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