 \sigma g_1) \otimes (f_2 \sigma g_2 )\)</div><div><br /></div><div>For the mathematically inclined reader we have constructed what it is called a <a href="https://en.wikipedia.org/wiki/Coalgebra" target="_blank">coalgebra </a>where the operation is called a coproduct: \(\Delta : C \rightarrow C\otimes C\). <b>In category theory a coproduct is obtained from a product by reversing the arrows.</b></div><div><b><br /></b></div><div>Now the task is to see if we can say something about the coproduct parameters: \(a_{11},..., b_{22}\). In general nothing can constrain their values, but in our case we do have an additional relation: Leibniz identity which arises out the functoriality of time evolution. This will be enough to fully determine the products \(\alpha\) and \(\sigma\), and from them the formalism of quantum mechanics. Please stay tuned.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-17661262499437536722017-03-26T18:30:00.000-04:002017-04-09T17:53:59.273-04:00<h2 style="text-align: center;">Time as a continous functor</h2><div><br /></div><div>To recall from prior posts, a functor maps objects to objects and arrows to arrows between two categories. In other words, it is structure preserving. In the case of a monoidal category, suppose there is an arrow * from \(C\times C \rightarrow C\). Then a functor T makes the diagram below commute:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-CUUpUxprXSA/WNgTFMIrt-I/AAAAAAAABLA/PwSgLfe0WyckeG022zqPP7r13Fg0HMEfgCLcB/s1600/time.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="196" src="https://3.bp.blogspot.com/-CUUpUxprXSA/WNgTFMIrt-I/AAAAAAAABLA/PwSgLfe0WyckeG022zqPP7r13Fg0HMEfgCLcB/s320/time.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">This is all fancy abstract math which has a simple physical interpretation when T corresponds to time evolution: <b>the laws of physics do not change in time. </b>Moreover it can be shown with a bit of effort and knowledge of C* algebras that <b>Time as a functor = unitarity</b>.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">But what can we derive from the commutative diagram above? With the additional help of two more very simple and natural ingredients we will be able to reconstruct the complete formalism of quantum mechanics!!! Today I will introduce the first one: time is a continuous parameter. Just like in group theory adding continuity results in the theory of Lie groups we will consider <b><u>continous functors</u> and we will investigate what happens in the neighborhood of the identity element.</b></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">In the limit of time evolution going to zero T becomes the identity. For infinitesimal time evolution we can then write:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">\(T = I + \epsilon D\)</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">We plug this back into the diagram commutativity condition \(T(A)*T(B) = T(A*B)\) and we obtain in first order the chain rule of differentiation:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">\(D(A*B) = D(A)*B + A*D(B)\)</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">There is not a single kind of time evolution and \(D\) is not unique (think of various hamiltonians). There is a natural transformation between different time evolution functors and we can express D as an operation like this: \(D_A = A\alpha\) where \((\cdot \alpha \cdot)\) is a product.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">\(\alpha : C\times C \rightarrow C\)</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Then we obtain the Leibniz identity:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">\(A\alpha (B * C) = (A\alpha B) * C + B * (A \alpha C)\)</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">This is extremely powerful, as it is unitarity in disguise. Next time we'll use the tensor product and the second ingredient to obtain many more mathematical consequences. Please stay tuned.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-88643292734778322672017-03-19T19:13:00.000-04:002017-03-19T19:13:50.992-04:00<h2 style="text-align: center;">Monoidal categories and the tensor product</h2><div><br /></div><div><br /></div><div>Last time we discussed the category theory product which forms another category from two categories. Suppose now that we start with one category \(C\) and form the product with itself \(C\times C\). It is natural to see if there is a functor from \(C\times C\) to \(C\). If such a functor exists and moreover it respects associativity and unit elements, then the category \(C\) is called a <b>monoidal category. </b>By abuse of notation, the functor above is called the tensor product, but this is not the usual tensor product of vector space. The tensor product of vector space is only one concrete example of a monoidal product. To get to the ordinary tensor product we need to inject physics into the problem. </div><div><br /></div><div>The category \(C\) we are interested in is that of physical systems where the objects are physical systems, and arrows are compositions of physical systems. The key physical concepts needed are that of <b>time</b> and <b>dynamical degree of freedom </b>inside <b>Hamiltonian formalism.</b></div><div><br /></div><div>Time plays an distinguished role in quantum mechanics both in terms of formalism (remember that there is no time operator) and in how quantum mechanics can be reconstructed. </div><div><br /></div><div>The space in Hamiltonian formalism is a Poisson manifold which is not necessarily a vector space but because the Hilbert space \(L^2 (R^3\times R^3)\) is isomorphic to \(L^2 (R^3 ) \otimes L^2 (R^3 )\) let's discuss monoidal categories for vector spaces obeying an equivalence relationship. Hilbert spaces form a category of their own and there is a functor mapping physical systems into Hilbert spaces. This is usually presented as the first quantum mechanics postulate: each physical system is associated with a complex Hilbert space H.