# What is the energy of a superposition of energy eigenstates?

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?

• Note that after you measure and get one energy or the other, the state collapses and now has that definite energy. I think you should be asking the general question of what the energy of a superposition is. – Ryan Unger Feb 9 '15 at 14:48
• Yea that is precisely my question! The system could have collapsed in any one of the energy states. Then how come does it have exactly the energy corresp. to that state. Also by energy conservation one could say that the system had precisely that much energy to begin with i.e it was never in the other state. – Physicsaholic Feb 9 '15 at 15:26

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.

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.

• energy is not conserved because there is no fix energy in the system to be measured. – Sofia Feb 9 '15 at 16:45
• i sort of agree, though i find that misleading. i'd say that energy is conserved for every possible consistent history. – innisfree Feb 9 '15 at 16:46
• in the frame of standard QM the wave-function is whatever we know. A constant of motion is the value of that operator whose eigenvalue is our wave-function. If there is no such operator, our wave-function admits no constant of motion. Do you want examples? Coherent light. Consistent history, or Bohm interpretation, or many-worlds (there are many others) are not what the asker asked. – Sofia Feb 9 '15 at 17:02
• i'm honestly not sure what your asking/telling me? you're right, though, that the OP didn't ask for an explanation in terms of a particular interpretation, but consistent histories is useful for this kind of question. – innisfree Feb 9 '15 at 17:10
• consistent histories is an interpretation. Let me say otherwise, in the frame of the standard quantum theory did you learn of consistent histories? I don't remember this interpretation exactly, but I know that it is an interpretation. – Sofia Feb 9 '15 at 17:23

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.

• Let me rephrase what I think OP's question is. Suppose we have some superposition oscillator, not an ensemble. We can get one of two energies by measuring the energy, with equal probabilities. Suppose we get the lower of the two. Where does the energy go that would have allowed it to be in the higher energy state? – Ryan Unger Feb 9 '15 at 15:37
• as I said the measurement procedure in quantum mechanics is of statistical nature. measuring something just once bears no information so it is meaningless. Hence why you need ensembles. – Phoenix87 Feb 9 '15 at 15:39
• are you saying that energy is (clearly) conserved if one considers correlation between the measured state and the "system" from which the state was produced? – innisfree Feb 9 '15 at 15:51
• i think the problem of conservation of energy in this case is even immaterial. Once you have created a state you might want to know which state you have created, therefore you measure its energy. Which other energy is there to compare this measure with? – Phoenix87 Feb 9 '15 at 15:54
• i prepare a superposition (i know my state is a superposition of energy eigenstates), then take a measurement of its energy. i can compare this energy to other energies I could have obtained from my measurement. – innisfree Feb 9 '15 at 16:01

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 $E_1$.