## Are Einstein's Boxes an argument for nonlocality?

### (an experimental proposal)

Today I want to discuss a topic from an excellent book by Jean Bricmont: Making Sense of Quantum Mechanics which presents the best arguments for the Bohmian interpretation. Although I do not agree with this approach I appreciate the clarity of the arguments and I want to present my counter argument.

On page 112 there is the following statement: "...

*the conclusion of his [Bell] argument, combined with the EPR argument is rather that there are nonlocal physical effects (and not just correlations between distant events) in Nature*".
To simplify the argument to its bare essentials, a thought experiment is presented in section 4.2: Einstein's boxes. Here is how the argument goes: start with a box B and a particle in the box, then cut the box into two half-boxes B1 and B2. If the original state is \(|B\rangle\), after cutting the state it becomes:

\(\frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)\)

Then the two halves are spatially separated and one box is opened. Of course the expected thing happens: the particle is always found in one of the half-boxes. Now suppose we find the particle in B2. Here is the dilemma: either there is action at a distance in nature (opening B1 changes the situation at B2), or the particle was in B2 all along and quantum mechanics is incomplete because \(\frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)\) does not describe what is going on.

**My take on this is that the dilemma is incorrect. Splitting the box amounts to a measurement regardless if you look inside the boxes or not and the particle will be in either B1 or B2.**

Here is an experimental proposal to prove that after cutting the box the state is not \(\frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)\):

split the box and connect the two halves to two arms of a Mach-Zehnder interferometer (bypassing the first beam splitter).

**Do you get interference or not? I say you will not get any interference because by weighing the boxes before releasing the particle inside the interferometer gives you the which way information.**

If we do not physically split the box, then indeed \(|B\rangle = \frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)\), but if we do physically split it \(|B\rangle \neq \frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)\). There is a hidden assumption in Einstein's boxes argument: realism which demands non-contextuality.

**Nature and quantum mechanics is contextual: w****hen we do introduce the divider the experimental context changes.**
Bohmian's supporters will argue that always \(|B\rangle = \frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)\). There is a simple way to convince me I am wrong: do the experiment above and show you can tune the M-Z interferometer in such a way that there is destructive interference preventing the particle to exit at one detector.

Florin,

ReplyDeleteI think that your point about weighing the boxes is irrelevant. After all the boxes are not required. It is enough that you have a source that could send a particle with 0.5 probability along two directions to Alice and Bob. If Alice detects the particle it means that either the particle was on that path all along, or that her detection changes the physics at Bob's location from a probability 0.5 to 0.

This experiment is also similar with the EPR-Bohm one (the one we discussed at length in the past) involving spin measurements along the same axis.

The dilemma remains and marks the death of non-realism as a serious scientific option.

Andrei

Andrei, you are missing the point. Physics is an experimental science. Prove experimentally that the state after the box cut it is what it is claimed to be. Do that and I concede my argument. The rest is speculation.

DeleteFlorin,

DeleteIf you rely want the experiment be done with boxes, fine. Instead of using two independent boxes that can be independently moved and weighed use a single, rigid box containing a mobile barrier in the middle. Closing the barrier should give you the state you want. In order to get the spatial separation just use a thick barrier.

Andrei

Florin,

DeleteI would also insist on a better version of this experiment that does not use boxes but has the same physical implications. Place a half-silvered mirror between Paris and London and send a photon towards it. If you detect the photon in London you know it will not be detected in Paris and vice-versa. Unlike the "boxes" experiment this is easily doable in practice and it has been done.

Andrei

You're right that those dilemmas etc. are wrong but all your comments are wrong as well. Dividing a box to two surely doesn't make any measurement automatic. A measurement is a process when an observer actually learns the value of a physical observable. If there's no such learning, there's no measurement.

ReplyDeleteDividing a box to two is just a situation when the Hamiltonian for the single particle in the box is time-dependent, and a wall grows in between. The wave function will remain constant when the change of the Hamiltonian is abrupt, and then it will evolve according to the new Hamiltonian - with the potential barrier in the middle etc. None of these events necessitates any collapse of the wave function.

I wrote a blog post explaining in some detail why you're absolutely wrong and how much I am offended by crooks like you who have no clue about (even undergraduate) quantum mechanics yet pretend to be experts.

ReplyDeleteThere is a similar problem in Griffiths book for undergraduate quantum mechanics. There an infinite square well of length a that is suddenly reduced in length and the problem is to find the new ground state wave function.The wall does essentially the same thing. Not only is this not a dilemma it is a homework problem

ReplyDeleteDoes an inanimate computing machine with memory, "learning" the value of a physical observable by recording a definite value into its memory, count as an "observer"?

ReplyDeleteIn my opinion yes. The role of the observer is essential, but the observer does not need to be alive to count as an observer. The observer is a witness which stores the result of measurement.

DeleteSuppose instead of splitting the box exactly in half, you split it into thirds, so that one new box is 1/3 and the other is 2/3. Would that be a measurement?

DeleteSuppose instead of splitting the box exactly in half, you split it into hundredths, so that one new box is 1/100 and the other is 99/100 the original. Would that be a measurement?

Suppose instead of splitting the box exactly in half, you split it into trillionths, so that one new box is 1/1,000,000,000,0000 and the other is the rest of the original volume. Would that be a measurement?

If so, then taking the limit of 1/x as x approaches infinity, one can perform a measurement without altering the volume of a box in any appreciable manner.

Hi Elliot,

DeleteYes that would also be a measurement. How you cut the box gives you the probability of the particle to be found in the left or the right box. Cut it very small on one side and the chance you'll get the particle there is very small, while the chance you get the particle on the other side is very large.

I don't see your reasoning. You can gain no new information and make no measurement by dividing the box. Until you observe the electron you haven't measured anything.

Deletep.s. my name is Dr. Elliot McGucken

ReplyDelete