Generally, Gamma matrices could be constructed based on the Clifford algebra. \begin{equation} \gamma^{i}\gamma^{j}+\gamma^{j}\gamma^{i}=2h^{ij}, \end{equation}

My question is how to generally construct the charge conjugation matrix to raise one spinor index in the gamma matrix.

In even dimensions (D=2m), consider complex Grassmann algebra $\Lambda_{m}[\alpha^{1},...,\alpha^{m}]$ with generators $\alpha^{1},...,\alpha^{m}.$) Namely, we define $\widehat{\alpha }^{i}$ and $\widehat{\beta}_{i}$ as multiplication and differentiation operators: \begin{equation} \widehat{\alpha}^{i}\psi=\alpha^{i}\psi, \end{equation} \begin{equation} \widehat{\beta}_{i}\psi=\frac{\partial}{\partial\alpha^{i}}\psi. \end{equation}

According to the Grassmann algebra, we have \begin{equation} \widehat{\alpha}^{i}\widehat{\alpha}^{j}+\widehat{\alpha}^{j}\widehat{\alpha }^{i}=0, \end{equation} \begin{equation} \widehat{\beta}_{i}\widehat{\beta}_{j}+\widehat{\beta}_{j}\widehat{\beta}% _{i}=0 \end{equation} \begin{equation} \widehat{\alpha}^{i}\widehat{\beta}_{j}+\widehat{\beta}_{j}\widehat{\alpha }^{i}=\delta_{j}^{i}. \end{equation} This means that $\widehat{\alpha}^{1},...,\widehat{\alpha}^{m},\widehat{\beta }_{1},...,\widehat{\beta}_{m}$ specify a representation of Clifford algebra for some choice of $h$ (namely, for $h$ corresponding to quadratic form $\frac{1}{2}(x^{1}x^{m+1}+x^{2}x^{m+2}+...+x^{m}x^{2m})$). It follows that operators \begin{equation} \Gamma^{j}=\widehat{\alpha}^{j}+\widehat{\beta}_{j},1\leq j\leq m, \end{equation} \begin{equation} \Gamma^{j}=\widehat{\alpha}^{j-m}-\widehat{\beta}_{j-m},m<j\leq2m, \end{equation} determine a representation of $Cl(m,m,\mathbb{C})$.

For example, in $D=4$, we can obtain $$\Gamma^{1}=\begin{pmatrix}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\ \end{pmatrix}$$, $$\Gamma^{2}=\begin{pmatrix}0& 0& 0& 1\\ 0& 0& {-1}& 0\\ 0& {-1}& 0& 0\\ 1& 0& 0& 0\\ \end{pmatrix}$$, $$\Gamma^{3}=\begin{pmatrix}0& {-1}& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& {-1}& 0\\ \end{pmatrix}$$, $$\Gamma^{4}=\begin{pmatrix}0& 0& 0& {-1}\\ 0& 0& 1& 0\\ 0& {-1}& 0& 0\\ 1& 0& 0& 0\\ \end{pmatrix}.$$

My question is how to generally construct the charge conjugation matrix C, so that we could have $$C\Gamma C^{-1}=\pm\Gamma^T$$

  • $\begingroup$ Lecture notes Supersymmetry summer term 2010 by Maximilian Kreuzer hep.itp.tuwien.ac.at/~kreuzer/inc/susy.pdf Here, in page 8 there a general answer to your question. Thanks for sharing that interesting representation in terms of Grassmann operators! $\endgroup$ – user77990 Apr 16 '15 at 20:27

Explicit expressions for the Euclidian signature are given in the following Hitoshi Murayama lecture notes (Section 1.3). The expressions are given in the Pauli matrix tensor product basis.

