In quantum mechanics, the position operator is the operator that corresponds to the position observable of a particle.

When the position operator is considered with a wide enough domain (e.g. the space of tempered distributions), its eigenvalues are the possible position vectors of the particle.^[1]

In one dimension, if by the symbol

$${\displaystyle |x\rangle }$$

${\displaystyle x}$

${\displaystyle |x\rangle }$

${\displaystyle x}$

Therefore, denoting the position operator by the symbol ${\displaystyle X}$ – in the literature we find also other symbols for the position operator, for instance ${\displaystyle Q}$ (from Lagrangian mechanics), ${\displaystyle {\hat {\mathrm {x} ))}$ and so on – we can write

$${\displaystyle X|x\rangle =x|x\rangle ,}$$

${\displaystyle x}$

One possible realization of the unitary state with position ${\displaystyle x}$ is the Dirac delta (function) distribution centered at the position ${\displaystyle x}$, often denoted by ${\displaystyle \delta _{x))$.

In quantum mechanics, the ordered (continuous) family of all Dirac distributions, i.e. the family

$${\displaystyle \delta =(\delta _{x})_{x\in \mathbb {R} },}$$

${\displaystyle X}$

It is fundamental to observe that there exists only one linear continuous endomorphism ${\displaystyle X}$ on the space of tempered distributions such that

$${\displaystyle X(\delta _{x})=x\delta _{x},}$$

${\displaystyle x}$

$${\displaystyle X(\psi )=\mathrm {x} \psi ,}$$

${\displaystyle \psi }$

${\displaystyle \mathrm {x} }$

$${\displaystyle \mathrm {x} :\mathbb {R} \to \mathbb {C} :x\mapsto x.}$$

Introduction

In one dimension – for a particle confined into a straight line – the square modulus

$${\displaystyle |\psi |^{2}=\psi ^{*}\psi ,}$$

$${\displaystyle \psi :\mathbb {R} \to \mathbb {C} ,}$$

${\displaystyle x}$

In other terms, if – at a certain instant of time – the particle is in the state represented by a square integrable wave function ${\displaystyle \psi }$ and assuming the wave function ${\displaystyle \psi }$ be of ${\displaystyle L^{2))$-norm equal 1,

$${\displaystyle \|\psi \|^{2}=\int _{-\infty }^{+\infty }|\psi |^{2}d\mathrm {x} =1,}$$

${\displaystyle [a,b]}$

$${\displaystyle \pi _{X}(\psi )([a,b])=\int _{a}^{b}|\psi |^{2}d\mathrm {x} .}$$

Hence the expected value of a measurement of the position ${\displaystyle X}$ for the particle is the value

$${\displaystyle \langle X\rangle _{\psi }=\int _{\mathbb {R} }\mathrm {x} |\psi |^{2}d\mathrm {x} =\int _{\mathbb {R} }\psi ^{*}\mathrm {x} \psi \,d\mathrm {x} ,}$$

1. the particle is assumed to be in the state ${\displaystyle \psi }$;
2. the function ${\displaystyle \mathrm {x} |\psi |^{2))$ is supposed integrable, i.e. of class ${\displaystyle L^{1))$;
3. we indicate by ${\displaystyle \mathrm {x} }$ the coordinate function of the position axis.

Additionally, the quantum mechanical operator corresponding to the observable position ${\displaystyle X}$ is denoted also by

$${\displaystyle X={\hat {\mathrm {x} )),}$$

$${\displaystyle \left({\hat {\mathrm {x} ))\psi \right)(x)=x\psi (x),}$$

${\displaystyle \psi }$

${\displaystyle x}$

The circumflex over the function ${\displaystyle \mathrm {x} }$ on the left side indicates the presence of an operator, so that this equation may be read:

The result of the position operator ${\displaystyle X}$ acting on any wave function ${\displaystyle \psi }$ equals the coordinate function ${\displaystyle \mathrm {x} }$ multiplied by the wave-function ${\displaystyle \psi }$.

Or more simply:

The operator ${\displaystyle X}$ multiplies any wave-function ${\displaystyle \psi }$ by the coordinate function ${\displaystyle \mathrm {x} }$.

Note 1. To be more explicit, we have introduced the coordinate function

$${\displaystyle \mathrm {x} :\mathbb {R} \to \mathbb {C} :x\mapsto x,}$$

Note 2. The expected value of the position operator, upon a wave function (state) ${\displaystyle \psi }$ can be reinterpreted as a scalar product:

$${\displaystyle \langle X\rangle _{\psi }=\int _{\mathbb {R} }\mathrm {x} |\psi |^{2}d\mathrm {x} =\int _{\mathbb {R} }\psi ^{*}(\mathrm {x} \psi )\,d\mathrm {x} =\langle \psi |X(\psi )\rangle ,}$$

${\displaystyle \psi \in L^{2))$

${\displaystyle \mathrm {x} \psi }$

${\displaystyle L^{2))$

${\displaystyle \mathrm {x} |\psi |^{2))$

${\displaystyle L^{1))$

Note 3. Strictly speaking, the observable position ${\displaystyle X}$ can be point-wisely defined as

$${\displaystyle \left({\hat {\mathrm {x} ))\psi \right)(x)=x\psi (x),}$$

${\displaystyle \psi }$

${\displaystyle x}$

${\displaystyle \psi \in L^{2))$

$${\displaystyle {\hat {\mathrm {x} ))\psi =\mathrm {x} \psi ,}$$

${\displaystyle \psi \in L^{2))$

Basic properties

In the above definition, as the careful reader can immediately remark, does not exist any clear specification of domain and co-domain for the position operator (in the case of a particle confined upon a line). In literature, more or less explicitly, we find essentially three main directions for this fundamental issue.

1. The position operator is defined on the subspace ${\displaystyle D_{X))$ of ${\displaystyle L^{2))$ formed by those equivalence classes ${\displaystyle \psi }$ whose product by the imbedding ${\displaystyle \mathrm {x} }$ lives in the space ${\displaystyle L^{2))$ as well. In this case the position operator
   $${\displaystyle X:D_{X}\to L^{2}:\psi \mapsto \mathrm {x} \psi }$$

   
   reveals not continuous (unbounded with respect to the topology induced by the canonical scalar product of ${\displaystyle L^{2))$), with no eigenvectors, no eigenvalues, consequently with empty eigenspectrum (collection of its eigenvalues).
2. The position operator is defined on the space ${\displaystyle {\mathcal {S))_{1))$ of complex valued Schwartz functions (smooth complex functions defined upon the real-line and rapidly decreasing at infinity with all their derivatives). The product of a Schwartz function by the imbedding ${\displaystyle \mathrm {x} }$ lives always in the space ${\displaystyle {\mathcal {S))_{1))$, which is a subset of ${\displaystyle L^{2))$. In this case the position operator
   $${\displaystyle X:{\mathcal {S))_{1}\to {\mathcal {S))_{1}:\psi \mapsto \mathrm {x} \psi }$$

   
   reveals continuous (with respect to the canonical topology of ${\displaystyle {\mathcal {S))_{1))$), injective, with no eigenvectors, no eigenvalues, consequently with void eigenspectrum (collection of its eigenvalues). It is (fully) self-adjoint with respect to the scalar product of ${\displaystyle L^{2))$ in the sense that
   $${\displaystyle \langle X(\psi )|\phi \rangle =\langle \psi |X(\phi )\rangle ,}$$

   
   for every ${\displaystyle \psi }$ and ${\displaystyle \phi }$ belonging to its domain ${\displaystyle {\mathcal {S))_{1))$.
3. This is, in practice, the most widely adopted choice in Quantum Mechanics literature, although never explicitly underlined. The position operator is defined on the space ${\displaystyle {\mathcal {S))'_{1))$ of complex valued tempered distributions (topological dual of the Schwartz function space ${\displaystyle {\mathcal {S))_{1))$). The product of a temperate distribution by the imbedding ${\displaystyle \mathrm {x} }$ lives always in the space ${\displaystyle {\mathcal {S))'_{1))$, which contains ${\displaystyle L^{2))$. In this case the position operator
   $${\displaystyle X:{\mathcal {S))'_{1}\to {\mathcal {S))'_{1}:\psi \mapsto \mathrm {x} \psi }$$

