What is the eigenvalue of angular momentum? (Zeeman)

[unscientific]

Homework Statement

In the calculation of the Zeeman Effect, the most important calculation is

$$\langle L_z + 2S_z \rangle = \langle J_z + S_z\rangle$$

Suppose we want to find the Zeeman Effect for $(2p)^2$, meaning $l=1$.

In Sakurai's book,

My question is, what is $m$? They say that $m$ is the eigenvalue of $J_z$, meaning $J_z = l + m_s = \frac{3}{2}$ in this case.

Homework Equations

The Attempt at a Solution

Using $m = \frac{3}{2}$, it gives the wrong Gordon-Clebsch coefficients.

I have worked out the correct form, which is adding orbital angular momentum $(l=1)$ to its spin $(\pm\frac{1}{2})$

$$|l + s, m_l + m_s\rangle = |\frac{3}{2},\frac{1}{2}\rangle = \sqrt{\frac{1}{3}}|1,1\rangle|-\rangle + \sqrt{\frac{2}{3}}|1,0\rangle|+\rangle$$

 

[vela]

Homework Statement

In the calculation of the Zeeman Effect, the most important calculation is

$$\langle L_z + 2S_z \rangle = \langle J_z + S_z\rangle$$

Suppose we want to find the Zeeman Effect for $(2p)^2$, meaning $l=1$.

In Sakurai's book,

My question is, what is $m$? They say that $m$ is the eigenvalue of $J_z$, meaning $J_z = l + m_s = \frac{3}{2}$ in this case.

Why do you assume $m_l=l$?

Homework Equations

The Attempt at a Solution

Using $m = \frac{3}{2}$, it gives the wrong Gordon-Clebsch coefficients.

I have worked out the correct form, which is adding orbital angular momentum $(l=1)$ to its spin $(\pm\frac{1}{2})$

$$|l + s, m_l + m_s\rangle = |\frac{3}{2},\frac{1}{2}\rangle = \sqrt{\frac{1}{3}}|1,1\rangle|-\rangle + \sqrt{\frac{2}{3}}|1,0\rangle|+\rangle$$

 

[unscientific]

Why do you assume $m_l=l$?

my biggest problem is, they say that the eigenvalue of $J_z$ is $m$, but then $m = \frac{1}{2}$ gives the correct gordon-clebsch coefficients. Does this mean that $J_z = \frac{1}{2}$? But isn't $J_z$ the TOTAL angular momentum? Shouldn't it be $J_z = l + s = \frac{3}{2}$?

 

[vela]

$J_z$ is the z-component of the total angular momentum. It can take on any value from $j$ to $-j$. According to the rules about adding angular momenta, $j$ can take on any value from $l+s$ to $\lvert l-s \rvert$.

 

[dauto]

you seem to be confusing l with m_l and s with m_s. If l = 1 than m_l = -1, 0, 1 are all possible, and if s = 1/2 than m_s = +1/2 and -1/2 are possible. m = m_l + m_s

 

[unscientific]

you seem to be confusing l with m_l and s with m_s. If l = 1 than m_l = -1, 0, 1 are all possible, and if s = 1/2 than m_s = +1/2 and -1/2 are possible. m = m_l + m_s

If that's the case, then for $l=1$ and $m_s = 1$:

$$|l + s, m_l + m_s\rangle = |\frac{3}{2},\frac{1}{2}\rangle = \sqrt{\frac{1 - (1 + \frac{1}{2}) + \frac{1}{2}}{3}}|1\rangle|-\rangle + \sqrt{\frac{1 + (1 + \frac{1}{2}) + \frac{1}{2}}{3}}|0\rangle|+\rangle$$

These coefficients are wrong.

 

[dauto]

For j = 3/2, l = 1, and m = 1/2 I'm getting

|j,m> = |3/2,1/2> = |1+1/2,1/2> = [(l+m+1/2)/(2l+1)]^1/2|m-1/2,1/2> + [(l-m+1/2)/(2l+1)]^1/2|m+1/2,-1/2>
|j,m> = [(1+1/2+1/2)/(2*1+1)]^1/2|1/2-1/2,1/2> + [(1-1/2+1/2)/(2*1+1)]^1/2|1/2+1/2,-1/2>
|j,m> = (2/3)^1/2|0,1/2> + (1/3)^1/2|1,-1/2>
|3/2,1/2> = (2/3)^1/2|1,0>|+> + (1/3)^1/2|1,1>|->

 

[vela]

If that's the case, then for $l=1$ and $m_s = 1$:

$$|l + s, m_l + m_s\rangle = |\frac{3}{2},\frac{1}{2}\rangle = \sqrt{\frac{1 - (1 + \frac{1}{2}) + \frac{1}{2}}{3}}|1\rangle|-\rangle + \sqrt{\frac{1 + (1 + \frac{1}{2}) + \frac{1}{2}}{3}}|0\rangle|+\rangle$$

These coefficients are wrong.

For the state $\left\lvert \frac{3}{2},\frac{1}{2} \right\rangle$, you have $j=3/2$ and $m=1/2$. In your expression for the coefficients, you're using $m=3/2$, which is why they're not coming out right.

 

[unscientific]

For the state $\left\lvert \frac{3}{2},\frac{1}{2} \right\rangle$, you have $j=3/2$ and $m=1/2$. In your expression for the coefficients, you're using $m=3/2$, which is why they're not coming out right.

Damn, I should have been more careful. Thanks a lot!