Performing a projective measurement, is the resulting expectation value $\langle \Psi|M|\Psi\rangle$ bounded between $+1$ and $-1$?

Question

Suppose we have a quantum state $|\Psi\rangle = \alpha|0\rangle + \beta|1\rangle$.According to a measurement operator M, the projective measurement of $|\Psi\rangle$ is given by $\langle\Psi|M|\Psi\rangle$. Is this measurement always bounded between +1 and -1? If it is, why is that?

Answer 1

The measurement operators $\{M_i\}$ obey two conditions, firstly they are positive operators, $M_i\geq 0$, which is $\forall |\psi \rangle,\, \langle \psi | M_i |\psi \rangle \geq 0$.

Secondly, they must also obey $\sum_i M_i = I$. We can then deduce that $\langle \psi | M_j |\psi \rangle \leq \sum_i \langle \psi | M_i |\psi \rangle =\langle \psi | \sum_i M_i |\psi \rangle = \langle\psi | I|\psi \rangle =1$. Where the inequality comes from the fact that the sum contains the term on the left and additional non-negative terms.

Putting this all together, we have $0 \leq \langle \psi | M_i |\psi \rangle \leq 1$.

Answer 2

The expectation value of an observable is not constrained to be between +1 and -1.

You can define $M$ to be any observable - i.e. a Hermitian operator. See https://physics.stackexchange.com/questions/27038/what-hermitian-operators-can-be-observables for example. If I have the eigendecomposition $M=\sum_\lambda \lambda |\lambda\rangle \langle \lambda|$, a single measurement value can be any of the $\lambda$ eigenvalues, that will be real due to $M$ being Hermitian.

$\langle \psi | M | \psi \rangle$ will give you the "expectation value" of these eigenvalues, which is a sum of each $\lambda$ weighted with the overlap of the eigenstate $|\lambda \rangle$ and $|\psi\rangle$.

Examples:

1. Think of a Hamiltonian, which has energy levels as the eigenvalues. Energy values are not restricted to be between +1 and -1.
2. You could measure the observable 2Z - it is a perfectly fine observable, and has eigenvalues +2 and -2.