Tag Info

New answers tagged

2

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 + ...


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.) ...



Top 50 recent answers are included