yes I have. I think I understand
If I write
H=\hbar\omega(a^\dagger a\ +1/2) \
or \ H= \hbar\omega(aa^\dagger\ -1/2)
And then act upon an eigenvector
|\psi_n>
, which returns
E_n = (n+1/2)\hbar\omega
|\psi_n>
I...
So would I be right in saying that this works because any constant multiple of an eigenstate must also be an eigenstate.
i.e. if the constant of proportionality is beta then this works because
H (a^\dagger \left| \psi _n \right\rangle) = (E + \hbar\omega) (a^\dagger \left| \psi...
I can't seem to find information regarding this anywhere.
I understand why when the ladder operators act upon an energy eigenstate of energy E it produces another eigenstate of energy E \mp\hbar \omega. What I don't understand is why the following is true:
\ a \left| \psi _n \right\rangle...