# Are the eigenvalues of an observable always -1 and 1?

What are the necessary & sufficient conditions for a matrix to be an observable, and what is the proof that any such matrix has eigenvalues -1 and 1 (if indeed that is the case)? I ask because in the standard Bell experimental setup the measurement outputs are always -1 or 1.

Possibly related: in a previous question I asked whether the squared absolute values of the eigenvalues of a unitary matrix are always 1 (they are).

## 1 Answer

An observable only needs to be Hermitian, and can have any real eigenvalues. They don't even need to be distinct eigenvalues: if there are repeated eigenvalues, we say that the eigenspace for that eigenvalue is degenerate.

(In the case of observables on a qubit, having a repeated eigenvalue makes the observable rather uninteresting, because absolutely all pure states are eigenstates in that case; I'd be tempted to call such an observable 'degenerate' in an informal sense as well in that case — though it is on occasion useful to include $$\mathbf 1$$ in an analysis of things to do with single-qubit observables.)

In the analysis of Bell's Theorem, the reason why the observables are taken to be ones with eigenvalues $$\pm1$$ are conventional. It makes them analogous to Pauli spin operators in particular, and it makes a perfectly mixed state have expectation value $$0$$. Having eigenvalues of $$\pm1$$ also allows the expectation values of the operators to describe a bias towards one of two outcomes, and for expectation values of tensor products to be straightforwardly interpreted as a correlation coefficient of outcomes. You could prove versions of Bell's Theorem for observables with other eigenvalues, but those versions could be derived from Bell's Theorem as its usually stated.

• I'm not talking about projectors, I'm talking about projective measurement operators like $\sigma_z$ and $\sigma_x$. – ahelwer Nov 16 '18 at 20:47
• This is a use of the term 'projective measurement operator' that I'm unfamiliar with. (Which reference are you using?) These are certainly observables, and in measuring an observable one may consider the state to be 'projected' onto an eigenstate, but (a) there is nothing in that description that requires that the eigenvalues be $\pm1$, and (b) one usually describes the action of 'projecting' onto an eigenstate terms of a projector. – Niel de Beaudrap Nov 16 '18 at 20:57
• All right, then I meant observable. I will change the question! – ahelwer Nov 16 '18 at 20:59
• No, they don't have to be. (Why should they?) Examples of non-unitary observables are the spin-component operators (eigenvalues $\pm\tfrac12$ for spin-$\tfrac12$ particles; eigenvalues $-1,0,+1$ for spin-$1$ particles; etc.) and usually Hamiltonian operators (whose eigenvalues represent the energy levels of a physical system). – Niel de Beaudrap Nov 16 '18 at 21:31
• Your comment about repeated eigenvalues are the reason why I make my remark on repeated eigenvalues on qubits. The identity operator is in effect a measurement of 'do I have a physical system at all', which is not something physically interesting but is a part of the general mathematical theory. – Niel de Beaudrap Nov 16 '18 at 21:36