Recent content by CrimsonFlash

Why does the problem only occur if we have the same subshell? Also, how do we know that these linear combinations that your wrote are eigenstates of the total orbital angular momentum? Thanks!

When you add two angular momentum states together, you get states which have exchange symmetry i.e. the highest total angular momentum states (L = l1 + l2) will be symmetric under the interchange of the two particles, (L = l1 + l2 - 1) would be anti-symmetric...and the symmetry under exchange...

l is the step size here.

Hey! I've been doing some research on random walks. From what I have gathered, a random walker in 1-D will have: <x> = [FONT="Georgia"]N l (2 p - 1) σ =[FONT="Georgia"] 2 l sqrt[N p (1 - p) ] Here, N is the number of steps, p is the probability to take a step to the right and l is the step...

The dirac delta is actually supposed to be all the way to the left. Sorry, that's the only picture I could find.

What are the "matrix elements" of the angular momentum operator? Hello, I just recently learned about angular momentum operator. So far, I liked expressing my operators in this way: http://upload.wikimedia.org/math/8/2/6/826d794e3ca9681934aea7588961cafe.png I like it this way because it...

Ok thanks.

bump?

Homework Statement It is simply the same as the one for lnz i.e. does it go from 0 to ∞? Also, is there any proper way to figure out branch points of a function? Homework Equations The Attempt at a Solution

Never mind, I got what I was looking for. Thanks anyway.

Ok, thanks everyone. Also, can you help me with the following: f(z) = sqrt(z)/(1+z^2) = sqrt(z)/[(i+z)(z-i)] The poles of this are at i and - i. The residues I calculate are: For i := sqrt(i)/(2i) = exp(πi/4)/(2i) For -i := sqrt(-i)/(-2i) = exp(-πi/4)/(-2i) Bur WolframAlpha says otherwise...

Homework Statement What is the integral of e-1/z around a unit circle centered at z = 0? Homework Equations - The Attempt at a Solution The Laurent expansion of this function gives : 1 - 1/z + 1/(2 z^2) - 1/(3! z^3) + . . . . . The residue of the pole inside is -1. So the integral...

Max Energy: 0.5 C (V/ωRC)2 Energy lost per cycle: V2/2R x 2π/ω ...and this gives the correct value for Q. Thank you.

I'll post the answers here : Q from the 1st bit: 1/R x (L/C)^(1/2) Q from my try: R x (C/L)^(1/2)

Latex...y u no work?