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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?

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    $\begingroup$ 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,...." $\endgroup$ – user198207 Jun 20 '18 at 15:33
  • $\begingroup$ 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. $\endgroup$ – Jack G Jun 20 '18 at 15:37
  • $\begingroup$ Which part is confusing? $\endgroup$ – Grayscale Jun 20 '18 at 16:07
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    $\begingroup$ Hi @Grayscale, it just took a little while to get my head around the wording. The response by Jahan Claes has clarified this. Thanks! $\endgroup$ – Jack G Jun 20 '18 at 18:23
  • $\begingroup$ 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 $\endgroup$ – user198207 Jun 20 '18 at 19:05
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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 $a^\dagger a$.

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  • $\begingroup$ Thanks for the additional proof! This has cleared things up. $\endgroup$ – Jack G Jun 20 '18 at 18:24

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