Calculating the expectation value of an observable over a finite region rather than an an entire distribution

Question

Given an operator $\hat{Q}(x)$, the expectation value of the observable $\langle Q \rangle$ is calculated by integrating the probability distribution over the entire space as

\begin{equation} \langle Q \rangle = \int_{-\infty}^{\infty} \psi^{*}(x) \hat{Q}(x) \psi(x) dx \end{equation}

Is it possible to calculate the expectation value $\langle Q \rangle_{ab}$ over just a finite range $x \in [a,b]$ such that

\begin{equation} \langle Q \rangle_{ab} = \int_{a}^{b} \psi^{*}(x) \hat{Q}(x) \psi(x) dx. \end{equation}

It would seem that the answer is yes. For example, a detector in the lab usually takes measurements only in a finite space rather than over an infinite space. Furthermore, as as example, it seems that for a piece-wise potential that gives three separate wave functions $\psi_{1}(x), \psi_{2}(x) \text{ and } \psi_{3}(x)$, the above calculation would be performed due to

\begin{equation} \langle Q \rangle = \int_{-\infty}^{a} \psi_{1}^{*}(x) \hat{Q}(x) \psi_{1}(x) dx + \int_{a}^{b} \psi_{2}^{*}(x) \hat{Q}(x) \psi_{2}(x) dx + \int_{b}^{\infty} \psi_{3}^{*}(x) \hat{Q}(x) \psi_{3}(x) dx. \end{equation}

Answer 1

The first question has positive answer. The second one has again positive answer if $Q$ preserves pairwise orthogonality of the three functions $\psi_j$. Roughly speaking, it happens if $Q$ commutes with the position operator.

Answer 2

The below seems to long to add as a comment, so, just trying to mathematically formalize the @Valter Moretti 's answer, namely, there is a useful approach consisting using Heaviside function. Precisely, the piece-wise your wave function is \begin{equation} \psi(x) = \psi_1(x) H(a-x) + \psi_2 ( H(x-a) + H(x-b)) + \psi_3 H(x-b) \end{equation} when calculating the expected value of $\hat{Q}$, it is seen that:

\begin{align} &\int\limits_{-\infty}^{\infty} \psi(x) \hat{Q} \psi(x) \mathrm{d} x = \int\limits_{-\infty}^{\infty} \left(\psi_1(x) H(a-x) + \psi_2 ( H(x-a) + H(x-b)) + \psi_3 H(x-b) \right) \hat{Q} \psi(x) \mathrm{d} x = \\ & = \int\limits_{-\infty}^{\infty} \psi_1(x) H(a-x) \hat{Q} \psi(x) \mathrm{d} x + \int\limits_{-\infty}^{\infty} \psi_2 ( H(x-a) + H(x-b)) \hat{Q} \psi(x) \mathrm{d} x \\ & + \int\limits_{-\infty}^{\infty} \psi_3 H(x-b) \hat{Q} \psi(x) \mathrm{d} x = \\ & = \int\limits_{-\infty}^{a} \psi_1(x) \hat{Q} \psi(x) \mathrm{d} x + \int\limits_{a}^{b} \psi_2 \hat{Q} \psi(x) \mathrm{d} x + \int\limits_{b}^{\infty} \psi_3 \hat{Q} \psi(x) \mathrm{d} x \end{align} Next, the $\hat{Q}$ operates somehow on the Heaviside function so if the function on the right commutes with the operator you can write the equation at the end of the question.