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

Suppose we have a many-particle system described via a many-particle wavefunction that involves single-particle states $\lvert\lambda_{a}\rangle$, $\lvert\lambda_{b}\rangle$, $\lvert\lambda_{c}\rangle$. 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 $$\lvert\psi\rangle = \lvert\lambda_{a}\rangle_1 \otimes \lvert\mu_{a}\rangle_2$$ 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 $\lvert\lambda\rangle$ and $\lvert\mu\rangle$ are the eigenstates of the one-particle Hamiltonians, then the wavefunction $\lvert\psi\rangle$ is an eigenstate of the single-particle Hamiltonian in the two-particle Hilbert space since $$H(1)\lvert\psi\rangle = H_{1}\lvert\lambda_{a}\rangle_{1} \otimes \mathbb{I}_{2}\lvert\mu_{a}\rangle_{2}= \lambda_{a} \lvert\lambda_{a}\rangle_1 \otimes |\mu_{a}\rangle_2$$ 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 $$\lvert\psi\rangle = \frac{1}{\sqrt{2}}\Bigl(\lvert\lambda_{a}\rangle_{1} \otimes \lvert\lambda_{b}\rangle_2 + \lvert\lambda_{b}\rangle_{1} \otimes \lvert\lambda_{a}\rangle_2\Bigr)$$ However, this wavefunction is not an eigenstate of $H(1)$. Furthermore, $\lvert\psi(1)\rangle = \lvert\lambda_{a}\rangle_1 \otimes \lvert\lambda_{b}\rangle_2$, 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?

• @AccidentalFourierTransform for stuff like that I'd say just fix it - which I did, here. AUT, I think you left out a subscript $a$ in $\lvert \mu_a\rangle_2$ at one point, so I edited it in, but if that omission was intentional, please feel free to change it. Jan 8, 2016 at 19:10
• Yes, one can measure the energy of one "particle", what that measurement does to the entire wavefunction... that's a good question, though. I admit that I lack a good intuitive picture for that. Jan 8, 2016 at 19:16

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.

• Thanks for this answer. But I did not assume that the particles were interacting. Suppose I have two identical bosons in an external harmonic oscillator potential, so the single-particle $\lvert\lambda\rangle$ would be the usual $\lvert n\rangle$ energy eigenstates. Even if they don't interact, the two-particle wavefunction would still need to be symmetrised and would thus be a superposition of $\lvert n\rangle$ states, right? Also, what about the case of non-identical particles? Would you be able to measure the energy of one of them (as the operator description above suggests)?
– AUT
Jan 8, 2016 at 22:04
• If they are nonidentical and noninteracting then you should be able to measure the energy of one without disturbing the others. If they are identical the act of symmetrising (for bosons) or antisymmetrising (for fermions) entangles them and my answer still applies. In the Hartree-Fock procedure this entanglement is called the exchange interaction. Jan 9, 2016 at 3:38
• Generalising this argument, does this mean that if the $\lvert \lambda\rangle$ are the eigenstates of a general operator $O$, one cannot perform a measurement on just one particle of the observable corresponding to this operator in a system of many identical particles? In particular, it would not be possible to affect the state of one of a pair of entangled electrons by measuring the spin of the other?
– AUT
Jan 9, 2016 at 19:09
• I believe you meant to leave out the "not" in your last sentence. If that is the case then the answer is yes. This is the point that gave Einstein so much heartburn (EPR paradox) and led Feynman to say "anyone who thinks he understands quantum mechanics doesn't understand quantum mechanics" or words to that effect. Jan 11, 2016 at 19:28
• Actually, this was intentional. I have explained my reasoning in a new question, which can be found here. @LewisMiller
– AUT
Jan 12, 2016 at 13:28

The definition of an entangled state is a state $$\left|\psi\right>$$ which cannot be factorized.
In your case you have $$$$\left|\psi\right>=\left|A\right>_1\otimes\left|B\right>_2$$$$ This is by definition a factorized state, so it's not an entangled state and you can measure the single particle energy as you suggested, with $$H_1\otimes\mathbb{I}_2$$ or vice versa.
This wouldn't be true if, e.g., you had a state $$\left|\phi\right>$$ where $$$$\left|\phi\right>=\left|\eta_1\eta_2\right>\ne\left|\eta_1\right>\otimes\left|\eta_2\right>$$$$ How can you even define a single particle operator here?