Responsible party: John Galt, Can

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J = L + S

When l = 1 and s = 1/2, j has possible values of 1/2, 3/2

In Bra and Ket notation, we write:
<math> |j, j_z> = |l, l_z>|s, s_z> </math>

and in the case of the board example:
<math> |3/2, 3/2> = |1, 1>|1/2, 1/2> </math>

but for the purpose of finding Clebsch-Gordon Coefficients using the tables, we can leave out l and s because they are constants, simply writing: <math> |j, j_z> = |l_z>|s_z> </math>

Yuichi put up a projection of the Clebsch Gordon Tables and asked the class to find the coefficients for |j,j_z> = |3/2,1/2>, at which point Joe asked for clarification on how to know when to look at the rows or the columns of the table, and for what information, since they are a little tricky to work with.

Around this point in the lecture, Yuichi wrote Spin-Orbit Coupling on the board, but the reason why remains a little unclear to me and I do not

Another valid question was raised by Kirby and Kate when they noticed negative representations of probability coefficients in the table and Yuichi showed us that they do not correspond to:

with the negative inside the square root,

but:

with the negative outside.

Yuichi then asked us to calculate the coefficients of the different states if we had used a plus rather than a minus… show

(IN PROGRESS)

The eigenvector (and eigenstate, I believe) of S_x is <math>\S = \begin{pmatrix} 1
1\end{pmatrix}</math>
and yields and eigenvalue of 1/2ħ with 100% probability on the first measurement and if S_y is then measured, it yields a 50% mix of probability for spin up and spin down.

The eigenvector (and eigenstate, I believe) of S_y is <math>\S = \begin{pmatrix} 1
i\end{pmatrix}</math>
and yields and eigenvalue of 1/2ħ with 100% probability on the first measurement and if S_x is then measured, it yields a 50 mixed probability of spin up and spin down.

(In this section, the code proved difficult, so fulls words are used to replace simpler notation.

<math>|1></math> and <math> |0> </math> can be replaced by the spherical harmonic functions (<math>Y_{1}^1 (Theta Psi)</math>) and (<math>Y_{1}^{0}(Theta Psi)</math>), respectively.

(<math>P(Theta Psi)=|PSI|^2</math>)

And using these, Yuichi said the probability at any angle can be found, and I think he was referring to position probability, but this is still a little nuclear to me, and he continued:

(<math>P(Theta Psi)dOmega)=P(Theta Psi)dCos(Theta)dPsi</math>)
(<math>P'(Theta)=INTEGRATE(P(Theta Psi)d(Psi)d(Cos(Theta))</math>)
(<math>P'(Theta')dTheta=(P(Theta,Psi))^2Sin(Theta)</math>)

Only material covered up to today, December 4th, will be covered on the third midterm and the final. Chapter 5 is still an interesting chapter to look in to, but its material will appear on neither the third midterm nor the final.

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