Vacuum vector and basis defined by anti-commuting operators

Question

Let $\mathcal{H}$ be a finite-dimensional inner product space over $\mathbb{C}$. Suppose $A_{1},...,A_{N}$ are linear operators on $\mathcal{H}$ such that: $$\{A_{i},A_{j}\} = 0 \quad \mbox{and} \quad \{A_{i}^{*},A_{j}\} = \delta_{ij}1$$ with $1$ being the identity matrix, $A_{i}^{*}$ being the adjoint of $A_{i}$ and $\{A,B\} := AB+BA$. It is known that, in this case, $\mathcal{H}$ must have a dimension which is a multiple of $2^{N}$ and each $A_{i}$ can be represented as a matrix operator on $\mathbb{C}^{2^{N}}$.

In physics, one postulates the existence of a unique vector (vacuum vector) $\Omega \in \mathcal{H}$ such that: $$A_{i} \Omega = 0 $$ for every $i=1,...,N$ and such that $\Omega$, $\Omega_{i}$, $\Omega_{i_{1}, i_{2}}$ ($i_{1}<i_{2}$),..., $\Omega_{i_{1},...,i_{N}}$ $(i_{1}<\cdots < i_{n}$) form a basis for $\mathcal{H}$, with: $$\Omega_{i_{1},...,i_{k}} = A_{i_{1}}^{*}\cdots A_{i_{k}}^{*}\Omega.$$

Question: Why does $\Omega$ exists and is unique? What about $\Omega = 0$? And why is it so evident that those $\Omega_{i_{1},...,i_{k}}$ as defined above form a basis for $\mathcal{H}$? I know these are $2^{N}$ vectors, so it would suffice to prove linear independence, and this could (in principle, assuming neither of these vectors are the zero vector) be proved because these vectors are all pairwise orthogonal, but how to garantee that neither of them are the zero vector?

Answer 1

1. There is a general algebraic result which states that the abstractly defined associative $\mathbb{R}$-algebra with generators $X_1,\ldots,X_n$, $Y_1,\ldots,Y_n$ and relations $$ X_iX_j+X_jX_i=0, \quad Y_iY_j+Y_jY_i=0,\quad X_iY_j+Y_jX_i=\delta_{i,j}1 $$ for all $i,j=1,\ldots,n$, is of dimension $2^{2n}$ and is isomorphic to the algebra of $2^n\times 2^n$-matrices. In fact, one can do better: those matrices are matrices of all possible linear transformations of the Grassmann algebra $\Lambda(Y_1,\ldots,Y_n)$, on which $Y_i$ acts by multiplication, and $X_i$ acts as the odd derivation $\frac{\partial\phantom{Y_i}}{\partial Y_i}$.

To prove the latter point, one can argue as follows:

A) the operators of multiplication by $Y_i$ and odd derivatives $\frac{\partial\phantom{Y_i}}{\partial Y_j}$ satisfy these commuting relations, this is an easy calculation, so there is a surjective map from the abstract algebra A with the commutation relations $$ X_iX_j+X_jX_i=0, \quad Y_iY_j+Y_jY_i=0,\quad X_iY_j+Y_jX_i=\delta_{i,j}1 $$ to the algebra of "differential operators" on the Grassmann algebra (algebra generated by multiplications by $Y_i$ and odd derivatives $\frac{\partial\phantom{Y_i}}{\partial Y_j}$).

B) every endomorphism of the matrix algebra can be represented by a differential operator, since it is easy to express each matrix unit sending $Y_I=Y_{i_1}\wedge\cdots\wedge Y_{i_k}$ to $Y_J=Y_{j_1}\wedge\cdots\wedge Y_{j_p}$ and others to zero: first multiply by all the $Y$'s not in $I$, then take derivative with respect to each $Y_i$ once, then multiply by all the $Y$'s from $J$.

C) Thus, we have a surjective map from our algebra to the matrix algebra, but our commutation relation allow to put all $X$'s before $Y$'s and order them, which shows that the dimension of our algebra is at most $2^{2n}$, and therefore the surjective map we constructed is an isomorphism.

2. Since we are working with the matrix algebra, every module over it is a direct sum of several copies of the standard module. In particular, every module of dimension $2^n$ is isomorphic to the one given by the action on the Grassmann algebra. In particular, there is, up to rescaling, just one vector annihilated by all $X_i$.

3. All the questions in your post are addressed by this. Roughly speaking, you define a representation $\rho$ of this algebra by putting $\rho(X_i)=A_i$, $\rho(Y_i)=A_i^*$. The only thing to take care of is that you work over $\mathbb{C}$, not over $\mathbb{R}$, but this you can take care of by separating real and imaginary parts of your operators.

Answer 2

Proof of the existence of $\Omega$ :

It is easy to see that the number operators $N_i = A_i^* A_i$ commute. Therefore there exists an orthonomal basis of common eigenvectors. Now let $\Omega$ one of these vectors which is a ground state of $N = \sum_i {N_i}$ . If there would be an $i$ such that $N_i \Omega \neq 0$ than $A_i \Omega$ would be an eigenvector of $N$ with smaller eigenvalue than the eigenvalue of $\Omega$ . Contradiction !

Answer 3

The easiest way to construct the vacuum state and the basis states is to consider the fermionic Fock space associated to the $2^N$ dimensional Hilbert space. The $2^N$ basis states of Fock space are states $|n_1,n_2,\ldots n_N\rangle$ described by $N$ binary occupation numbers $n_i\in\{0,1\}$. If the operator $A_i$ acts on a basis state two outcomes are possible: If $n_i=0$ the state is annihilated, meaning the outcome is the zero vector, while if $n_i=1$ that occupation number is lowered to $0$.

The vacuum state is then the state $\Omega=|0,0,\ldots 0\rangle$ with all $n_i$'s equal to zero, since that is the unique state that is annihilated by all $A_i$'s.

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Further questions in the OP:

"What about 0?" --- that vector is not allowed as a vacuum state, because by definition a state must be normalizable.

Construction of basis states: Since $\{A_i^\ast,A_j\}=\delta_{ij}$, it follows that the operator $A_i^\ast$ increases an occupation number from 0 to 1, so the state $\Omega_{i_{1},...,i_{k}} = A_{i_{1}}^{*}\cdots A_{i_{k}}^{*}\Omega$ is the basis state with all occupation numbers 0 except $n_{i_1},\ldots n_{i_k}$ equal to 1.