The Lie algebra of generators

Last time we derived the fundamental composition relationship which is the root cause of why complex numbers play a  key role in quantum mechanics. However the pattern:

Real          = real * real           -  imaginary * imaginary

Imaginary = imaginary * real + real * imaginary

can also be understood as:

Symmetric       = symmetric * symmetric       -  antisymmetric * antisymmetric

Antisymmetric = antisymmetric * symmetric + symmetric * antisymmetric

and when we look at it from this point of view we arrive at a Lie algebra.

Now let me start by giving a high level overview of Lie algebras. First, the name comes from Sophus Lie, a Norwegian mathematician (no, he was not Chinese :) ) who pursued to do in partial differential equation what Galois did for polynomial equations.

Sophus Lie

Galois was using the permutation of solutions to determine if a particular polynomial equation is solvable. Instead of discrete symmetries, Lie was using continuous symmetries to simplify partial differential equations. From this came the idea of a continuous group (like the rotation group). In physics one encounters the gauge groups of the Standard Model as examples of Lie groups. In special relativity one encounters SO(1,3).

Linearizing the Lie group around the identity gives rise to what is called a Lie algebra. The elements of a Lie algebra are sometimes called generators. From the Lie algebra one may reconstruct its Lie group by exponentiation, but this does not always work as there can be distinct Lie groups with the same Lie algebra.

Now back on $\alpha$, is this a Lie algebra? It respects bilinearity and the Leibniz identity, but what about antisymmetry and Jacobi identity? It turns out that all we need is antisymmetry! We get the Jacobi identity for free because Leibniz + antisymmetry = Jacobi:

$A \alpha (B\alpha C) = (A\alpha B) \alpha C + B\alpha (A \alpha C)$

by antisymmetry:

$A \alpha (B\alpha C) +C \alpha (A\alpha B)  - B\alpha (A \alpha C)= 0$
$A \alpha (B\alpha C) +C \alpha (A\alpha B)  + B\alpha (C \alpha A)= 0$ q.e.d.

But how can we prove antisymmetry? By itself $\alpha$ is only a Loday algebra. However, invariance under composition comes to the rescue! So how can we do it? First , let's observe that in the Leibniz identity above the order of $A$ and $B$ terms occurs as $AB$ and as $BA$ and we want to take advantage of this. Second, by invariance under composition we can define the bipartite Leibniz identity:

$(A_1 \otimes A_2) \alpha_{12} ((B_1 \otimes B_2) \alpha_{12} (C_1 \otimes C_2)) = ((A_1 \otimes A_2 ) \alpha_{12} (B_1 \otimes B_2)) \alpha_{12} (C_1 \otimes C_2)$
$+ (B_1 \otimes B_2 ) \alpha_{12} ((A_1 \otimes A_2) \alpha_{12} (C_1 \otimes C_2))$

and now we want to use the fundamental composition relationship we derived last time:

$\alpha_{12} = \alpha \otimes \sigma + \sigma \otimes \alpha$

to expand the mess above. However, a great simplification occurs if we pick: $B_1 = C_2 = 1$ :

$0 = (A_1 \alpha C_1) \otimes (A_2 \alpha B_2 + B_2 \alpha A_2)$

which is valid for any $A_1 \alpha C_1$ Do it as an exercise to convince yourself it is true. Let me only show that the left hand side is zero:

$((B_1 \otimes B_2) \alpha_{12} (C_1 \otimes C_2)) = B_1 \alpha C_1 \otimes B_2 \sigma C_2 + B_1 \sigma C_1 \otimes B_2 \alpha C_2$

but because $B_1 = 1$: $B_1 \alpha C_1 = 0$ and because  $C_2 = 1$: $B_2 \alpha C_2 = 0$  and the LHS is zero.

Hence:

$A\alpha B = - B\alpha A$

So $\alpha$ is a Lie algebra. Moreover by the fundamental composition relationship $\sigma$ must be a symmetric product to preserve the antisymmetry of $\alpha$ under composition.

Now we are getting somewhere. We know that quantum mechanics is described by C* algebras which can be decomposed into Lie and Jordan algebras. We got the Lie part and we are almost there for the Jordan side. Please stay tuned.

Where are the complex numbers coming from in Quantum Mechanics?

So last time we produced the ingredients used to build quantum mechanics and today we will show how they combine just like Lego. And the main building pattern will turn out to be nothing but to complex number multiplication.

The mathematical structure will be that of a trigonometric coalgebra. 

