Edit: I have corrected some conceptual errors in the question. I actually intend to ask about a situation in which potential changes with position.

When we need to solve Schrödinger's equation for a particle, let's say, undergoing harmonic oscillations, the first approach is usually to assume the wave function as a product of two functions each depending only on space or on time, respectively.

Solving for the time part, we get a solution which describes a particle oscillating with constant energy. So what remains is finding the spatial part of the solution to get the complete wave function.

But there is a problem in this approach. If the particle is in a potential whose value is changing with position, then as it moves, let's say, down the potential, it should gain energy. So the temporal part of the wave function should oscillate faster and faster for spatial positions down the potential. In other words, given a non constant potential (with respect to spatial coordinates), the temporal part of the wave function can't be independent of space. So, it is wrong to assume that a valid solution for the wave function can be had by making such an assumption.

For example, in Griffith's book, the in order to solve the Schroedinger's equation for a harmonic oscillator using the variable separable method, the author concludes that we need to solve the following equation:


But my argument is that this method should not even be applicable in a situation where the potential changes with position.

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    $\begingroup$ if it's a constant potential, how can it move "down" this potential? $\endgroup$ Apr 18, 2017 at 6:39
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    $\begingroup$ Constant potential means the potential energy doesn't change as the particle moves. I think you have confused this with a constant field i.e. $dV/dx$ is constant. $\endgroup$ Apr 18, 2017 at 8:07
  • $\begingroup$ The separation of variables approach leads to stationary state solutions. These are not the only possible solutions. Other solutions like those you envision may be obtained by taking linear combinations of stationary state solutions. $\endgroup$ Apr 18, 2017 at 14:27
  • $\begingroup$ Sorry! Yes, I had indeed confused potential with constant force field. I could not edit earlier because I was traveling. $\endgroup$
    – Prem
    Apr 19, 2017 at 11:43

1 Answer 1


If the particle is in a constant [I assume you mean a function of position, constant in time] potential, then as it moves, let's say, down the potential, it should gain energy.

No. What you call energy when solving the time-independent Shroedinger equation is an eigenvalue of the Hamiltonian, i.e. of the total energy. This is conserved in time.

  • $\begingroup$ Thank you! If what I call as kinetic energy is actually the total energy, then it will indeed be conserved in time. $\endgroup$
    – Prem
    Apr 19, 2017 at 11:45
  • $\begingroup$ I don't know, what do you call kinetic energy? $\endgroup$
    – Martino
    Apr 19, 2017 at 16:19
  • $\begingroup$ I was assuming that the frequency of oscillation of the wave function corresponds to the kinetic energy of the particle. I now of course realize that i have no reason to assume that this is so. (This is why i was talking about the (kinetic) energy of the particle changing as it moves down the force field.) $\endgroup$
    – Prem
    Apr 20, 2017 at 1:07

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