How Do Raising Operators Work in Quantum Mechanics?

[Chronos000]

I don't understand the following step:

using \hat{}a*\hat{}a = (\hat{}H/\hbarw ) -1/2

<n|\hat{}a*\hat{}a|n> = n<n|n>.

my first thoughts were to use a|n> = sqrt n | n-1> but I don't think that's relevant

if you sub in a*a and separate it into two expressions I don't see what good that would do

 

[fzero]

I don't understand the following step:

using \hat{}a*\hat{}a = (\hat{}H/\hbarw ) -1/2

<n|\hat{}a*\hat{}a|n> = n<n|n>.

my first thoughts were to use a|n> = sqrt n | n-1> but I don't think that's relevant

if you sub in a*a and separate it into two expressions I don't see what good that would do

You can indeed obtain \langle n|\hat{a}^\dagger \hat{a} |n\rangle by using your formula for \hat{a} |n\rangle as well as the corresponding formula for \hat{a}^\dagger |n-1\rangle.

 

[Chronos000]

I shouldn't actually have mentioned that, as my notes use a|n> initially to get a constant out. But then they provide the above step in order to obtain the value of that constant which turns out to be sqrt n