I have the Schrödinger equation:

$$\dfrac{-\hbar^2}{2m} \nabla^2 \Psi + V \Psi = i \hbar \dfrac{\partial{\Psi}}{\partial{t}},$$

where $m$ is the particle's mass, $V$ is the potential energy operator, and $(-\hbar^2/2m) \nabla^2$ is the kinetic energy operator ($p^2/2m$).

The state function can be expressed as the product of space-dependent and time-dependent factors, $\Psi = \psi(r) \psi(t)$. If we substitute these into the above equation and divide by $\psi w$, we obtain a function on the left that depends on $r$ and a function on the right that only depends on $t$:

$$\dfrac{-\hbar^2}{2m} \dfrac{\nabla^2 \psi}{\psi} + V = \dfrac{i \hbar}{w} \dfrac{\partial{w}}{\partial{t}}.$$

Therefore, to be valid for all $r$ and $t$, each side must equal a constant, $E$:

$$\dfrac{-\hbar^2}{2m} \dfrac{\nabla^2 \psi}{\psi} + V = \dfrac{i \hbar}{w} \dfrac{\partial{w}}{\partial{t}} = E.$$

From this, we immediately have

$$w(t) = Ce^{-i(E/\hbar)t},$$

from which we can identify that $E = \hbar \omega$, where $\omega$ is the angular frequency.

I am told that $E$ is the energy of the particle in question. As a novice, I find this difficult to come to terms with, since it is not obvious to me what the connection between the reduced Planck constant and the angular frequency of a particle is with its energy. Perhaps this is a flaw that stems from thinking about these things in "classical" terms. Even so, I brushed this aside as something that I would get used to, and decided not to ask about it.

I was then watching this Youtube video by Sabine Hossenfelder, where she said that

$$E \Rightarrow \Delta x = \dfrac{\hbar c}{E}.$$

I recognized this immediately, because it is connected to a question I recently asked about the uncertainty principle. Given this, we also have that

$$E = \dfrac{\hbar c}{\Delta x}.$$

I was then watching this other Youtube video by Sabine Hossenfelder, where she says that energies are proportional to frequencies. In other words, assuming the "frequencies" in question are angular frequencies, we have that

$$E \propto \omega,$$

which agrees with $E = \hbar \omega$ from above. But it is not clear how this agrees with

$$E = \dfrac{\hbar c}{\Delta x}.$$

How are all of these definitions of energy connected, and why would they describe the energy of particle (that is, what would the energy of a particle have to do with concepts such as the reduced Planck constant and angular frequency)? I suspect that there are some physical intuitions/understandings that physicists have that make such concepts clear to them, and that I am missing these intuitions/understandings. Or perhaps this is just one of those "that's just the reality of how quantum mechanics is, and you need to get used to it" things.


1 Answer 1


I am told that E is the energy of the particle in question. As a novice, I find this difficult to come to terms with, since it is not obvious to me what the connection between the reduced Planck constant and the angular frequency of a particle is with its energy.

It seems your try to show "Schrödinger-Eq. $\Rightarrow E=\hbar\omega$", but whats going on is "S.Eq$ \Leftarrow E=\hbar\omega$"

It's called energy there, because that was the way the Schrödinger Equation was designed in the first place:

De Broglie ($p = \hbar k $) and Planck ($E=\hbar\omega$) precede the Schrödinger Equation, which I would say are somewhat empirical observations or postulates if you like.

The Schrödinger equation is constructed in such a way, that it describes a wave with the Dispersion relation of a classical free particle $E=\frac{p^2}{2m}$ with the above mentioned relations to incorporate the actual wave quantities $\lambda$ and $\omega$.

What you have done is just going backwards again there: you went from the Schroedinger equation back to $E=\hbar\omega = \frac{\hbar^2k^2}{2m}$ which is $E=\frac{p^2}{2m}$ because this relation (Planck & De-Broglie) were used to construct the Schrödinger Equation.

the second part of the question (with the youtube video) is about a somewhat independent relationship (Heisenberg uncertainty relation is not the same as Energy-Frequency-relation) Also, the equation in your second part is for a relativistic situation (i think). The Schrödinger Equation does not describe a relativistic situation (See Energy-momentum relation I wrote above)

Here is how you can arrive at the second equation: A relativistic particle has a dispersion relation $E=\sqrt{p^2 c^2 + m^2 c^4}$ so when the momentum gets very large (like in a particle collider) you can neglect the restmass $m$ and you end up with $p\approx E/c$. Now you might have a particle coming from the left and one from the right so you can handwavy set $\Delta p = 2E/c$, put this into Heisenberg and you end up with the equation $\Delta x = \hbar c/E$

  • $\begingroup$ Thanks for the answer. Keep in mind that the $\Rightarrow$ part is not from me -- it is copied from the video. $\endgroup$ Jun 2, 2020 at 12:17
  • $\begingroup$ I haven't watched the video :P. But with that I meant that you try to conclude from the Schroedinger Equation that $E=\hbar \omega$. $\endgroup$
    – hagebutte
    Jun 2, 2020 at 12:18
  • $\begingroup$ I linked directly to the relevant times, so there's no need to watch the entire thing. The videos were mainly posted as evidence/reference for the parts that I'm referring to. $\endgroup$ Jun 2, 2020 at 12:19
  • $\begingroup$ Yes, I watched a few seconds, but there I couldn't see that she was talking about the Schrödinger equation. $\endgroup$
    – hagebutte
    Jun 2, 2020 at 12:20
  • $\begingroup$ I've read your answer a number of times, but it isn't clear to me that it's answering my question. In particular, it isn't clear which, if any, of my points it specifically addresses. It seems like it might be tangential to my actual question. $\endgroup$ Jun 2, 2020 at 12:46

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