Is the interference pattern of an entangled particle affected by the measurement of it's pair far enough away to take into account special relativity?

Question

I've been considering a theoretical setup for a quantum experiment where 2 particles are prepared with entangled spin. They are then moved far away from one another where particle B is next to Bob who has set up a double slit experimental setup where the slit it goes through is determined by its spin.

While I know that what happens next shouldn't affect the result of the experiment, I do not understand how that is the case beyond "reality just must be consistent". (Also assume that every time an interference pattern is measured the entire setup is recreated for each particle in the final distribution)

Say I am in a reference frame in which Alice measures particle A to be spin up and then Bob preforms the double slit experiment with particle B. Now if Alice learns that Bob did not measure an interference pattern, that seems to be the expected scenario since Alice would know which slit each of Bobs particles went through and would "collapse" the wave function.

However if I am in a reference frame in which Bob preforms the double slit experiment before Alice measures her particle, then shouldn't he see an interference pattern? From his point of view, isn't the particle in a superposition of states, or does interference just not work with entanglement in that way?

Thanks for any clarification you can provide.

Answer 1

a. Neither Alice nor Bob see any discernible difference in outcomes regardless of which of them measures first.  Note that it is possible to set experiments up so that Alice and Bob agree as to who is measuring first (i.e. the ordering is the same in all reference frames).

b. Further, even when comparing outcomes from Alice and Bob: there is no discernible difference regardless of which of Alice or Bob measures first.  The rule is that time ordering is not a factor in the quantum mechanical expectation value in entangled pair scenarios.

c. The idea of using a double slit test by Bob to probe whether Alice made one type of measurement or another is a good one.  (The idea being that you could use Bob's outcome to predict Alice's which-slit information before her particle arrives at her measurement apparatus.)

The problem is that entangled particles (such as photons) are not coherent sources.  They will NOT produce an interference pattern in a double slit setup.  The explanation of this is complicated, but a single slit must precede the double slit in order to get sufficient coherence to produce the classic interference pattern. That single slit "collapses" the entanglement of the relevant observable, and you get nothing useful from the subsequent measurement.

Reference paper by Zeilinger, Figure 2: http://courses.washington.edu/ega/more_papers/zeilinger.pdf

Answer 2

In a typical double slit experiment, the state of the particle passing through the slits is something like $\lvert\text{left}\rangle+\lvert\text{right}\rangle$, meaning that the interference pattern will have the form $$I(x) = \lvert\psi(x)\rvert^2 = \left\lvert\langle x|\text{left}\rangle+\langle x|\text{right}\rangle\right\rvert^2.$$ In this case, the double slit interference pattern comes from the fact that we can directly add $\langle x|\text{left}\rangle$ and $\langle x|\text{right}\rangle$, when projecting the state onto the position basis. However, if the slit the particle goes through is determined by its spin, this means that the spatial component of the particle's state becomes entangled with its spin state. In this case, the state of the particle passing through the slits will be something like $\lvert\text{left}\rangle\otimes\lvert\uparrow\rangle+\lvert\text{right}\rangle\otimes\lvert\downarrow\rangle$, meaning that if we want to compute the interference pattern, it's not sufficient to just apply $\langle x\rvert$; we also need to project the state onto a spin basis as well, giving us $$\lvert\psi_\updownarrow(x)\rvert^2 = \lvert\langle x|\text{left}\rangle\langle\updownarrow|\uparrow\rangle+\langle x|\text{right}\rangle\langle\updownarrow|\downarrow\rangle\rvert^2,$$ where I use $\langle\updownarrow\rvert$ to denote quantifying over $\{\langle\uparrow\rvert,\langle\downarrow\rvert\}$. However, since the interference pattern on the screen is really only showing us the position measurement, we can add the patterns for the two spins separately, giving us $$I(x) = \lvert\psi_\uparrow(x)\rvert^2+\lvert\psi_\downarrow(x)\rvert^2 = \lvert\langle x|\text{left}\rangle\rvert^2+\lvert\langle x|\text{right}\rangle\rvert^2.$$ Thus, we see that Bob will observe a sum of two single slit diffraction patterns, offset based on the slit spacing, rather than a double slit pattern, regardless of whatever Alice does. Of course, if they compare results, they will find correlations between Alice's spin measurements and which single slit diffraction pattern Bob observes. However, it should be noted that, as @Aiden pointed out in a comment, the experiment must be repeated many times for the pattern to appear, since each particle only produces a single blip on the screen.

