# Questions tagged [superalgebra]

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### What does it exactly mean by right and left functional derivatives?

In BV formalism of the gauge theory, we need to compute the right / left functional derivatives of the actions that include fermions. I do not quite see what it means by that. For example, let us ...
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### Complex valued Grassmann variables $(\theta \eta)^*$, $(\theta \eta)^T$ and $(\theta \eta)^\dagger$

Since hermitian conjugation and complex conjugation are serious issues in a QFT lagrangian with Grassmann variables, see here and here. Let us try to go to the bottom. We start by accepting the ...
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1 vote
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### Odd or Even symplectic structure in BV formalism

I am studying Batalin-Vilkovisky formalism. I am a little bit confused on what an odd (or even) symplectic structure is (i know what the degree of the underlying 2-form is). I can not find a clear ...
1 vote
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### Logarithm of Grassmann numbers

What is $\log \theta$ where $\theta$ is a Grassmann number such that $\theta^2 = 0$. How does one then look at its logarithm? Does it even make sense?
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### Integration with complex Grassmann numbers

I have a question about a convention from Peskin & Schroeder, namely that $$\int d\theta^{*}\, d\theta \, (\theta \theta^*) = 1,$$ where $\theta$ and $\theta^*$ are independent Grassmann numbers. ...
1 vote
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### Superbundles and spin in Dirac equation

The Dirac operator $D$ of a Clifford bundle $S$ is a first order differential operator on $C^{\infty}(S)$. The Clifford bundles considered are $\mathbb{Z} / 2 \text { -graded }$ (superbundles). This ...
1 vote
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### Ward identity of QED - whether the fields are all $c$-number fields

I am following Sidney Coleman's lectures of Quantum Field Theory. At the end of ch.32, he derived the Ward identity for the 1PI generating functional $\Gamma[\psi,\bar{\psi},A_{\mu}]$ for QED: \begin{...
331 views

### Right vs Left Derivatives

Let $\theta$ be a fermionic quantity and $f(\theta)=f(0)+\theta\frac{\partial f}{\partial\theta}=f(0)+\frac{\partial_r f}{\partial\theta}\theta$. Under a variation $\theta\mapsto\theta+\delta\theta$ ...
1 vote
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### Jacobi Identity in de Sitter Superalgebra

In the book "Supergravity" (by D. Freedman and A. Van Proeyen) they talk about why $\mathcal{N}=1$ de-sitter superalgebra is impossible to construct (Section 12.6.1). Basically de-Sitter algebra is a ...
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### What is the meaning of a Grassmann variable?

I don't seem to understand the concept of a Grassmann-variable. When studying superspaces and superfields I am told that the coordinates being used are $$(x^\mu, \theta_\alpha, \bar{\theta}_\alpha)$$ ...
1 vote
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### Are there cases where the use of the Grassmann variables simplifies computations in the usual bosonic analysis?

When one introduces complex numbers and complex analysis one can then use the new machinery to solve some real-analysis problems. A lamppost example is computing integrals via residues. I think I've ...
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### Is the expectation value of a Fermi field operator a Grassmann number?

It's often noted that Bosonic fields result from quantizing classical field theories defined on a regular numbers, whereas Fermionic fields arise when quantizing a classical field theory defined on ...
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### Unitarity/Hermiticity condition for $osp(m,n|\mathbb{C})$ superalgebra

According to Dictionary on Lie Superalgebras (page 82), the compact form of $OSP(m,n|\mathbb C)$ Lie superalgebra must satisfy $M^{\text st}H\,M=1$ and $M^{\ddagger}M=1$ (is this the unitarity ...
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### What exactly are "Grassmann-valued fields"?

Peskin & Schroeder define a Grassmann field $\psi(x)$ as a function whose values are anticommuting numbers, that can be written as : [p.301 eq. 9.71] $$\psi(x) = \sum\psi_i \phi_i(x),\tag{9.71}$$ ...
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### Expand superspace function into component form

In 2D (1,1) superconformal field theory, the invariant "distance" between two points $Z_1=(z_1,\theta_1)$ and $Z_1=(z_1,\theta_1)$ in superspace is $$Z_{12}=z_1-z_2-\theta_1\theta_2.$$ My question ...
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### One question about BRST symmetry in reading Srednicki’s book: Why should the BRST charge $Q_B$ be nilpotent?

In p.453, Srednicki claims that since the BRST transformation of a BRST transformation is zero, $Q_B$, the BRST charge, must be nilpotent: $$Q_{B}^{2}=0.\tag{74.32}$$ I don't know why.
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