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

[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

 

[eljose79]

by Daniel Bes, Springer

Hello,

Thank you for your question. I understand your confusion and will try to explain the reasoning behind the statement in your text.

Firstly, let's consider the commutator [H,a^+]. From the given equations, we know that this commutator should equal 2\hbar. Now, let's expand this commutator using the definition of the commutator:

[H,a^+] = Ha^+ - a^+H

Substituting the expressions for H and a^+, we get:

[H,a^+] = (x^2+p^2)(x+ip) - (x+ip)(x^2+p^2)

= x^3 + ix^2p + px^2 + ip^2x + x^2p + ip^3 - x^3 - ix^2p - px^2 - ip^2x

= 2ip^2x

= 2\hbar

Since the commutator equals 2\hbar, we can conclude that the matrix element <i|a^+|j> must be non-zero only when the difference between the energies of the two eigenstates is 2\hbar.

Now, let's consider the statement that the text makes about sequentially ordering the eigenstates connected by a^+. This means that we can assign an integer number n to each eigenstate, starting from the ground state (n=0) and increasing by 1 for each consecutive eigenstate. This is because, as we saw earlier, the commutator [H,a^+] only has non-zero matrix elements when the difference between the energies of the two eigenstates is 2\hbar. This means that the eigenstates connected by a^+ have energies that differ by multiples of 2\hbar. Therefore, we can assign an integer number n to each eigenstate, with the ground state having n=0 and the next eigenstate having n=1, and so on.

I hope this explanation helps you understand why the text states that the matrix element <i|a^+|j> vanishes unless the difference between the energies of the two eigenstates is 2\hbar, and how this fact allows us to assign integer numbers to each eigenstate. Please let me know if you need any further clarification.