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Delta function for grassmann numbers?

  1. Dec 19, 2009 #1
    Claim: if [tex]\psi[/tex] is a variable grassmann number, then [tex]\delta(\psi)=\psi[/tex] is a Dirac delta function for integrals over [tex]\psi[/tex].

    I'm not seeing this.

    A general function of a grassmann number can be written [tex]f(\psi)=a+b\psi[/tex] because anti-commutativity requires [tex]\psi^2=0[/tex]. According to the wikipedia article (which doesn't elaborate on why), integrals satisfy

    [tex]\int 1 d\psi=0[/tex]
    [tex]\int \psi d\psi=1[/tex]

    Ok. So let's check.

    [tex]\int\delta(\psi-\psi')f(\psi)d\psi[/tex]
    [tex]=\int(\psi-\psi')f(\psi)d\psi[/tex]
    [tex]=\int(\psi-\psi')(a+b\psi)d\psi[/tex]
    [tex]=a\int \psi d\psi + 0 -a\psi'\int d\psi-b\psi'\int\psi d\psi[/tex]
    [tex]=a - 0 -b\psi'[/tex]
    [tex]=f(-\psi')[/tex]
    [tex]\neq f(\psi')[/tex]

    Are one of my assumptions wrong?
     
  2. jcsd
  3. Dec 19, 2009 #2

    Hurkyl

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    The definition of [itex]\delta[/itex] is [itex]\int_{-\infty}^{+\infty} \delta(x) f(x) \, dx = f(0)[/itex].

    Does that change of variable make sense for such integrals? Does it make sense for f?
     
  4. Dec 19, 2009 #3
    I don't know, Hurkyl. I'm not even sure what integration itself means here. I am just following the steps in a formal sort of way. Warren Siegel on page 48 of Fields has that [tex]\int d\psi'(\psi'-\psi)f(\psi')=f(\psi)[/tex] (the prime-unprimed reverse of what I have above.) But I get [tex]f(-\psi)[/tex] .

    Never mind if we can call it the delta function. Siegel himself puts "anti-commuting delta function" in quotes.
     
  5. Dec 19, 2009 #4

    Hurkyl

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    I remember trying to read through that book; I stopped after finding that section extremely off-putting. :frown:

    Anyways, one thing Siegel says in that section is if we write [itex]f(\psi) = a + b \psi[/itex], then a and b aren't necessarily complex scalars -- they can be commuting numbers or anticommuting numbers or whatever.


    Assuming that we really do have
    [tex]\int d\psi (a + b \psi) = b[/tex]
    whenever a and b don't involve psi, then if I take special care not to pass any variable through another I get

    [tex]\int d\psi' (\psi' - \psi) f(\psi') = -\psi b + \int d\psi' \psi' a[/tex]

    which can be rearranged to

    [tex]= f(\psi) - \{\psi, b\} + \int d\psi' [\psi', a][/tex]

    of course, it can be rearranged into other things -- such as

    [tex]= f(-\psi) - [\psi, b] + \int d\psi' [\psi', a][/tex]
     
    Last edited: Dec 19, 2009
  6. Dec 20, 2009 #5
    I think I'm with you about that section. I'll just let it go and press on.

    Thanks!!
     
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