Levene's test

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.

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

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

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

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

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:

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

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 Ω_Q is given by

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

i.e. Ω_Q 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).

See also

• Position and momentum space
• Momentum operator

References

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