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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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