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We know that the energy values for an one-dimensional infinite square well potential is given by $$E_n = \frac{n^2{\pi}^2{\hbar}^2}{2ma^2}$$ where $a$ is the width of the well. Now, as we can see that the difference in energy of two consecutive levels is given by $$E_{n+1} - E_n = (2n+1)E_1.$$

In my PhD interview, it was asked that how can we make these energy levels equispaced?

I know that the quantum harmonic oscillator energy levels are equispaced, but how can we make the levels equispaced for one-dimensional infinite square well potential? Is this related to Bohr Correspondence Principle somehow?

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They just wanted you to use the formula for the energy and adapt it very likely. This means if you know the formula for $E_n$, you can see that a cuadratic dependence on $n$ will not lead to equispaced energy levels. So which parameter could you play with so that $E_n$ does not scale cuadratically with $n$?

You might think of $m$ or $a$ but, let us say the particle you are given has a fixed mass, so the only thing remaining is $a$, you could then require that $a$ is not constant and you make it such that $a^2 \propto n$ then $$\Delta E_n = E_{n+1}-E_n = \frac{\pi^2\hbar^2}{2m}.$$ This turns out is achieved by the potential not being a square well, but a parabolic well, i.e. a harmonic oscillator.

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  • $\begingroup$ $a$ and $n$ are not related. I don't think you can make one proportional to the other, even if you allow variable width. At a fixed width, there are infinitely many possible states, each labeled by an integer $n$. If you have a variable width, the wavelength of the allowed states changes, but they are still infinitely many of them. The width at any moment is not proportional to anything. $\endgroup$
    – EM_1
    Sep 12, 2023 at 12:00
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the question is not very well defined because it is not clear what we are allowed to change. One thing I can think of is if we make the particle massless, then the energy disperssion becomes linear. We don't have $E_p \propto p^2$ but rather $E_p \propto p$. So if we have $H=\sqrt{c^2p^2 + m^2c^4} + V(x)$ and set $m=0$ then we get the same eigenstates of the standard version (sine and cosine) but the energy is $E \propto n$, making it linear and equally spaced.

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  • $\begingroup$ You are right. The question is not very well defined. I was also confused at that time. So, it seems, there are two possibilities as one marked by @ohneVal and other by you. So, these are the only two possibilities? $\endgroup$
    – Neutralino
    Nov 8, 2021 at 17:30

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