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## Main Question or Discussion Point

Hey,

If [tex] \xi^a [/tex] and [tex] \bar{\xi^a} [/tex] are some finitie number of complex Grassman variables.

i.e. if [tex]\theta, \eta [/tex] are two real grassman numbers, then [tex] \xi^a=\frac{1}{\sqrt{2}} \left( \theta+i\eta \right) [/tex] and [tex] \bar{\xi^a}=\frac{1}{\sqrt{2}} \left( \theta-i\eta \right) [/tex].We then have the anticommutation relations:

[tex] \{ \xi^a,\xi^b \}=0 [/tex], [tex] \{ \bar{\xi^a},\bar{\xi^b} \}=0 [/tex], [tex] \{ \xi^a,\bar{\xi^b} \}=0 [/tex]

Now the question is, if I have the sum (where f is antisym on first two and last two indices: [tex] f_{abcd}=-f_{bacd}=-f_{abdc} [/tex]):

[tex] \sum_{abcd} f_{abcd} \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}=\left( \sum_{ab} k_{ab}\bar{\xi^a}\xi^b \right)^2 [/tex]

How does one now go about writing [tex] f_{abcd} [/tex] in terms of [tex] k_{ab} [/tex]? I think I've managed something but I'm not convinced it's correct.

RHS:

[tex] \left( \sum_{ab} k_{ab}\bar{\xi^a}\xi^b \right)^2= \left( \sum_{ab} k_{ab}\bar{\xi^a}\xi^b \right)\left( \sum_{cd} k_{cd}\bar{\xi^c}\xi^d \right)

=\left( \sum_{abcd} k_{ab}k_{cd}\bar{\xi^a}\xi^b\bar{\xi^c}\xi^d \right)[/tex]

Now focussing on LHS to bring the correct ordering of the [tex] \xi[/tex]'s:

[tex] \sum_{abcd} f_{abcd} \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}=\sum_{abcd} -f_{abcd} \xi^a\bar{\xi^c}\xi^b\bar{\xi^d}=\sum_{abcd} f_{abcd} \bar{\xi^c}\xi^a\xi^b\bar{\xi^d}= \sum_{abcd} -f_{abcd} \bar{\xi^c}\xi^a\bar{\xi^d}\xi^b[/tex]

Comparing LHS and RHS now, and relabelling dummies from RHS:

[tex] \sum_{abcd} -f_{abcd} \bar{\xi^c}\xi^a\bar{\xi^d}\xi^b= \sum_{abcd} k_{ca}k_{db}\bar{\xi^c}\xi^a\bar{\xi^d}\xi^b [/tex]

I'm not sure at this point if we can simply equate [tex] -f_{abcd}=k_{ca}k_{db} [/tex] however. Since the product of [tex] \xi\xi\xi\xi [/tex] is antisymetric on exchange of c and a, and antisymmetric on exchange of any two indices (c and a, d and b, a and d but also c and d, a and b, and c and b). I'm thinking it must be the antisymmetric part of [tex]f_{abcd} [/tex] and [tex]k_{ca}k_{db} [/tex] that we can equate with each other.

I'm not sure how to find the antisymmtric part on all four indices however, if that is indeed the right track.

Appreciate any help at all, thanks alot

If [tex] \xi^a [/tex] and [tex] \bar{\xi^a} [/tex] are some finitie number of complex Grassman variables.

i.e. if [tex]\theta, \eta [/tex] are two real grassman numbers, then [tex] \xi^a=\frac{1}{\sqrt{2}} \left( \theta+i\eta \right) [/tex] and [tex] \bar{\xi^a}=\frac{1}{\sqrt{2}} \left( \theta-i\eta \right) [/tex].We then have the anticommutation relations:

[tex] \{ \xi^a,\xi^b \}=0 [/tex], [tex] \{ \bar{\xi^a},\bar{\xi^b} \}=0 [/tex], [tex] \{ \xi^a,\bar{\xi^b} \}=0 [/tex]

Now the question is, if I have the sum (where f is antisym on first two and last two indices: [tex] f_{abcd}=-f_{bacd}=-f_{abdc} [/tex]):

[tex] \sum_{abcd} f_{abcd} \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}=\left( \sum_{ab} k_{ab}\bar{\xi^a}\xi^b \right)^2 [/tex]

How does one now go about writing [tex] f_{abcd} [/tex] in terms of [tex] k_{ab} [/tex]? I think I've managed something but I'm not convinced it's correct.

RHS:

[tex] \left( \sum_{ab} k_{ab}\bar{\xi^a}\xi^b \right)^2= \left( \sum_{ab} k_{ab}\bar{\xi^a}\xi^b \right)\left( \sum_{cd} k_{cd}\bar{\xi^c}\xi^d \right)

=\left( \sum_{abcd} k_{ab}k_{cd}\bar{\xi^a}\xi^b\bar{\xi^c}\xi^d \right)[/tex]

Now focussing on LHS to bring the correct ordering of the [tex] \xi[/tex]'s:

[tex] \sum_{abcd} f_{abcd} \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}=\sum_{abcd} -f_{abcd} \xi^a\bar{\xi^c}\xi^b\bar{\xi^d}=\sum_{abcd} f_{abcd} \bar{\xi^c}\xi^a\xi^b\bar{\xi^d}= \sum_{abcd} -f_{abcd} \bar{\xi^c}\xi^a\bar{\xi^d}\xi^b[/tex]

Comparing LHS and RHS now, and relabelling dummies from RHS:

[tex] \sum_{abcd} -f_{abcd} \bar{\xi^c}\xi^a\bar{\xi^d}\xi^b= \sum_{abcd} k_{ca}k_{db}\bar{\xi^c}\xi^a\bar{\xi^d}\xi^b [/tex]

I'm not sure at this point if we can simply equate [tex] -f_{abcd}=k_{ca}k_{db} [/tex] however. Since the product of [tex] \xi\xi\xi\xi [/tex] is antisymetric on exchange of c and a, and antisymmetric on exchange of any two indices (c and a, d and b, a and d but also c and d, a and b, and c and b). I'm thinking it must be the antisymmetric part of [tex]f_{abcd} [/tex] and [tex]k_{ca}k_{db} [/tex] that we can equate with each other.

I'm not sure how to find the antisymmtric part on all four indices however, if that is indeed the right track.

Appreciate any help at all, thanks alot