How to manipulate indices when Grassmannian numbers are present?

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Homework Help Overview

The discussion revolves around manipulating indices in expressions involving Grassmannian numbers, specifically focusing on the evaluation of derivatives with respect to these variables. The original poster presents a term involving anticommuting variables and seeks clarification on the implications of their properties during differentiation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to differentiate a term involving Grassmann variables and questions the validity of their result, suspecting it may be zero due to the properties of the involved quantities.
  • Some participants question the reasoning behind the original poster's conclusion, pointing out potential errors in index handling and the need for clarity on the definition of derivatives in this context.
  • There are inquiries about the distinction between left and right derivatives, indicating a deeper exploration of the mathematical framework involved.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's calculations and raising important questions about the definitions and properties of derivatives in the context of Grassmann algebras. There is no explicit consensus yet, as various interpretations and clarifications are being explored.

Contextual Notes

Participants are navigating the complexities of Grassmann numbers, including their anticommutative properties and the implications for differentiation. The original poster's assumptions and the notation used are under scrutiny, highlighting the need for precision in this mathematical framework.

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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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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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What is the definition of derivative here? I.e. upper- vs lower case index
 
Also you must distinguish between "left" and "right" derivatives!
 
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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?
 
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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