Ladder Operators I want to show that the following formula for the ground state $\psi_0$ of the harmonic oscillator is valid:$$<\psi_0,\hat x^{2n}\psi_0>=\frac{(2n)!}{2^{2n}n!}(\frac{h}{m \omega})^n$$Ok
I want to use ladder operators to do this: $$\hat a_+={\frac{1}{\sqrt{2hm\omega}}}(-i\hat p+m\omega\hat x)$$ "raising operator"$$\hat a_-={\frac{1}{\sqrt{2hm\omega}}}(i\hat p+m\omega\hat x)$$  "lowering operator" I used this to express the operator $\hat x$ as follows:$$\hat x=\frac{\sqrt{2hm\omega}}{2m\omega}(\hat a_++\hat a_-)$$Then I tried to use induction over n to prove the formula, however I can´t show the step n-->n+1Can someone help me with this please? (If you want see more details of my attempt just ask.)
 A: Induction won't work very well for this problem. 
A better way is to start with a variant of the Baker-Campbell-Hausdorff formula, a.k.a the Zassenhaus formula:
$$
e^{\lambda({\hat A}+{\hat B})} = e^{\lambda {\hat A}}e^{\lambda {\hat B}}e^{-\frac{\lambda^2}{2}[ {\hat A}, {\hat B}]}\cdot \dots
$$ 
Factors not shown become $\hat I$ when $[ {\hat A}, {\hat B}] \sim {\hat I}$ and will not be needed in the following. For the present case take 
$$
\lambda({\hat A}+{\hat B}) = {\hat x} \equiv \sqrt{\frac{h}{2m\omega}}\left( {\hat a}^\dagger + {\hat a}\right)
$$
with $[{\hat a}, {\hat a}^\dagger] = {\hat I}$. This gives immediately
$$
e^{\lambda\left( {\hat a}^\dagger + {\hat a}\right)} = e^{\lambda {\hat a}^\dagger}e^{\lambda {\hat a}}e^{\frac{\lambda^2}{2}}
$$
The idea is now to take the ground state matrix element on both sides, calculate the tractable right hand side, then expand on $\lambda$ and identify coefficients of corresponding powers. So, we have successively,
$$
\langle 0 | e^{\lambda\left( {\hat a}^\dagger + {\hat a}\right)} | 0 \rangle = \langle 0 | e^{\lambda {\hat a}^\dagger}e^{\lambda {\hat a}} | 0 \rangle e^{\frac{\lambda^2}{2}} = e^{\frac{\lambda^2}{2}}
\\
\Rightarrow\;\; \; \langle 0 |\left( {\hat a}^\dagger + {\hat a}\right)^{2n+1}|0\rangle = 0
\\
\Rightarrow \;\;\; \langle 0 |\left( {\hat a}^\dagger + {\hat a}\right)^{2n}|0\rangle = \frac{(2n)!}{2^n n!}
\\
\langle 0 |{\hat x}^{2n}|0\rangle \equiv \langle 0 |\left[ \sqrt{\frac{h}{2m\omega}}\left( {\hat a}^\dagger + {\hat a}\right)\right]^{2n}|0\rangle =  \frac{(2n)!}{2^{2n} n!}\left( \frac{h}{m\omega} \right)^n
$$