   
   reveals continuous (with respect to the canonical topology of ${\displaystyle {\mathcal {S))'_{1))$), surjective, endowed with complete families of eigenvectors, real eigenvalues, and with eigenspectrum (collection of its eigenvalues) equal to the real line. It is self-adjoint with respect to the scalar product of ${\displaystyle L^{2))$ in the sense that its transpose operator
   $${\displaystyle {}^{t}X:{\mathcal {S))_{1}\to {\mathcal {S))_{1}:\phi \mapsto \mathrm {x} \phi ,}$$

   
   which is the position operator on the Schwartz function space, is self-adjoint:
   $${\displaystyle \left\langle \left.\,{}^{t}X(\phi )\right|\psi \right\rangle =\left\langle \phi |\,{}^{t}X(\psi )\right\rangle ,}$$

   
   for every (test) function ${\displaystyle \phi }$ and ${\displaystyle \psi }$ belonging to the space ${\displaystyle {\mathcal {S))_{1))$.

Eigenstates

The eigenfunctions of the position operator (on the space of tempered distributions), represented in position space, are Dirac delta functions.

Informal proof. To show that possible eigenvectors of the position operator should necessarily be Dirac delta distributions, suppose that ${\displaystyle \psi }$ is an eigenstate of the position operator with eigenvalue ${\displaystyle x_{0))$. We write the eigenvalue equation in position coordinates,

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

recalling that ${\displaystyle {\hat {\mathrm {x} ))}$ simply multiplies the wave-functions by the function ${\displaystyle \mathrm {x} }$, in the position representation. Since the function ${\displaystyle \mathrm {x} }$ is variable while ${\displaystyle x_{0))$ is a constant, ${\displaystyle \psi }$ must be zero everywhere except at the point ${\displaystyle x_{0))$. Clearly, no continuous function satisfies such properties, and we cannot simply define the wave-function to be a complex number at that point because its ${\displaystyle L^{2))$-norm would be 0 and not 1. This suggest the need of a "functional object" concentrated at the point ${\displaystyle x_{0))$ and with integral different from 0: any multiple of the Dirac delta centered at ${\displaystyle x_{0))$. Q.E.D.

The normalized solution to the equation

$${\displaystyle \mathrm {x} \psi =x_{0}\psi }$$

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

$${\displaystyle \psi =\delta _{x_{0)).}$$

Proof. Here we prove rigorously that

$${\displaystyle \mathrm {x} \delta _{x_{0))=x_{0}\delta _{x_{0)).}$$

Indeed, recalling that the product of any function by the Dirac distribution centered at a point is the value of the function at that point times the Dirac distribution itself, we obtain immediately

$${\displaystyle \mathrm {x} \delta _{x_{0))=\mathrm {x} (x_{0})\delta _{x_{0))=x_{0}\delta _{x_{0)).}$$

Meaning of the Dirac delta wave. Although such Dirac states are physically unrealizable and, strictly speaking, they are not functions, Dirac distribution centered at ${\displaystyle x_{0))$ 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 space-time wavefunction is now ${\displaystyle \psi (\mathbf {r} ,t)}$ and the expectation value of the position operator ${\displaystyle {\hat {\mathbf {r} ))}$ at the state ${\displaystyle \psi }$ is

$${\displaystyle \left\langle {\hat {\mathbf {r} ))\right\rangle _{\psi }=\int \mathbf {r} |\psi |^{2}d^{3}\mathbf {r} }$$

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

Momentum space

Usually, in quantum mechanics, by representation in the momentum space we intend the representation of states and observables with respect to the canonical unitary momentum basis

$${\displaystyle \eta =\left(\left[(2\pi \hbar )^{-{\frac {1}{2))}e^{(\iota /\hbar )(\mathrm {x} |p)}\right]\right)_{p\in \mathbb {R} }.}$$

