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Supersymmetry generators are Spinors are by definition.


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Formally, the properties "bosonic" and "fermionic" are captured by the relevant objects being part of a $\mathbb{Z}_2 = \{1,-1\}$-graded algebra (a superalgebra). The even degrees (with grading $1$) of that algebra are called "bosonic", the odd degrees (with grading $-1$) are called "fermionic". It follows from the grading that if $xy$ is even for $y$ odd, ...


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Let's start from $$H = \hbar \omega \left(f^\dagger f - \frac{1}{2}\right),$$ with $\{f, f^\dagger\}=1$, $\{f, f\} = 0$ and define fermionic position and momentum coordinates by $$ \psi_1 = \sqrt{\frac{\hbar}{2}} \left(f + f^\dagger\right) \\ \psi_2 = i\sqrt{\frac{\hbar}{2}} \left(f - f^\dagger\right) $$ with the following anticommutation relations: $$ ...


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Fermions are strange beasts in many ways. The first problem you will encounter, and which will make it impossible to write an harmonic oscillator for fermions is the following: The fermion ladder operators $f$ and $f^\dagger$ require that $\{f,f^\dagger\}=1$. Translated to $X$ and $P$ this means that $\{X,P\}=i\hbar$. But is also means that $\{X,X\}=0$ and ...


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Assuming that $X=X^\dagger$, $P=P^\dagger$ and $[X,P] = i\hbar$, let me try $$f = \sqrt{\frac{m\omega}{2\hbar}}\left( \alpha X + \frac{\beta}{m\ \omega } P \right) $$ where $\alpha$ and $\beta$ are complex numbers of modulus one. From this follows that $$ \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + ...


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The bracket you have written is of the form $$[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu).$$ As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a ...



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