We are supposed to give a matrix representation of $L\cdot S$ for an electron with $l=1$ and $s=\frac{1}{2}$.

I read $L\cdot S$ as $L \otimes S$. Is this correct? Then we would have e.g. for

$L\otimes S (|1,1\rangle \otimes |1/2,1/2\rangle) = L |1,1\rangle \otimes S|1/2,1/2\rangle $ $= \sqrt{2} \hbar |1,1\rangle \otimes \sqrt{\frac{3}{4}} \hbar |1/2,1/2 \rangle = \sqrt{\frac{3}{2}}\hbar^2 |1,1\rangle \otimes |1/2,1/2\rangle $.

Is this correction correct? In that case should I proceed in this way with all the other basis vectors and write the eigenvalues down the diagonal in a matrix?

  • 3
    $\begingroup$ ${\bf L}$ and ${\bf S}$ are vectors. ${\bf L} \cdot {\bf S}$ is a scalar. What is $L\otimes S$? If you want to use the tensor product of operators, the correct expression is ${\bf L} \cdot {\bf S} = \sum_{k=1}^3 L_i \otimes S_i$. $\endgroup$ Jan 17, 2014 at 22:03
  • $\begingroup$ so you are saying that this is nonsense? well, do you know how to get a matrix representation for $L \cdot S$? $\endgroup$
    – Xin Wang
    Jan 17, 2014 at 22:07
  • $\begingroup$ Yes: replace each operator $L_i$ and $S_i$ for the corresponding matrix. Use the fact that $l=1$ and $s=1/2$ to find these matrices in textbooks. $\endgroup$ Jan 17, 2014 at 22:08
  • $\begingroup$ I don't understand this. If we say that $J=L+S$, does this not mean that $J=L \otimes Id + Id \otimes S$? I think I have not understood the whole concept of addition of angular momenta. What is the correct mathematical notation for $J=L+S$? $\endgroup$
    – Xin Wang
    Jan 17, 2014 at 22:11
  • 2
    $\begingroup$ ${\bf J}= {\bf L} + {\bf S}$ actually means: $J_k = L_k \otimes I_{spin} + I_{orb}\otimes S_k$ for $k=1,2,3$. $\endgroup$ Jan 17, 2014 at 22:14

1 Answer 1


There are two problems to deal with which must be disentangled to solve problems like these.

  • Both angular momentum operators are vector operators, so in some sense they "take values" in $\mathbb R^3$; you are being asked for their dot product, which should be taken within that copy of $\mathbb R^3$. You would have the same problem if you were asked to calculate the dot product $\mathbf r\cdot\mathbf p$ for a single particle without spin.

  • The orbital and spin angular momentum operators act on the two different factors of a tensor product of Hilbet spaces. Thus any (operator) product of a scalar orbital operator with a scalar spin operator should be interpreted as a tensor product. You would have the same problem if you were asked to calculate the product $L^2S^2$, which would need to be interpreted as $L^2\otimes S^2$.

Thus, in your case, you must read $L\cdot S$ as $$ \mathbf{L}\cdot \mathbf{S}=\sum_{i=1}^3L_iS_i=\sum_{i=1}^3L_i\otimes S_i. $$ To compute the matrix representation of this, you should begin with the matrix representation of each $L_i$ and $S_i$. You then compute the tensor product matrices $L_i\otimes S_i$. Finally, you add all of those matrices together to get the final result.

This is all much clearer with an example. The $z$ component, for example, is easy, since each matrix is given by $$ L_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \quad\text{and}\quad S_z=\frac\hbar 2 \begin{pmatrix}1&0\\0&-1\end{pmatrix}, $$ in the bases $\{|1\rangle,|0\rangle,|-1\rangle\}$ and $\{|\tfrac12\rangle,|-\tfrac12\rangle\}$ respectively. The tensor product matrix, then, in the basis $\{|1\rangle\otimes|\tfrac12\rangle ,|0\rangle\otimes|\tfrac12\rangle ,|-1\rangle\otimes|\tfrac12\rangle , |1\rangle\otimes|-\tfrac12\rangle ,|0\rangle\otimes|-\tfrac12\rangle ,|-1\rangle\otimes|-\tfrac12\rangle \}$, is given by $$ L_z\otimes S_z=\frac{\hbar^2} 2 \begin{pmatrix} 1\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} & 0\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \\ 0\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} & -1\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} \end{pmatrix} =\frac{\hbar^2} 2 \begin{pmatrix} 1&0&0& 0&0&0\\0&0&0&0&0&0\\0&0&-1&0&0&0\\ 0&0&0&-1&0&0\\0&0&0&0&0&0\\0&0& 0&0&0&1 \end{pmatrix}. $$ This procedure should be repeated with both the $x$ and the $y$ components. Each of those will yield a six-by-six matrix (in this case). To get your final answer you should add all three matrices.


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