# Are entangled particles indistinguishable?

No, entangled particles need not be indistinguishable.

Consider two particles $$A$$ and $$B$$, which correspond to single-particle Hilbert spaces $$\mathcal H_A$$ and $$\mathcal H_B$$. The Hilbert space underlying the composite system is the tensor product space $$\mathcal H = \mathcal H_A \otimes \mathcal H_B$$.

Question: Can all states in $$\mathcal H$$ be written in the form $$\alpha \otimes \psi$$, where $$\alpha \in \mathcal H_A$$ and $$\psi \in \mathcal H_B$$? The answer is no - while this may be possible for some states, in general elements of $$\mathcal H$$ cannot be written in this way.

Consider for example the state $$\Psi = \alpha \otimes \psi + \beta \otimes \psi + \alpha \otimes \phi + \beta \otimes \phi$$ where $$\alpha,\beta \in \mathcal H_A$$ and $$\psi,\phi\in\mathcal H_B$$. It isn't difficult to see that this state can be "factored" $$\Psi= (\alpha + \beta) \otimes (\psi + \phi)$$

and therefore falls into the above category. However, states such as

$$\Phi = \alpha \otimes \psi + \beta \otimes \phi$$

cannot be written in this way. States such as $$\Phi$$ which cannot be factored into a single tensor product between an element of $$\mathcal H_A$$ and an element of $$\mathcal H_B$$ are called entangled.

Nowhere in the above description of entanglement did I imply or require that $$\mathcal H_A$$ and $$\mathcal H_B$$ correspond to identical particles. They need not even be copies of the same Hilbert space - we could have $$\mathcal H_A = L^2(\mathbb R^3)\otimes \mathbb C^2$$ (corresponding to a spin-1/2 particle) and $$\mathcal H_B = L^2(\mathbb R^3)$$ (corresponding to a spin-0 particle). But even if $$\mathcal H_A = \mathcal H_B$$, the particles need not be indistinguishable.

That two particles are indistinguishable means (1) that $$\mathcal H_A = \mathcal H_B$$, and (2) that every state $$\Psi \in \mathcal H$$ is mapped to $$\Psi' = e^{i\theta} \Psi$$ under particle interchange (in which $$\alpha\otimes\psi \mapsto \psi \otimes \alpha$$). In most cases, $$\theta= \{0,\pi\}$$ corresponding to bosons and fermions, respectively (though this does not always hold in 2D, where we can have more exotic possibilities).

In any case, this added constraint of indistinguishability is downstream of the notion of entanglement, which requires only that the Hilbert space be written as a tensor product of simpler Hilbert spaces.

Have you any idea on what John Rennie might have meant by the two particles become mixed up?

The two particles become "mixed up" in the sense that it is no longer meaningful to speak about the state of either particle independently of the other.

Consider two (not necessarily indistinguishable) non-interacting particles in a box of length $$L$$. One possible state for this composite system is

$$\Psi = \left(\frac{\psi_1+\psi_2}{\sqrt{2}}\right) \otimes \left(\frac{\psi_3 + 2\psi_4}{\sqrt{5}}\right)$$

where $$\psi_n$$ is the $$n^{th}$$ energy eigenstate for the single particle in a box.

In this state, measurements of the energies of the two particles are independent in the sense of statistical independence. Physically, this means that a measurement of the first particle gives me no information whatsoever about the second particle; in this sense, we can think of them as two separate particles (upon which we can perform two separate measurements) going about their business.

On the other hand, consider an entangled state

$$\Phi = \frac{1}{\sqrt{5}} \left[ \psi_1 \otimes \psi_3 + 2\psi_2 \otimes \psi_4 \right]$$

In this state, the measurements are no longer independent. A measurement of the state of the first particle constitutes a measurement of the state of the second as well; if I find the first particle to have energy $$E_1$$, then I will find the energy of the second particle to be $$E_3$$ with 100% probability.

In this sense, it is no longer meaningful to talk about the state of one particle or the other. We can only talk about the state of the system of two particles.

Incidentally, this is the origin of much confusion among laymen and non-laymen alike. People intuitively feel that if you measure the first particle by itself, then some superluminal influence must propagate out to the second, but this is wrong. In truth, there is no such thing as a measurement of the first particle alone.

Quick Review of what Exactly entanglement is:

In general we know quantum amplitudes are assigned to possible configuration states. I think most people understand what this means when you start with one object and you consider it in multiple states. But when you have two objects, the configurations of the TOTAL possible configurations increase the amount of possible states.

For instance, if I flip a coin, you have heads & tails as possible configuration states for a single coin, but if I flip 2 coins, now I have 4 possible configuration states: HH, HT, TH, and TT.

Now in the "Quantum coin" case, these possibilities are EACH assigned an INDEPENDENT probability amplitude. This is unlike the "boring case" in normal probability, in which if you have coins are independent the probabilities are:

$$P(HH) = P(\text{Coin}_1 = H)*P(\text{Coin}_2 = H)\\ P(HT) = P(\text{Coin}_1 = H)*P(\text{Coin}_2 = T)\\ P(TH) = P(\text{Coin}_1 = T)*P(\text{Coin}_2 = H)\\ P(TT) = P(\text{Coin}_1 = T)*P(\text{Coin}_2 = T)\\$$

Now in the classical case, we can say that sometimes things aren't that simple, and the coins are not independent anymore. That is, the outcome of one of the coins depends on the outcome of the other. Is there a quantum equivalent of "states that aren't independent"? Yes, that's what an entangled state is! And to figure it out, it's as simple as checking if your individual states are acting independently.

The "formal way" of identifying nonindependent states in QM are seeing that their combination state is the same as just multiplying the individual probability amplitudes.

$$(p_{1H}, p_{1T}) \otimes (p_{2H}, p_{2T})$$

If you see that your coefficients for $$c1|HH\rangle+ c2|HT\rangle +c3|TH\rangle +c4|TT\rangle$$ are the same as $$(p_{1H} \langle H | + p_{1T} \langle T |) \otimes ((p_{2H} \langle H | + p_{2T} \langle T |))$$

Then you can see that our state is not independent, and isn't as simple as some independent coin flips.

Now, I explained the concept of entanglement first because I think it will clarify some confusion here. Entanglement is basically quantum non-Independence, and it doesn't have to do at all with distinguishability or indistinguishability. Things that are distinguishable or indistinguishable can sometimes cause entanglement in different ways, but I don't think that's so fundamental to entanglement itself.

For instance, I can take an entangled pair of photons (entangled in polarization), and have one of the pairs get absorbed by an atomic system (where it goes in a different state depending on its polarization). Now I have a photon entangled to an atom. Distinguishability is not really a factor here that matters.

It's not really particles that become entangled, it's degrees of freedom that become entangled.

But in any case, it seems to me that the existing answers make this sound a lot more mysterious and complicated than it needs to be.

Are entangled particles indistinguishable?

No.

Let's say that a nucleus with spin 0 undergoes fission into two fragments which are not identical, but each of which has spin 1. (In real fission, there would also be neutrons emitted, but let's simplify by saying that doesn't happen.) Then the two angular momenta of the fragments have to add up to 0. That means they're entangled.