# How can we make the energy levels of an one-dimensional infinite square well potential equispaced?

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?

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.
• $a$ 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.
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.