What is the energy of a superposition of energy eigenstates?

Question

Suppose I have a system say SHO in a superposition of energy eigenstates $|{n_1}\rangle$ and $|{n_2}\rangle $ given by $|{\psi}\rangle = \frac{1}{\sqrt{2}}|{n_1}\rangle + \frac{1}{\sqrt{2}}|{n_2}\rangle $.

If I measure the energy of system there is equal probability of getting $(n_1 + \frac{1}{2})\hbar\omega$ or $(n_2 + \frac{1}{2})\hbar\omega$.

The question then arises where does rest of the energy go or come from? On one good day I measure system to be in state $|{n_1}\rangle$ while on other I measure it to be $|{n_2}\rangle$. It seems to me that it violates the energy conservation. Does this have to do with the measuring device? If yes then why do we never talk about it in formulating states?

Answer 1

First, the energy expectation value of the superposition state you have written down is

$$ \left(\frac{n_1 + n_2}{2} + \frac{1}{2}\right)\hbar\omega $$

and one might naively conclude that therefore the energy of the state lies in between the energy of its constituents.

This naive concept doesn't work, though - the "energy" of a state that is not an energy eigenstate is not well-defined, just as the spin of a state that is not a spin eigenstate is not well-defined - all you have is the expectation value, which tells you what you would get averaging over many measurements. The spin superposition $|{\uparrow}\rangle + |{\downarrow}\rangle$ hasn't got a definite spin, and it is certainly not zero, although its expectation value is.

Therefore, the question "Where does the energy come from/go?" is simply ill-posed. A state that is not an eigenstate has no well-defined property "energy".

You might ask how conservation of energy is realised here, and the answer is simple and unsatisfactory at first: Classical conservation laws are realized on the quantum level as operator laws, or, in this case, as the conservation of energy expectation value

$$ \frac{\mathrm{d}}{\mathrm{d}t}\langle \psi \vert H |{\psi}\rangle = 0$$

in the course of the usual time evolution, which, by Ehrenfest's theorem, is always true for time-independent Hamiltonians. Thus, energy is indeed conserved.

The measurement process itself constitutes an interaction with the state $\psi$, and is in particular not a unitary (time) evolution on the system of the state. There is hence no reason to demand that the energy expectation of a state after measurement be the same as before.

Answer 2

If energy is conserved, how so that a measurement of the energy of a state, such as $$ \psi = N (\phi_1 + \phi_2), $$ could result in two different energies? I think your question is easiest to tackle with the consistent histories interpretation of QM (though you'll reach similar conclusions with any other interpretation).

We must remember that energy can be exchanged between between the state $\psi$, the "subsystem" that made that state, and the "subsystem" (i.e. measuring apparatus) with which you propose to measure the state.

If we consider the whole system, state, the original process, and the measuring apparatus, energy is always conserved. Each possible (consistent) history of the state conserves energy - if the state has "more" energy upon measurement, it is because the process that made that state gave it more energy (though until measurement, the history of the state is undetermined).

In general, conservation laws, such as the conservation of energy, are obeyed by consistent histories for closed systems.

Answer 3

Suppose you prepare a state and you want to determine which state you have created. Of course a single measure tells you nothing unless you know a priori that you have created an eigenstate of the Hamiltonian. So from a practical point of view, a measurement is a repetition of the same experimental operations on a large enough ensemble of identical copies of the state you are creating. If you find that half of the times your state has energy $E_1$ and the other half of the times it has energy $E_2$, you can declare that the state you are producing is a superposition of state $\psi_1$ and state $\psi_2$, the coefficients of the superposition coming from the measured statistics.

Formally, the energy of this state is then the statistical averages of all the outcomes, which in this simple case is just $\frac{E_1 + E_2}2$, but in general is just the expectation value of the state on the Hamiltonian.

Answer 4

As you write the state $|\psi\rangle$ it seems that your system is in a superposition of $n_1$ oscillators and $n_2$ oscillators, each one of energy $\hbar \omega$. So, in the state $|\psi\rangle$ the number of oscillators in not fixed, and with it the energy is not fixed, it is not a constant of motion. In this case the question where go the other oscillators is meaningless.

The situation is analogue with the 2slit experiment. The path is not a constant of motion, i.e. if you put detectors near the slits you can find in one trial that the particle passed through one slit, and in another trial you can find that is passed through the other slit.

Energy makes no exception, it may be no constant of motion as any other variable. That doesn't mean that if the measurement gave us $E_1$ before the measurement the energy was $E_1$. No, if the energy is not a constant of motion is wasn't $E_1$.