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This question is an extension of a question I asked earlier about the possibility of measuring the energy of just one particle in a system of many identical particles, which can be found here. One answer confirmed what I already suspected in my question, namely that one cannot measure the energy of a single-particle constituent of an many-particle system. I assume this is because the eigenstates of the single-particle operators are not acceptable correctly-symmetrised many-particle wavefunctions.

If we now extend the formalism of my earlier question to other observables (and thus other Hermitian operators), I am led to believe that one cannot measure an observable of just one particle in a system of many identical particles. Is this correct? In particular, consider the case of a system of two particles (non-identical or identical) in the spin-singlet state: $$\lvert\psi\rangle = \frac{1}{\sqrt{2}}(\lvert\uparrow\rangle_{1} \lvert\downarrow\rangle_{2} - \lvert\downarrow\rangle_{1} \lvert\uparrow\rangle_{2})$$. In Bohm's version of the EPR argument, a system of an electron and a positron (which are non-identical) starts out in this state. Then a measurement of $S_{z}$ is performed on the electron and it is, for example found in state $\lvert\uparrow\rangle_{1}$. The state of the two particles after the measurement is thus $\lvert\psi'\rangle = \lvert\uparrow\rangle_{1} \lvert\downarrow\rangle_{2}$, which is an acceptable state for two non-identical particles. Therefore, the state of the positron has been affected by a measurement on the electron.

However, if we now take two electrons and prepare them in the singlet state, this reasoning can no longer apply. This is because, as mentioned above, one cannot measure the spin of just one of the two identical electrons and $\lvert\psi'\rangle = \lvert\uparrow\rangle_{1} \lvert\downarrow\rangle_{2}$ is not an acceptable wavefunction for two identical particles. Does this mean that one cannot perform the EPR experiment with two identical particles?

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  • $\begingroup$ Entanglement is a statistical selection, not an interaction term. Phenomenologically one can measure the energy of individual particles of a multi-particle system just fine, as long as it doesn't interfere with the effective potentials in the system, which you haven't specified. It does not look like you even care about them... even though it's the effective potentials that make the physical difference. If they are weak, it makes sense to do measurements on individual particles. If they are not, it may or may not make sense, depending on the details. $\endgroup$ – CuriousOne Jan 12 '16 at 14:25
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Of course you can! There are several important points to consider.

  1. The (anti)symmetrization for identical particles involves only the number of the particle but not e.g. it's position. So if you have two detectors $A$ and $B$ then spin states for them are different states! So you can consider, $$\vert\psi\rangle=\frac{1}{\sqrt{2}}\left(\vert\uparrow,A\rangle_1\vert\downarrow,B\rangle_2-\vert\downarrow,B\rangle_1\vert\uparrow,A\rangle_2\right)$$ but not $$\frac{1}{\sqrt{2}}\left(\vert\uparrow,A\rangle_1\vert\downarrow,B\rangle_2-\vert\downarrow,A\rangle_1\vert\uparrow,B\rangle_2\right)$$ which is what you would expect the EPR state to be. Actual EPR state for identical particles looks like

\begin{align} \vert\chi\rangle=\frac{1}{\sqrt{2}}\Biggl[\frac{1}{\sqrt{2}}\Bigl(\vert\uparrow,A\rangle_1\vert\downarrow,B\rangle_2-\vert\downarrow,B\rangle_1\vert\uparrow,A\rangle_2\Bigr)\\ -\frac{1}{\sqrt{2}}\Bigl(\vert\uparrow,B\rangle_1\vert\downarrow,A\rangle_2-\vert\downarrow,A\rangle_1\vert\uparrow,B\rangle_2\Bigr) \Biggr] \end{align}

  1. The essense of the entanglement is not really in the state non-factorization but in the non-factorization of the expectation values i.e. if you have two commuting observables $\hat{O}_A$ and $\hat{O}_B$ their entanglement means that $\langle\chi|\hat{O}_A\hat{O}_B|\chi\rangle\neq \langle\chi|\hat{O}_A|\chi\rangle\langle\chi\hat{O}_B|\chi\rangle$

Now, for identical particles the observable should commute with any permutation operator $\mathcal{P}_\sigma$. That means that its eigenstates can be appropriately (anti)symmetrized and it conserves the (anti)symmetrization of the states it acting on. We can get such an operator by using (anti)symmetrization operator, $$\mathcal{S}=\sum_{\sigma}\mathcal{P}_\sigma,\quad\mathcal{A}=\sum_{\sigma}\epsilon(\sigma)\mathcal{P}_\sigma$$ where $\epsilon(\sigma)$ is a sign of the permutation $\sigma$. For two fermions it's simply, $$\mathcal{A}=1-\mathcal{P}$$

Then taking local (acting only on the particles in the detector $A$) single-particle operator $O_{A}$ and the projector $P_{B}$ on the states in the detector $A$ we can construct the operator, $$\hat{O}_A=\mathcal{A}\Bigl(\hat{O}_{A,1}\otimes P_{B,2}\Bigr)\mathcal{A}$$ And similarly operator $\hat{O}_B$ for the second detector. You can check that it commutes with any permutation simply because $\mathcal{P}_\sigma\mathcal{A}=\epsilon(\sigma)\mathcal{A}$.

It's eigenstates are antisymmetrized like if $\hat{O}_{A}$ is a spin operator, the $|\psi\rangle$ is its eigenstate.

You can check that then the expectation values of antisymmetrized operators $\hat{O}_A$ and $\hat{O}_B$ factorize for $|\psi\rangle$ and don't factorize for $|\chi\rangle$. So we can say that for $|\psi\rangle$ the spins of the particles in detectors $A$ and $B$ (we don't identify them as particles #1 and #2!) are not entangled (it's analog of the separable state) whereas for $|\chi\rangle$ they are (it's EPR state for identical particles)

So the point is you can observe the spin of the particle going to detector $A$, but not the spin of the first particle!

We can "distinguish" the particle at some region by using local operators at that region but actually we are measuring the (anti)symmetrized wavefunction of all particles of this type at once so it's not a real "label". So in the EPR experiment with identical particles the entanglement should be stated not in terms of the particles #1 and #2 but in terms of the observables for detectors $A$ and $B$.

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  • $\begingroup$ Sorry, the formula I gave for symmetrized observables is obviously incorrect (while everything else is fine). In short their eigenstates (i.e. the states after the observation) should be (anti)symmetrized. $|\chi\rangle$ is actually an eigenstate of the spin operator in $A$ detector. I'll try now to rewrite the wrong part. $\endgroup$ – OON Jan 13 '16 at 18:26
  • $\begingroup$ Corrected and expanded a bit $\endgroup$ – OON Jan 13 '16 at 19:57

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