Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1120
Q&A for the next lecture: Q_A_1125

If you want to see lecture notes, click lec_notes

Main class wiki page: home

In the proof that [L^2,Lx] = 0 on page 161, how does Griffiths get from the first line to the second line?

He uses the commutation relation [AB,C] = A[B,C] + [A,C]B. The first term [Lx^2,Lx], is zero since Lx commutes with itself. Then using the commutation relation I introduced you expands [Ly^2,Lx] and [Lz^2,Lx]. Lets just expand [Ly^2,Lx], just to illustrate how the commutation relation works. In this case, A=Ly, B=Ly, and C=Lx. Using the relation we have [Ly*Ly,Lx] = Ly[Ly,Lx] + [Ly,Lx]Ly. He similar does this with [Lz^2,Lx].

Where does the minus sign come from when you operate Sz on <math>\chi-</math> in eq. 4.144 (pg 174)?

When you multiply the definition matrices for each value you have: <math> S_z \chi_- = \frac{\hbar}{2} \left(\begin{array}{cc} 1 & 0
0 & -1 \end{array} \right) \left(\begin{array}{c} 0
1 \end{array} \right) = \frac{\hbar}{2} \left(\begin{array}{c} 1*0+0*1
0*0+(-1)*1 \end{array} \right) = \frac{\hbar}{2} \left(\begin{array}{c} 0
-1 \end{array} \right) = \frac{-\hbar}{2} \left(\begin{array}{c} 0
1 \end{array} \right) = \frac{-\hbar}{2}\chi_-</math>

Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1120
Q&A for the next lecture: Q_A_1125