I'm solving problem 3.D in H. Georgi Lie Algebra etc for fun where one is to compute the matrix elements of the direct product $\sigma_2\otimes\eta_1$ where $[\sigma_2]_{ij}\text{ and }[\eta_1]_{xy}$ are two different Pauli matrices in two different two dimensional spaces.

Defining the basis in our four dimensional tensor product space $$\tag{1}\left|1\right\rangle = \left|i=1\right\rangle\left|x=1\right\rangle\\ \left|2\right\rangle = \left|i=1\right\rangle\left|x=2\right\rangle\\ \left|3\right\rangle = \left|i=2\right\rangle\left|x=1\right\rangle\\ \left|4\right\rangle = \left|i=2\right\rangle\left|x=2\right\rangle$$

Now we know that when we multiply representations, the generators add in the sense of

$$\tag{2}[J_a^{1\otimes2}(g)]_{jyix} = [J_a^1]_{ji}\delta_{yx} +\delta_{ji}[J_a^2]_{yx}, $$ where the $J$s are the generators corresponding to the different representations $D_1$ and $D_2$ ($g$ stands for the group elements).

Using all of this I find that in the basis of $(1)$ the matrix representation of the tensor product is given by

$$\tag{3}\sigma_2\otimes\eta_1 = \begin{pmatrix} 0 & \mathbf{1} & -i & 0 \\ 1 & 0 & 0 & -i \\ i & 0 & 0 & 1 \\ 0 & i & 1 & 0 \end{pmatrix}$$

(The bold $\mathbf{1}$ is just notation, see below!) I am not asking you to redo the calculations for me but does $(3)$ make sense?

Appendix. My calculations were done in the following fashion [using equation $(2)$]: $$\tag{4}\langle 1| \sigma_2\otimes \eta_1 |1\rangle = \\ \langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=1\rangle \\ = [\sigma_2]_{11}\delta_{11}+\delta_{11}[\eta_1]_{11} \\ = 0.$$ Similarly for eg $$\tag{5} \langle 1| \sigma_2\otimes \eta_1 |2\rangle = \\ \langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=2\rangle \\ = [\sigma_2]_{11}\delta_{12}+\delta_{11}[\eta_1]_{12} \\ = 1. $$ This is how the bold $\mathbf{1}$ was obtained.

So are my calculations $(4), (5)$ totally wrong?

The Pauli matrices $$\begin{align} \sigma_1 &= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix} \\ \sigma_2 &= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix} \\ \sigma_3 &= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \,. \end{align} $$


3 Answers 3


I think it is easier to compute direct products when you write the matrices in component form; basically, you just have to multiply each element of the first matrix by the whole second matrix: $$ \mathbf{A}\otimes\mathbf{B} = \begin{bmatrix} A_{11} \mathbf{B} & \cdots & A_{1n} \mathbf{B} \\ \vdots & \ddots & \vdots \\ A_{n1} \mathbf{B} & \cdots & A_{nn} \mathbf{B}\end{bmatrix} $$ In your case, using the Pauli matrices $\boldsymbol{\sigma}_2$ and $\boldsymbol{\eta}_1$, we get: $$ \boldsymbol{\sigma}_2 \otimes \boldsymbol{\eta}_1 = \begin{bmatrix} 0 & -i\boldsymbol{\eta}_1 \\ i\boldsymbol{\eta}_1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ i & 0 & 0 & 0\end{bmatrix} $$

  • $\begingroup$ So my equation 5 is wrong? $\endgroup$ Commented Feb 1, 2015 at 14:34
  • $\begingroup$ The direct product $\boldsymbol{\sigma}_2\otimes\boldsymbol{\eta}_1$ should be the same no matter which method you use to calculate it. Thus I believe that your equation (3) is wrong, which would imply that (5) is also wrong. Unfortunately, I do not know much about formal Lie theory, so I am unable to pinpoint exactly what went wrong. $\endgroup$
    – jabirali
    Commented Feb 1, 2015 at 14:46
  • $\begingroup$ Are you e.g. sure that you got eq. (2) right? $\endgroup$
    – jabirali
    Commented Feb 1, 2015 at 14:47
  • $\begingroup$ Yes eq 2 comes from the book and the derivation looks valid. $\endgroup$ Commented Feb 1, 2015 at 17:28
  • $\begingroup$ Remember that your answer should be written in the basis (1). $\endgroup$ Commented Feb 2, 2015 at 15:15

Your equation (2) is right, in principle: it is the standard coproduct of Lie algebras, but it is irrelevant, and should have never been used for anything here. The language confused you. It should read $$ \boldsymbol{J^a} = \boldsymbol{j^a} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{j^a} .$$ If you wished to apply it to two doublet reps, you should have used the same Pauli matrix $\sigma^a$ for both $j^a$s, and multiplying by the same angle and exponentiating you would have seen how nicely the group elements tensor-factor in the respective subspaces 1 and 2: $\exp (i\theta^a \boldsymbol{J}^a)=$ $\exp(i\theta^a(\boldsymbol{j^a} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{j^a}))=\exp(i\theta^a(\boldsymbol{j^a} \otimes 1\!\!1)) \exp(i\theta^a (1\!\!1\otimes \boldsymbol{j^a}))= \exp(i\theta^a \boldsymbol{j^a}) \otimes\exp(i\theta^a\boldsymbol{j^a} )$.

But, instead, your assignment asked you to simply mechanically evaluate the tensor product of two different matrices, to see if you understand the rules @jabirali correctly applied to get the correct answer you were meant to find. So, your equation (3) is magnificently wrong: you evaluated $\boldsymbol{\sigma_2} \otimes 1\!\!1 +1\!\!1\otimes \boldsymbol{\sigma_1} $. jabirali is actually using your conventions, basis (1), exactly.

As a further exploratory excursion, you might use his and your rules, "right matrix into entries of left matrix", "left-coarse, right-fine" to compute (2) for a common matrix, e.g. $\sigma^2$, and then C-G rotate/reduce the 4x4 matrix to find $J_2$ in the triplet representation (3x3 block) and a singlet (0! in the remaining 1x1 block).


Each Pauli matrix has two non-zero elements. Therefore, direct product of Pauli matrices will have four non-zero elements. Your answer, unfortunately, has eight.

  • $\begingroup$ So my equation 5 is wrong? $\endgroup$ Commented Feb 1, 2015 at 14:34
  • $\begingroup$ No, it is just a wrong equation to apply. Addition of generators is just like addition of angular momenta. I haven't read the Georgi's book, but this exercise looks like something intended to get you familiarized with the direct product space. A typical QM book would then explain how this product space can be represented as a direct sum of spin-0 and spin-1 spaces. $\endgroup$
    – mho
    Commented Feb 1, 2015 at 20:55
  • $\begingroup$ I'm asking for the answer in basis $(1)$. $\endgroup$ Commented Feb 2, 2015 at 14:29

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