- #1
Osiris
- 19
- 0
Generally, Gamma matrices with one lower and one upper indices 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 gamma matrices with two lower indices. There should be at least two ways.
1). construct and use the charge conjugation matrix to lower one spinor index in the gamma matrix;
2). use inner product to directly obtain the gamma matrices with two lower spinor indices, something like $$<\Gamma e_{\alpha}, e_{\beta}>=\Gamma_{\alpha,\beta}$$, where $$e_{\alpha}$$ are the basis element.
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
$$C\Gamma C^{-1}=+/-\Gamma^T$$
\begin{equation}
\gamma^{i}\gamma^{j}+\gamma^{j}\gamma^{i}=2h^{ij},
\end{equation}
My question is how to generally construct gamma matrices with two lower indices. There should be at least two ways.
1). construct and use the charge conjugation matrix to lower one spinor index in the gamma matrix;
2). use inner product to directly obtain the gamma matrices with two lower spinor indices, something like $$<\Gamma e_{\alpha}, e_{\beta}>=\Gamma_{\alpha,\beta}$$, where $$e_{\alpha}$$ are the basis element.
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
$$C\Gamma C^{-1}=+/-\Gamma^T$$
