Can we measure the energy of one of several identical particles?

Question

Suppose we have a many-particle system described via a many-particle wavefunction that involves single-particle states $lvertlambda_{a}rangle$, $lvertlambda_{b}rangle$, $lvertlambda_{c}rangle$. In the following, I'll assume a two-particle system, but I think the argument generalizes.

In the case of non-identical particles, the wavefunction does not need to have any particular symmetry, so
$$lvertpsirangle = lvertlambda_{a}rangle_1 otimes lvertmu_{a}rangle_2$$
is an acceptable state. (I'm using $lambda$ and $mu$ since the single-particle states for two non-identical particles might be different.) Now, suppose I want to measure the energy of just one of the two particles. This requires the eigenstates of the operator $H(1) = H_{1} otimes mathbb{I}_{2}$. If we assume that the $lvertlambdarangle$ and $lvertmurangle$ are the eigenstates of the one-particle Hamiltonians, then the wavefunction $lvertpsirangle$ is an eigenstate of the single-particle Hamiltonian in the two-particle Hilbert space since
$$H(1)lvertpsirangle = H_{1}lvertlambda_{a}rangle_{1} otimes mathbb{I}_{2}lvertmu_{a}rangle_{2}= lambda_{a} lvertlambda_{a}rangle_1 otimes |mu_{a}rangle_2$$
correct? This means that I can measure the energy of just one of the two non-identical particles, right?

Now to the case of two identical particles, say two bosons. The two-particle wavefunction has to be symmetric, so take for example
$$lvertpsirangle = frac{1}{sqrt{2}}Bigl(lvertlambda_{a}rangle_{1} otimes lvertlambda_{b}rangle_2 + lvertlambda_{b}rangle_{1} otimes lvertlambda_{a}rangle_2Bigr)$$
However, this wavefunction is not an eigenstate of $H(1)$. Furthermore, $lvertpsi(1)rangle = lvertlambda_{a}rangle_1 otimes lvertlambda_{b}rangle_2$, which would be an eigenstate of $H(1)$, is not a possible (correctly symmetrised) wavefunction. I know that one cannot actually attach labels to identical particles, so I guess we can't measure the energy of, say, only particle 1. But does this, quite generally, mean that one cannot possibly measure the energy of any one particle only in a many-particle system?

Lewis Miller · Answer

You cannot measure the energy of a single particle constituent of a system of N interacting (bound) particles.  It is entangled with all the other constituents and you cannot measure (disturb) one of them without changing the others.  You can measure its removal energy but this is not an energy associated with it alone.

Theoretically you can come close to assigning an energy to one constituent however.  You can formulate a Hartree-Fock problem and find an eigenvalue associated with one constituent (or at least one state).  This is a Schrodinger calculation where the single particle potential results from a self-consistent field approximation.  These HF eigenvalues are frequently compared to the experimental removal energies and usually the agreement is satisfactory for the least bound states.

Birrabenzina · Answer

The definition of an entangled state is a state $left|psiright>$ which cannot be factorized.
In your case you have
begin{equation}
left|psiright>=left|Aright>_1otimesleft|Bright>_2
end{equation}
This is by definition a factorized state, so it's not an entangled state and you can measure the single particle energy as you suggested, with $H_1otimesmathbb{I}_2$ or vice versa.
This wouldn't be true if, e.g., you had a state $left|phiright>$ where
begin{equation}
left|phiright>=left|eta_1eta_2right>neleft|eta_1right>otimesleft|eta_2right>
end{equation}
How can you even define a single particle operator here?

Can we measure the energy of one of several identical particles?

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