# Orbital and Spin angular momentum

1. Apr 5, 2007

### stunner5000pt

1. The problem statement, all variables and given/known data
Consider the states for which l=4 and s=1/2.For the state wit hteh largest j, and the largest mj, calculate
a) the nagle between L and S
the angle between $\mu_{l}$ and $\mu_{s}$
c) the angle between J and the +z axis

2. Relevant equations
j=l+/-s

3. The attempt at a solution
the largest possible j is 4 +1/2 which is 9/2.
the largest mj is also 9/2
since mj ranges from -j to +j.

now my prof said that this could be done gemetrically that is
since l=4, then the value of value of L is $\sqrt{4(4+1)}\hbar=3\hbar$
the projection on the z axis, that is ml may be -4,-3,...,3,4

the spin s=1/2, ms=1/2
the magnitude of the spin angular momentum vector is $\sqrt{3}/2 \hbar$

usig the projection of both we can draw the vectors and we can find the nagles between them

3. Apr 5, 2007

### cgw

Someone correct me if I am wrong but:
The lengths of the vectors (L, S, J) are proportional to sqrt{l(l+1)}, sqrt{s(s+1)}, sqrt{j(j+1)}
Then maybe the angle between L and J can be had by the cosine law (which to me looks similar to what nrqed has above).
BTW check your sqrt{4(4+1)}.

Last edited: Apr 5, 2007
4. Apr 5, 2007

### stunner5000pt

here we are looking for the largest possible j so

j = 9/2
so
$$L \cdot S = \frac{\hbar^2}{2} (j(j+1) - l(l+1) -s(s+1)) = \frac{\hbar^2}{2} \left(\frac{9}{2}\left(\frac{9}{2} +1\right) - 4(4+1) - \frac{1}{2}\left(\frac{1}{2} +1\right) \right) = \hbar^2$$

$$|L| = \sqrt{20} \hbar$$
$$|S| = \sqrt{3}/2 \hbar$$

are u sure it should be cosine law or dot product??

5. Apr 6, 2007

### nrqed

Well, since S, L and J are really quantum operators, one cannot really see them as ordinary vectors. So it does not really make sense to talk about the angle between two operators. But people define the angle between ${\vec L}$ and ${\vec S}$ in a certain state of definite L,S and J as the ratio of the expectation values $$\frac{ < {\vec L } \cdot {\vec S} >} { \sqrt{ < {\vec L}^2 >} \sqrt{ < {\vec S}^2>}}$$

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