Recent content by Werbel22

Got it, thank you very much!

Hello, this is just a general question, how is <x^2> evaluated, if <x> = triple integral of psi*(r,t).x.psi(r,t).dr (this is the expectation value of position of wavepacket) Is it possible to square a triple integral? Is <x^2> the same as <x>^2 ? I'm only wondering how the squared works...

So then what is the value of the levi-civita symbol then? I only know it to be -1, 1, 0, and I thought you need to know about j and k as well to find it's value?

Hello, I am having a little difficulty understanding what exactly the Levi-Civita symbol is about. In the past I believed that it was equal to 1, -1 and 0, depending on the number of permutations of i,j,k. I had just accepted that to be the extent of it. However, now I am seeing things...

I'm still stuck. I understand that last post, and it has helped. But regardless if I take that into account, I don't see how having a f'(x) helps me, I tried doing it that way, and seeing I end up with a combination of f(x) and f'(x) in my final answer I don't see how that will give 1 :(

Yeah, but I thought in the example d/dx was operating on x, just to give 1. When I put it in, in my original post, it didn't make a difference. I know operators don't commute, but when I sub IN the actual operator (say into xp - px) it isn't an operator anymore, it's just some...

[X,P] = xp - px ? When I sub in the equations for those now, I get zero for [x,p], which gives zero for [a+,a-]. I got down to: [x,p] = (-i(d/dx)(x)) - (-i(d/dx))x = -i + i = 0 :S I don't get why I can't get the answer.

Homework Statement Show [a+,a-] = -1, Where a+ = 1/((2)^0.5)(X-iP) and a- = 1/((2)^0.5)(X+iP) and X = ((mw/hbar)^0.5)x P = (-i(hbar/mw)^0.5)(d/dx)2. The attempt at a solution It would take forever to write it all up, but in summary: I said: [a+,a-] = (a+a- - a-a+) then...