TITLE: Will the Hilbert space get expanded when measuring the entangled state?
QUESTION [1 upvotes]: If I have two electrons that are indistinguishable, as I understand, they are entangled. And the Hilbert space that describes the behavior of this pair of electrons is a subspace of the Tensor product of Hilbert spaces for each electron. So the wave function should be: $$\frac1{\sqrt2}(|0\rangle_a|1\rangle_b+|1\rangle_a|0\rangle_b)$$
Assuming I measure the state of a and get $|0\rangle_a$, the wave function would collapse to $|0\rangle_a|1\rangle_b$, so now the Hilbert space that describes the behavior of this pair of electrons becomes into the full space of the Tensor product of Hilbert spaces for each electron.
This seems different from the single-particle case, where the wave function only collapses to the eigenvector of its Hilbert space, there won't be any changes on the Space itself.
My question is: When measuring one particle of the entangled pairs, will the Hilbert space get expanded as I said above? I don't see this in the five principles of quantum mechanics.

REPLY [6 votes]: If the two particles are indistinguishable, the Hilbert space is the antisymmetrized tensor product of two 1-particle spaces, both before and after measurement.
If you want to only consider the spins, then the 1-particle Hilbert space is 2 dimensional : $\mathfrak h = \mathbb C^2$. Then, there is only one 2-particle state, which is :
$$\frac{1}{\sqrt 2}\Big(|1\rangle_a|0\rangle_b - |0\rangle_a|1\rangle_b\Big) $$
Then you cannot actually measure the 'first' particle, because they are indistinguishable. The measurement gives you "one of the two particles is in the state $|1\rangle$, and the state remains unchanged.
If you want to say, the particle $a$ is in the state $|1\rangle$", then you need both particles to be distinguishable, in which case the Hilbert space is the full tensor product both before and after measurement.

If you want to describe the situation where you have two particles in two different boxes, with two possible internal states, then you need a $4$-dimensional $1$-particle state. For example, let the state $|A1\rangle$ be "the particle is in the box $A$ with internal state $1$ and the other states of the basis be $|A0\rangle,|B1\rangle,|B0\rangle$.
Then, the initial state "there is a particle in each box and their internal states are different" is :
$$|i\rangle = \frac{1}{2}\Big(|A1\rangle|B0\rangle - |B0\rangle |A1\rangle \Big) + \frac{e^{i\varphi}}{2}\Big(|A0\rangle|B1\rangle - |B1\rangle|A0\rangle\Big)$$
"Is there a particle in the box $A$ with state $1$" is a valid measurement. The associated operator is the projector :
$$\hat P = |A1\rangle\langle A1|\otimes\mathbf 1 +\mathbf 1\otimes|A1\rangle\langle A1|$$
If you measure $|i\rangle$ and find a particle in state $1$ and in box $A$, the state after measurement is :
$$|f\rangle = \frac{1}{\sqrt 2}\Big(|A1\rangle|B0\rangle - |B0\rangle|A1\rangle\Big)$$
Here, you see that both states are antisymmetric, as is required for indistinguishable fermions. But you can also notice that the presence of the degrees of freedom given by the two boxes allow us to distinguish the two particles. Hence, you could safely take them to be distinguishable. This is often done in discussions of entanglement.
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