Simple question about anticommuting numbers

  • Thread starter pellman
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  • #1
pellman
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... I hope.

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

http://en.wikipedia.org/wiki/Grassmann_number

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?
 

Answers and Replies

  • #2
samalkhaiat
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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?

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

and

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


sam
 
  • #3
pellman
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[tex]\int d \theta = 0, \ \ \int d \theta \ \theta = 1[/tex]

and

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


sam

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]

?
 
  • #4
Avodyne
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[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.
 
  • #5
pellman
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Thank you.
 

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