- #1

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- Thread starter hokhani
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- #1

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- #2

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##J_z = \hbar

\begin{pmatrix}

\frac{3}{2} & 0 & 0 & 0 \\

0 & \frac{1}{2} & 0 & 0 \\

0 & 0 & -\frac{1}{2} & 0 \\

0 & 0 & 0 & -\frac{3}{2}

\end{pmatrix}

##

P.S.: i'm studying this right now, so maybe i could probably make some mistakes

- #3

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$$\vec{J}=\vec{L}+\vec{S}.$$

Then according to representation theory of the rotation group, if the orbital angular angular momentum has definite ##l## and spin ##s##, then the total angular momentum has ##j \in \{|l-s|,|l-s|+1,\ldots,l+s \}##, i.e., in your case you have ##j \in \{1/2,3/2 \}##.

Indeed the total-angular momentum space is ##(2l+1)(2s+1)##, i.e., in your case ##3 \cdot 2=6## dimensional. The possible values for ##j## imply the same dimension, i.e., ##2+4=6##.

For the eigenvalues of ##j_z=l_z+s_z##. You should look this up in a good textbook. Look for Clebsch-Gordan coefficients.

- #4

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- #5

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You use the Kronecker product.

- #6

Khashishi

Science Advisor

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##J^2 = (L + S)^2 = L^2 + S^2 + 2LS##

##S## is a constant, so the result depends on ##L^2## and ##J^2##. With ##l=1## fixed, you can write out the basis states of ##j## in terms of the basis states of ##s##.

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