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authorJaron Kent-Dobias <jaron@kent-dobias.com>2024-10-29 10:19:16 +0100
committerJaron Kent-Dobias <jaron@kent-dobias.com>2024-10-29 10:19:16 +0100
commit405f6727a6915c61e09160fba52dd8832c2207e3 (patch)
tree8fd5536f30cc3a067d5ef76cc73308aaeb07d37d
parent49e34257f5974cf63ab925f260457a1d5a7be079 (diff)
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Slightly modified convention for superbases.
-rw-r--r--marginal.tex18
1 files changed, 9 insertions, 9 deletions
diff --git a/marginal.tex b/marginal.tex
index 442e7e2..c8d0b8b 100644
--- a/marginal.tex
+++ b/marginal.tex
@@ -1804,26 +1804,26 @@ where the usual role of the determinant is replaced by the superdeterminant.
The superdeterminant can be defined using the ordinary determinant by writing a
block version of the matrix $M$. If $\mathbf e(1)=\{1,i\bar\theta_1\theta_1\}$ is
the basis vector of the even subspace of the superspace and $\mathbf
-f(1)=\{i\bar\theta_1,i\theta_1\}$ is that of the odd subspace, dual bases $\mathbf e^\dagger(1)=\{i\bar\theta_1\theta_1,1\}$ and $\mathbf f^\dagger(1)=\{\theta_1,-\bar\theta_1\}$ can be defined by the requirement that
+f(1)=\{i\bar\theta_1,i\theta_1\}$ is that of the odd subspace, dual bases $\mathbf e^\dagger(1)=\{i\bar\theta_1\theta_1,1\}$ and $\mathbf f^\dagger(1)=\{-\theta_1,\bar\theta_1\}$ can be defined by the requirement that
\begin{align}
- \int d1\,\mathbf e(1)\mathbf e^\dagger(1)=iI
+ &\int d1\,e_i^\dagger(1)e_j(1)=i\delta_{ij}
&&
- \int d1\,\mathbf f(1)\mathbf f^\dagger(1)=iI \\
- \int d1\,\mathbf e(1)\mathbf f^\dagger(1)=0
+ \int d1\,f_i^\dagger(1)f_j(1)=i\delta_{ij} \\
+ &\int d1\,e_i^\dagger(1)f_j(1)=0
&&
- \int d1\,\mathbf f(1)\mathbf e^\dagger(1)=0
+ \int d1\,f_i^\dagger(1)e_j(1)=0
\end{align}
With such bases and dual bases defined, we can form a
block representation of $M$ in analogy to the matrix form of an operator in quantum mechanics by
\begin{equation}
\int d1\,d2\,\begin{bmatrix}
- \mathbf e(1)M(1,2)\mathbf e^\dagger(2)
+ \mathbf e^\dagger(1)M(1,2)\mathbf e(2)
&
- \mathbf e(1)M(1,2)\mathbf f^\dagger(2)
+ \mathbf e^\dagger(1)M(1,2)\mathbf f(2)
\\
- \mathbf f(1)M(1,2)\mathbf e^\dagger(2)
+ \mathbf f^\dagger(1)M(1,2)\mathbf e(2)
&
- \mathbf f(1)M(1,2)\mathbf f^\dagger(2)
+ \mathbf f^\dagger(1)M(1,2)\mathbf f(2)
\end{bmatrix}
=i\begin{bmatrix}
A & B \\ C & D