Is it possible to write the fermionic quantum harmonic oscillator using $P$ and $X$?

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 $f^{\dagger}$?

• One problem you have is that $f^\dagger$ as you define it is not the hermitian conjugate of $f$, because you require that $X^\dagger=X$ and $P^\dagger = P$. – sagittarius_a May 19 '15 at 9:08
• And I think you miss a factor of $m$ right in front of $P$ in your definition of creation operators. – sagittarius_a May 19 '15 at 9:40
• @sagittarius_a I am working in units where $m=1$ – Yossarian May 19 '15 at 9:43

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.

• $[f,f^{\dagger}]=-1$ I think – Yossarian May 19 '15 at 9:58
• Yes you are right :) – sagittarius_a May 19 '15 at 10:00
• why don't you try with $\{X,P\}=i\hbar$ instead of $[X,P]=i\hbar$ ? – Yossarian May 19 '15 at 10:01
• This might work. But its not true if $X$ and $P$ are the position and momentum operator as usually defined in quantum mechanics. – sagittarius_a May 19 '15 at 10:02

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.

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.

• You can typeset braces in MathJax using the command \{ – Mark Mitchison May 19 '15 at 10:08
• how do you conclude that $\{f,f^{\dagger}\}=1$ implies $\{X,P\}=i\hbar$? what formulas are you using for $f$ and $f^{\dagger}$? – Yossarian May 19 '15 at 10:12
• Hmm. I am not convinced about $\left\lbrace X,X\right\rbrace=0$. Can you elaborate your point? From my understanding $X$ and $P$ are still ordinary one-particle operators. To have a ladder operator that is fermionic, we just need to have a quantum systems with two possible states: the ground state and the first excited state. – sagittarius_a May 19 '15 at 10:25
• @silvrfück. I'm not using any formulas here. I'm simply arguing that if $f$ and $f^\dagger$ are fermionic operators, then so are $X$ and $P$. – Mikael Fremling May 19 '15 at 10:26
• The Majorana fermion operators $x \sim (f+f^\dagger)/\sqrt{2}$ and $p \sim i(f-f^\dagger)/\sqrt{2}$ do not satisfy your requirement $\{x,x\} = 0$ etc. Actually $\{x,x\} = 1$. So it is not the case that linear combinations of fermion operators are always nilpotent. – Mark Mitchison May 19 '15 at 12:05