</div><div><br /></div><div>For complete generality of the definition of the tensor product we consider two distinct vector space V and W for which we first consider the category theory product (in this case the Cartesian product) but for which we make the following identifications:</div><div><ul><li>\((v_1, w)+(v_2, w) = (v_1 + v_2, w)\)</li><li>\((v, w_1)+(v, w_2) = (v, w_1 + w_2)\)</li><li>\(c(v,w) = (cv, w) = (v, cw)\)</li></ul><div>For physical justification think of V and W as one dimensional vector spaces corresponding to distinct dynamical degrees of freedom. Linearity is a property of vector spaces and we expect this property to be preserved if vector spaces are to describe nature. Bilinearity in the equivalence relationship above arises because the degrees of freedom are independent.</div></div><div><br /></div><div>Now a Cartesian product of vector spaces respecting the above relationships is a new mathematical object: a tensor product.</div><div><br /></div><div>The tensor product is unique up to isomorphism and respects the following <a href="https://en.wikipedia.org/wiki/Universal_property" target="_blank">universal property</a>:</div><div><br /></div><div>There is a bilinear map \(\phi : V\times W \rightarrow V\otimes W\) such that given <b>any</b> other vector space Z and a bilinear map \(h: V\times W \rightarrow Z\) there is a unique linear map \(h^{'}: V\otimes W \rightarrow Z\) such that the diagram below commutes.</div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-5hwrfRw5nIM/WM8KagnQ8wI/AAAAAAAABKo/iWiQyIfEqqU3GXb9epQIqqooEqTokZKuwCLcB/s1600/tensor.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-5hwrfRw5nIM/WM8KagnQ8wI/AAAAAAAABKo/iWiQyIfEqqU3GXb9epQIqqooEqTokZKuwCLcB/s1600/tensor.png" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">This universal property is very strong and several mathematical facts follows from it: the tensor product is unique up to isomorphism (instead of Z consider another tensor product \(V\otimes^{'}W\) ), the tensor product is associative, and there is a natural isomorphism between \(V\otimes W\) and \(W\otimes V\) making the tensor product an example of a <a href="https://en.wikipedia.org/wiki/Symmetric_monoidal_category" target="_blank">symmetric monoidal category</a>, just like the category of physical systems under composition.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">This may look like an insignificant trivial observation, but it is extremely powerful and it is the starting point of quantum mechanics reconstruction. <b>On one hand we have composition of physical systems and theories of nature describing physical systems. On the other hand we have dynamical degrees of freedom and the rules of quantum mechanics. The two things are actually identical and each one can be derived from the other. </b>To do this we need one additional ingredient: time viewed as a functor. Please stay tuned.</div><div><br /></div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-78499928397017430542017-03-13T00:21:00.000-04:002017-03-13T00:21:36.966-04:00<h2 style="text-align: center;">Category Theory Product</h2><div><br /></div><div>Before we discuss this week's topic, I want to make two remarks from the prior posts content. First, why we need natural transformations in algebraic topology? Associating groups to topological spaces (which incidentally describe the hole structure of the space) is done by the use of functors. Different (co)homology theories are basically different functors, and their equivalence is the same as proving the existence of a natural transformation. Second, the logic used in category theory is intuitionistic logic where truth is proved constructively. Since this is mapped into computer science by the Curry-Howard isomorphism, the fact that some statements have no constructive proof is equivalent with a computation running forever. In computation theory one encounters <a href="https://en.wikipedia.org/wiki/Halting_problem" target="_blank">the halting problem</a>. If the halting problem were decidable then category theory would have been mapped to ordinary logic instead of intuitionistic logic.</div><div><br /></div><div>Now back to the topic of the day. We are still in pure math domain and we are looking at mathematical objects from 10,000 feet disregarding their nature and observing only their existence and their relationships (objects and arrows). The first question one asks is how to construct new categories from existing ones? One way is to simply reverse the direction of all arrows and the resulting category is unsurprisingly called the opposite category (or the dual). Another way is to combine two category into a new one. Enter the concept of a product of two categories: \(\mathbf{C}\times \mathbf{D}\). In set theory this would correspond with to the Cartesian product of two sets. However we need to give a definition which is independent of the nature of the elements. Moreover we want to give it in a way which guarantees uniqueness up to isomorphism. </div><div><br /></div><div>The basic idea is that of a projection from the elements of \(\mathbf{C}\times \mathbf{D}\) back to the elements of \(\mathbf{C}\) and \(\mathbf{D}\). So how do we know that those projections and the product is unique up to isomorphism? Suppose that there is another category \(\mathbf{Y}\) with maps \(f_C\) and \(f_D\). Then there is a unique map \(f\) such that the diagram below commutes</div><div><br /></div><div><br /></div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-RNLjwZadnjs/WMYT7wKhrFI/AAAAAAAABJE/IhaFCcX3po4pGii2Nm0iFq8cBfKcIt3VwCLcB/s1600/Prod.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="120" src="https://1.bp.blogspot.com/-RNLjwZadnjs/WMYT7wKhrFI/AAAAAAAABJE/IhaFCcX3po4pGii2Nm0iFq8cBfKcIt3VwCLcB/s320/Prod.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">This diagram has to commute for all categories \(\mathbf{Y}\) and their maps \(f_C\) and \(f_D\). From this definition, can you prove uniqueness of the product up to isomorphism? It is a simple matter of "diagram reasoning". Just pretend that Y is now the "true incarnation" of the product. You need to find a morphisms f from Y to CxD and a morphism g from CxD to Y such that \(g\circ f =1_{C\times D}\), \(g\circ f = 1_Y\). See? Category theory is really easy and not harder than linear algebra.