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

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

---