Suppose I have a quantum system governed by a time-independent Hamiltonian $H$. Its eigenvectors $\{|\varphi_n\rangle\}_{n\in\mathbb{N}}$ form a complete orthonormal set (or basis) for the Hilbert space $\mathscr{H}$ hosting the state $|\psi\rangle$ of the system.

If ${\displaystyle \left|\psi (t)\right\rangle }$ is a state at time $t$, then

$${\displaystyle H\left|\psi (t)\right\rangle =i\hbar {\partial \over \partial t}\left|\psi (t)\right\rangle .}$$

The eigenvectors of $H$ at a given time $t_0$ are $\{|\varphi_n\rangle\}_{n\in\mathbb{N}}$ (assuming discrete, non-degenerate spectrum). Then $$ |\varphi_n(t)\rangle = \exp\left(-\frac{i}{\hbar}E_nt\right)|\varphi_n\rangle \qquad \forall\;t>t_0 $$

Since $\{|\varphi_n\rangle\}_{n\in\mathbb{N}}$ form a complete orthonormal set (or basis) for the Hilbert space $\mathscr{H}$ any state $|\psi(t_0)\rangle \in \mathscr{H}$ at the time $t_0$ can be expressed as

$$ |\psi(t_0)\rangle= \sum_{n\in\mathbb{N}}c_n|\varphi_n\rangle \qquad \operatorname{at} \ t=t_0$$

What will the general expression for $|\psi(t)\rangle$ at time $t>t_0$ be? Well, evidently $\{|\varphi_n(t)\rangle\}_{n\in\mathbb{N}}$ is still a basis for $\mathscr{H}$, at any time $t$. Therefore

$$ |\psi(t)\rangle= \sum_{n\in\mathbb{N}}c_n(t)|\varphi_n(t)\rangle \qquad \operatorname{at} \ t>t_0$$

But actually what turns out is slightly different $$ |\psi(t)\rangle= \sum_{n\in\mathbb{N}}c_n\exp\left(-\frac{i}{\hbar}E_n t\right)|\varphi_n\rangle \qquad \operatorname{at} \ t>t_0$$ In other words $c_n(t)=c_n$. Why do the coefficients remain the same?


1 Answer 1


Write $$ \vert\Psi(t)\rangle= \sum_n c_n(t)e^{-iE_nt/\hbar}\vert\varphi_n\rangle $$ with $H\vert\varphi_n\rangle=E_n\vert\varphi_n\rangle$. Insert this into the TDSE: $$ i\hbar\sum_n \dot{c}_n(t)e^{-iE_nt/\hbar}\vert\varphi_n\rangle +i\hbar \sum_n \frac{-iE_n}{\hbar}c_n(t)e^{-iE_nt/\hbar}\vert\varphi_n\rangle = \sum_n c_n(t) E_n e^{-iE_nt/\hbar}\vert\varphi_n\rangle\, , $$ where $\dot{c}_m(t)=\frac{d}{dt}c_m(t)$ and where $H\vert\varphi_n\rangle=E_n\vert\varphi_n\rangle$ has been used.

Close with $\langle\varphi_m\vert$ and use orthogonality: $$ i\hbar\dot{c}_m(t)e^{-iE_mt/\hbar} +i\hbar \frac{-iE_m}{\hbar}c_m(t)e^{-iE_mt/\hbar} = c_m(t) E_m e^{-iE_mt/\hbar}\, . $$ The terms in $c_m(t)$ cancel out and you're left with $$ i\hbar\dot{c}_m(t)e^{-iE_mt/\hbar}=0 $$ which implies $c_m$ is constant.


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