My textbook says that for a simple harmonic oscillator the Hamiltonian can be expressed in the "energy basis" in this way:
$$\hat H=\hbar\omega\bigg(\hat a^{\dagger}\hat a + {1\over 2}\bigg).$$
I know that $\hat a^{\dagger}$ and $\hat a$ are the raising and lowering operators, and that they can be written in terms of $\hat p_x$ and $\hat x$, but how is this the "energy" basis? What does that even mean?
Best Answer
Any of our observable operators in their own eigenbasis are diagonal, where the diagonal entries are the eigenvalues.$^*$
We can see this is true. Let $|\psi_i\rangle$ be the eigenvector such that $H|\psi_i\rangle=E_i|\psi\rangle$. Then the Hamiltonian in its own eigenbasis is: $$[H]_{m,n}=\langle\psi_m|H|\psi_n\rangle=\langle\psi_m|E_n|\psi_n\rangle=E_n\langle\psi_m|\psi_n\rangle$$
Since the eigenvectors are orthonormal: $$[H]_{m,n}=\delta_{m,n}E_n$$
Which means that the Hamiltonian is diagonal in its own eigenbasis basis.
Notice how this doesn't depend on what $H$ actually is. If you want to work with your specific example (I'll leave the work to you): $$\langle\psi_m|\hbar\omega\left(a^\dagger a+\frac 12\right)|\psi_n\rangle=\delta_{m,n}\hbar\omega\left(n+\frac12\right)=\delta_{m,n}E_n$$
Therefore, the expression you give must be the Hamiltonian is it's own eigenbasis.
$^*$In treating our operators like matrices, in general an operator in some basis tells us the following information. Each column of the operator tells us how the corresponding basis vector transforms upon multiplication by that operator. Therefore, it makes sense that an operator in its own eigenbasis is diagonal, because the eigenvectors are the basis vectors, and the resulting transformation of each basis vector corresponds to just multiplying them by the corresponding eigenvalue.