Recent content by sean_mp

You need to use one, two and three dimensional dirac delta functions for your charge densities for the first three parts. Check out page 4 of these notes: http://theory.uchicago.edu/~sethi/Teaching/P221-F2008/DiracandGreenNotes(08).pdf

Sorry, I misunderstood your last question, so I'm editing my post. I'm going to have to come back to your #2 point later, since it's getting late where I am. The following was written when I misunderstood your question, so I'm just going to post it rather than waste it. If we're being honest...

Ok, for the first part, all you need to know is that the states are orthogonal and normalized. This means \big< \Psi_{n'l'm'} \big\lvert \, \Psi_{nlm} \big> =\big<n'l'm' \big\lvert \,nlm\big> = 0~~ \text{if}~~nlm=n'l'm';~1~~\text{if}~~nlm \ne n'l'm' So for your state, we have \big< \Psi...

You are definitely on the right track. Your expression is going to be ugly. Now, set z\equiv 2 \hbar at/m and use \lvert \psi \lvert^2= \sqrt{ \frac{2a}{ \pi}} \frac{1}{ \sqrt{1+ z^2}} e^{-l^2/2a}e^{a\big\{ \frac{(ix+l/2a)^2}{(l+iz)}+ \frac{(-ix +l/2a)^2}{(1-iz)}\big\}} Expand the term in...

God, do I know it. Do yourself a favor (I tell this to everybody) and buy a copy of Zettili's book. I've been in grad school for over two years, but I've had this book since the end of my undergrad. It is absolutely shredded. I still use it all the time, even though I never do...

Yes! That's the amazing thing about it! Take the particle in a box, or infinite square well, with V=0,~~0<x<a; V= \infty,~~x<0,~x>a Now plug \psi(x)= \sin\bigg(\frac{ \pi x}{a}\bigg) into both methods. With straight-forward way, you'd get -\hbar^2 \int^a_0 dx\, \sin\bigg(\frac{ \pi...

Ok, great. You just want to take the derivative of the ket (not the conjugate), don't sweat the bra. If you were just doing this problem straight out, you'd do \big< \psi \big\lvert \, \hat p^2 \big\lvert \psi \big> = \int dx \, \psi^*(x) \, \hat p^2 \, \psi(x) = -\hbar^2 \int dx \...

Well, your bra should just be the Hermitian conjugate of your ket, right? I'm not sure what your level of quantum is, so I'm not really sure if I'm dumbing things down or making them complicated.

I can't tell what you're asking. Are you saying \int \psi^* d^3 \psi dx^3 dx = \int d^2 \psi dx^2 dx or \int \psi^* \frac{d^3 \psi}{ dx^3} dx= \int \frac{d^2 \psi}{dx^2} dx? I'm guessing it's the latter.

Have you checked http://en.wikipedia.org/wiki/Fine_structure ?

Here's one thing. For \psi(x,t)= \frac{1}{ \sqrt{2 \pi \hbar}} \int dp \, \phi(p)e^{i(kx- \omega t)}, \psi(x,t)^*= \frac{1}{ \sqrt{2 \pi \hbar}} \int dp \, \phi(p)^*e^{-i(kx- \omega t)} \big< \psi \big\lvert \,p^2 \big\lvert \psi \big> = \frac{1}{2 \pi \hbar} \int \phi(p) \,p^2...

If you have something like \psi (p) = \frac{1}{(2 \pi \hbar)^{1/2}} \int dx \, \phi(x) e^{-ipx}, it is legitimate to use \psi^* (p) = \frac{1}{(2 \pi \hbar)^{1/2}} \int dx \, \phi(x)^* e^{ipx}, In other words, you can take the complex conjugate of an integral before integrating (at...

Homework Statement I'm dealing with a dirac particle in an attractive spherical square well. I've solved for the transcendental equation to find energy, found the normalized wave function, and now I'm trying to explain what happens when the well becomes very deep, when V0 ≥ 2mc2. If I plug...

Wow... I'm embarrassed that I didn't get this myself... I must have had a typo in my equation or something. Thanks a lot, you're a lifesaver.

Homework Statement I'm trying to verify equations 16 and 17 on the attached paper, but I'm just not getting the same values they are. I've used mathematica and it just isn't happening. For one thing, there should be a factor of 1/2 in front of equation 16, but I'm really not seeing how they...