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## On tensor product generators of the gamma matrices.

Posted by peeterjoot on June 20, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

In [1] he writes

\begin{aligned}\gamma^0 &=\begin{bmatrix}I & 0 \\ 0 & -I\end{bmatrix}=I \otimes \tau_3 \\ \gamma^i &=\begin{bmatrix}0 & \sigma^i \\ \sigma^i & 0\end{bmatrix}=\sigma^i \otimes i \tau_2 \\ \gamma^5 &=\begin{bmatrix}0 & I \\ I & 0\end{bmatrix}=I \otimes \tau_1\end{aligned}

The Pauli matrices $\sigma^i$ I had seen, but not the $\tau_i$ matrices, nor the $\otimes$ notation. Strangerep in physicsforums points out that the $\otimes$ is a Kronecker matrix product, a special kind of tensor product [2]. Let’s do the exersize of reverse engineering the $\tau$ matrices as suggested.

# Guts

Let’s start with $\gamma^5$. We want

\begin{aligned}\gamma^5 = I \otimes \tau_1 =\begin{bmatrix}I \tau_{11} & I \tau_{12} \\ I \tau_{21} & I \tau_{22} \\ \end{bmatrix}= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.1)

By inspection we must have

\begin{aligned}\tau_1 = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}= \sigma^1\end{aligned} \hspace{\stretch{1}}(2.2)

Thus $\tau_1 = \sigma^1$. How about $\tau_2$? For that matrix we have

\begin{aligned}\gamma^i = \sigma^i \otimes i \tau_2 =\begin{bmatrix}\sigma^i \tau_{11} & \sigma^i \tau_{12} \\ \sigma^i \tau_{21} & \sigma^i \tau_{22} \\ \end{bmatrix}= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(2.3)

Again by inspection we must have

\begin{aligned}i \tau_2 = \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.4)

so

\begin{aligned}\tau_2 = \begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}= \sigma^2.\end{aligned} \hspace{\stretch{1}}(2.5)

This one is also just the Pauli matrix. For the last we have

\begin{aligned}\gamma^0 = I \otimes \tau_3 =\begin{bmatrix}I \tau_{11} & I \tau_{12} \\ I \tau_{21} & I \tau_{22} \\ \end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.6)

Our last tau matrix is thus

\begin{aligned}\tau_3 = \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}= \sigma^3.\end{aligned} \hspace{\stretch{1}}(2.7)

Curious that there are two notations used in the same page for exactly the same thing? It appears that I wasn’t the only person confused about this.

# The bivector expansion

Zee writes his wedge products with the commutator, adding a complex factor

\begin{aligned}\sigma^{\mu\nu} = \frac{i}{2} \left[{\gamma^\mu},{\gamma^\nu}\right]\end{aligned} \hspace{\stretch{1}}(3.8)

Let’s try the direct product notation to expand $\sigma^{0 i}$ and $\sigma^{ij}$. That first is

\begin{aligned}\sigma^{0 i} &= \frac{i}{2} \left( \gamma^0 \gamma^i - \gamma^i \gamma^0 \right) \\ &= i \gamma^0 \gamma^i \\ &= i (I \otimes \tau_3)(\sigma^i \otimes i \tau_2) \\ &= i^2 \sigma^i \otimes \tau_3\tau_2 \\ &= - \sigma^i \otimes (-i \tau_1) \\ &= i \sigma^i \otimes \tau_1 \\ &= i \begin{bmatrix}0 & \sigma^i \\ \sigma^i & 0\end{bmatrix},\end{aligned}

which is what was expected. The second bivector, for $i=j$ is zero, and for $i\ne j$ is

\begin{aligned}\sigma^{i j} &= i \gamma^i \gamma^j \\ &= i (\sigma^i \otimes i \tau_2) (\sigma^j \otimes i \tau_2) \\ &= i^3 (\sigma^i \sigma^j) \otimes I \\ &= i^4 (\epsilon_{ijk} \sigma^k) \otimes I \\ &= \epsilon_{ijk} \begin{bmatrix}\sigma^k & 0 \\ 0 & \sigma^k\end{bmatrix}.\end{aligned}

# References

[1] A. Zee. Quantum field theory in a nutshell. Universities Press, 2005.

[2] Wikipedia. Tensor product — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 21-June-2011]. http://en.wikipedia.org/w/index.php?title=Tensor_product&oldid=418002023.