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Where do fermionic coherent states live?

Physics Asked on May 28, 2021

Although there have been a couple of questions on fermionic coherent states, I don’t think any has answered the question "on what space do fermionic coherent states live?", or at least not to my understanding. Hopefully, someone with more knowledge can clarify the situation.

The usual "explanation" is that coherent states $|psirangle = exp(-psi a^*) |0rangle$ live on a space "larger" than the usual fermionic Fock space $mathscr{F}$ (or the exterior algebra of the single-particle Hilbert space $mathscr{H}$), where Grassman variables $psi$ are the coefficients.

I understand that you can define Grassmann variables as elements in the exterior algebra $mathscr{G}$ of an infinite-dim vector space $V$ and whether the variable is fermionic or bosonic depends on whether $psi in mathscr{G}_-=oplus Lambda^{2k+1}(V)$ or $in mathscr{G}_+=oplus Lambda^{2k}(V)$. You can then define the Grassmann integral via an abstract algebraic generalization of the Lebesgue/Riemann integral. A rigorous explanation can be found Tao’s blog.

However, my problem is how would you define/construct this "larger space"? It can’t just be the tensor product $mathscr{G} otimes mathscr{F}$ since we require that a fermionic Grassman variable $psiin mathscr{G}_-$ anti-commute with $mathscr{F}_- = oplus Lambda^{2k+1}(mathscr{H})$ and commute with $mathscr{F}_+ = oplus Lambda^{2k}(mathscr{H})$ so that $psi$ anti-commutes with the ladder operators $a,a^*$. It should also have a well-defined "inner product", in the sense that, it is a sesqui-linear map on this "larger space" and maps into the Grassmann variable $mathscr{G}$.

One attempt would be to think of fermionic coherent states as anti-linear maps on the fermionic Fock space which satisfy
$$
langle m|psi rangle=psi_{i_M} cdots psi_{i_1}
$$

where $|mrangle = (a_{i_1}^*)cdots(a_{i_M}^*)$, and similarly think of $langle psi|$ as linear maps which satisfy $langle psi|a^* = langle psi|psi$ and $langle psi|0rangle = 1$. However, I’m not sure if this is the right way to think of this problem.

Best attempt so far. After further thought, it’s possible that the "larger space" is the exterior algebra $Lambda$ of the direct sum $Voplus mathscr{H}$, so that $mathscr{F},mathscr{G}subseteq Lambda$. Also notice that as vector spaces, $Lambda$ is isomorphic to $mathscr{G}otimes mathscr{F}$, which is easily seen if we were to "push" all the Grassmann variables to the left and Fock space states to the right based on the (anti)-commutation relation. Therefore, the ladder operators $a,a^*$ are well-defined on $Lambda$ as $Iotimes a, Iotimes a^*$ on $Lambda cong mathscr{G}otimes mathscr{F}$. We can then proceed to define an "inner product" based on the anti-commutation rules. I haven’t yet worked out the details though.

One Answer

  1. Consider a fermionic Fock space ${cal F}$, which is a $mathbb{Z}_2$-graded $mathbb{C}$-Hilbert space.

    Example: Given a single Grassmann-odd creation operator $a^{dagger}$, the fermionic Fock space is $${cal F}~=~mathbb{C}|Omegarangleoplusmathbb{C}a^{dagger}|Omegarangle. $$

  2. The most pedestrian approach is probably to consider a (generalized) $mathbb{Z}_2$-graded $R$-Hilbert space $$ {cal H}~=~Rotimes_{mathbb{C}}{cal F}, $$ where $$R~=~mathbb{C}^{1|0}oplusmathbb{C}^{0|1}$$ is the $mathbb{Z}_2$-graded ring of supernumbers. E.g. the inner product $$langlecdot,cdotrangle: {cal H} times {cal H}~to ~R$$ should be $R$-sesquilinear.

  3. Returning to OP's title question, the fermionic coherent states then belongs to ${cal H}$.

  4. See also this Phys.SE post and links therein.

Correct answer by Qmechanic on May 28, 2021

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