9.4: Normalisation and Hermitean conjugates

If you look at the expression $\int_{-\infty}^{\infty} f(y)^* \hat{a}^{\dagger} g(y) d y$ and use the explicit form $\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}\left(y-\frac{d}{d y}\right)$, you may guess that we can use partial integration to get the operator acting on $f$,

$$\begin{aligned} \int_{-\infty}^{\infty} & f(y)^* \hat{a}^{\dagger} g(y) d y \\ = & \int_{-\infty}^{\infty} f(y)^* \frac{1}{\sqrt{2}}\left(y-\frac{d}{d y}\right) g(y) d y \\ = & \int_{-\infty}^{\infty} \frac{1}{\sqrt{2}}\left(y+\frac{d}{d y}\right) f(y)^* g(y) d y \\ = & \int_{-\infty}^{\infty}[\hat{a} f(y)]^* g(y) d y \end{aligned}$$

This is the first example of an operator that is clearly not Hermitean, but we see that $\hat{a}$ and $\hat{a}^{\dagger}$ are related by "Hermitean conjugation". We can actually use this to normalise the wave function! Let us look at

$$\begin{aligned} O_n & =\int_{-\infty}^{\infty}\left[\left(\hat{a}^{\dagger}\right)^n e^{-y^2 / 2}\right]^*\left(\hat{a}^{\dagger}\right)^n e^{-y^2 / 2} d y \\ & =\int_{-\infty}^{\infty}\left[\hat{a}\left(\hat{a}^{\dagger}\right)^n e^{-y^2 / 2}\right]^*\left(\hat{a}^{\dagger}\right)^{n-1} e^{-y^2 / 2} d y \end{aligned}$$

If we now use $\hat{a} \hat{a}^{\dagger}=\hat{a}^{\dagger} \hat{a}+\hat{1}$ repeatedly until the operator $\hat{a}$ acts on $u_0(y)$, we find

$$O_n=n O_{n-1}$$

Since $O_0=\sqrt{\pi}$, we find that

$$u_n(y)=\frac{1}{\sqrt{n!\sqrt{\pi}}}\left(\hat{a}^{+}\right)^n e^{-y^2 / 2}$$

Question: Show that this agrees with the normalisation proposed in the previous study of the harmonic oscillator!

Question: Show that the states $u_n$ for different $n$ are orthogonal, using the techniques sketched above.