EDIT: added (assuming $\lambda$ to be non-degenerate). Based on the specifics of the question, we don't in fact know whether this is the case, so it may be that $\left|\lambda\right>$ is not an energy eigenstate...

Consider a particle which at time $t=0$ is in a state $\left|\Psi,t=0\right> =\left|\lambda\right>$, which is an eigenstate of the operator $\hat{A}$, such that $\hat{A}\left|\lambda\right>=\lambda\left|\lambda\right>$, where $\lambda$ is the eigenvalue of $\left|\lambda\right>$. Further, suppose that $[\hat{H},\hat{A}]=0$, and that $\hat{H}$ is dependent on time.

How can it be shown that for all $t$ the particle exists in state $\left|\lambda\right>$ with eigenvalue $\lambda$?

What follows is the approach I've taken so far.

$$ [\hat{H},\hat{A}]=\hat{H}\hat{A}-\hat{A}\hat{H}=0. $$ This obviously means that: $$ \hat{H}\hat{A}\left|\lambda\right>=\hat{A}\hat{H}\left|\lambda\right>=\lambda\hat{H}\left|\lambda\right>, $$ which implies that $\left|\lambda\right>$ is an eigenstate of $\hat{H}$ at $t=0$ (assuming $\lambda$ to be non-degenerate).

It is after this point that I don't know how to proceed. What I think I have to show next is that the more general state when $t>0$, $\left|\Psi,t\right>$, is always an eigenstate of $\hat{H}$, but with the general form of $\left|\Psi,t\right>$being such that: $$ \left|\Psi,t\right>=\sum_{n}C_n(t)\left|n\right>, $$ I am not sure how to go about doing this. I am thinking that I have to solve the general Schrodinger equation: $$ i\hbar\frac{\partial}{\partial t}\left|\Psi,t\right>=\hat{H}\left|\Psi,t\right>, $$ where: $$ \hat{H}=\hat{H_0}+\hat{V}(t), $$ with $\hat{H_0}$ being independent of time. Is this correct? And, if so, how can I show that all $C_n(t)=0$ except for the one in corresponding to the initial state?


1 Answer 1


My understanding of the question is that $\hat H = \hat H(t)$ is time-dependent, but $[\hat H(t), \hat A] = 0$ at all times. Then your argumentation shows that $|\lambda\rangle$ is actually an eigenstate of $\hat H(t)$ at all times $t$, only the respective eigenvalue can change: $$ \hat H(t) |\lambda\rangle = E_\lambda(t) |\lambda\rangle . $$

To find the solution $|\psi(t)\rangle$ of the Schrödinger equation (so that $|\psi(0)\rangle = |\lambda\rangle$), we can simply make the ansatz that $$ |\psi(t)\rangle = C(t) |\lambda\rangle . \tag{1} $$ Plugging that in yields $\mathrm i\hbar\, \dot C(t) = E_\lambda(t) C(t)$ which is an ordinary differential equation that can be easily solved. This shows that our ansatz was correct and (1) indeed solves the Schrödinger equation.

Another way of seeing this is that the solution of the Schrödinger equation can be generally written as $$ |\psi(t)\rangle = \mathcal T\! \mathrm e^{-\mathrm i/\hbar\, \int_0^t \hat H(\tau) \mathrm d\tau} |\psi(0)\rangle , $$ where $\mathcal T\! \mathrm e$ is the time-ordered exponential. The time-ordered exponential can be defined in the following way: $$ |\psi(t)\rangle = \lim_{N \to \infty} \mathrm e^{-\mathrm i \hat H(t_N) \Delta t / \hbar} \cdots \mathrm e^{-\mathrm i \hat H(t_0) \Delta t / \hbar} |\lambda\rangle , $$ where $\Delta t = t / N$ and $t_k = (k/N) \Delta t$. Now note that $|\lambda\rangle$ is an eigenstate of every "slice" $\mathrm e^{-\mathrm i \hat H(t_k) \Delta t / \hbar}$, and therefore an eigenstate of the whole time-evolution operator.


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