In quantum mechanics, the momentum operator is the operator associated with the linear momentum. The momentum operator is, in the position representation, an example of a differential operator. For the case of one particle in one spatial dimension, the definition is:

$${\displaystyle {\hat {p))=-i\hbar {\frac {\partial }{\partial x))}$$

${\displaystyle \partial /\partial x}$

$${\displaystyle {\hat {p))\psi =-i\hbar {\frac {\partial \psi }{\partial x))}$$

In a basis of Hilbert space consisting of momentum eigenstates expressed in the momentum representation, the action of the operator is simply multiplication by p, i.e. it is a multiplication operator, just as the position operator is a multiplication operator in the position representation. Note that the definition above is the canonical momentum, which is not gauge invariant and not a measurable physical quantity for charged particles in an electromagnetic field. In that case, the canonical momentum is not equal to the kinetic momentum.

At the time quantum mechanics was developed in the 1920s, the momentum operator was found by many theoretical physicists, including Niels Bohr, Arnold Sommerfeld, Erwin Schrödinger, and Eugene Wigner. Its existence and form is sometimes taken as one of the foundational postulates of quantum mechanics.

Origin from de Broglie plane waves

The momentum and energy operators can be constructed in the following way.^[1]

One dimension

Starting in one dimension, using the plane wave solution to Schrödinger's equation of a single free particle,

$${\displaystyle \psi (x,t)=e^((\frac {i}{\hbar ))(px-Et)},}$$

$${\displaystyle {\frac {\partial \psi (x,t)}{\partial x))={\frac {ip}{\hbar ))e^((\frac {i}{\hbar ))(px-Et)}={\frac {ip}{\hbar ))\psi .}$$

This suggests the operator equivalence

$${\displaystyle {\hat {p))=-i\hbar {\frac {\partial }{\partial x))}$$

Since the partial derivative is a linear operator, the momentum operator is also linear, and because any wave function can be expressed as a superposition of other states, when this momentum operator acts on the entire superimposed wave, it yields the momentum eigenvalues for each plane wave component. These new components then superimpose to form the new state, in general not a multiple of the old wave function.

Three dimensions

The derivation in three dimensions is the same, except the gradient operator del is used instead of one partial derivative. In three dimensions, the plane wave solution to Schrödinger's equation is:

$${\displaystyle \psi =e^((\frac {i}{\hbar ))(\mathbf {p} \cdot \mathbf {r} -Et)))$$

$${\displaystyle {\begin{aligned}\nabla \psi &=\mathbf {e} _{x}{\frac {\partial \psi }{\partial x))+\mathbf {e} _{y}{\frac {\partial \psi }{\partial y))+\mathbf {e} _{z}{\frac {\partial \psi }{\partial z))\\&={\frac {i}{\hbar ))\left(p_{x}\mathbf {e} _{x}+p_{y}\mathbf {e} _{y}+p_{z}\mathbf {e} _{z}\right)\psi \\&={\frac {i}{\hbar ))\mathbf {p} \psi \end{aligned))}$$

$${\displaystyle \mathbf {\hat {p)) =-i\hbar \nabla }$$

This momentum operator is in position space because the partial derivatives were taken with respect to the spatial variables.

Definition (position space)

For a single particle with no electric charge and no spin, the momentum operator can be written in the position basis as:^[2]

$${\displaystyle \mathbf {\hat {p)) =-i\hbar \nabla }$$

In one spatial dimension, this becomes^[3]

$${\displaystyle {\hat {p))={\hat {p))_{x}=-i\hbar {\partial \over \partial x}.}$$

This is the expression for the canonical momentum. For a charged particle q in an electromagnetic field, during a gauge transformation, the position space wave function undergoes a local U(1) group transformation,^[4] and ${\textstyle {\hat {p))\psi =-i\hbar {\frac {\partial \psi }{\partial x))}$ will change its value. Therefore, the canonical momentum is not gauge invariant, and hence not a measurable physical quantity.

The kinetic momentum, a gauge invariant physical quantity, can be expressed in terms of the canonical momentum, the scalar potential φ and vector potential A:^[5]

$${\displaystyle \mathbf {\hat {P)) =-i\hbar \nabla -q\mathbf {A} }$$

The expression above is called minimal coupling. For electrically neutral particles, the canonical momentum is equal to the kinetic momentum.

Properties

Hermiticity

The momentum operator is always a Hermitian operator (more technically, in math terminology a "self-adjoint operator") when it acts on physical (in particular, normalizable) quantum states.^[6]

(In certain artificial situations, such as the quantum states on the semi-infinite interval [0, ∞), there is no way to make the momentum operator Hermitian.^[7] This is closely related to the fact that a semi-infinite interval cannot have translational symmetry—more specifically, it does not have unitary translation operators. See below.)

