The Hamiltonian of the quantum harmonic oscillator is $$\mathcal{H}=\frac{P^2}{2m}+\frac{1}{2}m\omega^2X^2$$ and we can define creation and annihilation operators $$b=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}b^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(X-\frac{i}{\omega}P)$$ where the following commutation relations are fulfilled $$[X,P]=i\hbar\qquad{}[b,b^{\dagger}]=1$$ and the Hamoltonian can be written $$\cal{H}=\hbar\omega\left(b^{\dagger}b+\frac{1}{2}\right).$$ Now, it is also known that we can define a fermionic quantum harmonic oscillator with the Hamiltonian $$\cal{H}=\hbar\omega\left(f^{\dagger}f-\frac{1}{2}\right)$$ where $f$ and $f^{\dagger}$ satisty the following anticommutation relation $$\{f,f^{\dagger}\}=1.$$ What I am trying to get is a Hamiltonian for the fermionic harmonic oscillator using $P$ and $X$. I have tried defining $$f=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{\omega}P)\qquad{}f^{\dagger}=\sqrt{\frac{m\omega}{2\hbar}}(-X-\frac{i}{\omega}P)$$ because after imposing the anticommutation relation $\{X,P\}=i\hbar$ for $X$ and $P$ (as I guess would suit a fermionic system) these definitions of $f$ and $f^{\dagger}$ imply $\{f,f^{\dagger}\}=1$. Nonetheless, for the Hamiltonian I get $$\mathcal{H}=\frac{P^2}{2m}-\frac{1}{2}m\omega^2X^2$$ where I get an undesired minus sign. My question is then the following: is it possible (with an appropriate definition of $f$ and $f^{\dagger}$ in terms of $X$ and $P$) to obtain the first hamiltonian I have written from the fermionic oscillator Hamiltonian written in terms of $f$ and $f^{\dagger}$?
3 Answers
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: $$ \{\psi_i, \psi_j\} = \hbar \delta_{ij}.$$ So the operators anticommute with each other and square to $\hbar/2$.
We the find the Hamiltonian formulated in the new coordinates $$H = -i \omega \psi_1 \psi_2,$$
which clearly gives rise to oscillatory motion, as can be seen by calculating the Heisenberg equations of motion: $$\dot \psi_1 = -\omega \psi_2 \\ \dot \psi_2 = +\omega\psi_1. $$
This doesn't have the form you expected it to have, but that just shows the weirdness of fermionic degrees of freedom.
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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
$$ \hbar \omega \left( f^\dagger f - \frac{1}{2} \right) = \frac{P^2}{2m}+ \frac{1}{2} m \omega^2 X^2 + \hbar \omega \left(\alpha^\ast\beta \frac{XP}{2\hbar} + \alpha\beta^\ast \frac{PX}{2\hbar}- \frac{1}{2} \right) $$
You see now why I chose $\alpha$ and $\beta$ the way I did. We recover the original Hamiltonian if
$$ i\alpha^\ast\beta\ XP + i\alpha\beta^\ast\ PX \overset{!}{=} i\hbar = [X,P] $$
is fulfilled. Thus, we are led to the conclusion $\alpha\beta^\ast = i$. Two complex numbers of modulus one that fulfill this equation are $\alpha = i$ and $\beta = 1$ and therefore
$$f = \sqrt{\frac{m\omega}{2\hbar}}\left( i X + \frac{1}{m\ \omega } P \right) $$
could be a possible canditate. So remarkably we get $f = i b^\dagger$. We can check the result by inserting this relation
$$f^\dagger f - \frac{1}{2} = (-i b)(+i b^\dagger)- \frac{1}{2} = bb^\dagger - \frac{1}{2} = b^\dagger b + \frac{1}{2}$$
where the last step follows from $[b,b^\dagger] = 1$. But unfortunately
$$\left\lbrace f,f^\dagger\right\rbrace = f f^\dagger + f^\dagger f = b^\dagger b + b b^\dagger \neq 1$$
and $[f,f^\dagger] = -1$. You will always get a boson operator. Which makes perfectly sense if you think about it. A fermionic ladder operator would imply that your system suddenly has only two states left while you found infinitely many before. If you want to have a fermionic oscillator something has to happen with the Hamiltonian and the assumptions have to be altered.
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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 $\{P,P\}=0$ since they are now fermionic operators. As a result the Hamiltonian can at most have bilinear terms in $X$ and $P$.
Especially the terms $X^2$ and $P^2$ are forbidden, so no "Harmonic oscillator"-style Hamiltonian exists.
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