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

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

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

$*$ Or perhaps you know the answer already.

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

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

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

I am not asking you to redo the calculations$^*$ for me but does $(3)$ make sense?

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

$*$ Or perhaps you know the answer already.

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

$*$ Or perhaps you know the answer already.