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Now what happens if we flip all arrows in the diagram above? We obtain a coproduct category \(\mathbf{C}\oplus \mathbf{D}\) and the projections maps become injection maps. </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">OK, time for concrete examples:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"></div><ul><li>sets: product = Cartesian product, coproduct = disjoint union</li><li>partial order sets: product = greatest lower bounds (meets), coproduct = least upper bounds (joins)</li></ul><div>So where are we now? The concept of the product is very simple, but we need it as a stepping stone to the concept of tensor product and (symmetric) monoidal category. Why? Because <b>physical systems form a symmetric monoidal category</b>. Using categorical arguments we can derive the complete mathematical properties of any theory of nature describing a symmetric monoidal category. And the answer will turn out to be: quantum mechanics. Please stay tuned.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-87634074684798928382017-03-04T23:08:00.000-05:002017-03-05T00:44:53.316-05:00<h2 style="text-align: center;">The Curry–Howard isomorphism</h2><div><br /></div><div>Category theory may seem vary abstract and intimidating, but in fact it is extremely easy to understand. In category theory we look at concrete objects from far away without any regard for the internal structure. This is similar with Bohr's position on physics: physics is about what we can say about nature, and not decide what nature is. Surprisingly, a lot of information about the objects in category theory is derivable from the behavior of the objects and this is where I am ultimately heading with this series on category theory.</div><div><br /></div><div>Last time I mentioned the origin of category theory as the formalism to clarify when two homology theories are equivalent. But category theory can be started from two other directions as well, and those alternative viewpoints help provide the intuition needed to navigate the abstractions of category theory. One thread of discussion starts with the idea of computability and the work of <a href="https://en.wikipedia.org/wiki/Alonzo_Church" target="_blank">Alonzo Church</a> and <a href="https://en.wikipedia.org/wiki/Alan_Turing" target="_blank">Alan Turing.</a> Turing was Church's student and each started an essential line of research: <a href="https://en.wikipedia.org/wiki/Lambda_calculus" target="_blank">lambda calculus</a> and <a href="https://en.wikipedia.org/wiki/Universal_Turing_machine" target="_blank">universal Turing machines</a>. Those later grew into one hand functional languages like Java script, and the other hand into object oriented languages like C++. What one can do with lambda calculus can be achieved with universal Turing machines, and the other way around. The essential idea of computer programming is to build complex structures out of simpler building blocks. Object oriented programming starts from the idea of packaging together actions and states. An object is a "black box" containing actions (functions performing computations) and information (the internal state of the object). Functional programming on the other hand lacks the concept of an internal state and you deal only with functions which take an input, crunch the numbers, and then produce an output. The typical example is FORTRAN: FORmula TRANslation (from higher level human understandable syntax into zeroes and ones understandable by a machine).<br /><br />The second direction one can start category theory is <a href="https://en.wikipedia.org/wiki/Intuitionistic_logic" target="_blank">intuitionistic logic</a> and the foundation of set theory. The problem of naive set theory is that one can create paradoxes like <a href="https://en.wikipedia.org/wiki/Russell's_paradox" target="_blank">Russel's paradox</a>: the set of all sets which are not members of themselves. The solution Russel proposed was <a href="https://en.wikipedia.org/wiki/Type_theory" target="_blank">type theory</a>. Types introduce structure to set theory preventing self-referential constructions. In computer programming, types are semantic rules which tell us how to understand various sequences of zeros and ones from computer memory as integers, boolean variables, etc.<br /><br />In intuitionistic logic statements are not true by simply disproving their falsehood, but they are true by providing an explicit construction. Truth must be computed and the parallel with computer programming is obvious. There is a name for this relationship, the <a href="https://en.wikipedia.org/wiki/Curry%E2%80%93Howard_correspondence" target="_blank">Curry-Howard isomorphism</a>. <b>The mathematical formalism needed to rigorously spell out this correspondence is category theory. </b>At high level:<br /><ul><li>proofs are programs</li><li>propositions are types</li></ul><div>More important is that we can attach logical and programming meaning to category theory constructions which helps dramatically reduce the difficulty of category theory to that of elementary linear algebra. </div><div><br /></div>There are two additional key points I want to make. First category theory ignores the structure of the objects: they can be sets, topological spaces, posets, even physical systems. As such <b>uniqueness is relaxed in category theory and things are unique up to isomorphisms. </b>Second, <b>we are strengthening uniqueness by seeking </b><a href="https://en.wikipedia.org/wiki/Universal_property" style="font-weight: bold;" target="_blank">universal properties</a><b>. </b>This gives category theory its abstract flavour:<b> </b>the generalization of standard mathematical concepts in category theory involve diagrams which must commute. The typical definition is something like: "if there is an "impostor" which claims to have the same properties as the concept being defined, then there exist a so and so isomorphism such that a certain