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Grassman numbers

  1. Mar 11, 2010 #1

    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.

    [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
  2. jcsd
  3. Mar 11, 2010 #2
    I think I finally solved this if anyone is interested....

    We can express RHS as:

    \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)
    =\left( \sum_{abcd} k_{ab}k_{cd}\xi^b\bar{\xi^a}\xi^d \bar{\xi^c} \right)
    =\left( \sum_{abcd} -k_{ab}k_{cd}\xi^b\xi^d\bar{\xi^a}\bar{\xi^c} \right)
    Relabel dummies:
    [tex] =\left( \sum_{abcd} -k_{ca}k_{db}\xi^a\xi^b\bar{\xi^c}\bar{\xi^d} \right)

    and LHS we had:

    \sum_{abcd} f_{abcd} \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}
    So now the coefficents are hitting the same components of the tensor. Adopting Einstein convention, we have [tex] f_{abcd} \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}=-k_{ca}k_{db}\xi^a\xi^b\bar{\xi^c}\bar{\xi^d} [/tex]

    You can think of the object [tex] \xi^a\xi^b\bar{\xi^c}\bar{\xi^d} [/tex] as a tensor in a way. But the tricky thing is to realize how its symmetry actually works. In a sense it is antisymmetric under every index pair being swapped (due to the anticom of the grassmans). e.g. [tex] \xi^a\xi^b\bar{\xi^c}\bar{\xi^d}=-\xi^a\bar{\xi^c} \xi^b\bar{\xi^d} [/tex]. But then this symmetry isn't just minus the same object (the object being the structure [tex] \xi\xi\bar{\xi}\bar{\xi} [/tex]). Where as swapping on the first two or last two indices does leave this form invariant.
    (It's easier to see this by explicitly expanding, e.g. if it was just [tex] f_{ab} \xi^a\bar{\xi^b}=g_{ab} \xi^a\bar{\xi^b} [/tex] and we only let the index run from 1,2. Then [tex] f_{12}\xi^1\bar{\xi^2}+f_{21}\xi^2\bar{\xi^1}=g_{12}\xi^1\bar{\xi^2}+g_{21}\xi^2\bar{\xi^1} [/tex]. The usual trick then would be to use the antisymmetry of the tensor to write: [tex] f_{12}\xi^1\bar{\xi^2}-f_{21}\bar{\xi^1}\xi^2=g_{12}\xi^1\bar{\xi^2}-g_{21}\bar{\xi^1}\xi^2 [/tex]. But then if it wasnt for the bars, we would be able to equate the antisymmetric parts of g and f, but the bars means, that if though of variables commute under every pair swap, the "tensor" is only antisymmetric under exchange of first two, or exchange of the second too. So actually in this case, using the arbitrary freedom of the variables, we we would have been able to deduce f=g completely.

    Thus returning to the original equation:

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

    But as noted in post one f is already antisymmetric under these exchanges, so we can equate the whole of f, with the part of the

    [tex] -k_{ca}k_{db} [/tex], which is anytisymmetrized on swapping a&b, and b&c ( not antisymmetrized on it's own first and last pair, but with the indices in accord with first and last pair of the "tensor" the coeff mutplies. )

    Thus we are finally led to:

    [tex] f_{abcd}=-\frac{1}{2^2}\left( k_{ca}k_{db}- k_{cb}k_{da} -k_{da}k_{cb}+k_{db}k_{ca} \right) [/tex]

    [tex] f_{abcd}=\frac{1}{2}\left(k_{cb}k_{da} -k_{db}k_{ca} \right) [/tex]
    Last edited: Mar 11, 2010
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