So let us start by recalling the two products from last time: $\alpha$ and $\sigma$. $\alpha$ respects the Leibniz identity (because it stems from infinitesimal time evolution) and apart from that we have no other information at this time about the two products. If we assume the real numbers to be the mathematical field corresponding to actual physical measurement values, the Lego operation of combining two physical systems into one is called a coproduct. Here is how we do it:

Let C be a R-space with {$\alpha$, $\sigma$} as a basis. We define the coproduct $\Delta : C\rightarrow C \otimes C$ as:

$\Delta (\alpha ) = a_{11} \alpha \otimes \alpha + a_{12} \alpha \otimes \sigma + a_{21} \sigma \otimes \alpha + a_{22} \sigma \otimes \sigma$

$\Delta (\sigma) = b_{11} \alpha \otimes \alpha + b_{12} \alpha \otimes \sigma + b_{21} \sigma \otimes \alpha + b_{22} \sigma \otimes \sigma$

There is also another operation called counit but that is only important for the mathematical point of view to complete what mathematicians call a coalgebra (S. Dascalescu, C. Nastasescu, and S. Raianu, Hopf Algebra: An Introduction, Chapman & Hall/CRC Pure and Applied Mathematics (Taylor & Francis, 2000)).

To get this abstraction more down to earth we need to see it in action and show what it means to construct the bi-partite products $\alpha_{12}$ and $\sigma_{12}$ from $\sigma$ and $\alpha$. If we have $f_1, g_1$ elements belonging to physical system 1, and $f_2, g_2$ elements belonging to physical system 2 we basically have the following:

$(f_1\otimes f_2) \alpha_{12} (g_1\otimes g_2) = a_{11} (f_1\alpha g_1) \otimes (f_2 \alpha g_2) + a_{12} (f_1 \alpha g_1) \otimes (f_2 \sigma g_2)$                                                       $+ a_{21}(f_1 \sigma g_1) \otimes (f_2 \alpha g_2) + a_{22} (f_1 \sigma g_1) \otimes (f_2 \sigma g_2)$

$(f_1\otimes f_2) \sigma_{12} (g_1\otimes g_2) = b_{11} (f_1\alpha g_1) \otimes (f_2 \alpha g_2) + b_{12} (f_1 \alpha g_1) \otimes (f_2 \sigma g_2)$                                                       $+ b_{21}(f_1 \sigma g_1) \otimes (f_2 \alpha g_2) + b_{22} (f_1 \sigma g_1) \otimes (f_2 \sigma g_2)$

Basically we take all possible combinations of the original products, so what does this lead us? We have seem to make no progress whatsoever. However, there is hope to determine the 4+4  constants $a$ and $b$!!! Why? Because $\alpha$ respects the Leibniz identity. 

Because $\alpha$ respects Leibniz identity (invariance of the laws of nature under infinitesimal time evolution) it is basically a derivation. And the derivation of the unit element (corresponding to "no physical system") is zero. So what if we take $f_1 = g_1 = 1$? [As a side note, because $\alpha$ is distinct from $\sigma$ (otherwise we find ourselves in the trivial case discussed last time), it can be normalized to respect: $1\sigma f = f\sigma 1 = f$]

So now let's plug in $f_1 = g_1 = 1$ first in the $\alpha_{12}$ equation above. On the left hand side we get:

$(1\otimes f_2) \alpha_{12} (1\otimes g_2) = f_2\alpha g_2$

and in the right hand side we get:

$a_{21}(f_2 \alpha g_2) + a_{22} (f_2 \sigma g_2)$

which demands $a_{21}= 1$ and $a_{22} = 0$.

We can play the same game with $f_2 = g_2 = 1$ and get $a_{12} = 1$ This cannot reveal anything about $a_{11}$ at this time, but at least we have trimmed the possible combinations.

Same game on $\sigma_{12}$ yields: $b_{21} = 0$, $b_{22} = 1$, and $b_{12} = 0$. Again nothing on $b_{11}$.

So now we have the reduced pattern:

$\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha + a_{11} \alpha \otimes \alpha$

and

$\Delta (\sigma) = \sigma\otimes \sigma +  b_{11} \alpha \otimes \alpha$

It turns out that $b_{11}$ is a free parameter which can be normalized to +1, 0 , -1 resulting in 3 composition classes: hyperbolic (hyperbolic quantum mechanics), parabolic (classical mechanics), and elliptic (quantum mechanics). But we can eliminate $a_{11}$ by applying Leibniz identity on the $\alpha_{12}$.  This is tedious and I will skip it here, but take my word on it because in the bipartite identity $\alpha_{12}$ will occur squared and by linearity of $\alpha$ it must vanish.

