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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: $$... 0 Fermions are strange beasts in many ways. The first problem you will encounter, and which will make it impossible to write an harmonic oscillator for fermions is the following: The fermion ladder operators f and f^\dagger require that \{f,f^\dagger\}=1. Translated to X and P this means that \{X,P\}=i\hbar. But is also means that \{X,X\}=0 and ... 1 Assuming that X=X^\dagger, P=P^\dagger and [X,P] = i\hbar, let me try$$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 + ...
The variation $\delta F$ for any field (or degree of freedom) $F$, given an infinitesimal transformation, is always calculated as the commutator $$\delta F = [ \bar\epsilon Q, F ]$$ where $\bar \epsilon$ is a parameter ("angle" or "shift" or some generalization) of the transformation and $Q$ is the generator. (Those may be replaced by other letters.) ...