diagram commutes which guarantees that the impostor is nothing but a restatement of the same concept up to isomorphism". Next time I will talk about the first key definition we need from category theory, that of a product, and by flipping the arrows that of a coproduct. </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-41329477995214228272017-02-28T00:16:00.000-05:002017-02-28T00:16:01.799-05:00<h2 style="text-align: center;">Objects and arrows</h2><div><br /></div><div>With one day delay, let's continue the discussion about category theory. One way to look at category theory is as a generalization of the notion of equivalence: <b>category theory = equivalence on steroids</b>. </div><div><br /></div><div>It is informative to look at the original motivation for category theory and also to look at a problem around 1900. Suppose you go back in time without knowing any modern math except group theory and you are aware of <a href="https://en.wikipedia.org/wiki/M%C3%B6bius_strip" target="_blank">Mobius strip</a>, and <a href="https://en.wikipedia.org/wiki/Klein_bottle" target="_blank">Klein bottle</a>. Your task is to try to figure out what else is possible? In other words, classify all two dimensional surfaces. Who can help you on this quest? Well, clothes are two dimensional surfaces made by tailors. How do they make them? By two operations: cutting and stitching. Knowing group theory, you realize cutting and stitching are opposite operations, and they do respect the axioms of a group. This is how <a href="https://en.wikipedia.org/wiki/Homology_(mathematics)" target="_blank">homology </a>theory actually came from: associating groups with topological spaces in order to classify them. Now fast forward to 1940s, several homology theories were known and the problem was why the groups involved in them are the same? How do we axiomatize homology theory and how do we know if two homologies are equivalent? The answer lies in the concept of <b>natural transformation</b> which requires the concept of <b>functor</b>, which in turn needs the idea of a <b>category.</b> </div><div><br /></div><div>So what is a category? A category consists of objects and morphisms (arrows) such that the morphisms can be composed. Here are some examples:</div><div><br /></div><div>-examples from math:</div><div><ul><li>sets and functions</li><li>groups and group homomorphisms</li><li>Hilbert space and operators</li><li>partial order sets and monotone functions</li><li>manifolds and cobordisms</li></ul><div>-examples from logic</div></div><div><ul><li>propositions and proofs</li></ul><div>-example from physics</div><div><ul><li>physical systems and physical processes</li></ul></div>-examples from computer science<br /><ul><li>data types and programs</li></ul><div>Now a <b>functor</b> maps a category to another category by mapping objects to objects and arrows to arrows in a way that preserves structure. This is how for example in algebraic topology we associate groups to topological spaces. </div></div><div><br /></div><div>A <b>natural transformation</b> is a arrow (morphism) between functors subject to some (natural) conditions. </div><div><br /></div><div>Apart of naturality, another key concept is <b>universality </b>which means a unique (up to an isomorphism) solution to problems of constructions. We will encounter that when we will express quantum mechanics in category formalism.</div><div><br /></div><div>Category theory reveals surprising relationships:</div><div><ul><li>Cartesian products of sets are like greater lower bounds of partial order sets</li><li>Proofs in logic are like programs in functional programming</li></ul><div>Back to quantum mechanics, unitary evolution preserves information and it should be no surprise that there quantum information can be represented in diagramatic fashion. However this is not the path I am going to take and I will make use of universality in deriving quantum mechanics from a simple principle - composition: <b>a theory T describing two physical systems A and B must described the composite system A+B as well. </b>This is very intuitive principle but in the formalism of category theory it has extremely powerful mathematical consequences, spelling out the complete internal details of the theory T. <b>Quantum mechanics comes out of this in its full detail. </b></div></div><div><br /></div><div>Please stay tuned...</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-117126206166693902017-02-19T19:54:00.000-05:002017-02-19T19:54:56.768-05:00<h2 style="text-align: center;">Monoids: the root of it all</h2><div><br /></div><div>Let's start talking about category theory. We will start from set theory and in the end try to get away from it. The first thing we need to discuss is <a href="https://en.wikipedia.org/wiki/Magma_(algebra)" target="_blank">magma</a>. Basically you have a binary operation on a set and that's all: \(M \times M\rightarrow M\). One problem with magmas is that there is no associativity. Now not all mathematical operations lacking associativity are inherently primitive. Think of Lie algebras: the operation is not associative. However there you have something else: the Jacobi identity. But a pure magma without any additional structure is a rather inert object. The other problem with magmas is the lack of a unit. Add associativity and a unital element and category theory comes alive. </div><div><b><br /></b></div><div><b>To link the discussion to physics, nature obeys the structure of a (commutative) monoid: <u>two physical systems can be composed into a larger physical system:</u></b></div><div><b>- </b>composition is the binary operation </div><div><b>- </b>associativity guarantees our ability to reason about physical systems regardless of how we split a physical system into subsystems: quantum mechanics is valid for both an electron or an atom containing an electron</div><div><b>-</b> the unital element is nothingness: composition with nothing leaves the original physical system intact.