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 $|\psi\rangle = \exp(-\psi a^*) |0\rangle$ 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 $\psi\in \mathscr{G}_-$ 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 $|m\rangle = (a_{i_1}^*)\cdots(a_{i_M}^*)$, and similarly think of $\langle \psi|$ as linear maps which satisfy $\langle \psi|a^* = \langle \psi|\psi$ and $\langle \psi|0\rangle = 1$. 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 $V\oplus \mathscr{H}$, 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 $I\otimes a, I\otimes a^*$ 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.


1 Answer 1

  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 $${\cal F}~=~\mathbb{C}|\Omega\rangle\oplus\mathbb{C}a^{\dagger}|\Omega\rangle. $$

  2. The most pedestrian approach is probably to consider a (generalized) $\mathbb{Z}_2$-graded $R$-Hilbert space $$ {\cal H}~=~R\otimes_{\mathbb{C}}{\cal F}, $$ where $$R~=~\mathbb{C}^{1|0}\oplus\mathbb{C}^{0|1}$$ is the $\mathbb{Z}_2$-graded ring of supernumbers. E.g. the inner product $$\langle\cdot,\cdot\rangle: {\cal H} \times {\cal H}~\to ~R$$ 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.

  • $\begingroup$ Thanks for the answer, but I don’t think your answer quite captures the entire picture. As explained in my (admittedly) very long question, just the tensor product of the ring of super numbers and the fermionic Fock space is not enough to capture the commutation relations between super numbers and Fock states, i.e., a fermionic Grassman variable anti-commutes with odd particle number states and commutes with even particle number states. Hence, my “best attempt so far”. $\endgroup$ Feb 5, 2021 at 21:08
  • $\begingroup$ NB: Appropriate $\mathbb{Z}_2$-graded generalizations of tensor products, etc, are implicitly understood. $\endgroup$
    – Qmechanic
    Feb 5, 2021 at 21:14
  • $\begingroup$ So H is a module instead of a vector space, since R is not a field, right? Doesn't this have any consequences on what we can do with this space? (e.g. taking eigenvalues,etc.). $\endgroup$
    – megaleo
    Mar 15, 2021 at 22:02

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