If we now consider only the quantum mechanics case of $b_{11} = -1$ we have:

$\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha$

and

$\Delta (\sigma) = \sigma\otimes \sigma - \alpha \otimes \alpha$

Does this remind you of something? How about:

Imaginary =  imaginary real  + real imaginary

Real          =  real real             - imaginary imaginary

This is how complex numbers start occurring naturally in quantum mechanics and this is why observables are Hermitean operators and generators are anti-Hermitean operators. The 1-to-1 map between observables and generators known as "dynamic correspondence" in literature is the 1-to-1 map between $\alpha$ and $\sigma$. It is this dynamic correspondence which is at the root of Noether theorem! Noether's theorem is baked in the quantum and classical formalism but you need to know where and how to look to uncover it.

So today we made good progress on the road to reconstruct quantum mechanics but there is quite a way to go. We do not know anything yet about the properties of the  $\alpha$ and $\sigma$ products, and without it we cannot hope to finds their concrete representation. But we'll get there. Please stay tuned.

Cooking quantum mechanics:

One product, two products?

As I was posting this a large terror attack was unfolding in Paris. I stand in  

solidarity with the victims of this barbarity. Je suis Parisienne.

Coming back to physics, I will now start a series of posts where I will rigorously derive quantum mechanics from physical principles. To prevent "abstraction indigestion", I will chop off this proof into easily manageable segments and today I want to discuss the minimum number of ingredients needed to cook the quantum soup.

The first physical principle needed is that of the invariance of the laws of nature under time evolution. This is a no-brainer, otherwise the search for physical laws would make no sense. But what can we get out of this? If there are algebraic operations which make sense in nature (and yes, there are plenty), those operations commute with time translation. Let us denote such operation with a generic symbol *, and let us use uppercase letters for the objects on which * operates. If we call T(A) the time translated version of the abstract object A, invariance under time evolution demands:

$T(A*B) = T(A)*T(B)$

So what? We did not get too far, unless... Unless we can do something more. Who says T has to be large? We can take as small time steps as we want. Infinitesimal steps. In other words we linearize $T$ into the identity and a small correction:

$T= I + \epsilon D$

So what do we get?

$(I + \epsilon D)(A*B) = ((I +\epsilon D)A)*((I+\epsilon D)B)$

which to first order yields the trivial $A*B=A*B$ but to first order in $\epsilon$ we get:

$D(A*B) = D(A)*B + A*D(B)$

Recognize this? It is the chain rule of differentiation. But we can do more. We can transform $D$ into a product which we will call $\alpha$:

$A D B  = A\alpha B$

To put this abstraction in perspective, $\alpha$ will later become either the Poisson bracket or the commutator: those are concrete representation of the abstract product.

What we can do now is to rewrite the chain rule from above as the Leibniz identity:

$A\alpha (B*C) = (A\alpha B)*C + B*(A\alpha C)$

We know that one of the products * is $\alpha$, so the simplest case possible is that all there is in nature is this one and only product $\alpha$. What kind of world would we get is there is only one product possible in nature?

What we can do now is to apply the Leibniz identity on itself.

$A\alpha (B\alpha C) = (A\alpha B)\alpha C + B\alpha(A\alpha C)$

So we seem to be stuck, unless... Unless we can tell something about the abstract objects $A, B, C, ...$. Here is where category theory of k-algebras comes to the rescue. Now I can either go for the abstract math, or I can present the physical interpretation. Let's go the physical route and anticipate that what we ultimately want to consider are compositons of physical systems. Such compositions have a unit element: the "compose with nothing" element. Let's denote it's corresponding uppercase letter "I" and see what we obtain from Leibniz in this case:

$I\alpha (I\alpha A) = (I\alpha I) \alpha A + I \alpha (I\alpha A)$

from which we get:

$(I \alpha I) \alpha A = 0$

for any A!!! Which means $I \alpha I = 0$ or "nothing comes from nothing".

It is now time for the second physical principle: the invariance of the laws of nature under composition. What does this mean? It means the laws of nature do not change when we add additional degrees of freedom. For a silly "what if" game, how would the world would look like if the laws of nature would depend on the number of degrees of freedom? Imagine going to the airport and boarding a plane. When enough passengers board this plane increasing its internal degrees of freedom, the plane would suddenly become a bird which would fly to its destination. There it would revert back to a plane when the passengers would disembark. Sorry Harry Potter, no quantum soup for you. Come back one year.