</div><div><br /></div><div>In later posts I will show how quantum mechanics is a logical consequence of the commutative monoid above. In other words, quantum mechanics is inescapable and nature is quantum all the way.</div><div><br /></div><div>Back on monoids, let's fall back on the usual example: composable functions: the image of a function is the domain of the next function. The link with programming is obvious: the output of one computation is plugged in as the input of another computation. As a side note, because of this functional programming is best explained in the language of category theory. When we talk function composition we usually write: \(f \circ g\) which means \(f(g(x))\). To jump in abstraction and eliminate the nature of the elements considerations, there is an elementary trick to help navigate complex composition chains: call \(\circ\): AFTER like this: \(f~composed~with~g = f\circ g = f~ AFTER~ g\)</div><div><br /></div><div>Now let's review the usual properties of injectivity and surjectivity:</div><div><br /></div><div>Injectivity: for any elements \(x, x^{'}\), a function is injective if \(f(x) = f(x^{'}) ~implies~ x=x^{'}\)</div><div>Surjectivity: for any \(y\) in the range, there is an \(x\) in the domain such that \(f(x)=y\)</div><div><br /></div><div>So how can we abstract this away and eliminate the talk about the elements? The corresponding category theory concepts are monic and epic:</div><div><br /></div><div>Monic: a morphism is monic if for any \(g, h\) \(f\circ g = f\circ h ~implies~g=h\)</div><div>Epic: a morphism is epic if for any \(g, h\) \(g\circ f = h\circ f ~implies~g=h\)</div><div><br /></div><div>Can you prove that if \(f:X\rightarrow Y\) then \(f\) is injective if and only if it is monic and it surjective if and only if it is epic? The proof can be found in many places but it is instructive to try to prove it yourself without looking it up first as this will help you better understand category theory. </div><div><br /></div><div>The last point I want to make today is that in category theory we move away from functions into abstract morphisms. <b>The key point of morphisms is that they preserve mathematical structures.</b> As such they can be used to jump between categories of very different nature. This is how category theory is a unifying structure of mathematics where the same patterns of reasoning can be replicated from logic to computer science, to algebraic topology, to quantum mechanics.</div><div><br /></div><div>To be continued...</div><div><br /></div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-28736351905012547962017-02-12T21:13:00.000-05:002017-02-12T22:13:13.150-05:00<h2 style="text-align: center;">A new way to look at mathematics</h2><div><br /></div><div>I want to start today a series of posts about <a href="https://en.wikipedia.org/wiki/Category_theory" target="_blank">category theory</a>. This is a vast area of mathematics which unifies logic, computer programming, combinatorics, cohomology, etc, and <b>quantum mechanics </b>into a cohesive paradigm. It also settles the problem of interpretation for quantum mechanics. By its very construction category theory has no need for any realism baggage. The entire mathematics can be expressed not in the language of sets (which are abstractions based on our classical intuition) but <b>in the language of categories free of any considerations about the nature of elements</b>. Regarding physics, the paradigm of category theory is best expressed by a famous Bohr quote:</div><br /><i>"It is wrong to think that the task of physics is to find out how Nature is. Physics concerns what we say about Nature."</i><br /><div><i><br /></i></div><div>Let me start slow. The usual usage of math is on the practical side to solve problems. How many times did we hear the lazy student complaint: why should we learn this? Math is not about memorization and math is very easy once we absorb its content. Learning math is a journey in mastering abstractions and general ways of reasoning. For example when you learn about Lie groups you can extract a lot of key result by elementary methods simply by studying matrices. However you hit a wall with octonions because they are not associative and do not admit a matrix representation for this very reason. In turn this precludes the proper understanding of exceptional Lie groups.</div><div><br /></div><div>Or consider a simpler example, topology. A lot of functional analysis can be done using the concept of distance and metric spaces. For example a space in \(R^n\) is compact iff it is closed and bounded. Then the metric spaces are generalized by the concept of <a href="https://en.wikipedia.org/wiki/Topological_space" target="_blank">topological space</a>s which are based on the idea of neighborhoods, unions, and intersections. In this case compactness is defined much more abstractly: a space is compact iff any open covering has a finite subcover. </div><div><br /></div><div>A similar thing happens in category theory. Patterns of reasoning in various mathematical domains are abstracted away in a formalism which does not care about the nature of the elements. On one end this is harder and to help navigate this in the beginning you hold on particular examples; the typical examples are functions. However at some point you let go of the examples just like in topology you let go the notion of distance. At that point you learn to reason properly in category theory and a lot can be achieved in this way. Then we can make the journey backwards from abstract to concrete. There is a big bonus in this: we have the flexibility to pick the concrete examples we want. And in our case we will pick quantum mechanics. <b>Quantum mechanics is best and most naturally expressed in the language of category theory. </b>Goodbye sets, goodbye classical realism, let the category journey begin. Please stay tuned. </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-71473466697640513722017-02-04T22:00:00.000-05:002017-02-04T22:43:01.570-05:00<h2 style="text-align: center;">Trump, DeVos and the fleecing of America</h2><div><br /></div><div>Once upon a time there was a Sputnik circling the Earth and the <a href="https://en.wikipedia.org/wiki/Sputnik_crisis" target="_blank">fear </a>it created spurred America to wake up and invest massively into education. Those days are long gone and now religious extremists (like late Jerry Falwell and his son Jerry Falwell Jr) wage war on science in United States. Sadly they are about to destroy the education system and the end result will be an <a href="https://en.wikipedia.org/wiki/Idiocracy" target="_blank">Idiocracy </a>society where we study only creationism and we water plans with sport drinks because they have electrolytes - what plants crave.