Functional Derivative with respect to Dirac Spinors

In summary, the functional derivative with respect to Dirac spinors involves finding the change in a functional caused by an infinitesimal variation in the spinor field. This is a crucial tool in quantum field theory and is used to derive equations of motion for spinor fields. By taking the functional derivative, one can find the variation of the action with respect to the spinor field and ultimately solve for the dynamics of the system. This concept is fundamental in understanding the behavior of fermionic particles in quantum systems and has many important applications in theoretical physics.
  • #1
ur22

Homework Statement



I am currently working on an exercise list where I need to calculate the second functional derivative with respect to Grassmann valued fields.

$$
\dfrac{\overrightarrow{\delta}}{\delta \psi_{\alpha} (-p)} \left( \int_{x} \widetilde{\bar{\psi}}_{\mu} (x) i \partial_{s}^{\mu \nu} \widetilde{\psi}_{\nu} (x) \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} \\
$$

where the ##\psi## are euclidean Grassmann valued Dirac spinors. The tilde denotes Spinors in position space and the index ##s## stands for Feynman slashes
$$
a_{s} = \gamma^{\mu} a_{\mu}
$$

Homework Equations



I used

$$
\int_{x} \equiv \int\mathrm{d}^{4}x \\
\int_{p} \equiv \int \dfrac{\mathrm{d}^{4}p}{(2 \pi)^{4}}
$$

The convolution is defined as:
$$
\left( F \ast G \right) (p) \equiv \int_{- \infty}^{+ \infty} F(\lambda) G(p - \lambda) \mathrm{d} \lambda
$$

The Attempt at a Solution



My calculation reads:

$$
\dfrac{\overrightarrow{\delta}}{\delta \psi_{\alpha} (-p)} \left( \int_{x} \widetilde{\bar{\psi}}_{\mu} (x) i \partial_{s}^{\mu \nu} \widetilde{\psi}_{\nu} (x) \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
\dfrac{\overrightarrow{\delta}}{\delta \psi_{\alpha} (-p)} \left( \int_{x} \int_{p'} e^{-ip'x} \left[ -\int d \lambda \, \bar{\psi}_{\mu} (\lambda) (p_{s}'^{\mu \nu} - \lambda_{s}^{\mu \nu}) \psi_{\nu}(p' - \lambda) \right] \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
\left( \int_{x} \int_{p'} e^{-ip'x} \left[ \int d \lambda \, \bar{\psi}_{\mu} (\lambda) (p_{s}'^{\mu \nu} - \lambda_{s}^{\mu \nu}) \delta^{\alpha}_{\nu} \delta (\lambda - (p' + p)) \right] \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
\left( \int_{x} \int_{p'} e^{-ip'x} \bar{\psi}_{\mu} (p' + p) (-p_{s}^{\mu \alpha}) \right) \dfrac{\overleftarrow{\delta}}{\delta \bar{\psi}_{\beta} (-q)} = \\
%
- \int_{x} \int_{p'} e^{-ip'x} \delta^{\beta}_{\mu} \delta (p'+p+q) \, p_{s}^{\mu \alpha} = \\
%
- \dfrac{1}{(2 \pi)^4} \int_{x} e^{i (p+q) x} \, p_{s}^{\beta \alpha} = \\
%
- p_{s}^{\beta \alpha} \delta (p+q) = \\
%
- (p_{s}^{T})^{\alpha \beta} \delta (p+q)
$$However, the solution of the exercise reads:

$$
- p_{s}^{\alpha \beta} (2 \pi)^{4} \delta (p-q)
$$

The factor ##(2 \pi)^{4}## might be due to different conventions regarding the Fourier Transform, but what's the deal with the different arguments in the delta function and the transposition of ##p_s##?Best regards
 
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  • #2
,

Thank you for sharing your approach and question. After reviewing your calculations, I have identified two potential issues that may have led to the discrepancy between your solution and the given solution.

Firstly, in your third line of calculation, when you use the delta function to eliminate the integral over ##\lambda##, you should also take into account the delta function in the second term of the action. This will result in an additional factor of ##\delta(p' - (p + q))## in your final answer.

Secondly, the transposition of ##p_s## in the given solution may be due to the definition of the Feynman slash operator, which is sometimes written as ##\gamma^{\mu} p_{\mu}## instead of ##p_{\mu} \gamma^{\mu}##. This would result in a transposition of the spinor indices in the final answer.

I hope this helps clarify the discrepancy in your solution. Keep up the good work!
 

1. What is the definition of a functional derivative with respect to Dirac spinors?

A functional derivative with respect to Dirac spinors is a mathematical concept used in quantum field theory to calculate the change in a functional (a function of a function) with respect to a small variation in a Dirac spinor field. It is represented by the symbol ∄.

2. How is a functional derivative with respect to Dirac spinors different from a regular derivative?

A functional derivative with respect to Dirac spinors is different from a regular derivative in that it operates on a function of a function instead of just a single variable. It is also a functional operator, meaning that it maps a function to another function.

3. What are some common applications of functional derivatives with respect to Dirac spinors?

Functional derivatives with respect to Dirac spinors are commonly used in quantum field theory to solve for the equations of motion of fermionic fields. They are also used in the calculation of scattering amplitudes and in the path integral formulation of quantum mechanics.

4. How do you compute a functional derivative with respect to Dirac spinors?

The computation of a functional derivative with respect to Dirac spinors involves taking the derivative of the action functional, which is a functional of the Dirac spinor field, with respect to small variations in the spinor field. This is typically done using the Euler-Lagrange equations.

5. What are the properties of functional derivatives with respect to Dirac spinors?

Functional derivatives with respect to Dirac spinors have several important properties, including linearity, chain rule, and integration by parts. They also satisfy a fundamental identity known as the Leibniz rule, which states that the derivative of a product is equal to the product of the derivatives.

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