Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple question about anticommuting numbers

  1. Apr 29, 2009 #1
    ... I hope.

    I wasn't sure which math forum to put this in to get an answer, but since the application is quantum, I figured this forum would be better.


    We see here that Grassman numbers have the property

    [tex]\int [ \frac{\partial}{\partial\theta}f(\theta)]d\theta=0[/tex]

    I don't see it. Suppose [tex]f(\theta)=a + b\theta[/tex].

    Then [tex]\frac{\partial}{\partial\theta}f(\theta)=b[/tex]

    And so

    [tex]\int [ \frac{\partial}{\partial\theta}f(\theta)]d\theta=b\theta + constant[/tex]

    right? At least that is what we would get with regular numbers. How does the anti-commutativity affect that?
  2. jcsd
  3. Apr 29, 2009 #2


    User Avatar
    Science Advisor

    [tex]\int d \theta = 0, \ \ \int d \theta \ \theta = 1[/tex]


    [tex]\int d \theta f(\theta) = \frac{d}{d \theta} f(\theta) = b[/tex]

  4. Apr 29, 2009 #3
    Well, now I am totally confused. Considering a single anti-commuting variable theta, the defining property is [tex]\theta^2=0[/tex] right? Is there anything else? If not, how does that get us

    [tex]\int d \theta = 0[/tex]

  5. Apr 29, 2009 #4


    User Avatar
    Science Advisor

    [itex]\int d\theta = 0[/itex] and [itex]\int d\theta\,\theta = 1[/itex] are just definitions. They are motivated by the following considerations. An integral over [itex]\theta[/itex] is supposed to be an analog of a definite integral over a real variable [itex]x[/itex] from [itex]-\infty[/itex] to [itex]+\infty[/itex]. So consider [itex]I=\int_{-\infty}^{+\infty}dx\,f(x)[/itex]. Assume the integral coverges. One key property is linearity: if we multiply [itex]f(x)[/itex] by a constant [itex]c[/itex], the result is [itex]cI[/itex]. Another is translation invariance: if we replace [itex]f(x)[/itex] with [itex]f(x+a)[/itex], the result is still [itex]I[/itex].

    Now, for a Grassmann variable, the most general function is [itex]f(\theta)=a+b\theta[/itex]. Let [itex]I=\int d\theta\,f(\theta)[/itex]. If we want [itex]I[/itex] to be both linear and translation invariant, we must define [itex]I=b[/itex], up to a possible numerical multiplicative constant.
  6. May 2, 2009 #5
    Thank you.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook