How to manipulate indices when Grassmannian numbers are present?

In summary, the conversation discusses evaluating the derivative of a term involving repeated indices and Grassmann variables. The calculation provided in the conversation appears to have a mistake in the second term, which should contain a free r-index instead of a. The correct calculation is given as C_{rb} \psi^b + \psi^{a}C_{ar}. The conversation also mentions the importance of distinguishing between left and right derivatives, and provides a resource for further reading on differentiation in Grassmann algebras.
  • #1
LCSphysicist
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Homework Statement
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Suppose i have a term like this one (repeated indices are being summed)

$$x = \psi^a C_{ab} \psi^b$$

Such that ##C_{ab} = - C_{ba}##, and ##\{\psi^a,\psi^b\}=0##. How do i evaluate the derivative of this term with respect to ##\psi_r##?

I mean, my attempt g oes to here

$$\frac{\partial x}{\partial \psi_r} = C_{rb} \psi^b + \psi^r C_{ar}$$

But, i think this is zero!!, isnt? Ok, maybe we can't change a and b in ##x## because we have the anticommuting property of psi, but since in the term above we have one psi for each term, i can't see a problem in change a to b, using the C antisymmetry property and getting 0.

What is wrong?
 
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  • #2
Why do you think it's zero?

Your calculation can't be right; look at the indices. Your second term should contain a free r-index, not a, and hitting the second psi with your derivative should give an addition minus sign.
 
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  • #3
What is the definition of derivative here? I.e. upper- vs lower case index
 
  • #4
Also you must distinguish between "left" and "right" derivatives!
 
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  • #5
haushofer said:
Why do you think it's zero?

Your calculation can't be right; look at the indices. Your second term should contain a free r-index, not a, and hitting the second psi with your derivative should give an addition minus sign.
Ups, just tiped it wrong.

So it should be

$$C_{rb} \psi^b + \psi^{a}C_{ar}$$

By the way

"Why do you think it's zero?"

Well,

$$C_{rb} \psi^b + \psi^a C_{ar} = C_{rb} \psi^b + \psi^b C_{br} = \psi^b ( C_{rb} - C_{rb} ) = 0$$

Where i have used that ##C## is anti-symmetric, and that since we are dealing only with indices, we "could" just say that ##A_{bc}x^{c} = x^{c}A_{bc}##.

malawi_glenn said:
What is the definition of derivative here? I.e. upper- vs lower case index
Ok, let's be more specific, derivating it with respesct to ##\psi^r##. So that the indices are fine.

vanhees71 said:
Also you must distinguish between "left" and "right" derivatives!
What do you mean?
 
  • #7
LCSphysicist said:
Ok, let's be more specific, derivating it with respect to ##\psi^r##. So that the indices are fine.
The word you're looking for is differentiating. :smile:
 
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