# (In)distinguishable particles

1. Aug 1, 2009

### neworder1

I have the following conceptual problem with the quantum effects of (in)distinguishability of particles:

Imagine the following experimental setup. I have two pairs of entangled electrons, A1, A2, B1, B2 (i.e. A1 is maximally entangled with A2, and so is B1 with B2). The spins of all electrons are random and I don't know them. I keep A1 and B1 and send A2, B2 to my partner.

My partner makes a measurement on his particles, say, measures the spin z-component of A2 and B2, but keeps the results to himself. So, for example, if he gets +1 on A2, he knows that A1 is now in the -1 eigenstate, since these particles were entangled, and likewise for the -1 result. Suppose that A2 and B2 now have opposite spins (if not, I simply repeat the experiment from the start).

Now imagine two experiments:
a) I collide A1 with B1, without knowing the results of my partner's measurements; since, to my knowledge, the particles are indistinguishable (I don't know anything about their spins before the collision), I get a "fermion-like" collision, i.e. interference of two possible collision scenarios; I assume that the interaction during the collision is such that no spin exchange occurs
b) I phone my partner to get the results of his measurements; now I know which of my electrons is which (beacuse I know that A1 has spin +1 and B1 has -1 or vice versa); the particles are therefore distinguishable, so the collision I'll get is the "distinguishalbe particles collsion", with no interference

The only difference between a) and b) is my telephone call, i.e. acquiring information about the identity of electrons A1 and B1. So, it seems that either my information can affect the physical collision, which is nonsense, or I don't get interference in a) (i.e. electrons are indistinguishable to me, but collide like distinguishable particles), because in principle I could know their identity.

Now, my question is: how, in this example, can you formalize the notion of "knowing in principle" or "being able, in principle, to acquire information about the system"? Maybe this can be done under the information-theoretic framework? Does it matter that the information about my particles is physically "encoded" somewhere (in this case, in my partner's measurement results) and so affects (how?) the outcome of my experiment? I know that questions like "how does the electron know whether you have the information or not?" are silly nonsense, however, I think that the notions of "information" and "in principle" here needs clarifying.

2. Aug 2, 2009

### lstellyl

I could be wrong here, as I only have a limited undergraduate knowledge of SR, but I think a and b will end in the same result.

The deterministic act is the act of measuring, not knowing...

3. Aug 2, 2009

### Hurkyl

Staff Emeritus
Given the setup, I think you do -- the (relative) state of A1 is the totally mixed state, and similarly for B1.

4. Aug 3, 2009

### neworder1

Correct. On purely computational level, gaining knowledge of my state turns my initial totally mixed state into some pure state.

However, I think I don't quite understand the physics here. There is no interaction with macroscopic devices or anything that would constitute a "measurement" (whatever that means), only my phoning the partner. Even if I decide not to phone him, I don't get the interference because my particles are still, in principle, distinguishable. Now suppose that my partner's lab is blown up, so the information about his results is irrecoverable. Are my particles still "distinguishable in principle"? If yes, what does this mean? If not, how can blowing up my partner's lab affect the behavior of my particles (after his measurements, my and his particles become disentangled)?

5. Aug 3, 2009

### Hurkyl

Staff Emeritus
If the joint A1-A2 state is maximally entangled, then the relative state of A1 alone is totally mixed.

Any experiment that involves A1 but not A2 has access only to the relative state of A1, and thus the distribution on the outcomes arises from a classical statistical mixture of the spin + and spin - states.

"Knowledge" and "measurement" are irrelevant here.

6. Aug 4, 2009

### neworder1

Sorry, I think I don't get something here.

After my partner's measurement, the joint state of the system of all four particles is either $$\vert 01 \rangle \otimes \vert 10 \rangle$$ or $$\vert 10 \rangle \otimes \vert 01 \rangle$$. If I don't know his measurement's results, I only know that with probability $$\frac{1}{2}$$ he obtained $$\vert 10 \rangle$$ and with $$\frac{1}{2}\vert 01 \rangle$$. Because the particles were pairwise entangled before the measurement, the partial density matrix of my subsystem is $$\rho_{1} = \frac{1}{2}\vert 01 \rangle \langle 01 \vert + \frac{1}{2}\vert 10 \rangle \langle 10 \vert$$.

If I get my partner's results, I know precisely the state of the whole system (agian, because of the initial entanglement - if he says he obtained $$\vert 10 \rangle$$, I know that the system is in the pure state $$\vert 01 \rangle \otimes \vert 10 \rangle$$. So now my partial density matrix is $$\rho_{2} = \vert 01 \rangle \langle 01 \vert$$.

$$\rho_{1}$$ and $$\rho_{2}$$ are clearly different density operators, so give different outcome statistics. However, the only thing that happened is obtaining the knowledge about my partner's subsystem. Perhaps I misunderstood something?

7. Aug 4, 2009

### Hurkyl

Staff Emeritus
The outcome statistics of p1 are computed as follows
1. Compute the distribution of outcomes you'd get if A1 is spin up and B1 is spin down
2. Compute the distribution of outcomes you'd get if A1 is spin down and B1 is spin up
3. The distribution of outcomes for p1 are the average of the results from step 1 and 2. (Each with weight 50%)​

Because it's a mixture, there can be no interference effects between the two cases -- formally, the question you're asking is exactly the same as the classical scenario where you and your partner each have a ball, one green and one red, but you don't know who has which. (And both cases are equally probable)

It might help to pay attention to the fact that 50% of the time you try this experiment, your partner's answer will answer "I got |10>", and 50% of the time, he will answer "I got |01>".

Interference effects can only show up in an experiment that involves both sets of particles.