# Interpretation of position projection operator

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?

• Feb 16, 2016 at 18:06

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 $[a,b]$?".

• Thanks, helpful to have the maths in a more rigorous form. But you haven't answered the question of what state the wavefunction collapses to upon performing this measurement.
– J.L.
Feb 16, 2016 at 18:28
• @J.L.: What "measurement"? In general, measuring gives you an eigenstate of the operator that is being measured, what about that is confusing you and what does it have to do with this projector? Feb 16, 2016 at 18:32
• That is what I am asking, does a projection operator in general correspond to a measurement? Or can they only be interpreted via the Born-Interpretation.
– J.L.
Feb 16, 2016 at 18:36
• @J.L.: Projectors, or, more precisely, projection valued-measures are closely related to measurements and needed to formally express such things as "probability to measure position in the interval $[a,b]$". I don't know what you mean by "does a projection operator in general correspond to a measurement". Feb 16, 2016 at 18:43
• Sorry I'm not being very clear. I mean that in elementary QM courses you are told that an operator such as $\int dp \, p | p\rangle \langle p |$ corresponds to a measurement observing momentum (p). You are told that the eigenvalues correspond to possible observations for a single measurement and the probability amplitude squared of the corresponding eigenstate tell you the probability of obtaining that eigenvalue in your measurement. I was wondering if projection operators can be interpreted in this way or not.
– J.L.
Feb 16, 2016 at 18:51

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.