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classes:2009:fall:phys4101.001:q_a_1211

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Dec 11 (Fri)

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Green Suit

I think there's a typo on the Quiz 4 practice sheet. On problem 4 second line, it reads: <math>f_l^{m+1}=AL_+f_l^m</math>.

According to equation [4.120] it should read: <math>Af_l^{m+1}=L_+f_l^m</math>.

Can someone verify?

poit0009 12/10 00:57

At this point, it is an undetermined constant. Whether it is on the left or right doesn't really matter.

Your fix does correspond to the formula in the book, though (eq. 4.120).

liux0756 12/10 22:00

I guess the A in the quiz is equal to 1/A in the textbook.

poit0009 12/10 12:35

On the first problem for the practice quiz, how is this normalized? Is there one factor that goes in front of the whole expression for chi? Or should there be a normalization factor in front of each component?

Edit: I think we find a specific value of theta. Does anybody know for sure?

Yuichi

My intention was that there should be an overall factor in front of chi.

Super Hot Guy

Here is where I am and I'm a bit confused because my normalization constant is in terms of <math>\theta

\chi=A\begin{pmatrix}1+\cos{\theta}
\sin{\theta}\end{pmatrix}

normalizing…

\frac{1}{A^2}=\begin{pmatrix}1+\cos{\theta} & \sin{\theta}\end{pmatrix}\begin{pmatrix}1+\cos{\theta}
\sin{\theta}\end{pmatrix}

\frac{1}{A^2}=1+2\cos{\theta}+\cos^2{\theta}+\sin^2{\theta}=2+2\cos{\theta}

A=\frac{1}{\sqrt{2(1+\cos{\theta})}}

using

\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\cos{\theta}}{2}}

A=\frac{1}{2\cos{\frac{\theta}{2}}}\\</math>

poit0009 10/12 16:43

I think it is okay to have the normalization include <math>\theta</math>. The idea was to not have a square root, which you accomplished.

Super Hot Guy

So do the probabilities have theta in them as well? We got <math>

\\P_{+\frac{\hbar}{2}}=\frac{1+2\cos{\theta}+\cos^2{\theta}}{4\cos^2{\frac{\theta}{2}}}
P_{-\frac{\hbar}{2}}=\frac{\sin^2{\theta}}{4\cos^2{\frac{\theta}{2}}}</math>

Andromeda 12/10 2PM

are the relations <math>S{x}=[{S+} +{S-}]/2 </math>and<math> S{y}=[{S+} -{S-}]/2i </math>correct for all spin particles or only spin 1/2 particles?

poit0009 12/10 14:25

That relationship should be good for any angular momentum. It comes from the definition of L+ and L- (eq. 4.105)

chap0326 12/10 14:16

I just had a general question from section 4.3 on angular momentum. When applying the raising operator to the ladder of angular momentum states, why is it that the “process cannot go on forever”? I guess I don't see why we would eventually reach a state for which the z-component exceeds the total.

poit0009 12/10 14:26

If the z-component keeps increasing, then it could definitely exceed the total angular momentum. Take the l=1 case. Here we can have <math>m_l</math> = -1,0,or 1. If you apply <math>L_+</math> to the -1 state, you return the zero state. If you apply it to the 0 state, you return the +1 state. If you apply it to the +1 state, you would end up with <math>L_z</math> being +2. This doesn't make any sense. The angular momentum in the z-direction would be greater than the particle's total angular momentum.

Super Hot Guy

Question 3a:

<math>J_+|\Psi>=\hbar\sqrt{2}|1,0>+\hbar|\frac{1}{2},\frac{1}{2}></math>

Or is there a different notation I should be using? And is this what other people are getting?

liux0756

I get <math>J_+|\Psi>=\hbar\sqrt{2}|1,0>|\frac{1}{2},-\frac{1}{2}>+\hbar|1,-1>|\frac{1}{2},\frac{1}{2}></math>


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classes/2009/fall/phys4101.001/q_a_1211.1260504438.txt.gz · Last modified: 2009/12/10 22:07 by x500_liux0756