Eigenvalues of Hamiltonian in Another Basis

Question

I am taking a quantum mechanics class and was assigned this problem:

Among other things, I am asked to find the eigenvalues of $H$ in terms of $a$, $b$ and $\sigma$. I'm sort of lost of even how to approach this.

Since $\hat{A}$ is Hermitian, I am assuming that $|a\rangle$ and $|b\rangle$ are orthonormal and complete. After that I'm pretty lost about even how to start. I have roughly written down that

$ \hat H | \lambda_n\rangle = \lambda_n|\lambda_n\rangle$ where $\lambda$ is an eigenvalue and $|\lambda_n\rangle$ is an eigenvector. What I don't grasp at all, is how to reconcile the outer product,

$|a\rangle \langle b| + |b\rangle \langle a| $ since that result is a matrix, right?

Answer 1

I think the easiest way to solve these problems when you first encounter them is to convert them into a matrix problem, where it is easier to proceed.

Since $|a\rangle$ and $|b\rangle$ form an orthonormal basis, one way of proceeding is to represent them as the column vectors: $$ |a\rangle := \begin{bmatrix} 1 \\ 0 \end{bmatrix} \ \ \text{and} \ \ |b\rangle := \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$ You can pick any vectors you like (as long as $|a\rangle$ and $|b\rangle$ are orthonormal), so we may as well pick the above simple case.

You can look up the rules for outer products on wikipedia, what you get is: $$ |a\rangle\langle b| = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \\ |b\rangle\langle a| = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \\ $$ which means that your Hamiltonian is the matrix: $$ H = \sigma \big( |a\rangle\langle b|+|b\rangle\langle a| \big) = \begin{bmatrix} 0 & \sigma \\ \sigma & 0 \end{bmatrix} $$ This is probably going to be useful for you, because it is less abstract and I assume you've diagonalized a $2\times 2$ matrix before. You should check this yourself, but the eigenvalues turn out to be $\lambda_{\pm} = \pm \sigma$ and the (normalized) eigenvectors are $| \lambda_{\pm} \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ \pm 1 \end{bmatrix}$.

What you do at the end of the day is notice that the eigenvectors can be writen in terms of $|a\rangle := \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $|b\rangle := \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, where we get: $$ | \lambda_{\pm} \rangle = \frac{1}{\sqrt{2}}|a\rangle \pm \frac{1}{\sqrt{2}}|b\rangle $$ You should double check that these satisfy $H |\lambda_{\pm}\rangle = \lambda_{\pm} |\lambda_{\pm}\rangle$ in terms of the abstract vectors (no longer written as column vectors).

The magic of the above is that you would have found exactly the same answer, even if you used a different representation of $|a\rangle$ and $|b\rangle$ (for example, if you want to give yourself a headache, try the same calculation with the different representation $|a\rangle := \frac{1}{\sqrt{17^2 + \pi^2}} \begin{bmatrix} 17 \\ \pi \end{bmatrix}$ and $|b\rangle := \frac{1}{\sqrt{17^2 + \pi^2}} \begin{bmatrix} \pi \\ - 17 \end{bmatrix}$. You'll get the same abstract answer $| \lambda_{\pm} \rangle = \frac{1}{\sqrt{2}}|a\rangle \pm \frac{1}{\sqrt{2}}|b\rangle$ with $\lambda_{\pm} = \pm \sigma$ at the end of the day!)

Answer 2

You can write down the matrix representation of $\hat{H}$ in the given basis of $|a\rangle$ and $|b\rangle$, and then diagonalize $\hat{H}$ to find its eigenvalues and eigenvectors. That's the definition of finding eigenvalues and eigenvectors in linear algebra.