Canonical commutation relation

By applying the commutator to an arbitrary state in either the position or momentum basis, one can easily show that:

$${\displaystyle \left[{\hat {x)),{\hat {p))\right]={\hat {x)){\hat {p))-{\hat {p)){\hat {x))=i\hbar .}$$

The Heisenberg uncertainty principle defines limits on how accurately the momentum and position of a single observable system can be known at once. In quantum mechanics, position and momentum are conjugate variables.

Fourier transform

The following discussion uses the bra–ket notation. One may write

$${\displaystyle \psi (x)=\langle x|\psi \rangle =\int \!\!dp~\langle x|p\rangle \langle p|\psi \rangle =\int \!\!dp~{e^{ixp/\hbar }{\tilde {\psi ))(p) \over {\sqrt {2\pi \hbar ))},}$$

$${\displaystyle {\hat {p))=\int \!\!dp~|p\rangle p\langle p|=-i\hbar \int \!\!dx~|x\rangle {\frac {d}{dx))\langle x|~,}$$

$${\displaystyle \langle x|{\hat {p))|\psi \rangle =-i\hbar {\frac {d}{dx))\psi (x).}$$

An analogous result applies for the position operator in the momentum basis,

$${\displaystyle \langle p|{\hat {x))|\psi \rangle =i\hbar {\frac {d}{dp))\psi (p),}$$

$${\displaystyle \langle p|{\hat {x))|p'\rangle =i\hbar {\frac {d}{dp))\delta (p-p'),}$$

$${\displaystyle \langle x|{\hat {p))|x'\rangle =-i\hbar {\frac {d}{dx))\delta (x-x'),}$$

Derivation from infinitesimal translations

The translation operator is denoted T(ε), where ε represents the length of the translation. It satisfies the following identity:

$${\displaystyle T(\varepsilon )|\psi \rangle =\int dxT(\varepsilon )|x\rangle \langle x|\psi \rangle }$$

$${\displaystyle \int dx|x+\varepsilon \rangle \langle x|\psi \rangle =\int dx|x\rangle \langle x-\varepsilon |\psi \rangle =\int dx|x\rangle \psi (x-\varepsilon )}$$

Assuming the function ψ to be analytic (i.e. differentiable in some domain of the complex plane), one may expand in a Taylor series about x:

$${\displaystyle \psi (x-\varepsilon )=\psi (x)-\varepsilon {\frac {d\psi }{dx))}$$

$${\displaystyle T(\varepsilon )=1-\varepsilon {d \over dx}=1-{i \over \hbar }\varepsilon \left(-i\hbar {d \over dx}\right)}$$

As it is known from classical mechanics, the momentum is the generator of translation, so the relation between translation and momentum operators is^[8]:^{[further explanation needed]}

$${\displaystyle T(\varepsilon )=1-{\frac {i}{\hbar ))\varepsilon {\hat {p))}$$

$${\displaystyle {\hat {p))=-i\hbar {\frac {d}{dx)).}$$

4-momentum operator

Inserting the 3d momentum operator above and the energy operator into the 4-momentum (as a 1-form with (+ − − −) metric signature):

$${\displaystyle P_{\mu }=\left({\frac {E}{c)),-\mathbf {p} \right)}$$

obtains the 4-momentum operator:

$${\displaystyle {\hat {P))_{\mu }=\left({\frac {1}{c)){\hat {E)),-\mathbf {\hat {p)) \right)=i\hbar \left({\frac {1}{c)){\frac {\partial }{\partial t)),\nabla \right)=i\hbar \partial _{\mu ))$$

where ∂_μ is the 4-gradient, and the −iħ  becomes +iħ  preceding the 3-momentum operator. This operator occurs in relativistic quantum field theory, such as the Dirac equation and other relativistic wave equations, since energy and momentum combine into the 4-momentum vector above, momentum and energy operators correspond to space and time derivatives, and they need to be first order partial derivatives for Lorentz covariance.

The Dirac operator and Dirac slash of the 4-momentum is given by contracting with the gamma matrices:

$${\displaystyle \gamma ^{\mu }{\hat {P))_{\mu }=i\hbar \gamma ^{\mu }\partial _{\mu }={\hat {P))=i\hbar \partial \!\!\!/}$$

If the signature was (− + + +), the operator would be

$${\displaystyle {\hat {P))_{\mu }=\left(-{\frac {1}{c)){\hat {E)),\mathbf {\hat {p)) \right)=-i\hbar \left({\frac {1}{c)){\frac {\partial }{\partial t)),\nabla \right)=-i\hbar \partial _{\mu ))$$

instead.