Matrix Tricks in QM

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In a previous article on this site “A Gentle Introduction to Computational Quantum Mechanics” we derived a matrix form for the Hamiltonian in the time-independent Schrödinger equation. Let’s solve that Hamiltonian analytically. No Numpy needed for the special case of a particle in a box!

The standard textbook answer for the eigenenergies of a particle in a box is

\[\begin{align} E_n = \frac{n^2\pi ^2}{2L^2}.\end{align}\]

Let’s try to see if we can figure this out from the matrix form

\[\begin{align} D = -\frac{1}{2(\Delta x)^2} \begin{bmatrix} -2 & 1 & 0 & \cdots \\ 1 & -2 & 1 & \ddots \\ 0 & 1 & -2 & \ddots \\ \vdots & \ddots & \ddots & \ddots \end{bmatrix}.\end{align}\]

It turns out that this sort of matrix is called a "tridiagonal Toeplitz matrix". We can simply take the eigenvalues from the literature; they are

\[\begin{align} E_n=a-2\sqrt{bc} \cos \bigg(\frac{n \pi }{N+1}\bigg)\end{align}\]

where $a$ are the diagonal entries, $b$ and $c$ are the offdiagonal entries. In our case, $a=1/(\Delta x)^2$ and $b=c=-0.5/(\Delta x)^2$ and $N\cdot N$ is the size of the matrix.

Presuming $N\gg 1$, we can approximate $N+1\approx N$ and take the Taylor series of $\cos$, to wit

\[\begin{align} \cos \bigg( \frac{n\pi }{N+1} \bigg) \approx \cos \bigg( \frac{n\pi }{N} \bigg) \approx 1 - \frac{n^2 \pi ^2 }{2N^2}.\end{align}\]

Now, since our matrix comes from discretizing the space in to N+1 intervals with N gridpoints, evidently $N = L/\Delta x$. Hence

\[\begin{align} 1 - \frac{n^2 \pi ^2 }{2N^2} = 1 - \frac{n^2 \pi ^2 (\Delta x)^2 }{2L^2}.\label{eq:finalcos}\end{align}\]

Plugging in $\eqref{eq:finalcos}$ and the values for $a,b$ and $c$, we get

\[\begin{align} E_n =\frac{1}{(\Delta x)^2} - \frac{1}{(\Delta x)^2}\bigg( 1-\frac{n^2 \pi ^2 (\Delta x)^2 }{2L^2}\bigg) = \frac{n^2\pi ^2 }{2L ^2}\end{align}\]

which is precisely what we would get by standard methods.

Well, you say, it’s not a very impressive trick – this is a very special form of matrix. If we added a potential, then all the entries on the diagonal wouldn’t be the same and we wouldn’t get such a pretty result.

True! There is one more exception: if we add 1 to the top right and bottom left corners – indicating periodic boundary conditions – the matrix is still solvable (it is now a circulant matrix). It is also possible to apply the perturbation theory of matrices¹ to get more analytical results, even for more unusual matrix types.

This technique is not very useful in practice, but it was a funny find anyway. If you find more quantum systems solvable in this way, I would be interested – contact me.

1. Kato’s book "Perturbation theory of linear operators" is the definitive resource for this. ↩