In this answer, a state, $\psi(t)$ is said to be stationary if

$$ \begin{equation*} |\psi(t)|^2=|\psi(0)|^2. \end{equation*} $$

That answer then concludes that a state can only be stationary if it is an eigenstate of the Hamiltonian.

But, I believe it's possible to satisfy $|\psi(t)|^2=|\psi(0)|^2$ for a state that is not a Hamiltonian eigenstate. Consider, for example, the state $|\psi(t)\rangle=\sum_j \alpha_j(0)e^{-iE_jt/\hbar}|j\rangle$. In this case,

$$ \begin{align*} |\psi(t)|^2 &= \langle\psi(t)|\psi(t)\rangle,\\ &= \left(\sum_k \alpha_k^*(0) e^{iE_kt/\hbar}\langle k|\right) \left(\sum_j \alpha_j(0) e^{-iE_jt/\hbar}|j\rangle\right),\\ &= \sum_{j,k} \alpha_k^*(0) \alpha_j(0) \exp[i(E_k-E_j)t/\hbar] \langle k|j\rangle,\\ &= \sum_{j,k} \alpha_k^*(0) \alpha_j(0) \exp[i(E_k-E_j)t/\hbar] \delta_{jk},\\ &= \sum_j \alpha_j^*(0) \alpha_j(0) \exp[i(E_j-E_j)t/\hbar],\\ &= \sum_j |\alpha_j(0)|^2,\\ &= |\psi(0)|^2. \end{align*} $$

So, if we use $|\psi(t)|^2=|\psi(0)|^2$ as the definition of a stationary state, wouldn't it also be possible for a stationary state to be a superposition of Hamiltonian eigenstates? However, this other answer says that a superposition of stationary states is not a stationary state.

Which is correct? It seems to me like this comes down to how stationary state is defined (unless, of course, I made a mistake in my proof). If so, what's the definition of a stationary state?


1 Answer 1


I think there is a muddle of notation here.

If $$ | \psi(t) \rangle $$ is a quantum state then one might define the notation $$ | \psi(t) |^2 \equiv \langle \psi(t) | \psi(t) \rangle . \tag{1} $$

On the other hand, if $\psi(x,y,z,t)$ is a wavefunction then one might define the notation $$ |\psi(t)|^2 $$ as a shorthand for $$ |\psi(x,y,z,t)|^2 . \tag{2} $$

For a normalized state, version (1) will give 1 no matter how the state evolves, while version (2) will give the modulus-squared wavefunction which can change with time.

To relate the wavefunction to the Dirac notation, use $$ \psi(x,y,z,t) = \langle (x,y,z) | \psi(t) \rangle $$ where the bra part of the bracket on the right refers to an eigenstate of position.

  • $\begingroup$ In the original Dirac's notation, the bra would be written as $< x,y,z |$, without parentheses. $\endgroup$ Mar 18 at 15:53

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