</div><div><br /></div><div>So what is president Trump's policy? <b>He has only one policy: to continuously demonstrate he has the largest dick.</b> </div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-RA2xjHuCz54/WJZWV5yiX_I/AAAAAAAABIo/IINHx_gWz6kZ5CwFnaNY7QYbSu7CMlGZwCLcB/s1600/camacho.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="154" src="https://1.bp.blogspot.com/-RA2xjHuCz54/WJZWV5yiX_I/AAAAAAAABIo/IINHx_gWz6kZ5CwFnaNY7QYbSu7CMlGZwCLcB/s320/camacho.jpg" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">From inauguration crowd size to popular vote size, it is all about how he is "yuuuge". Help him masturbate his ego in public and you get away with anything. One such person is Betsy DeVos. </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><a href="https://en.wikipedia.org/wiki/Betsy_DeVos" target="_blank">Betsy DeVos</a> is a billionaire who made her money with <a href="https://en.wikipedia.org/wiki/Amway" target="_blank">Amway</a> and she bought her way into the Trump administration by donating to the republican party something to the tune of 200 million dollars. <b>On the recent confirmation hearings she said she is supports guns in schools to protect students from grizzly bears!!! </b>Also she did not understand the difference between the value of something and the rate of increase of that value. Her intellectual level is that of a moronic imbecile who repeatedly failed to complete 3rd grade. Honestly, she deserves a prize for managing to beat Sarah Palin in stupidity: a really really hard thing to achieve.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">So why does she want to lead the education department? Because <b>the education budget is over 140 billion dollars. Cha-ching! By refusing to hold both public and private schools accountable <u>to the same standards</u> she opens to door to scams like her boss' "Trump University" And it is all paid for by us, the taxpayers. </b>What? Did you believe those 200 million dollars were donated out of the goodness of her hearth?</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Now maybe Amway is a legit business and she is not a nutcase. Do you know how Amway works? It is a pyramid scheme going under the name "<a href="https://en.wikipedia.org/wiki/Multi-level_marketing" target="_blank">multi-level marketing</a>": you buy their soap and you sell it to say 5 of your friends and if you sign them up as Amway distributors you get a cut from their sale as well. Now you don't get rich by selling 3 dollars worth of soap in a month, but by signing up many more people and they in turn do the same. The end result is a pyramid of losers. Most of them end up broke and with a garage full of soap inventory. They loose money on books and brainwashing cult-like company seminars. So why is this not illegal? Because the Federal Trade Commission guys are crooks too: since there is an actual product flowing (which keeps the scam going) they merely get their cut of the scam by huge fines (which would have been impossible from traditional Ponzi schemes because in that case when the scheme crashes the money stops flowing). Now Amway is not alone in MLM. Here is a nice video about the dangers of multi-level marketing:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div style="text-align: center;"><iframe allowfullscreen="" class="YOUTUBE-iframe-video" data-thumbnail-src="https://i.ytimg.com/vi/s6MwGeOm8iI/0.jpg" frameborder="0" height="266" src="https://www.youtube.com/embed/s6MwGeOm8iI?feature=player_embedded" width="320"></iframe></div><br /><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Back to DeVos. She is promoted by Trump to buy republican support for keeping him in power, but she is a pure republican creation. <b>Both republicans and Trump are willing to fleece America, their only difference is how they sell it to their base:</b> republicans trick them by appealing to freedom, self-reliance, and independence, while Trump promises to make their dicks great again. <b>Trump's ideology is nothing but the fascist utopia: "you are the best, screw everyone else" </b>(like mexicans, muslims, and recently australians)<b>.</b></div><div class="separator" style="clear: both; text-align: left;"><b><br /></b></div><div class="separator" style="clear: both; text-align: left;">I don't want to end on a dark note so here is a nice video: </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div style="text-align: center;"><iframe allowfullscreen="" class="YOUTUBE-iframe-video" data-thumbnail-src="https://i.ytimg.com/vi/9Sq-VPDtNK4/0.jpg" frameborder="0" height="266" src="https://www.youtube.com/embed/9Sq-VPDtNK4?feature=player_embedded" width="320"></iframe></div><div style="text-align: center;"><br /></div><div style="text-align: left;">because the best way to deal with Trump is by making fun of him. There are many more videos like this for various countries in the world. Have fun watching them all.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com0tag:blogger.com,1999:blog-3832136017893749497.post-52145621259178352032017-01-29T22:50:00.000-05:002017-01-29T22:50:51.501-05:00<h2 style="text-align: center;">Surreal Trajectories: the main argument against Bohmian ontology</h2><div><br /></div><div>I was extremely busy for the past two weeks and I simply did not have any time to write the weekly post. But without any more delays, as promised, here is the argument against Bohmian interpretation. The argument comes from a famous paper by Englert, Scully, Süssmann, and Walther: <i>Surrealistic Bohm Trajectories</i>.