# Ansatz for wavefunction infinite square well with linear perturbation $\alpha \cdot x$

Suppose we have an infinite square well extending from $$0 and a particle in its ground state. However, the infinite square well contains a linear perturbation $$\alpha \cdot x$$. The potential is accordingly: $$V(x) = \alpha \cdot x$$ if $$0 and $$V(x)= \infty$$ otherwise. We want to use the variational method to approximate the energy of the particle. How do we know what ansatz to use for the "guess" function to use with this method? We were thinking about a gaussian, however, if you want to normalize this by integrating from zero to $$L$$, this gives us an error function.

• One idea is to let the test function be a lin. comb. of the 2 lowest modes of the unperturbed problem. Commented Dec 19, 2023 at 11:08
• Maybe thinking about en.wikipedia.org/wiki/Airy_function might also give some insight. Commented Dec 20, 2023 at 14:31

I'll set $$L=1$$ in the following.
The issue with the Gaussian is that it won't cancel at the boundaries. This will give you an infinite kinetic energy so you cannot apply the variational principle. One way to amend this is to use the method of images to make it 2-periodic with odd parity (related to the Jacobi theta function when you 1-periodise it): $$e^{-(x-\mu)^2/2\sigma^2} \to \sum_{n\in\mathbb Z}e^{-(x-\mu-2n)^2/2\sigma^2}-e^{-(-x-\mu-2n)^2/2\sigma^2}$$ It's easier to see this in the unperturbed basis: $$e^{-(x-\mu)^2/2\sigma^2} \to \sqrt\pi\sigma\sum_{n=1}^\infty e^{-k_n^2\sigma^2/2}2\sin(k_n\mu)\sin(k_nx) \\ k_n = \pi n$$ Things might be complicated analytically.
As Qmechanics proposed, you can try to write your ansatz as a combination of eigenmodes of the original system. You can even check the solution with the perturbative method with $$\alpha\to0$$.
Another way would be to use ansatz of the form (setting $$L=1$$): $$\psi = \sqrt{\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}}$$ by choosing large values of $$\alpha,\beta$$, you can get a concentrated wave function like for a gaussian. To have a finite kinetic energy, you'll still need to take $$\alpha,\beta >1$$ and if one of them is $$1$$, then $$\alpha=\beta=1$$.