Why do these raising and lowering operator combinations give the following answers?

$\langle n^{(0)}\mid\eta x^4\mid n^{(0)}\rangle$

$ = \eta\left(\frac{\hbar}{2m\omega}\right)^2\langle n^{(0)}\mid\left(a^\dagger aaa^\dagger+aa^\dagger aa^\dagger+aaa^\dagger a^\dagger+a^\dagger a^\dagger aa+a^\dagger aa^\dagger a+aa^\dagger a^\dagger a\right)\mid n^{(0)}\rangle$


I was under the impression, from my textbook, that the following was true:

The left hand side of this equation is $n\mid n\rangle$ and we therefore conclude that $$a^\dagger\mid n\rangle = \sqrt{n+1}\mid n+1\rangle,$$ which should be compared with $$a\mid n\rangle = \sqrt{n}\mid n-1\rangle.$$

So how is it possible to get $(n-1)$ or $(n+2)$ with those definitions?

Essentially I'm asking for someone to show me, step by step, how to get from, for example:

$$a†aaa† → n(n+1)$$

Am I using the wrong definitions or just reading into this wrong? Because I don't see how it's possible to get an $n-1$ if you're only using $\sqrt{n}$ and $\sqrt{n+1}$.

Essentially, I am confused on what the formulae are actually asking me to do. If n is the same throughout the entire calculation, how does this work out?

  • 3
    $\begingroup$ Please do not post formulae as screenshots, but use MathJax instead. $\endgroup$
    – ACuriousMind
    Jul 24, 2017 at 8:09
  • $\begingroup$ I cover some of it in my answer to this related question $\endgroup$
    – Kyle Kanos
    Jul 24, 2017 at 11:51
  • $\begingroup$ You should check Chapter 2 of Sakurai's Modern Quantum Mechanics. It shows very clearly how the creation/annihilation operators work. $\endgroup$ Jul 24, 2017 at 16:07

1 Answer 1


When you apply the raising operator, it raises the value of $\left|n\right>$ to $\left|n+1\right>$ and multiplies by $\sqrt{n+1}$. You then apply the next operator to the new value of $\left|n+1\right>$. For example, $a^\dagger aaa^\dagger\left|n\right>=\sqrt{n+1}a^\dagger aa\left|n+1\right>=(n+1)a^\dagger a\left|n\right>=(n+1)\sqrt na^\dagger\left|n-1\right>=n(n+1)\left|n\right>$. I assume you can work out the rest.

  • $\begingroup$ Right, but in the particular example I gave above, I don't see how that works because the ⟨n(0)∣ and ∣n(0)⟩ disappear (I would assume that's because ⟨n(0)∣n(0)⟩=1). If only the eigenstate can be ∣n+1⟩ or ∣n-1⟩ but the eigenvalues can only be changed by n and n+1... How would one get n-1 or n-2? $\endgroup$
    – q-compute
    Jul 24, 2017 at 17:52
  • $\begingroup$ \left\n\right> is not an eigenstate of the raising and lowering operators. That's why they become a different state when you apply the operators. $\endgroup$ Jul 24, 2017 at 18:18

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