I want to show that the following formula for the ground state $\psi_0$ of the harmonic oscillator is valid:
$$<\psi_0,\hat x^{2n}\psi_0>=\frac{(2n)!}{2^{2n}n!}(\frac{h}{m \omega})^n$$


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+1
Can someone help me with this please?
(If you want see more details of my attempt just ask.)


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, $$ \langle 0 | e^{\lambda\left( {\hat a}^\dagger + {\hat a}\right)} | 0 \rangle = \langle 0 | e^{\lambda {\hat a}^\dagger}e^{\lambda {\hat a}} | 0 \rangle e^{\frac{\lambda^2}{2}} = e^{\frac{\lambda^2}{2}} \\ \Rightarrow\;\; \; \langle 0 |\left( {\hat a}^\dagger + {\hat a}\right)^{2n+1}|0\rangle = 0 \\ \Rightarrow \;\;\; \langle 0 |\left( {\hat a}^\dagger + {\hat a}\right)^{2n}|0\rangle = \frac{(2n)!}{2^n n!} \\ \langle 0 |{\hat x}^{2n}|0\rangle \equiv \langle 0 |\left[ \sqrt{\frac{h}{2m\omega}}\left( {\hat a}^\dagger + {\hat a}\right)\right]^{2n}|0\rangle = \frac{(2n)!}{2^{2n} n!}\left( \frac{h}{m\omega} \right)^n $$

  • $\begingroup$ I like this solution $\endgroup$
    – Prahar
    Apr 2 '16 at 17:38
  • $\begingroup$ @udrv Thank you for your help. I did not think on these identities. Many thanks ! $\endgroup$
    – user326049
    Apr 2 '16 at 21:01
  • $\begingroup$ @user326049 Welcome :) $\endgroup$
    – udrv
    Apr 2 '16 at 22:10

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