Recent content by Godmar02

ahhhhhh you legend. Makes so much sense now, sorry for being a bit dense. Thanks so much!

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...

Any help would be greatly appreciated, I am sure I am missing something simple. Thanks

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...