</div><div><br /></div><div>For clarity, here are the original <a href="https://www.degruyter.com/downloadpdf/j/zna.1992.47.issue-12/zna-1992-1201/zna-1992-1201.xml" target="_blank">paper</a>, the <a href="https://www.degruyter.com/downloadpdf/j/zna.1993.48.issue-12/zna-1993-1219/zna-1993-1219.xml" target="_blank">rebuttal</a>, and the <a href="https://www.degruyter.com/downloadpdf/j/zna.1993.48.issue-12/zna-1993-1220/zna-1993-1220.xml" target="_blank">response</a>.</div><div><br /></div><div>The argument is simple: in a double slit experiment with a which way detector present before the slit (which incidentally kills the interference pattern), the Bohmian trajectories do not cross the axis of symmetry. </div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-qtSXrVmbuDE/WI6vy6Vis4I/AAAAAAAABIM/w4O6dRpqHq8R8PHnF8BRAFfXE2VpFRtkACLcB/s1600/DoubleSlit.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="296" src="https://1.bp.blogspot.com/-qtSXrVmbuDE/WI6vy6Vis4I/AAAAAAAABIM/w4O6dRpqHq8R8PHnF8BRAFfXE2VpFRtkACLcB/s320/DoubleSlit.png" width="320" /></a></div><div><br /></div><div><br /></div><div>However the wavefunction is:</div><div><br /></div><div>\(\Psi = \psi_>|detect~up\rangle + \psi_<|detect~down\rangle\)</div><div><br /></div><div>and \(\psi_>\) does not vanish in the bottom part and \(\psi_<\) does not vanish in the top part. <b>As such, the particle can be found in the down section while the particle was detected earlier by the upper which way detector. <u>But this is at odds with Bohmian trajectories which by symmetry considerations do not connect the up with the down.</u></b></div><div><b><u><br /></u></b></div><div>The conclusion is that Bohmian trajectories do not <i>always</i> have a correspondence in reality. The issue is not whether Bohmian quantum mechanics does or does not make the same prediction as standard quantum mechanics as the rebuttal seems to imply, but <b>the issue is the ontology of Bohmian trajectories</b>. The claimed advantage of Bohmian mechanics is its clarity rooted in realism, but <b>if Bohmian trajectories are at odds with experiments, what is the value of Bohmian interpretation?</b> <b>Remember that in Bohmian interpretation the only thing "real" is the particle trajectory</b>. I could not find a valid answer from the Bohmian community to the surrealistic paper challenge and <b><u>in my opinion this paper it is a decisive clear cut argument against Bohmian interpretation.</u></b> </div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com7tag:blogger.com,1999:blog-3832136017893749497.post-19638246834488980892017-01-16T01:04:00.000-05:002017-01-17T21:54:04.351-05:00<h2 style="text-align: center;">Book Review: "Making Sense of Quantum Mechanics"</h2><div><br /></div><div>After much delay I had found the time to finish reading Jean Bricmont's "Making Sense of Quantum Mechanics" <a href="http://www.springer.com/us/book/9783319258874" target="_blank">book</a>.</div><div><br /></div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-u-tgekEgpoo/WHw00O_XgpI/AAAAAAAABHs/g5mVi8MWfcEONS1QK3MCgY9BhZlvMNuDACLcB/s1600/MakingSense.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://4.bp.blogspot.com/-u-tgekEgpoo/WHw00O_XgpI/AAAAAAAABHs/g5mVi8MWfcEONS1QK3MCgY9BhZlvMNuDACLcB/s320/MakingSense.jpg" width="212" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div><b>The book is the best presentation of Bohmian interpretation</b> I have ever read. It masterly combines the philosophical ideas with a bit of math, famous quotes, and some historical perspective. </div><div><br /></div><div>After preliminary topics in chapter one, chapter two discusses the first quantum "mystery": superposition, while chapter four discuses the second "mystery": nonlocality. It was chapter three, a philosophical "intermetzzo" which took me a very loooong time to read and prevented me to write this review much sooner: one one end I could not write this post without reading it, and on the other end I was loosing interest very quickly in it after a couple of pages of historical review. Then Bricmont proceeds into presenting Bohniam mechanics - the heart of the book. </div><div><br /></div><div>Let's dig a bit deeper into it. Chapter two is a very well written introduction into why quantum mechanics is counter-intuitive. This is presented in the style of modern quantum foundation undergrad classes. Chapter four main idea is this: to many physicists Bell's result proved the impossibility of non-contextual hidden variables (or local realism) while Bell should be understood in conjunction with EPR: EPR+Bell = nonlocality. But what does nonlocality mean? Is it just higher than expected correlations? Here Bricmont makes a very bold and provocative claim: </div><div><br /></div><div>"<i>the conclusion of his </i>[Bell's]<i> argument, combined with the EPR argument is rather that there are nonlocal physical effects (and not just correlations between events) in Nature</i>."</div><div><br /></div><div>To support this chapter 4.2 discusses "Einstein's boxes" [I had a series of posts discussing <u>why in my opinion they do not</u> represent an argument for nonlocality. What EPR+Bell shows is that the composition of two physical systems into a larger physical system does not respect the rules of classical physics - parabolic composability but new rules - elliptic composability. Nature is not "nonlocal" but "non-parabolic composable"]. </div><div><br /></div><div>Onto the main topic, the presentation of Bohmian mechanics is standard and what it is surprising is the degree on which the underdetermination issue is addressed: there are an infinite number of alternative theories (like Nelson's stochastic theory) which are in the same realistic vein and which make the same predictions as Bohmian mechanics. Chapter three discussion is invoked here but I feel the argument is very weak (not even a handwaving).</div><div><br /></div><div>Then the book talks about alternative approaches to Bohmian mechanics courageously taking (some well deserved some not) shots at alternative interpretations (like GRW, MWI, CH, Qbism), and wraps up with historical topics and sociological arguments.