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 $f^{\dagger}$?

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    $\begingroup$ 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 $P^\dagger = P$. $\endgroup$ – sagittarius_a May 19 '15 at 9:08
  • $\begingroup$ And I think you miss a factor of $m$ right in front of $P$ in your definition of creation operators. $\endgroup$ – sagittarius_a May 19 '15 at 9:40
  • $\begingroup$ @sagittarius_a I am working in units where $m=1$ $\endgroup$ – Yossarian May 19 '15 at 9:43

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.


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.

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

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.

  • $\begingroup$ You can typeset braces in MathJax using the command \{ $\endgroup$ – Mark Mitchison May 19 '15 at 10:08
  • $\begingroup$ how do you conclude that $\{f,f^{\dagger}\}=1$ implies $\{X,P\}=i\hbar$? what formulas are you using for $f$ and $f^{\dagger}$? $\endgroup$ – Yossarian May 19 '15 at 10:12
  • $\begingroup$ 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. $\endgroup$ – sagittarius_a May 19 '15 at 10:25
  • $\begingroup$ @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 $P$. $\endgroup$ – Mikael Fremling May 19 '15 at 10:26
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    $\begingroup$ 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. $\endgroup$ – Mark Mitchison May 19 '15 at 12:05

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