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.


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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