| cite | improve this answer | |
  • $\begingroup$ Thank you. But they used a different method to construct the gamma matrices. Any idea how to construct the charge conjugation matrix for the gamma matrices given in my original post? $\endgroup$ – Osiris Xu Oct 4 '12 at 5:01
  • $\begingroup$ I can show you how to construct in this particular case, but can you please check your calculations because in your expressions $(\Gamma^1)^2=(\Gamma^2)^2=1$ while $(\Gamma^3)^2=(\Gamma^4)^2=-1$, i.e., you are working in a signature $(1, 1, -1, -1)$. Is this really the case you need. $\endgroup$ – David Bar Moshe Oct 4 '12 at 16:06
  • $\begingroup$ Dear David Yes, I am working with the general signatures of (1..1,-1..-1), where there are m +1s, and m -1s, for the 2m dimensional spacetime. $\endgroup$ – Osiris Xu Oct 4 '12 at 21:41

The charge conjugation matrix will depend on the choice of the basis you are representing the Dirac matrix. This is so because you want to satisfy:

$$ C\Gamma^{m}C^{-1}=\pm(\Gamma^{m})^{T} $$

Two charge conjugation matrices of different choices of basis will be related by

$$ C\rightarrow U^{T}CU $$

where the Dirac matrices of different basis will be related by

$$ \Gamma^{m}\rightarrow U\Gamma^{m} U^{-1} $$

Now, you need to fix a basis and find the charge conjugation matrix for this basis.

There is a very convenient basis obtained by splitting the representations of $SO(2n)$ into $U(n)$. This is obtained by grouping the gamma matrices as follows:

$$ \Gamma_{a}=\frac{1}{2}\left(\Gamma^{a}+i\Gamma^{a+n}\right) $$

$$ \Gamma_{\bar{a}}=\frac{1}{2}\left(\Gamma^{a}-i\Gamma^{a+n}\right) $$

Note that this new index $a$ labels the fundamental representations of the $U(n)$ subgroup of $SO(2n)$, while $\bar{a}$ labels the anti-fundamental representation. Raising the anti-fundamental $U(n)$ index by the $U(n)$ metric we have the following algebra:

$$ \{\Gamma_{a},\Gamma_{b}\}=0,\,\,\,\,\,\,\{\Gamma^{a},\Gamma^{b}\}=0,\,\,\,\,\,\,\{\Gamma_{a},\Gamma^{b}\}=\delta_{a}^{b} $$

This is the well know algebra of a fermionic quantum oscillator. A representation is built by fixing a ground state that is annihilated by all the $\Gamma^{a}=\Gamma_{\bar{a}}$ annihilation operators, and others states can then be obtained by exciting this ground state with the raising operators $\Gamma_{a}$.

Example, for $d=10$, we have $n=5$, then we have the ground state

$$ |0\rangle=|-----\rangle $$

that is annihilated by all the $\Gamma^{a}$, and for instance:

$$ \Gamma^{3}|-----\rangle=|--+--\rangle $$

The charge conjugation matrix $C$ will be the one that conjugate all the $U(1)$ charges, and then switching fundamental representations to the anti-fundamental and vice versa:

$$ C|-----\rangle=|+++++\rangle $$

while $C\Gamma_{a}=\pm\Gamma^{a}C$ and $C\Gamma^{a}=\pm\Gamma_{a}C$. Explicitly, this matrix can be written as

$$ C=(\Gamma)\Gamma^{a+1}...\Gamma^{a+n} $$

Where now, this indices are the $m$ kind of index, the $SO(2n)$ index. The $\Gamma$ inside the parenthesis is optional, and it is fixed if we fixed the signs of $C\Gamma_{a}=\pm\Gamma^{a}C$ and $C\Gamma^{a}=\pm\Gamma_{a}C$

In this notation, chirality is simply defined by the number of $+$ signs. If it is even, it is called Weyl or Chiral, if it is odd, it is called anti-Weyl or anti-Chiral. Then you see that depending on the dimension, the charge conjugation matrix can switch the chirality or not. More precisely, if $n$ is even then the chirality is preserved by the charge conjugation matrix, while if it is odd it will switch the chirality.

There are maybe some signs that I am missing here, but the idea is this.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.