In momentum space, the position operator in one dimension is represented by the following differential operator

$${\displaystyle \left({\hat {\mathrm {x} ))\right)_{P}=i\hbar {\frac {d}{d\mathrm {p} ))=i{\frac {d}{d\mathrm {k} )),}$$

where:

• the representation of the position operator in the momentum basis is naturally defined by ${\displaystyle \left({\hat {\mathrm {x} ))\right)_{P}(\psi )_{P}=\left({\hat {\mathrm {x} ))\psi \right)_{P))$, for every wave function (tempered distribution) ${\displaystyle \psi }$;
• ${\displaystyle \mathrm {p} }$ represents the coordinate function on the momentum line and the wave-vector function ${\displaystyle \mathrm {k} }$ is defined by ${\displaystyle \mathrm {k} =\mathrm {p} /\hbar }$.

Formalism in L²(R, C)

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 contains the L²-space (Hilbert space) ${\displaystyle L^{2}(\mathbb {R} ,\mathbb {C} )}$ of complex-valued and square-integrable (with respect to the Lebesgue measure) functions on the real line.

The position operator in ${\displaystyle L^{2}(\mathbb {R} ,\mathbb {C} )}$,

$${\displaystyle Q:D_{Q}\to L^{2}(\mathbb {R} ,\mathbb {C} ):\psi \mapsto \mathrm {q} \psi ,}$$

$${\displaystyle Q(\psi )(x)=x\psi (x)=\mathrm {q} (x)\psi (x),}$$

${\displaystyle \psi \in D_{Q))$

$${\displaystyle D_{Q}=\left\{\psi \in L^{2}(\mathbb {R} )\mid \mathrm {q} \psi \in L^{2}(\mathbb {R} )\right\},}$$

${\displaystyle \mathrm {q} :\mathbb {R} \to \mathbb {C} }$

${\displaystyle x\in \mathbb {R} }$

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 theory in L²(R, C)

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

$${\displaystyle X:D_{X}\to L^{2}(\mathbb {R} ,\mathbb {C} ):\psi \mapsto \mathrm {x} \psi }$$

$${\displaystyle X=\int _{\mathbb {R} }\lambda \,d\mu _{X}(\lambda )=\int _{\mathbb {R} }\mathrm {x} \,\mu _{X}=\mu _{X}(\mathrm {x} ),}$$

${\displaystyle \mu _{X))$

Since the operator of ${\displaystyle X}$ is just the multiplication operator by the embedding function ${\displaystyle \mathrm {x} }$, its spectral resolution is simple.

For a Borel subset ${\displaystyle B}$ of the real line, let ${\displaystyle \chi _{B))$ denote the indicator function of ${\displaystyle B}$. We see that the projection-valued measure

$${\displaystyle \mu _{X}:{\mathcal {B))(\mathbb {R} )\to \mathrm {Pr} ^{\perp }\left(L^{2}(\mathbb {R} ,\mathbb {C} )\right)}$$

$${\displaystyle \mu _{X}(B)(\psi )=\chi _{B}\psi ,}$$

${\displaystyle \mu _{X}(B)}$

${\displaystyle B}$

Therefore, if the system is prepared in a state ${\displaystyle \psi }$, then the probability of the measured position of the particle belonging to a Borel set ${\displaystyle B}$ is

$${\displaystyle \|\mu _{X}(B)(\psi )\|^{2}=\|\chi _{B}\psi \|^{2}=\int _{B}|\psi |^{2}\ \mu =\pi _{X}(\psi )(B),}$$

${\displaystyle \mu }$

After any measurement aiming to detect the particle within the subset B, the wave function collapses to either

$${\displaystyle {\frac {\mu _{X}(B)\psi }{\|\mu _{X}(B)\psi \|))={\frac {\chi _{B}\psi }{\|\chi _{B}\psi \|))}$$

$${\displaystyle {\frac {(1-\chi _{B})\psi }{\|(1-\chi _{B})\psi \|)),}$$

${\displaystyle \|\cdot \|}$

${\displaystyle L^{2}(\mathbb {R} ,\mathbb {C} )}$