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I've been trying to work the following problem:

If a system has a time-independent Hamiltonian with spectrum $\{E_n\}$, prove that the probability of measuring the energy $E_k$ is also time-independent.

To approach this, I differentiated the amplitude $\langle E_k\lvert\psi\rangle$ with the intent of showing it was zero, winding up with:

$$\begin{align} i\hbar\frac{d}{dt}\langle E_k\lvert\psi\rangle &= i\hbar\left(\frac{d\langle E_k\rvert}{dt}\lvert\psi\rangle +\langle E_k\rvert\frac{d\lvert\psi\rangle }{dt}\right)\\ &=i\hbar\left(\left[\langle\psi \rvert\frac{d\lvert E_k\rangle}{dt}\right]^* +\langle E_k\rvert\frac{d\lvert\psi\rangle }{dt}\right)\\ &=\left[-i\hbar\langle\psi \rvert\frac{d\lvert E_k\rangle}{dt}\right]^* +i\hbar\langle E_k\rvert\frac{d\lvert\psi\rangle }{dt}\\ &\stackrel{?}{=}-\langle\psi \rvert\hat H \lvert E_k\rangle^* +\langle E_k\rvert\hat H\lvert\psi\rangle\\ &=-\langle E_k\rvert\hat H\lvert\psi\rangle+\langle E_k\rvert\hat H\lvert\psi\rangle = 0\,\, \square \end{align}$$

The last equality follows because $\hat H$ is Hermitian. I have two questions on this somewhat sketchy looking derivation:

  1. I seem not to have used the fact that $\frac{\partial\hat H}{\partial t} = 0$ anywhere in this proof. Have I used it implicitly?

  2. Does the second-to-last equality follow? In particular, can I apply the TDSE to the state $\lvert E_k\rangle$?

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  • $\begingroup$ It's not obviously constant in time (without this argument or equivalent) as far as I'm aware. $\endgroup$ – theage Oct 27 '14 at 1:31
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Your derivation is not correct, since $c_k=\langle E_k|\psi\rangle$ is the coefficient of the expansion of $|\psi\rangle$ in the basis $\{|E_k\rangle\}$: $$|\psi\rangle=\sum_kc_k|E_k\rangle$$ and your derivation is implying that $\dfrac{dc_k}{dt}=0$, that is, $|\psi\rangle$ is a stationary state. But that cannot be true, since in general $$|\psi(t)\rangle=\sum_kc_k(t)|E_k\rangle=\sum_kc_k(0)e^{-iE_kt/\hbar}|E_k\rangle$$ The problem here is that you are considering the basis and the state to be time dependent, but to study the time evolution, you must have a fixed basis, not a time-evolving one. A correct derivation would be \begin{align}\frac{d}{dt}p_k(t)&=\frac{d}{dt}|\langle E_k|\psi(t)\rangle|^2\\ &=\frac{d}{dt}|c_k(t)|^2 \\ &=\frac{d}{dt}|c_k(0)e^{-iE_kt/\hbar}|^2\\ &=\frac{d}{dt}|c_k(0)|^2 \\ &=0 \end{align} So, the probability $p_k(t)$ is constant, although the coefficients $c_k(t)$ are not.

Realize that you can apply the TDSE equation to a state $|E_k\rangle$, but to study the time evolution, your basis is fixed and wou would have $$i\hbar\frac{d}{dt}|E_k(t)\rangle=H|E_k(t)\rangle \implies\\ |E_k(t)\rangle=e^{-iE_kt/\hbar}|E_k(0)\rangle$$ which is just a trivial evolution, since you ongly get a phase, but the basis is fixed at $t=0$ and does not evolve in time.

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  • $\begingroup$ Thanks a ton. I never thought to use the solution to the TDSE. $\endgroup$ – theage Oct 27 '14 at 20:47
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  1. As far as I can tell, you haven't used $\partial \hat{H}/\partial t = 0$. If you know that the statement in the question should not hold for a time-dependent Hamiltonian, that's a big clue that this proof isn't valid. Otherwise you could apply the proof to any arbitrary, time-dependent system and show that $\langle E_k\lvert\psi\rangle$ is constant even if the Hamiltonian changes in time.
  2. Yes, it is valid to apply the TDSE to $\lvert E_k\rangle$ - after all, it's a quantum state, it evolves like any other quantum state.

    $$i\hbar\frac{\partial}{\partial t}\lvert E_k\rangle = \hat{H}\lvert E_k\rangle$$

By the way, the TDSE has a partial time derivative, not a total time derivative as written in your question, but that's irrelevant to the question.

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  • $\begingroup$ Is the second statement concerning the second term? If so, am I not applying TDSE to $|\psi\rangle$, not $\langle E_k|$ (with the $E_k$ staying on the outside)? $\endgroup$ – theage Oct 27 '14 at 1:43
  • $\begingroup$ Oh, actually you're right. I guess I must just be too tired, I misread the question. (I'm going to edit this; if you want to unaccept it, that's fine.) $\endgroup$ – David Z Oct 27 '14 at 1:57

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