Is it wrong to say energy is the expectation value of Hamiltonian? Or should I say energy is the eigenvalue of Hamiltonian?

  • 3
    $\begingroup$ From a sufficiently advanced perspective the answer is very easy and exactly the same as the answer in the classical mechanics. Are you familiar with Noether's theorem? $\endgroup$ Feb 8, 2019 at 5:59
  • $\begingroup$ 1. Are you just asking a language question about whether "the energy" is used to denote the expectation value or an eigenvalue? 2. The Hamiltonian is not always equal to the energy even in classical mechanics, see e.g. physics.stackexchange.com/q/194772/50583 and its linked questions. $\endgroup$
    – ACuriousMind
    Feb 8, 2019 at 10:48

1 Answer 1


You must be a bit more explicit in your language than in the classical case. Either could be correct but I lean towards the eigenvalue description and I'll explain why.

First of all, the Hamiltonian $\hat{H}$ is something which "belongs" to system. Energy is something which belongs to a state. So for example if a state $|\psi_n\rangle$ has

$$ \hat{H}|\psi_n\rangle = E_n |\psi_n\rangle $$

We can unambiguously say "this state has energy $E_n$" because every time you measure it you will get energy $E_n$ and also the average energy of this state, $E_{avg}$ is $E_n$.

However, as was just pointed out in Superposition principle forbids quantisation? we can consider states which are superpositions of energy eigenstates such as $a|\psi_n\rangle + b|\psi_m\rangle$ which can have average energy anywhere between $E_n$ and $E_m$. One could say it is a state with this new energy, $E_{avg} = |a|^2E_n + |b|^2E_m$ but I think that would be misleading because if you do a measurement on this state you will never measure energy $E_{avg}$, you will always get either $E_n$ or $E_m$.

The language I would use is either that it is a state which is in a superposition of energy states or (more daringly) it is a state which both has energy $E_n$ and $E_m$.


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