In one of the questions about simple harmonic motion in quantum, it was required to find psi (x, t) then this question came to my mind,does stationary state in simple harmonic oscillator in quantum mechanics as stationary state in finite well?

  • 2
    $\begingroup$ Sorry, not clear what you are asking. Do you mean, are the states the same? $\endgroup$
    – Dan
    Jan 9, 2022 at 21:09
  • $\begingroup$ yes, that's what I mean $\endgroup$
    – A.M.PH2001
    Jan 9, 2022 at 21:13

1 Answer 1


No the states in a finite square well and a simple harmonic oscilator are not the same.

The square well is quite simple. Inisde the well it is a trig function. From the edge of the well outward it decays exponentially.

The harmonic oscilator is quite a bit more complicated, but still exactly solvable. The solutioins involve Hermite functions.

The energy levels in the square well may have a set of bound states. After that there are unbound states. The harmonic oscilator is idealized as increasing arbitrarily, so it has arbitrarily many bound states.

If you take an infinite square well to correspond to the infinite harmonic oscilator well, then the infinite square well is still trig functions. And the energy levels are given by $n$ times a constant, where $n$ is an integer greater than or equal to 1. The harmonic oscilator energy levels are given by $(n+\frac{1}{2})$ times a different constant. So the energy levels are in a different sequence.


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