Is the total energy of a quantum system the sum of the particle energies?

Question

I really should know this off by heart (this is my field...) but I never really grasped the difference between the total wavefunction of a system and the wavefunctions of particles within it, so it only just dawned on me that perhaps the total energy of a system was simply the sum of the energies of the individual particles.

It's true, isn't it?

By 'energy' here, I really mean $\langle E \rangle$. So would the expectation value of the total energy equal the sum of the expectation values of all the component particles? Or is there some conditionality to it? I.e. in a quantum computer, if the states of two qubits are opposites, then the expectation value of the total energy would be the sum of the component energies for each possible scenario $|0\rangle|1\rangle$ and $|1\rangle|0\rangle$, multiplied by the probability of each.

Took me a long time to learn to ask the stupid questions.

[Edit: brief mathsing for two qubits:

$ \hat{H} \Psi = E \Psi $

$ \Psi = \psi_m \psi_n $

$ \psi_m = a_m |0\rangle + b_m |1\rangle \quad\quad\mathrm{(ditto}~~\psi_n \mathrm{)} $

$ \Psi = a_m a_n |0\rangle |0\rangle + a_m b_n |0\rangle |1\rangle \dots $

$ \langle E \rangle = a_m a_n E_{00} + a_m b_n E_{01} \dots $

or time-independently

$ \langle E(t) \rangle = a_m(t) a_n(t) E_{00}(t) + a_m(t) b_n(t) E_{01}(t) $ ]

Answer 1

The expectation value of the total energy of a quantum system is not always the sum of the expectation values of the energy of each particle. It's certainly not true if you can't write the total wavefunction as the product of two wavefunctions corresponding to each particle. You also need to be able to write the energy as a sum of the energies of each particle. That is, if $|\Psi\rangle {}= |\psi_1\rangle\ldots|\psi_N\rangle$ and $\hat{E} = \hat{E}_{p1} + \ldots + \hat{E}_{pN}$ then you get $\langle\hat{E}\rangle {}={} \langle\psi_1|\hat{E}_{p1}|\psi_1\rangle+\ldots+\langle\psi_N|\hat{E}_{pN}|\psi_N\rangle$. I don't think you can get that result otherwise.

In your example, however, you do something different. You are expressing your state as a linear combination on a base of eigenstates, and then get $\langle\hat{E}\rangle$ as the weighted mean of the energies of each vector in your base. And of course, that's how quantum mechanics calculates expectation values. In the same notation as the previous paragraph, if you now express your state in the base of eigenstates ($\hat{E}|\psi_i\rangle{}=E_i|\psi_i\rangle$) you have $|\Psi\rangle {}= a_1|\phi_1\rangle+\ldots+a_N|\phi_N\rangle$ and now the energy can be written as $\langle\hat{E}\rangle {}={} \langle\phi_1|E_1|\phi_1\rangle+\ldots+\langle\phi_N|E_N|\phi_N\rangle$. This result is true always, but it's very different from the previous one.

Notice that $\hat{E}_{pk}$ is the hamiltonian operator for particle $k$, while $E_i$ is the $i$-eth eigenvalue of the hamiltonian operator of the complete system.