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If you have a system $A$ determined by the wavefunction $|\Psi(x,t)\rangle$ that is a linear combination of stationary states $|+\rangle$ and $|-\rangle$ with energies $E_+$ and $E_-$, respectively, how does the energy of system $A$ vary in time? I was given to believe that if you have a wavefunction that is a linear combination of stationary states, the expectation value of the energy of the system would vary with time. That is, I thought that as time progressed, the relative probability of measuring $E_+$ and $E_-$ would change because of the factor $C_{+/-}e^{-iE_{+/-}t/\hbar}$ that each of the stationary states would be multiplied by.

However, I have been told that the energy of system $A$ is time-invariant.

1) Why is this the case?

2) Is it true that linear combinations of stationary states with the same energies will have an expectation value for energy that is time-invariant?

3) Is it possible that certain linear combinations of stationary states with different energies will coincidentally have a time-independent expectation value for energy, and that this is simply what happened in this case?

Thank you in advance for your help.

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The average, or expected value, of the energy will always be constant in time. To see this most straightforwardly, note that the probability amplitudes of your states $C_{\pm}e^{iE_\pm t}$ have a constant magnitude, only the complex phase varies. The probabilities of finding the system in either eigenstate at any time are just $\lvert C_{\pm}e^{-iE_\pm t}\rvert^2 = \lvert C_{\pm}\rvert^2$.

Fundamentally, this is because the energy is always conserved under Hamiltonian evolution (assuming that the Hamiltonian is constant in time). Any quantity represented by an operator $A$ will be conserved if it commutes with the Hamiltonian, i.e. if $[A,H] = 0$ then $\langle A\rangle$ is a constant in time. In particular, $[H,H]=0$ trivially, and so the energy $\langle H\rangle$ is conserved. To prove this in general, just take as your initial state an arbitrary superposition of energy eigenstates $\lvert k\rangle$: $$ \lvert \psi(0)\rangle = \sum_k c_k \lvert k\rangle. $$ The time evolved state is $$ \lvert \psi(t)\rangle = \sum_k c_k e^{-iE_kt}\lvert k\rangle. $$ Thus the average energy at any time is $$ \langle \psi(t)\lvert H\rvert \psi(t)\rangle = \sum_{k,l} c_k c_l^* e^{i(E_l-E_k)t} \langle l \lvert H\lvert k\rangle = \sum_{k,l} c_k c_l^* e^{i(E_l-E_k)t} E_k \langle l \lvert k\rangle = \sum_k \lvert c_k\rvert^2 E_k.$$ You should be able to see that the same argument will hold for any operator $A$ which has the same eigenstates as the Hamiltonian, which means that $[A,H]=0$.

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    $\begingroup$ Also assuming that H is independent of time. $\endgroup$ – mr blick May 8 '15 at 2:53
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    $\begingroup$ @mrblick Quite right, thanks for the clarification. $\endgroup$ – Mark Mitchison May 8 '15 at 2:54
  • $\begingroup$ Wow, that is an elegant proof indeed! Thank you, Mark. I had forgotten about the generalization of Ehrenfest's theorem to generic operators. $\endgroup$ – NewDogOldTricks May 8 '15 at 2:56

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