# Eigenvalues of Hamiltonian in Another Basis

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?

• Calculate H|a> and H|b>. The result will tell you if |a> and |b> are indeed eigenstates of the Hamiltonian (by inspection, it seems to me they are not since H|a>$\ne$ constant |a> and the same for |b>). It looks like H swaps the states around. Please at least have a shot then ask a question. Thanks Commented Feb 5, 2022 at 21:09

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!)
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.