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

+ 4 like - 0 dislike
143 views

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? 

asked Jul 18 in Theoretical Physics by half_degree_of_freedom (25 points) [ revision history ]
edited Jul 19 by half_degree_of_freedom

2 Answers

+ 3 like - 0 dislike

I think that the claimed equivalence just means that the unitary representations of SO(2N) created through the two approaches are isomorphic. The number of degrees of freedom in the parameterization of the Grassmann coherent states and the Perelomov coherent states is completely different.

This shows that there is no trivial relation between the two. Indeed, the literature on the two kinds of coherent states seems to be completely disjoint (apart from the mention Combescure's book).

On the other hand, since the representations defined are essentially the same, there should be a close connection between the two approaches. In a vague, informal sense, the Perelomov coherent states are to the Grassmann coherent states what the squeezed coherent stated are to the Glauber (or Schrödinger) coherent states.

I'd be interested myself in having such a connection made definite and spelled out somewhere - but currently there is no such somewhere.

answered Jul 22 by Arnold Neumaier (12,355 points) [ no revision ]
+ 2 like - 0 dislike

No, the two coherent states are not equivalent, exactly due to the reason mentioned by Arnold. They do not carry the same representation of $SO(2N)$.

The first coherent state $|\mathbf{\gamma} \rangle$ is a coherent state of the Berezin-Schwinger group $\mathfrak{bs} = \mathbb{R}^{0, 2n} \rtimes  \mathbb{R}^{1, 0}$, which is the fermionic counterpart of the Weyl-Heisenberg group $\mathfrak{hw}$.  It is built as the orbit of $ \mathfrak{bs} $ through the fermionic Fock vacuum isomorphic to the coset space $\mathbb{R}^{0, 2n} = \mathfrak{bs} / \mathbb{R}^{1, 0}$. This coherent state lives in a Hilbert superspace spanned by a $2^N$ dimensional basis.

The second coherent state is $|\mathbf{\xi} \rangle$ a coherent state of $Spin(2N)$, it is a coherent state representation of one of the spinor representations of this group. It is built as the orbit of $SO(2N)$  through the Fock vacuum  which is isomorphic to the coset space  $SO^*(2N)/P = SO(2N)/U(N)$ .

It is true that the dimensions of the representation spaces are the same. But the coherent states are not the same. The reason is that a coherent state irreducibly represents the group which acts on the vacuum. In the first it is the $ \mathfrak{bs} $ group and in the second it is $SO(2N)$.

It is not hard to see in the second case that the Cartan generators of SO(2N) are represented by:

$$H_i = a_i^{\dagger} a_i - \frac{1}{2}$$

Thus the weights of the vacuum are:

$$\begin{bmatrix}  -\frac{1}{2},&-\frac{1}{2} &...&-\frac{1}{2} \end{bmatrix}^t$$

Which are exactly the weights of the lowest weight of the spinor representation.

The group $SO(2N)$ acts on the coherent state manifold of the $ \mathfrak{bs} $ group as

$$\begin{bmatrix}  a& b\\ \bar{b}& \bar{a} \end{bmatrix} \begin{bmatrix} \mathbf{c}\\ \mathbf{\bar{c}} \end {bmatrix}$$

(The matrix representation is in a $U(2N)$ basis), and:

$$\mathbf{c} = \begin{bmatrix}  c_1,&... &c_N  \end{bmatrix}^t$$

Thus the action of $SO(2N)$ on the coherent state manifold is homogeneous, therefore the vacuum is invariant. Moreover, it is not hard to see that the action of $SO(2N)$ on the coherent states spanned by a partial subset of Grassmann coordinates, for example those in which only $\gamma_1$ and $\gamma_2$ are nonvanishing is closed, because it cannot change the ghost number of the terms.

 Thus the action decomposes to a direct sum

$$\bigoplus_{k=0}^N \bigwedge_k {N}$$

$N$ is the fundamental representation.

However, in spite what was said above, the two coherent state representations are not unrelated. There is a very interesting series of works by Berceanu in which the bosonic counterpart of:

$$|\mathbf{\gamma} \rangle \otimes |\mathbf{\xi} \rangle$$

was considered, please see for example the following

http://projecteuclid.org/download/pdf_1/euclid.jgsp/1495245699

article and the following works by him citing this work.

In the bosonic case, the combined coherent state represents the group

$$\mathfrak{hw} \rtimes Sp(N)$$

whose fermionic analogy is:

$$\mathfrak{bs} \rtimes SO(2N)$$

In the bosonic case, the $\mathfrak{hw} \rtimes Sp(N)$ coherent state describes a multimode squeezed  coherent state.

All the computations in the article can be repeated for the fermionic case with the obvious changes of signs when necessary.

The fermionic coherent state should describe the ground state of some fermionic superfluid.

answered Jul 23 by David Bar Moshe (3,845 points) [ revision history ]
edited Jul 23 by David Bar Moshe

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