# Equivalence between grassmann fermion states and $SO(2N,\mathbb{R})$ fermion coherent states

I am importing this question from https://www.physicsoverflow.org/39342/equivalence-between-grassmann-fermion-mathbb-fermion-coherent

Cahill and Glauber in the paper 'Density operators for Fermions' construct Fermionic coherent state as the displaced vacuum state

$$\vert\boldsymbol{\gamma}\rangle=D\left(\boldsymbol{\gamma}\right)\vert 0\rangle=\exp\left(\sum_{i}c_{i}^{\dagger}\gamma_{i}-\gamma_{i}^{\star}c_{i}\right)\vert 0\rangle$$

where $$\boldsymbol{\gamma}$$ is the set of the Grassmann variables $$\gamma_i$$, one for each fermionic mode $$c_i$$.

Perelomov in the book 'Generalized Coherent States and Their Applications' constructs the Fermionic coherent state in a different manner. He first shows that it is the group $$SO(2N,\mathbb{R})$$ that preserves the canonical anti-commutation relations obeyed by the fermionic modes $$c_i$$ and that the representation $$T(g)$$ of an arbitrary element of this group $$g\in SO(2N,\mathbb{R})$$ is

$$T\left(g\right)={\cal N}\exp\left(-\frac{1}{2}\xi_{ij}X^{ij}\right)\exp\left(\alpha_{k}^{l}X_{l}^{k}\right)\exp\left(-\frac{1}{2}\eta^{ij}X_{ij}\right)$$

where $$X_{ij}=a_{i}a_{j}$$, $$X^{ij}=a_{j}^{\dagger}a_{i}^{\dagger}$$ and $$X_{l}^{k}=\frac{1}{2}\left(a_{k}^{\dagger}a_{l}+a_{l}^{\dagger}a_{k}\right)$$. Coherent states are constructed by acting an element $$g$$ of this group on a vector $$\vert \phi_0 \rangle$$ that is annihilated by $$X_{ij}$$. For the Fock subspace of states with even number of fermionic modes occupied, this vector is just $$\vert 0 \rangle = \vert 0,0,....0 \rangle$$. And we get the coherent state to be

$$\vert\xi\rangle={\cal N}\exp\left(-\frac{1}{2}\xi_{ij}X^{ij}\right)\vert 0 \rangle.$$

In the book 'Coherent States and Applications in Mathematical Physics' by Monique Combescure and Didier Robert, it is claimed that the two points of view are equivalent but I am not sure where exactly they prove the equivalence. They also claim, perhaps in support of the same point, that the model of spin states (which, I believe, is used by Perelomov to construct the coherent states because it is analogous to the construction of spin coherent states) is unitarily equivalent to the fermionic Fock model using which they construct the Grassmann coherent states. Does this equivalence mean that there is a mapping between the Grassmann coherent states and the $$SO(2N,\mathbb{R})$$ coherent states? I can see that in the former, the displacement operation $$\exp\left(\sum_{i}c_{i}^{\dagger}\gamma_{i}-\gamma_{i}^{\star}c_{i}\right)\vert 0\rangle$$ has an exponent which is linear in the fermionic modes while in the latter, the operator $${\cal N}\exp\left(-\frac{1}{2}\xi_{ij}X^{ij}\right)$$ is quadratic in the fermionic modes. Mathematically, Grassmann variables are 'fermions', so even the former is quadratic in 'fermionic modes'. Is that the mapping between the two sets of coherent states? Or are the two sets of coherent states unitarily related? How would that be possible- since the former is a displaced vacuum state while the latter is constructed by the analog of squeezing and beam splitters in fermionic phase space?

I assume that you may already know the answer, but I just came across your post and wanted to give a relatively brief answer:

No, the two types of coherent states are not equivalent. Let \begin{align} \hat{\xi}^a\equiv(\hat{a}_1,\dots,\hat{a}_N,\hat{a}^\dagger,\dots,\hat{a}_N^\dagger) \end{align} a set of bosonic or fermionic creation/annihilation operators and $$|0\rangle$$ the associated bosonic or fermionic vacuum state.

We can define two types of transformations: \begin{align} \mathcal{D}(w)&=e^{i w_a\hat{\xi}^a}\,,\\ \mathcal{S}(h)&=e^{i h_{ab}\hat{\xi}^a\hat{\xi}^b}\,, \end{align} which are displacement and squeezing transformations, respectively. As both are unitary, we require $$w_a\hat{\xi}^a$$ and $$h_{ab}\hat{\xi}^a\hat{\xi}^b$$ to be Hermitian operators (we adopt Einstein's summation convention over repeated indices). Moreover, we require that $$w_a$$ is a Grassmann variable for fermions.

One can then show that the most general squeezed coherent state is given by \begin{align} |\psi\rangle=\mathcal{D}(w)\mathcal{S}(h)|0\rangle\,. \end{align} These states are also known as bosonic and fermionic Gaussian states.

Now to your questions:

• The Gilmore-Perelomov coherent states are also known as group theoretic coherent states and they are a generalization to arbitrary groups, which are usually required to be at least semi-simple and sometimes compact (like SO(2N)).
• The states resulting from pure displacement are called "regular coherent states" and the underlying group is the displacement group, which is not semi-simple. One can try to treat them with the Gilmore-Perelomov formalism, but it doesn't fit super well. The bosonic ones are heavily used to describe real physical systems, while the fermionic ones DO NOT DESCRIBE real physical states due to the Grassmann numbers contained (technically, these states are not even in Hilbert space, because we suddenly allow for multiplication with Grassmann numbers, which do not form a field, but an algebra, so I think this extended space is actually a module). Either way, as Hilbert space is still included as a subspace (sub-module), one can still use these fermionic coherent states for path-integrals and to get a resolution of the identity etc., but they do not describe physical states.
• It is the squeezing part, i.e., states with $$w=0$$, but $$h\neq0$$ that correspond to physical states for both bosons and fermions. They are called squeezed states and they are generated by semi-simple Lie groups, namely the $$\mathcal{S}(h)$$ form a unitary representation of the groups $$\mathrm{Sp}(2N,\mathbb{R})$$ for bosons and $$\mathrm{SO}(2N,\mathbb{R})$$ for fermions. It is true that Gilmore-Perelomov would refer to them as coherent states, but more precisely as the group-theoretic coherent states with respect to above Lie groups (and not the displacement group as regular coherent states).