# operator invertability and measurment

1. In Quantum mechanics, we define for each physical quantity an operator (for example - the position operator) whose eigenvalues are the possible outcomes of a measurement (different positions in space). So, if we have a system in the state - $$\left|\psi\right\rangle = \left| x_0 \right\rangle + \left|x_1\right\rangle$$ after measuring its postition we will get by random either $x_0$ or $x_1$. So measurement can't be an invertible operator since from the result $x_0$ we cannot deduce the original state was $\left|\psi\right\rangle$ and not just $\left|x_0\right\rangle$. But can a measurement be defined as an operator at all?
2. Another question is whether the position operator (or any other used in QM) is invertible? It is tempting to say it is not because it has $0$ as an eigenvalue, but it is not defined in a finite dimensional space, so i'm not​ familiar with this kind of operators.

ad 1: Measurement can not be defined as an operator on Hilbert space, because after the measurement the system can be in different states (with different classical probabilities) and this can not be encoded in the state vector $|\psi\rangle$.

I'll go into some more detail. Let's say we are measuring an observable corresponding to the self-adjoint operator $$A = \sum_n a_n P_n ,$$ where $P_n = | a_n \rangle\!\langle a_n |$ are the projectors on the eigenspaces.

• After the measurement, the system will be in one of the states $|a_n\rangle$. If we don't know which one, this can not be written as a state vector $|\psi\rangle$, as explained above. However, it can be written as a state matrix $\rho$ which includes classical probabilities. In fact, measurement of $A$ corresponds to a linear superoperator on the space of state matrices: $$\rho \mapsto \sum_n P_n \rho P_n .$$

• After the measurement under the condition that we know that the result was $a_k$, the system will be in the state $|\psi\rangle = |a_k\rangle$ (assuming no degeneration). The function on Hilbert space corresponding to this conditioned measurement, $$|\psi\rangle \mapsto |a_k\rangle$$ is not linear and therefore not an operator. The effect of conditioned measurement on the state matrix is also non-linear, $$\rho \mapsto \frac{P_k \rho P_k}{\operatorname{tr}(\rho P_k)} .$$

ad 2: Yes, you can define a (self-adjoint) operator $x^{-1}$. The tricky part is that $x$ and $x^{-1}$ are unbounded operators, their domain is not the entire Hilbert space. For each $\psi \in D(x)$ we get $(x^{-1} \circ x) \psi = \psi$ and for each $\psi' \in D(x^{-1})$, $(x \circ x^{-1}) \psi' = \psi'$. However note that $D(x) \neq D(x^{-1})$. You might find this question and the comments therein interesting.

In the simplest case, measurements are represented by projection operators $\vert \phi_k\rangle\langle \phi_k\vert$, where $\vert\phi_k\rangle$ is an eigenstate of Hermitian operator corresponding to the observable under measurement. The projectors must satisfy $$\hat 1=\sum_k \vert\phi_k\rangle\langle\phi_k\vert\, . \tag{1}$$ Each projector $\vert \phi_k\rangle\langle \phi_k\vert$ is idempotent and thus non-invertible, but an operator nonetheless.

For mixed states (or more generally in the density matrix formalism), three are also generalized measurements or POVMs, which also satisfy the sum rule of (1). An example of POVM used in state discrimination can be found on this wiki page.