Quantum Harmonic Oscillator Ladder operator propertie

Question

My problem is: I have to prove the next "property" of the quantum harmonic oscillator number operator:

Given

$\hat{N} = a^\dagger a$

Prove:

$\hat{N} | n \rangle = n | n \rangle$

My attempt: I already show this two properties:

$a^\dagger | n \rangle = \sqrt{n+1} | n+1 \rangle$

$a | n \rangle = \sqrt{n} | n-1 \rangle$

So, i think that i can "join" this two in order to have the one that i want, is that correct? I mean, it is possible? Or i have to try it in another way.

Can someone help me, please?

Thanks in advance.

Answer 1

Just apply the operators in the sequence they appear: \begin{align*} \hat{N} | n \rangle &= a^\dagger a | n \rangle\\ &= a^\dagger \sqrt{n} | n-1 \rangle\\ &= \sqrt{n} a^\dagger | n-1 \rangle\\ &= \sqrt{n}\sqrt{n} | n \rangle\\ &= n | n \rangle \end{align*}