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?

  • 1
    $\begingroup$ Eigenstates of $\hat a^\dagger \hat a$ are energy eigenstates, i.e. have definite energy if $\hat H$ is the Hamiltonian. $\endgroup$ – ZeroTheHero Nov 5 '18 at 0:06
  • $\begingroup$ @ZeroTheHero What do you mean by "if $\hat H$ is the Hamiltonian"? Isn't $\hat H$ always the Hamiltonian? $\endgroup$ – matryoshka Nov 5 '18 at 0:26
  • 2
    $\begingroup$ maybe I phrased it wrong. What I meant is that $\hat H$ doesn't necessarily have the form you gave. $\endgroup$ – ZeroTheHero Nov 5 '18 at 1:17

...how is this the "energy" basis? What does that even mean?

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.


It's the energy basis because the eigenstates of $\hat H$ are the number of excited particles in a given state. The first term in your equation is also known as $n$ and represents the total number of particles in the nth state.

So, $\hat H |n\rangle = \hbar\omega\left(\hat a^\dagger\hat a+\frac12\right) |n\rangle = \hbar\omega \left(n+\frac12\right) |n\rangle$


This is a good question, because what is going on here is a loose use of terminology. If you want strict terminology, then indeed when we write an operator such as $\hat{a}^\dagger \hat{a}$ we have not adopted any particular basis but simply written down the operator. The correct use of the phrase "in the energy basis" would mean writing out the matrix elements $\langle E_n | \hat{H} | E_m\rangle$. Then you would have a matrix representing the Hamiltonian in the basis $\{ | E_n \rangle \}$. The loose terminology here is drawing our attention to the fact that by manipulating raising and lowering operators it is possible to find out much that we might like to know, such as the energy levels and the effect of other operators on energy eigenstates, without needing to discover how the energy eigenstates can be written in terms of position or some other quantity such as momentum.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.