# How to show only multiples of unit energy are allowed in QHO by ladder operator

1. Sep 7, 2011

### HotMintea

I'd like to know how to do it without solving via Hermite polynomials, so that I can check by both methods when I solve other problems. I have tried to figure out myself but I need some help.

Let's say $H:=x^2+p^2$, and $a:= x-ip$.

So that $[a,a^+]=2\hbar$, $H a \varphi_i = (E_i - 2\hbar)\varphi_i$, and $H a^+ \varphi_i = (E_i + 2\hbar)\varphi_i$, where $\varphi_i$ is an eigenstate.

Further, $\langle i|[H,a^+]|j\rangle=(E_i-E_j)\langle i|a^+|j\rangle=2\hbar\langle i|a^+|j\rangle$ .

My text says "the matrix element $\langle i|a^+|j\rangle$ vanishes, unless the difference $E_i-E_j$ between the energies of the two eigenstates is the constant [ $2\hbar$ ]."
( I have put ' [ ] ' where I changed due to simplification of H. )

I understand the statement. My question is: How does it suggest that there won't be any energy eigenvalues between those multiples? ( Or is there any other way to show it? ) Note that the text doesn't explicitly say that, but it is followed by

"This fact implies that we may sequentially order the eigenstates connected by $a^+$, the diﬀerence between two consecutive energies being [ $2\hbar$ ]. Another consequence is that we may assign an integer number n to each eigenstate."

Any help would be appreciated.

Source: p.35, Quantum Mehcanics: A Modern and Concise Introductory Course

Last edited: Sep 7, 2011