So now consider a composite system  $1\otimes 2$. For this system we would have the bipartite product alpha: $\alpha_{1 \otimes 2}$

Now we need to construct this bipartite product just like Lego from the only ingredients we have available in this toy world: the ordinary products $\alpha$ for system 1 and 2:

$(A_1\otimes A_2 ) \alpha_{1\otimes 2} (B_1 \otimes B_2) = a (A_1 \alpha_1 B1)\otimes (A_2 \alpha_2 B_2)$

where $a$ is a generic proportionality constant. But what if we pick $A_2=B_2 = I$? In other words the second physical system is nothing.  On one hand we get:

$(A_1\otimes I ) \alpha_{1\otimes 2} (B_1 \otimes I) = a (A_1 \alpha_1 B_1)\otimes (I \alpha_2 I)$

which is zero because we showed earlier that: $I\alpha I = 0$. On the other hand:

$(A_1\otimes I ) = A_1$ and $(B_1\otimes I ) = B_1$ and we have:

 $(A_1\otimes I ) \alpha_{1\otimes 2} (B_1 \otimes I)  = A\alpha B$

So the product alpha by itself can only be trivial: $A\alpha B = 0$ for any A and B.

To get something non-trivial we need at least another product which we will call $\sigma$. For a rich quantum soup we need meat and potatoes. I mean the commutator and the Jordan product. We'll get there, please be patient. For now we only identified the soup ingredients:

• a product $\alpha$ which respects the Leibniz identity (due to invariance under time evolution)
• a second unspecified product $\sigma$

Can we consider more products? Of course, but it will turn out that we only need those ingredients. Invariance under composition will completely determine the properties of those two products. Please stay tuned for the next episode of cooking quantum mechanics.

The Martian

Today I want to stay in the realm of cosmos and talk about a light topic for a change: the very nice movie "The Martian". I saw it in 3D on an IMAX screen and this is the way to see it. However, I really needed to suspend disbelief to enjoy it and I want to talk about some blatant nonsense. So spoiler alert, if you did not see the movie and don't want me to ruin it for you do not scroll down below the pictures and please came back next week for the new (physics) post.

The plot of the movie is simple: caught in a Martian dust storm a crew of NASA astronauts are forced to leave the planet in a hurry but one of them is injured in the process and is presumed dead. However he is not dead and is forced to survive alone on Mars until the next NASA mission will arrive. He lives in a housing compound and here is the first problem. The airlock of the compound is way too big and this waists a lot of air and energy every time you go in and out. At some point in the movie the airlock malfunctions and blows up, and guess how the huge hole in the compound is repaired? With a thin plastic sheet attached with duct tape. Really? Duct tape?

Moving along. The rocket which brought the astronauts to Mars was supposed to be powered by some plutonium reactor. The plutonium core was buried next to the compound about 1 foot deep in the sand for "safe disposal" and the spot was marked with a flimsy flag which one usually finds on bicycle attachments for kids. The huge storm which stranded the main character on Mars was about to tip over the Mars departure rocket but it had no effect on the flag! And the astronaut finds this plutonium core later on and uses it as a heater for his rover. The radiation was apparently of no concern.

Now back on Earth, some sort of graduate student who sleeps in his office figures out a new trajectory for the main Mars ship to return by slingshot around the Earth. He needs to double check his computation on a supercomputer, and how does he do it? We see him on the floor in the supercomputer server room connecting his laptop with a USB cable directly to one of the servers blades. Apparently proper remote connections were yet to be discovered, and anyone can just stroll at will in server rooms. The supercomputer confirms his computations but when the final rescue is happening, the crew discoverers that they will miss the meeting point by some 32 kilometers. Some supercomputer calculation!

More blatant nonsense happens when the crew of the big spaceship attempts to slow down their ship and fix the 32 kilometer gap by blowing a hole in the front of the ship to use the escaping air as a speed break.  The person placing an improvised explosive device has to do a space walk outside the ship, and he is not tethered in any way. Any miss on gripping various parts of the ship as he flies around would have sent him to his death in outer space. But the captain of the ship who is exiting the ship to catch the stranded astronaut arriving from Mars is fully tethered, and her spacewalk suit is also equipped with small rockets for independent mobility. The perks of being the boss I presume.

And surprise: the unnecessary tether is too short. This would belong more in a Laurel and Hardy comedy than in this movie. And now for the grand finale and the icing on the cake, the stranded astronaut is cutting the finger of his spacesuit glove to use the escaping air to propel himself the missing distance. Never mind he should probably be dead from the suit decompression, he arrives quite healthy, smiling, and breathing normally in the arms of the captain of the mission: a nice romantic happy ending.

Now don't get me wrong, I do like this movie and I would enjoy seeing it again. It is just that common sense is abused too much and the movie starts resembling productions of Godzilla from the 60s where the monster is nothing but a series of poorly made still photographs of a plastic toy.