QM expectation value relation <x^n>, <p^n>

[sensou]

I need to calculate <x^n> and <p^n> for psi(x)=exp(-ax^2/2)
for n even.
For <x^n>:
<x^n>=integral(exp(-ax^2)*x^n )dx from -inf to +inf
then i use integration by parts to get an infinite series and i use a formula to find the finite sum of the series
=[exp(-ax^2)*x^(n+1)/((n+1-2a*(n+1)^2)] with limits -infinity to +infinity
i don't think this is sufficient as i believe the answer should result in a constant and when i use this to work out <x^2>, i can't get a finite answer. I am not sure of another way to work through this question.

For <p^n> i have shown that for n odd, this vanishes, but for n even i am not sure what the answer is. I would appreciate any help or suggestions on where to start with this question, thanks.

[dextercioby]

Here's a trick...What happens if n=odd integer for \langle \hat{x}^{n} \rangle ...?

Daniel.

P.S. I'm sure you did gaussian integrals in statistical physics seminar.

[nrqed]

I need to calculate <x^n> and <p^n> for psi(x)=exp(-ax^2/2)
for n even.
For <x^n>:
<x^n>=integral(exp(-ax^2)*x^n )dx from -inf to +inf
then i use integration by parts to get an infinite series and i use a formula to find the finite sum of the series
=[exp(-ax^2)*x^(n+1)/((n+1-2a*(n+1)^2)] with limits -infinity to +infinity
i don't think this is sufficient as i believe the answer should result in a constant and when i use this to work out <x^2>, i can't get a finite answer. I am not sure of another way to work through this question.

I don't see how you can get an *infinite* series!

First, write down the result for the case n=0. Then, consider the case with n=2. after doing integrations by parts, you may write the result in terms of the n=0 expression.

Now do it with n=4 which, after integrations by parts, may be written in terms of the result of n=2. At that point, the generalization should be obvious. The result is of the form "a" raised to a certain power times a factorial term.

[sensou]

Thanks for your quick replies. I am now able to get the answer :smile:
I should have done it that way from the start but for some reason that infinite series thing led me on a bit.
Thanks for all the help.

[sensou]

I just realised that i haven't been able to get the <p^n> part right where p=-i(h-bar)d/dx
i get <p^2>=(h-bar)^2 *a/2*sqrt(pi/a)
<p^4>=(-3/4)*(h-bar)^4 *a^2*sqrt(pi/a)
<p^6>=(-165/8)*(h-bar)^6 *a^4*sqrt(pi/a)
<p^8>=(4605/16)*(h-bar)^8 *a^6*sqrt(pi/a)
And this doesn't give me a nice relationship for <p^n>
i realise that p^n where n is odd =0.
I think i must have done something wrong with the maths somewhere, but i can't find where. And i am 90% sure that the <p^2> is correct but i don't know about the others.