Suppose there is an entangled state of two electrons, the spin part is $$| \downarrow \uparrow \rangle - | \uparrow \downarrow \rangle \tag{1} $$. If I add the spatial part of the wavefunction as two Gaussians, it should be something like $$ ( e^{- (r_1-R_a)^2- (r_2-R_b)^2} + e^{- (r_1-R_b)^2- (r_2-R_a)^2})( | \downarrow \uparrow \rangle - | \uparrow \downarrow \rangle ) \tag{2} $$.

Now I measure the spin of the electron at position $R_a$, and I get down result. The wavefunction should be $$ e^{- (r_1-R_a)^2- (r_2-R_b)^2} | \downarrow \uparrow \rangle - e^{- (r_1-R_b)^2- (r_2-R_a)^2} | \uparrow \downarrow \rangle \tag{3} $$ , which is still indistinguishable and antisymmetric. ( I cannot get just $ e^{- (r_1-R_a)^2- (r_2-R_b)^2} | \downarrow \uparrow \rangle $, which breaks the antisymmetry)

Since the entanglement is defined by anything more than simple product, both wavefunctions (2) and (3) are entangled. However, is wavefunction (3) too trivial to be called entangled? It looks antisymmetrization itself will automatically produce entanglement.


The morally correct answer is that the measurement of one spin in the EPR-entangled pair eliminates the entanglement as it picks a particular factorized basis vector for the measured spin, and the total wave function therefore has to factorize to $\psi_\text{just measured}\otimes \psi_{\rm something}$.

If you parameterize the multi-fermion states in terms of "fermion 1" and "fermion 2", you will have to antisymmetrize, so no multi-particle wave function will ever tensor factorize. (This is true even for bosons with the exception of the states where all the bosons are placed to the same one-particle state.)

However, as you say, this obstacle to factorization (in the form of the required antisymmetrization, and similarly symmetrization for bosons) is sort of "trivial". There is a technical way to support the claim that this state is really not entangled. If you write the "subsystems" not as "fermion 1" and "fermion 2" but as "region around $R_a$" and "region around $R_b$", using your notation, then you will see that the wave function(al) for the whole space is almost accurately tensor factorized to the wave functional that only depends on the fields near $R_a$ (where a spin-down electron excitation is added) and the fields around the point $R_b$ (where the spin-up electron excitation lives). $$ \Psi_{QFT} = c^\dagger_\downarrow ({\rm Gaussian}_{R_a}) \otimes c^\dagger_\uparrow ({\rm Gaussian}_{R_b}) $$ Of course, the formula above is just a suggestive way to show that the ordinary, correct state of a quantum field theory $$ c^\dagger_\downarrow ({\rm Gaussian}_{R_a}) c^\dagger_\uparrow ({\rm Gaussian}_{R_b}) |0\rangle $$ is really a simple tensor product of a sort.

For this reason, the notion of "entanglement" is usually modified for identical particles so that the unentangled states aren't just the states having the form of the simple tensor product but all states obtained as (anti)symmetrization of a tensor product.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.