3

The question is in the title. I have an algebra of Hermitian operators that satisfy: \begin{align} \{O_i, O_j\} = 2 \delta_{ij} \end{align} that means all of those operators have eigenvalues $\pm 1$, and I'm looking for a way to represent those operators. Because of the anticommutation, none of the operators can share an eigenbasis: If $|+, i\rangle$ is an eigenstate of $O_i$, then \begin{align} O_j O_i |+, i\rangle = O_j |+, i \rangle = ( O_j |+, i \rangle) \end{align}

But also:

\begin{align} O_j O_i |+, i\rangle = - O_i O_j |+, i\rangle = - O_i (O_j |+, i\rangle) \end{align} That means $O_j |+, i\rangle$ is an eigenstate of $O_i$ with eigenvalue $-1$. The action of any $O_j$ on an eigenvector of $O_i$ is turning it to an eigenvector of $O_i$ with the negative eigenvalue.

I find this particular system represented with the $O_i$ being the pauli matrices, acting on a two dimensional system, here we have 3 observables, acting on a two dimensional system. If I have $N$ observables, can I find (at least with certain $N$) a minimum of dimensions that is required?

To ask the reverse question: For a 2-state system, I can't have a fourth spin operator, which satisfies the same relations as the other two - 3 seems to be the maximum number of operators. Is there a general rule to that, for systems of higher dimensionality?

Qmechanic
  • 201,751
Quantumwhisp
  • 6,733
  • 2
  • 19
  • 50

1 Answers1

4

The answer $d= 2^{\lfloor N/2\rfloor}$ is found in many books as an application of fermionic Fock spaces and also here and here. I would give a more detailed account, but this sounds like a homewok question.

mike stone
  • 52,996
  • Perhaps you can expand the answer (e.g. give an explicit reference or be more detailed here)... – Tobias Fünke Nov 22 '22 at 13:44
  • Aren't the mentioned pauli matrices $\sigma_i$ already a counter example? 3 of them satisfy the mentioned relation, the dimension should then be not 2, but $2^{1,5}$. – Quantumwhisp Nov 22 '22 at 13:47
  • The floor function $\lfloor x \rfloor$ means the integer part, so $\lfloor 1.5 \rfloor=1$. – mike stone Nov 22 '22 at 13:49
  • @obias Fünke I would, but I suspect that this is a homework problem. – mike stone Nov 22 '22 at 13:50
  • @mikestone it is not. I try to get a deeper understanding of the the structure of hilbert spaces that represent such an operator algebra as the one mentioned, and trying to find the minimum number of base states / or alternatively the maximum number of observables seemed like a good starting point. – Quantumwhisp Nov 22 '22 at 13:55
  • 1
    I suggest, then, that you look at my arxiv paper arXiv:2009.00518 as an entry to the literature. Section two lays out the standard construction of the gamma matrices. – mike stone Nov 22 '22 at 14:02