Expectation Value of displacement to the $n$th power quantum harmonic oscillator? [closed]

Is there a closed-form expression for $$\langle x^n\rangle$$ for the ground state quantum harmonic oscillator, where $$n =$$ even integer $$>0$$? I am attempting to pursue this with rising and lowering operators but the foiling is getting out of hand.

I would like $$\langle0 | x^n |0\rangle = \underline{\qquad\qquad}\,, n={2,4,6,8,\dots}$$ Example: $$\langle0 | x^2 |0\rangle = \langle0|\left[\sqrt{\frac{\hbar}{2m\omega}}\left(a^{\dagger} + a\right)\right]^{2}| 0\rangle = \frac{1}{2\alpha},$$ where $$\alpha = m\omega/\hbar$$

closed as off-topic by DanielSank, Yashas, Jon Custer, Qmechanic♦May 3 at 6:25

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• There are closed form expressions for the WF of the harmonic oscillator, you can find them anywhere online. And as a hint the ground state is symmetric across the origin – Triatticus May 1 at 22:26
• Apologies, I edited to make the question more clear. I was looking for the expression for the average displacement^nth power – sealpancake May 1 at 23:30
• Have a look at physics.stackexchange.com/a/460206/135433 – Sunyam May 1 at 23:55
• Have you considered a solution using induction? – DanielSank May 2 at 1:57

You can get the answer without raising and lowering operators, by directly calculating the the overlap integral using the identity $$\int_{-\infty}^{\infty}dx \ x^{2n} \ e^{-\alpha x^2} = \frac{(2n-1)!!}{(2\alpha)^n} \sqrt{\frac{\pi}{\alpha}}\ ,$$ which you can prove by induction or just look up in a table of integrals such as Gradshteyn and Ryzhik.