# What is the relation between a measurement and an observable?

Observables are represented by Hermitian operators. First of all, it's a little strange (to me) that some measurable physical quantity is represented by a transformation (or linear map), given that I think of a linear map as a function and I don't think of physical quantities as functions. But this is not my doubt. I try to accept this definition.

The resulting quantum state to which a system collapses after a measurement is one of the eigenvectors of this Hermitian operator. The corresponding eigenvalue is the result of the measurement.

If I understood correctly, a measurement is performed by a transformation, that is, in linear algebraic terms, you multiply by a matrix.

Anyway, I am not fully understanding the relation and connection between the measurement and an observable (or the Hermitian operator that represents it). How are measurements and observables in quantum mechanics related?

• "If I understood correctly, a measurement is performed by a transformation, that is, in linear algebraic terms, you multiply by a matrix." - I'm not sure how this could be the case. Measurement leaves the system in an eigenstate of the observable but which eigenstate it will be in is not knowable beforehand (unless the system is known to be in an eigenstate beforehand). – Alfred Centauri Apr 22 '18 at 13:48

## 1 Answer

I give here only the simplest explanation. In quantum mechanics (Schrödinger representation), the state of a physical system is completely determined by its, in general, complex (normalized) wave function $\psi$. An observable $A$ is a physical quantity that can be measured. It is represented by a corresponding Hermitian operator $\hat A$ in the sense that results of measurements of this quantity can only be the (real) eigenvalues $a_j$ of this Hermitian operator. The corresponding (normalized) eigenfunctions $\psi_j$ form a complete orthogonal set of functions so that any wave function $\psi$ representing the state of a system can be written as a (possibly infinite) linear combination of these eigenfunctions $$\psi=\sum_j c_j\psi_j \tag1$$ where $c_j$ are complex numbers. The measurement of the observable $A$ on a system in state $\psi$ yields one of the eigenvalues $a_j$ given by $$\hat A\psi_j=a_j\psi_j \tag 2$$ of the operator $\hat A$ with probability $$P=|c_j|^2 \tag 3$$ The sum of all these probabilities is 1. After the measurement with result $a_j$, the system is in the state of the corresponding eigenfunction $\psi_j$.

The measurement is not performed by a transformation, but the results are given by the eigenvalues of the corresponding Hermitian operator. More sophisticated situation like continuous or degenerated eigenvalues and systems described by density matrices are described in many quantum mechanics textbooks.

• So, a measurement is not mathematically defined in quantum mechanics? – nbro Apr 22 '18 at 14:58
• @nbro -In QM a measurement of a system in state $\psi$ is mathematically defined in the sense that a resulting measurement value of a physical quantity has to be an eigenvalue of the corresponding operator with probability $|c_j|^2$. There is no certainty for a specific measurement result unless the state of the system is already an eigenfunction of the operator of the observable. – freecharly Apr 22 '18 at 15:09
• I understood that. But how the measurement is performed, that is, how we measure the observable (i.e. the physical quantity) is not mathematically defined, right? – nbro Apr 22 '18 at 15:11
• @nbro - You are right, how the measurement is performed is not described mathematically. – freecharly Apr 22 '18 at 15:16
• Given the explanations in your answer, it seems like a measurement is assumed to be performed when a system is in one of the possible states (i.e. one of the eigenvectors of the Hermitian representing the observable). Is this right? If this is right, does it make sense to talk about measurements without talking about the assumed state the system is in? – nbro Apr 22 '18 at 15:19