# Eigenfunctions of Hamiltonian (question about the book “Quantum Field Theory for the Gifted Amateur”)

In the book "Quantum Field Theory for the Gifted Amateur" by Blundell and Lancaster, (page 21) the Hamiltonian (when discussing the number operator) is given by

$$\hat{H} = \left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)\hbar\omega$$

which is presented as

If $\hat{a}^{\dagger}\hat{a}$ has an eigenstate $|n⟩$ with eigenvalue $n$, then $\hat{H}$ will also have an eigenstate $|n⟩$ with eigenvalue $\hbar\omega(n + \frac{1}{2})$, so that we have recovered the eigenvalues of a simple harmonic oscillator in the equation $E_n = \hbar\omega(n + \frac{1}{2})$.

I'm struggling to get my head around this sentence. Is anybody able to provide an alternative explanation to this?

• Quote from the book ..."†aˆ. If aˆ †aˆhas an eigenstate |ni with eigenvalue n, then Hˆwill also have an eigenstate |ni with eigenvalue ~ω( n + 1 2 ), so that we have recovered the eigenvalues of a simple harmonic oscillator in eqn 2.5. However, we need to prove that n takes the values 0, 1 , 2,...." – user198207 Jun 20 '18 at 15:33
• Thank you for correcting this — I miscopied the quote when I was inserting the equation (replacing "eqn 2.5"). I have updated the quote in the original question. – Jack G Jun 20 '18 at 15:37
• Which part is confusing? – Grayscale Jun 20 '18 at 16:07
• Hi @Grayscale, it just took a little while to get my head around the wording. The response by Jahan Claes has clarified this. Thanks! – Jack G Jun 20 '18 at 18:23
• No problem, thanks for asking the question, I am slogging through the same book myself, actually it's quite a good book imo. Good luck with it – user198207 Jun 20 '18 at 19:05

Say $a^\dagger a$ has an eigenstate $|n\rangle$ with eigenvalue $n$. By the definition of an eigenstate, this means that $a^\dagger a|n\rangle = n|n\rangle$. The author claims that $|n\rangle$ is ALSO an eigenstate of the Hamiltonian, $\hat H$, with eigenvalue $\hbar\omega(n+\frac{1}{2})$. Let's prove it:
$$\hat H |n\rangle = \hbar\omega(a^\dagger a+\frac{1}{2})|n\rangle= \hbar\omega a^\dagger a|n\rangle + \hbar\omega\frac{1}{2}|n\rangle = \hbar\omega n|n\rangle+\hbar\omega\frac{1}{2}|n\rangle=\hbar\omega(n+\frac{1}{2})|n\rangle$$
So to find the eigenstates of $\hat H$, you just find the eigenstates of $a^\dagger a$.