2
$\begingroup$

Let $V$ be a finite-dimensional inner product space. Suppose $A_{1},...,A_{N}$ are anti-commuting operators, meaning that these are linear operators on $V$ that satisfy: $$A_{i}A_{j}+A_{j}A_{i} = A_{i}^{*}A_{j}^{*}+A_{j}^{*}A_{i}^{*} = 0$$ and: $$A_{i}A_{j}^{*}+A_{j}^{*}A_{i} = \delta_{ij} I$$ for every $i,j=1,...,N$. Here $I$ denotes the identity operator and $A^{*}$ is the adjoint of $A$.

The following facts can be proved to be true:

  • For each $i =1,...,N$, $A_{i}$ is an isometry from $\operatorname{Ker}(A_{i}^{*})$ to $\operatorname{Ker}(A_{i})$ and $A_{i}^{*}$ is an isometry from $\operatorname{Ker}(A_{i})$ to $\operatorname{Ker}(A_{i}^{*})$.
  • $V$ can be decomposed as a direct sum: $$V = \bigoplus_{i=1}^{N}\operatorname{Ker}(A_{i})\oplus \operatorname{Ker}(A_{i}^{*})$$

Now each $\operatorname{Ker}(A_{i})$ and $\operatorname{Ker}(A_{i}^{*})$ has dimension at least one. Hence, $V$ has dimension at least $2^{N}$.

Suppose $V$ has dimension $2^{N}$. Then, there exists a nonzero vector $\Omega \in V$, called the vacuum, such that $A_{i}\Omega = 0$ for every $i$. Moreover, the vectors: $$\Omega_{i_{1},...,i_{k}} := A_{i_{1}}^{*}\cdots A_{i_{k}}^{*}\Omega \tag{1}\label{1}$$ with $1 \le i_{1} < i_{2} < \cdots < i_{k} \le N$ form a basis for $V$.

There is something puzzling me, and this is my question. Suppose more generally that $V$ has a dimension which is a multiple of $2^{N}$. Then, the vectors given by (\ref{1}) cannot form a basis for $V$, because there are $2^{N}$ such vectors $\Omega_{i_{1},...,i_{k}}$, $1 \le i_{1} < i_{2} < \cdots < i_{k} \le N$. What should be the basis of $V$, then?

Motivation: In some particular cases, e.g. fermionic Fock spaces, the operators $A_{i}$ and $A_{i}^{*}$ are annihilation and creation operators, respectively, and the vectors (\ref{1}) form a basis of the underlying Fock space. In the abstract setting I proposed above, the dimension of $V$ need not to be exactly $2^{N}$, but more generally a multiple of this number. However, it is somehow surprising that vectors of the form (\ref{1}) do not form a basis of $V$ because these are just the usual eigenvectors of the number operators $n_{i} = A_{i}^{*}A_{i}$. I don't see how to reconcile these two scenarios.

Improve this question
$\endgroup$
3
  • $\begingroup$ I would think that if the dimension is $2^{N+M}$ you can find an additional set of $M$ anticommuting operators to construct the basis; $\endgroup$
    – Carlo Beenakker
    Commented Apr 4 at 8:32
  • $\begingroup$ @CarloBeenakker the way I proved it, the only thing I know is that the dimension of $V$ is a multiple of $2^{N}$. In principle it could be, say, $m2^{N}$ for $m$ positive integer. It is not even clear to me that the dimension should be of the form $2^{k}$, for some $k$. $\endgroup$
    – MathMath
    Commented Apr 4 at 8:58
  • $\begingroup$ You may want to check your decomposition of $V$ as a direct sum; it isn't correct (and would imply that ${\rm dim}(V) \geq 2N$, not $2^N$). But anyway the simple answer to your question is that the number operator could have repeated eigenvalues. $\endgroup$
    – Nik Weaver
    Commented Apr 4 at 11:22

0

Reset to default

You must log in to answer this question.