# Double double-slit experiment

Suppose I have a double double slit experiment. That is, I have an electron gun in the center, that shoots entangled pairs of electrons in opposite directions, one to each double slit. I tried to portray the situation in the following image: Since the electrons are entangled, determining the slit through which the electron on the right goes, fixes the slit through which the electron on the left went. From what I know of quantum mechanics, I think that placing a measuring device on the slits on the right, destroys the interference pattern on both double-slits. I find this very unintuitive, even more so than the classical EPR thought experiment, because there is a directly observable effect. You can see the interference pattern disappear on the left after placing the detectors on the right.

Am I correct? Does determining the slit through which the electron on the right went destroy the interference pattern on the left?

• Interesting- not sure whether there's an electron quantum parameter which can be entangled this way. Also see recent experiments which seem to show that a particle's quantum state can be "found" in one slit while the particle is "seen" in the other. Mar 13, 2014 at 19:01
• @CarlWitthoft "see recent experiments which seem to show that a particle's quantum state can be "found" in one slit while the particle is "seen" in the other." .... can you provide some references?
– a06e
Mar 13, 2014 at 20:03
• wish I could find it :-( . There were a few articles on this within the last 6 months. Mar 13, 2014 at 21:00

I think the experiment you are proposing is not possible in the way you want it.

Let us say we produce two photons in an electron-positron-annihilation with total momentum zero. (Since I don't see an easy way to produce entangled electrons I will talk about photons here, but I think it is not important for the argument). Those two photons are of course entangled in momentum: if one has momentum $\vec p$ the other one has momentum $-\vec p$.

But in order to make this statement you have to make a moemntum measurement on the initial state, i.e. know that the total momentum is zero with a certain $\Delta p$. But then, by means of the uncertainty relation, you only know the position where the photons were emitted with an uncertainty $\Delta x \propto (\Delta p)^{-1}$.

Now you can have two scenarios: Either your double-slit is small enough and far enough away that due to the uncertainties $\Delta p$ and $\Delta x$ you do not know through which slit your photon goes. Or you still can tell (with some certainty).

In the second case there will never be an interference pattern. So no need for entanglement to destroy it.

But in the first case, due to the uncertainty $\Delta x$, measuring the position (by determining which slit your photon takes) does not give you an answer about the entangled photons position that is certain enough to tell which slit it will go through. Therefore you will see interference on both sides.

So an EPR like measurement is not possible in the experimental setup you propose. I would assume that in general you need commuting observables, like spin and position in the Stern-Gerlach experiment, in order to measure EPR. But I didn't think that through yet.

Forget about the second photon for a while. The first photon starts in a position state which is a Gaussian distribution around $\vec x_0$ and a momentum state which is a Gaussian around $\vec p_0$. After some time $t$ its position has evolved into a Gaussian of $\mu$ times the width around $\vec x_0 + \vec p_0 t$ (mass set equal to 1) while the momentum state is now $1/\mu$ times the width around $\vec p_0$. So while your spatial superposition gets larger - and thus better to measure with a double slit - the superposition in the momentum state, in which you have entanglement, gets smaller. You don't gain anything from entanglement, since your momentum wave-function is so narrow, that you know the momentum anyways.

It is actually not important to have space and momentum for this. Just take any non-commuting observables A and B, say with eigenstates A+, A-, B+, B-, and take two states S1 and S2 that are entangled in A. So measuring S1 in A+ implies S2 in A- and vice versa. But what you want is measure if S1 is in B+ or B- and from this conclude if S2 is in B+ or B-. And since A and B do not commute, measuring B with some certainty gives you a high uncertainty on A, meaning, for knowing if S1 is in B+ or B- you completely loose the information if it is in A+ or A-. So you cannot say anything about S2. On the other hand, as long as you are still in an eigenstate of A and know what to expect for the A measurement of S2, you don't know anything about the result of the B measurement.

So in order to do an EPR experiment you need entanglement in the observable you measure or an observable that commutes with it.

Please tell me if my thoughts are wrong.

• +1 Good point. But I think that in the first case there should be only one interference pattern, not both. After determining through which slit the measured photon went, I'll still be ignorant as to the slit used by the second photon. So in this case I destroy the interference pattern of the measured photon (because I know which slit it used), but I still see the other interference pattern (because $\Delta x$ doesn't let me determine the slit of the non-measured photon from the slit of the measured photon).
– a06e
Mar 18, 2014 at 17:38
• Let me phrase it differently: you will only see an interference pattern if there is a spatial superposition of photon states of which one goes through the first slit and the other through the second. In the EPR this superposition exists directly after the entangled states are created. But in your case the spatial superposition is built up during the evolution of your photons. That way you lose the entanglement. You cannot think of it as a two-state system |slit1>, |slit2> where each photon is in some constant superposition of both. Mar 18, 2014 at 19:10
• I don't understand your argument. Why you say that the spatial superposition is built up during the evolution of the photons, and why this loses the entanglement? The fact that there's an uncertainty $\Delta x$ means that the second photon (the one you didn't measure) is in a superposition of states spreading through some spatial region on the order of $\Delta x$. If this spatial region is large enough as to include both slits, you have an interference pattern.
– a06e
Mar 18, 2014 at 20:02
• As for the photon that you do measure, it loses the entanglement with the other photon after the measurement. And you lose the interference pattern after determining the slit through which it went.
– a06e
Mar 18, 2014 at 20:08
• I edited my answer hoping that this makes it a bit clearer. I am not sure if this is the best possible explanation. I am quite certain though that you cannot determine if a measurement was made on the one double slit by looking at the interference pattern of the other, since this would allow for superluminar signaling. Mar 18, 2014 at 23:28

I addressed this question at length in this paper and this presentation. The TL;DR is that entanglement and measurement are the same physical phenomenon. When you send an entangled particle through a two-slit experiment it already has had its interference destroyed in exactly the same way and by exactly the same physical mechanism as measuring the particle would destroy the interference.

• LPT: use Latex instead of Word. Aug 17, 2017 at 18:26