# Position operator

In quantum mechanics, the position operator is the operator that corresponds to the position observable of a particle. The eigenvalue of the operator is the position vector of the particle.[1]

## Introduction

In one dimension, the wave function ${\displaystyle \psi }$ represents the probability density of finding the particle at position ${\displaystyle x}$. Hence the expected value of a measurement of the position of the particle is

${\displaystyle \langle x\rangle =\int _{-\infty }^{+\infty }x|\psi |^{2}dx=\int _{-\infty }^{+\infty }\psi ^{*}x\psi dx}$

Accordingly, the quantum mechanical operator corresponding to position is ${\displaystyle {\hat {x}}}$, where

${\displaystyle ({\hat {x}}\psi )(x)=x\psi (x)}$

## Eigenstates

The eigenfunctions of the position operator, represented in position basis, are dirac delta functions. To show this, suppose ${\displaystyle \psi }$ is an eigenstate of the position operator with eigenvalue ${\displaystyle x_{0}}$. We write the eigenvalue equation in position coordinates,

${\displaystyle {\hat {x}}\psi (x)=x\psi (x)=x_{0}\psi (x)}$

recalling that ${\displaystyle {\hat {x}}}$ simply multiplies the function by ${\displaystyle x}$ in position representation. Since ${\displaystyle x}$ is a variable while ${\displaystyle x_{0}}$ is a constant, ${\displaystyle \psi }$ must be zero everywhere except at ${\displaystyle x=x_{0}}$. The normalized solution to this is

${\displaystyle \psi (x)=\delta (x-x_{0})}$

Although such a state is physically unrealizable and, strictly speaking, not a function, it can be thought of as an "ideal state" whose position is known exactly (any measurement of the position always returns the eigenvalue ${\displaystyle x_{0}}$). Hence, by the uncertainty principle, nothing is known about the momentum of such a state.

## Three dimensions

The generalisation to three dimensions is straightforward. The wavefunction is now ${\displaystyle \psi ({\mathbf {r} },t)}$ and the expectation value of the position is

${\displaystyle \langle {\mathbf {r} }\rangle =\int {\mathbf {r} }|\psi |^{2}d^{3}{\mathbf {r} }}$

where the integral is taken over all space. The position operator is

${\displaystyle {\mathbf {\hat {r}} }\psi ={\mathbf {r} }\psi }$

## Momentum space

In momentum space, the position operator in one dimension is

${\displaystyle {\hat {x}}=i\hbar {\frac {d}{dp}}}$

## Formalism

Consider, for example, the case of a spinless particle moving in one spatial dimension (i.e. in a line). The state space for such a particle is L2(R), the Hilbert space of complex-valued and square-integrable (with respect to the Lebesgue measure) functions on the real line. The position operator, Q, is then defined by:[2][3]

${\displaystyle Q(\psi )(x)=x\psi (x)}$

with domain

${\displaystyle D(Q)=\{\psi \in L^{2}({\mathbf {R} })\,|\,Q\psi \in L^{2}({\mathbf {R} })\}.}$

Since all continuous functions with compact support lie in D(Q), Q is densely defined. Q, being simply multiplication by x, is a self adjoint operator, thus satisfying the requirement of a quantum mechanical observable. Immediately from the definition we can deduce that the spectrum consists of the entire real line and that Q has purely continuous spectrum, therefore no discrete eigenvalues. The three-dimensional case is defined analogously. We shall keep the one-dimensional assumption in the following discussion.

## Measurement

As with any quantum mechanical observable, in order to discuss measurement, we need to calculate the spectral resolution of Q:

${\displaystyle Q=\int \lambda d\Omega _{Q}(\lambda ).}$

Since Q is just multiplication by x, its spectral resolution is simple. For a Borel subset B of the real line, let ${\displaystyle \chi _{B}}$ denote the indicator function of B. We see that the projection-valued measure ΩQ is given by

${\displaystyle \Omega _{Q}(B)\psi =\chi _{B}\psi ,}$

i.e. ΩQ is multiplication by the indicator function of B. Therefore, if the system is prepared in state ψ, then the probability of the measured position of the particle being in a Borel set B is

${\displaystyle |\Omega _{Q}(B)\psi |^{2}=|\chi _{B}\psi |^{2}=\int _{B}|\psi |^{2}d\mu ,}$

where μ is the Lebesgue measure. After the measurement, the wave function collapses to either

or