# Position operator

(Redirected from Position (quantum mechanics))

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 $\psi$ represents the probability density of finding the particle at position $x$. Hence the expected value of a measurement of the position of the particle is

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

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

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

## Eigenstates

The eigenfunctions of the position operator, represented in position basis, are dirac delta functions.

To show this, suppose $\psi$ is an eigenstate of the position operator with eigenvalue $x_0$. We write the eigenvalue equation in position coordinates,

$\hat{x}\psi(x) = x \psi(x) = x_0 \psi(x)$

recalling that $\hat{x}$ simply multiplies the function by $x$ in position representation. Since $x$ is a variable while $x_0$ is a constant, $\psi$ must be zero everywhere except at $x = x_0$. The normalized solution to this is

$\psi(x) = \delta(x - x_0)$

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 $x_0$). 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 $\psi(\bold{r},t)$ and the expectation value of the position is

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

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

$\bold{\hat{r}}\psi=\bold{r}\psi$

## Momentum space

In momentum space, the position operator in one dimension is

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

## 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]

$Q (\psi)(x) = x \psi (x)$

with domain

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

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:

$Q = \int \lambda d \Omega_Q(\lambda).$

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

$\Omega_Q(B) \psi = \chi _B \psi ,$

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

$|\Omega_Q(B) \psi |^2 = | \chi _B \psi |^2 = \int _B |\psi|^2 d \mu ,$

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

$\frac{\Omega_Q(B) \psi}{ \|\Omega_Q(B) \psi \|}$

or

$\frac{(1-\chi _B) \psi}{ \|(1-\chi _B) \psi \|}$, where $\| \cdots \|$ is the Hilbert space norm on L2(R).