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I was reading through Griffiths and in there it states that if we apply the raising and lowering operators (in that order) to the nth stationary state of the quantum harmonic oscillator, we get back the original nth stationary state times an (n+1) factor. If we apply it the other way around, we get the original nth stationary state times a factor of n. Why is this? For example, if I go from the 54th stationary state to the 55th and then back down, I'll get my original wavefunction times 55.

$$\hat{a}_+\hat{a_-}\psi_n=n\psi_n,\hat{a}_-\hat{a_+}\psi_n=(n+1)\psi_n \tag{2.66}$$

Interstellar
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KneeGash
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2 Answers2

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The simplest answer is that, using the definitions of $a_\pm$, we have $$ H=\left(a_+a_-+\frac{\mathbb{1}}{2}\right)\hbar\omega \tag{1} $$ so that $$ H\psi_n=\left(n+\frac12\right)\hbar\omega\psi_n\, . $$ This holds true by construction of the $a_\pm$. It follows immediately from this that $a_+a_-\psi_n=n\psi_n$.

The rest is standard algebra of commutator. Since $[H,a_+]=a_+$, it must be that $a_+\psi_n$ is proportional to the (normalized) state $\psi_{n+1}$: $$ a_+\psi_n=\alpha_n\psi_{n+1} $$ Then use $$ \vert\alpha_n\vert^2 =\int dx \psi^*_n a_-a_+\psi_n =\int dx \psi^*_n \left(a_-a_+-a_+a_-+a_+a_-\right)\psi_n =n+1 $$ using the commutator $[a_-,a_+]$ and the action of $a_+a_-$ on $\psi_n$, so $\alpha_n=\sqrt{n+1}$. Similar manipulations yield $a_-\psi_n=\sqrt{n}\psi_{n-1}$.

Of course you can "remove" the square roots but then you loose the connection with Eq.(1), which is a large part of the motivation to introduce $a_\pm$ in the first place. Moreover, the definitions of $a_\pm$ do not depend on $n$ but only on $x$, $p$ and some constants whereas "removing" the square roots does not lead to such simple definitions in terms of $x$ and $p$.

ZeroTheHero
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The answer by @ZeroTheHero is perfectly good and gives an intuition to the why. Namely $a_-a_+$ has a factor of $n$ because it is a piece of the Hamiltonian and the energy is linear in the number of excitations.

I will show an alternative approach which is purely algebraic as it depends on the commutator. It was proposed in a comment, I will just spell it out in detail.

We have $$ [a_-, a_+] = 1\,. $$ By induction we can show $$ [a_-, a_+^m] = m a_+^{m-1} $$ Let's work through the steps

  1. Base case: $m=1$ is the commutation relation
  2. Inductive case: $$ [a_-,a_+^{m+1}] = [a_-,a_+^m] a_+ + a_+^m [a_-,a_+] = m\, a_+^m + a_+^m = (m+1)\,a_+^m\,. $$

Now that we proved this identity we can build the Hilbert space of the harmonic oscillator by applying powers of $a_+$ to the ground state $$ \psi_n = \frac{a_+^n}{\sqrt{n!}}\psi_0\,. $$ From this we get $$ \begin{aligned} a_+ a_- \,\psi_n &= \frac{a_+ a_- a_+^n}{\sqrt{n!}}\psi_0 \\ &= \frac{a_+ (a_- a_+^n - a_+^n a_-)}{\sqrt{n!}}\psi_0 \\ &= \frac{a_+ [a_-, a_+^n]}{\sqrt{n!}}\psi_0 \\ &= \frac{n\, a_+^n}{\sqrt{n!}}\psi_0 \\ &= n \,\psi_n \end{aligned} $$ The second line is due to $a_-\psi_0 = 0$.

You might also wonder why I defined the state with a $\sqrt{n!}$. I did not have to, but it's a convention to make them unit normalized $|\psi_n|^2 = 1$. This can also be proved using the commutator rules $$ \begin{aligned} |\psi_n|^2 &= \langle 0 | \frac{a_-^n a_+^n}{n!} | 0\rangle \\ &= \langle 0 | \frac{a_-^{n-1} [a_-, a_+^n]}{n!} | 0\rangle \\ &= n\, \langle 0 | \frac{a_-^{n-1} a_+^{n-1}}{n!} | 0\rangle \\ &= \langle 0 | \frac{a_-^{n-1} a_+^{n-1}}{(n-1)!} | 0\rangle \\ &= |\psi_{n-1}|^2 \end{aligned} $$ By induction this can be taken all the way down to $|\psi_0|^2$ which is unit normalized by definition.

MannyC
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