Answer 3

You're mixing up some concepts here. But I suspect your answer is something along the lines of this: Two particles are entangled then sent away from each other. They are separated by a large distance. Experimenters with each particle have a set of experiments they can choose from to run on their particle. The results of the experiments on the two particles are expected to be correlated in various ways depending on which pairs of experiments are chosen. When the respective particles get to the experimenters they (1) make a choice about which experiment to do and (2) do the experiment.

In special relativity we say that the events (1A) and (2A) are space-like separated from (1B) and (2B) so, observers in different reference may observe these events to happen in either order.

"But how can this make sense if, due to quantum collapse, the experimenters decision (1A) affects the result of experiment (2B) (which is space-like separated)? Different observers will see the events happen in a different order!"

This is the heart of the "spooky action at a distance" paradox. Part of the answer to the paradox is this: We expect correlation between experiment results (2A) and (2B) based on what choices were made at (1A) and (1B). But, the experimenters don't know if they saw the correlated results until they travel to meet eachother to compare notes on their experimental results at event (3). At this time they re-enter eachothers' light cones and we need not worry about causation at a distance.

In a "many worlds" sort of approach we would say there is a global wavefunction (yes, that contains spacelike separated components, quantum field theory can handle this, I think) that has various "branches" with all the different types of correlations. In this case nothing really funny happens because the particles get entangled at event (0), then travel out (staying within the light cone of event (0)) where they're measured, which maybe changes the waveform and includes the experimenters in the entanglement, and then come together again at point (3). But it's all just wavefunctions evolving. There's no collapse or causation to worry about.

Answer 4

Your question is a bit unclear. I'm going to clarify it by considering each of the sub experiments individually and then combining them.

Let's start with the interference experiment. Suppose that the state for going through the left slit is $|L\rangle$ and the state for going through the right slit is $|R\rangle$. An interference experiment will usually involve producing a superposition of $|L\rangle$ and $|R\rangle$ that are then combined to give a final state that wouldn't occur if the particle goes through just $|L\rangle$ or just $|R\rangle$. This will typically take the form of something like a change in the position of interference fringes depending on whether light goes through both slits or just one slit. It would also often be the case that if a particle goes through just one slit there will be interference fringes that will be a bit different from those you would see if it went through the other slit. In any case, let's assume that we're going to do an experiment where the $|L\rangle$ and $|R\rangle$ states would lead to different detected outcomes and we'll just call those final states $|L\rangle$ and $|R\rangle$ and say they're orthogonal so they can be distinguished perfectly. And let's suppose you're measuring the observable $$X=|L\rangle\langle L|-|R\rangle\langle R|$$ which has the value $+1$ for $|L\rangle$ and $-1$ for $|R\rangle$.

Now suppose the particles you put into this experiment have a spin entangled state like so $$|\Psi(0)\rangle=\tfrac{1}{\sqrt{2}}(|\uparrow\rangle_1|\uparrow\rangle_2+|\downarrow\rangle_1|\downarrow\rangle_2)$$ You then futz with each particle so that the up spin particles go left and the down spin particles go right: $$|\Psi(1)\rangle=\tfrac{1}{\sqrt{2}}(|\uparrow\rangle_1 |L\rangle_1|\uparrow\rangle_2 |L\rangle_2+|\downarrow\rangle_1 |R\rangle_1|\downarrow\rangle_2\ |R\rangle_2)$$

The expectation value of the observable $$X_1=|L\rangle_1\langle L|_1-|R\rangle_1\langle R|_1$$ is given by $\langle\Psi(1)|X_1|\Psi(1)\rangle = 0$ and likewise for the corresponding observable on the second particle. This means for each particle you will see a mix of both outcomes.

You talk about collapse in your question. Quantum mechanical equations of motion don't include collapse, but they do help to provide an explanation of entanglement correlations

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