Understanding the Raising and Lowering Operator: A Scientific Analysis

[VVS2000]

TL;DR Summary

I was reading up on linear harmonic oscillator and here when they define 2 new operators for solving the problem.
I still am finding it difficult how they ended up with the relation mentioned in the grey box, describing how the above mentioned operators act on given eigen state
Thanks in advance

 

[PeroK]

The proof is not immediately obvious. We start by noting that $$\hat a|n\rangle = k|n-1\rangle$$for some constant ##k##, where we take ##k## to be real and positive by convention. Note that eigenstates are defined only up to a complex phase factor, so we need this convention.

Next, using $$\hat a^{\dagger}\hat a |n\rangle = n|n\rangle$$ and the normalisation of ##|n\rangle## we see that$$\langle n|\hat a^{\dagger}\hat a |n\rangle = \langle n| n|n\rangle = n$$Finally, we evaluate the LHS a different way: $$\langle n|\hat a^{\dagger}\hat a |n\rangle = \langle n|\hat a^{\dagger}k|n-1\rangle = k\langle n-1|\hat a|n\rangle^* = k\langle n-1|k|n-1\rangle^* = k^2$$using the normalisation of ##|n-1\rangle## and the important identity: $$\langle \psi |\hat X |\phi \rangle = \langle \phi |\hat X^{\dagger} |\psi \rangle^*$$which applies for all vectors and operators. That leads to the required equation $$k^2 = n$$

 

[PeroK]

PS it would be a good exercise to do the proof for ##\hat a^{\dagger}## yourself.

 

[VVS2000]

PS it would be a good exercise to do the proof for ##\hat a^{\dagger}## yourself.

Yeah I will try it for sure,
Thanks!