Inspired by this question, How is the Schroedinger equation a wave equation?, I started playing around with the the Schrodinger equation.

Let's consider we're working in the energy basis, and so the time dependence of our wavefunctions will appear as $$\psi(\vec{r}, t) = e^{-iEt/\hbar}f(\vec{r}).$$

From the very simple differentiation of exponentials, we can write: $$\partial_t \psi(\vec{r}, t) = -\frac{E}{\hbar}i\psi(\vec{r}, t).$$

If we now turn to the Schrodinger equation, we see a factor of $i \psi(\vec{r}, t)$ appears on the lefthand side; $$\hbar\partial_t (i\psi(\vec{r}, t)) = -\frac{\hbar^2}{2m}\nabla^2\psi (\vec{r}, t).$$ Making our substitution, this yields: $$ -\hbar\partial_t\bigg(\frac{\hbar}{E}\partial_t \psi\bigg) = -\frac{\hbar^2}{2m}\nabla^2\psi (\vec{r}, t).$$ Rearranging, this gives: $$\partial_t^2 \psi = \frac{E}{2m}\nabla^2\psi,$$ which is just the wave equation, but with the relationship $v^2 = E/2m$, which would give us a velocity, $$v = \sqrt{\frac{E}{2m}}.$$ My question is, is there any physical interpretation of this result?

Edit: Have fixed a rather significant sign error

  • $\begingroup$ The second sentence is questionable; energy eigenstates have trivial time dependence but any state can be expressed on the energy eigenbasis and these states, in general, do not have trivial time dependence. $\endgroup$ Commented May 25, 2017 at 1:19
  • $\begingroup$ This is true, though everything in the derivation has remained linear, and so the result will apply to general wavefunctions also. That would then only make $v(E)$ explicitly a function of $E$. $\endgroup$
    – CDCM
    Commented May 25, 2017 at 1:31
  • $\begingroup$ Well, if you want to measure velocity you must have an hermitian operator related to it (this is one of the postulates of Non-relativistic quantum mechanics). I would say that you have only a mathematical expression... $\endgroup$ Commented May 25, 2017 at 3:25

1 Answer 1


You made a sign mistake right at the beginning; you should have $\psi({\bf r}, t) = e^{-i E t/\hbar} \psi({\bf r})$, where $\psi({\bf r})$ is an energy eigenstate wavefunction.

If I may simplify your argument: if $\psi$ is an eigenstate of the Hamiltonian, then

$$\begin{align*}\hat{H} \psi &= i \hbar \partial_t \psi \\ \hat{H}^2 \psi &= -\hbar^2 \partial_t^2 \psi \\ E \hat{H} \psi = -\frac{E \hbar^2}{2m} \nabla^2 \psi &= -\hbar^2 \partial_t^2 \psi \\ \nabla^2 \psi - \frac{2m}{E} \partial_t^2 \psi &= 0, \end{align*}$$ which is the wave equation with $v = \sqrt{E/(2m)}$. This almost makes sense: a nonrelativistic free particle has energy $E = \frac{1}{2} m v^2$, so we should have $v = \sqrt{2E/m} = 2 \sqrt{E/(2m)}$. Why are we only getting half the velocity that we should?

The answer is that we assumed that $\hat{H}$ took the nonrelativistic form $\hat{p}^2/(2m)$. What if we instead used the expression $\hat{H} = \hat{p} c$ for the energy of a photon? Then the third line above would become $$\hat{p}^2 c^2 \psi = -\hbar^2 \partial_t^2 \psi.$$ If we assume the usual representation $\hat{p} \to -i\hbar {\bf \nabla}$ in the position basis, then the $\hbar^2$'s cancel and we get $$\nabla^2 \psi - \frac{1}{c^2} \partial_t^2 \psi = 0,$$ which the the relativistic wave equation! If we were to allow for relativistic mass, then we'd use $\hat{H}^2 = \hat{p}^2 c^2 + m^2 c^4$, and we'd end up with the Klein-Gordon equation.

So to answer your question: the physical interpretation of your result is just the velocity of a classical nonrelativistic free particle of energy $E$, except off by a factor of two. But if you were to instead use the relativistic form of the energy, you'd get the Klein-Gordon equation, which is the exact quantum description of a massless scalar relativistic field rather than nonrelativistic particle.

  • 2
    $\begingroup$ Good answer, though it should be mentioned perhaps that there is also the group velocity here of all of the eigenstates, that would probably more likely correspond to the velocity of the particle. A similar thing happens in classical electromagnetics. When one solves for modes in a cavity it's very important to use group velocity $\endgroup$
    – R. Rankin
    Commented May 25, 2017 at 5:22
  • $\begingroup$ @R.Rankin Yes, for a massive scalar field the group and phase velocities differ, and the group velocity is the physically relevant one (in fact, the phase velocity is always strictly greater than $c$). For a massless scalar field, the group and phase velocities are both equal to $c$. $\endgroup$
    – tparker
    Commented May 25, 2017 at 5:44
  • $\begingroup$ Woo thank you. That makes a lot more sense and will learn me for doing late night physics. I will edit & alter question later to not lead others astray, then will accept your answer. $\endgroup$
    – CDCM
    Commented May 25, 2017 at 8:32

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