</div><div><b><br /></b></div><div><b>Now onto what the book covers poorly: surreal trajectories, and quantum field theory in Bohmian mechanics. </b>Surreal trajectories are mentioned in passing in a quote while they are <i><u>the</u> </i>major objection to the interpretation. As I said before, the very name "<b>surreal trajectory</b>" was a masterful catchy clever title for a paper but it backfired in the long term because it was attaching a stigma to Bohmian mechanics which in turned allowed Bohmian supporters to summarily and unfairly dismiss the argument. I will revisit the argument in next post. <b>The key point of surreal trajectories paper is that the particle is detected where Bohmian mechanics predicts it must not go, and since the only thing "real" in Bohmian mechanics is the position of the particle, it represents a fatal blow to the Bohmian ontology.</b> Currently, to my knowledge, there is no consensus inside the Bohmian community on the proper answer the surreal trajectory paper: some deny it is a problem at all while others acknowledge the problem and propose (faulty) ideas on how to deal with it. This is similar with the situation inside the MWI camp where the big pink elephant in the room there is the notion of probabilities: some in MWI disagree it is an issue while others attempt to solve it (but fail). </div><div><br /></div><div>Quantum field theory in Bohmian mechanics is another sore point which is not properly discussed. <b>My take on the topic is that a Bohmian quantum field theory is impossible to be constructed, and I want to be proven wrong by a consistent proposal: show me the money, show me the archive paper where the problem is comprehensibly solved.</b></div><div><b><br /></b></div><div>Bad points aside, overall I liked the book, I find it stimulating, and I enjoyed very much reading it (except chapter 3 which invariably succeeded putting me to sleep). <b>The book is a must read for any person seriously interested in the foundations of quantum mechanics.</b></div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com6tag:blogger.com,1999:blog-3832136017893749497.post-29365863491256703542017-01-09T00:01:00.000-05:002017-01-09T00:04:32.573-05:00<h2 style="text-align: center;">(The nonsense of) Joy Christian Reloaded</h2><div><br /></div><div>I was preparing the first physics posts of the year when I got some comments and a question on Joy Christian on an old blog <a href="http://fmoldove.blogspot.com/2015/07/joy-christians-program-of-achieving.html" target="_blank">post</a>. In the words of late <a href="https://en.wikipedia.org/wiki/Yogi_Berra" target="_blank">Yogi Berra</a>, this is "deja vu all over again". Probably the best description of Joy Christian is given by the Monty Python: The Dead Parrot sketch:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen="" class="YOUTUBE-iframe-video" data-thumbnail-src="https://i.ytimg.com/vi/4vuW6tQ0218/0.jpg" frameborder="0" height="266" src="https://www.youtube.com/embed/4vuW6tQ0218?feature=player_embedded" width="320"></iframe></div><div><br /></div><div>The question I got is the following:</div><div><br /></div><i>"I would like to understand whether the equations (67) - (75) in Joy Christian’s paper “Local Causality in a Friedmann-Robertson-Walker Spacetime” make any sense at all. I don't understand how the mathematical limes operation are carried out."</i><br /><div><i><br /></i></div><div>The paper which got past the referees by trickery is on the archive: <a href="https://arxiv.org/pdf/1405.2355v7.pdf" target="_blank">https://arxiv.org/pdf/1405.2355v7.pdf </a>and there you see the full derivation of the main faulty claim. Minus some obfuscation techniques, Eqs. 67-75 are nothing but the one-pager Joy preprint: <a href="https://arxiv.org/pdf/1103.1879v1.pdf" target="_blank">https://arxiv.org/pdf/1103.1879v1.pdf</a></div><div><br /></div><div><b>The main hand-waving trick in the "derivation" is a conversion inside of a sum of </b>\(\lambda^k\) <b>from a variable into an index</b> which amounts to adding apples to oranges and obtaining the incorrect result (see the bottom of page 8 on my preprint: <a href="https://arxiv.org/pdf/1109.0535v3.pdf" target="_blank">https://arxiv.org/pdf/1109.0535v3.pdf</a>). </div><div><br /></div><div>The mistake happens on the transition from Eq 73 to Eq 74 because the L's belong to two distinct kinds of algebras: let's call them apples and oranges. Ignoring the axb, the troubled sum term is something like this:</div><div><br /></div><div>\(L(\lambda^1) + L(\lambda^2)+L(\lambda^3)+L(\lambda^4)+L(\lambda^5)+...=\)</div><div>apple_1 + apple_2 + orange_3 +apple_4 + orange_5+...</div><div><br /></div><div>with \(\lambda^1 = +1, \lambda^2\ = +1, \lambda^3 = -1, \lambda^4 = +1, \lambda^5 = -1...\)</div><div><br /></div><div>and with the transformation rule: "apple = - oranges" when we convert to objects of the same kind (let's pick apples) we get:</div><div><br /></div><div>\(apple(\lambda^1) + apple(\lambda^2)-apple(\lambda^3)+apple(\lambda^4)-apple(\lambda^5)+...=\)</div><div>\(apple(+1) + apple(+1)-apple(-1)+apple(+1)-apple(-1)+...=\)</div><div>\(apple+ apple+apple+apple+apple+...=\)</div><div><br /></div><div><b>which no longer vanishes.</b></div><div><b><br /></b></div><div><b>The preparation for this trick is on Eq. 49 which encodes the two <u>distinct</u> algebras (of apples and oranges) into a common formula. </b>In my preprint you can double check this by trying out the matrix representations of the two algebras (eqs 53-56).</div><div><br /></div><div>Hopefully my explanation is clear enough. I know all Joy's mathematical tricks in all of his papers or in his blog debates, but I ran out of energy debunking his nonsense. <b>Kudos to Richard Gill for pursuing this further. </b>I was aware of the "<i>Causality in a Friedmann-Robertson-Walker Spacetime" </i>paper and it was on my to do list to write a rebuttal to it, but the journal withdrew it before I could get to it.</div>Florin Moldoveanuhttps://plus.google.com/117661996860849443251noreply@blogger.com2