Interpretation of position projection operator

Question

Does the operator:

$$\int_a^b \mathrm{d}x \, |x\rangle \langle x|$$

have physical meaning acting in a Hilbert space where it does not exactly correspond to the identity operator?

Does it correspond to a measurement that can be at least in theory be performed on the system, and based on the mathematical properties of the operator what can we say about the state of the system after such a measurement.

What is the measurement that this corresponds to in terms of a question that is asked of the system? (i.e. $\hat{x}$ would be interpreted as asking "What is the position of the particle?")

What state does the wavefunction collapse to on performing such a measurement?

Answer 1

If $X^b_a := \int^b_a\lvert x\rangle\langle x \rvert\mathrm{d}x$ is not the identity operator, that means your Hilbert space is $L^2([c,d])$ with $c<a,b<d$ (for $c,d$ possibly infinite). $X^b_a$ is simply the projector onto $L^2([a,b])\subset L^2([c,d])$ since for any $\lvert \psi \rangle$ we have that the wavefunction of $X^b_a\lvert \psi\rangle$ at $x\in[c,d]$ is given by $$ \langle x\vert X^b_a\vert \psi\rangle = \int^b_a\delta(x'-x)\langle x\vert\psi\rangle\mathrm{d}x$$ which is zero if $x\notin[a,b]$ and just the usual value of the wavefunction $\langle x\vert\psi\rangle$ if $x\in[a,b]$.

As it therefore is the projector onto the $[a,b]$ part of the spectrum of the position operator, its expectation value is the probability to measure the position as being in the interval $[a,b]$.

Since the projector is self-adjoint, you can think about introducing it as an observable and measuring it. It is a very boring observable: It has two eigenvalues, $0$ and $1$, and its eigenvectors are precisely the wavefunctions that vanish on $[a,b]$ (for $0$) respectively those that vanish outside $[a,b]$ (for $1$), where "vanishing" means they are zero except on a zero-measure subset.

Therefore, if you measure it, you either get "0" and the resulting state is the wavefunction you had before with its values on $[a,b]$ zeroed out or you get "1" and the resulting state is the wavefunction restricted to $[a,b]$ (in both cases suitably normalized, of course). The "question" this measurement corresponds to is very simply "Is the particle inside $[a,b]$?".

Answer 2

In simple terms I can write a state $|{\psi}\rangle = \alpha_1|x\in\{y|a<y<b\}(=A)\rangle (=|\alpha_1\rangle)+\alpha_2|x\notin A\rangle(=|\alpha_2\rangle)$

When you use operator $$\int_{a}^{b}|x\rangle\langle x|dx$$

Then it will get you simply $\alpha_1|\alpha_1\rangle$.

Physical meaning: You see your operator projects to a state and if you normalize the projected state, you will get a state which has probability of finding particle between a and b is 1.

Consider a measurement equipment that tells you if your particle position is between a and b. After measurement the particle is either in one of the states $|\alpha_1\rangle$ or $|\alpha_2\rangle$ with probability $|\alpha_1|^2$ or $|\alpha_2|^2$. (once you are in one of the states you can now remove the constants to normalize it). Now if our new state is $|\alpha_1\rangle$ then your operator is Identity operator to it.