Recent content by mrausum

sure: <H> = <(b/\pi).exp(-bx^2)|H|(b/\pi).exp(-bx^2)> <H> = b/\pi\int^{inf}_{-inf}exp(-2bx^2).\left(\frac{2h^2x}{m}+\frac{1}{2}mw^2x^2\right)dx

Wait..that still doesn't work. The norm.const = b^.5/(pi/2)^.5, which doesn't quite cancel with the b^3/2 in the denominator from the integral?

Yeah that's the one, thanks.

yeah, I know that from wolfram alpha too. But that can't be right, because then my upper limit for the ground energy would depend on 1/b^3/2, which would mean if b was very large it would violate the uncertainty principle.

Homework Statement Use trial wavefunction exp(-bx^2) to get an upper limit for the groundstate energy of the 1-d harmonic oscillator The Attempt at a Solution This is always going to give an integral of x^2*exp(-x^2). How do you do it? :/

That's absurd. Just because you haven't been to CalTech or MIT doesn't mean you can just go to any old university and still expect to get a decent job.

ok, I'll end the pretense: I don't really care about Physics. How good the department is for research doesn't interest me. I want the college that will look best on my CV. so, suggestions?

I'm going to spend a semester at an American University. Here is a list of allowed places: USA Arizona State University Tempe, Arizona www.asu.edu USA University of Arizona Tucson, Arizona studyabroad.arizona USA California State University Long Beach, California www.csulb.edu USA...

Homework Statement There are N point particles in a volume V. Find an equation for the mean spacing between the particles. The Attempt at a Solution distance = (V^1/3)/N would be my first guess?

For a string fixed at x=0 and free at x=l I know \frac{dy}{dx}(l,t)=0 (d's are meant to be partials) but what is the other boundry associated with the end of the string? Is the second derivative also equal to 0?

Nevermind, solved it. I was really close, I just made a stupid mistake with my algebra.

Homework Statement Show that: \frac{dc}{d\omega}=\frac{1}{k}\left(1-\frac{c}{v_{g}}\right) The Attempt at a Solution...

So I can't take the two inside the exp? I don't really see why not? So I'm guessing you're saying that I have to do the integral of phi multiplied by its conjugate? Thanks anyway, that's solved it :)

A^2.exp(2i\sqrt{\Lambda}\Phi)

The probability density function. So?