Momentum operator

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: $${\displaystyle \hat{\mathbf{p}}=-i\hslash \mathrm{\nabla }}$$ where ∇ is the gradient operator, ħ is the reduced Planck constant, and i is the imaginary unit.

In one spatial dimension, this becomes $${\displaystyle \hat{p}={\hat{p}}_x=-i\hslash \frac{\mathrm{\partial }}{\mathrm{\partial }x}.}$$

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, and $\hat{p}\psi =-i\hslash \frac{\mathrm{\partial }\psi }{\mathrm{\partial }x}$ 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: $${\displaystyle \hat{\mathbf{P}}=-i\hslash \mathrm{\nabla }-q\mathbf{A}}$$

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 can be described as a symmetric (i.e. Hermitian), unbounded operator acting on a dense subspace of the quantum state space. If the operator acts on a (normalizable) quantum state then the operator is self-adjoint. In physics the term Hermitian often refers to both symmetric and self-adjoint operators.

(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. 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: $${\displaystyle \left[\hat{x},\hat{p}\right]=\hat{x}\hat{p}-\hat{p}\hat{x}=i\hslash \mathbb{I},}$$ where ${\displaystyle \mathbb{I}}$ is the unit operator. 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 $${\displaystyle \psi (x)=⟨x|\psi ⟩=\int dp⟨x|p⟩⟨p|\psi ⟩=\int dp\frac{e^{ixp/\hslash }\tilde{\psi }(p)}{\sqrt{2\pi \hslash }},}$$ so the tilde represents the Fourier transform, in converting from coordinate space to momentum space. It then holds that $${\displaystyle \hat{p}=\int dp|p⟩p⟨p|=-i\hslash \int dx|x⟩\frac{d}{dx}⟨x|,}$$ that is, the momentum acting in coordinate space corresponds to spatial frequency, $${\displaystyle ⟨x|\hat{p}|\psi ⟩=-i\hslash \frac{d}{dx}\psi (x).}$$

An analogous result applies for the position operator in the momentum basis, $${\displaystyle ⟨p|\hat{x}|\psi ⟩=i\hslash \frac{d}{dp}\psi (p),}$$ leading to further useful relations, $${\displaystyle ⟨p|\hat{x}|p^{\prime }⟩=i\hslash \frac{d}{dp}\delta (p-p^{\prime }),}$$ $${\displaystyle ⟨x|\hat{p}|x^{\prime }⟩=-i\hslash \frac{d}{dx}\delta (x-x^{\prime }),}$$ where δ stands for Dirac's delta function.

Derivation from infinitesimal translations

The translation operator is denoted T(ε), where ε represents the length of the translation. It satisfies the following identity: $${\displaystyle T(\epsilon )|\psi ⟩=\int dxT(\epsilon )|x⟩⟨x|\psi ⟩}$$ that becomes $${\displaystyle \int dx|x+\epsilon ⟩⟨x|\psi ⟩=\int dx|x⟩⟨x-\epsilon |\psi ⟩=\int dx|x⟩\psi (x-\epsilon )}$$

Assuming the function ${\displaystyle \psi }$ to be analytic (i.e. differentiable in some domain of the complex plane), one may expand in a Taylor series about x: $${\displaystyle \psi (x-\epsilon )=\psi (x)-\epsilon \frac{d\psi }{dx}}$$ so for infinitesimal values of ε: $${\displaystyle T(\epsilon )=1-\epsilon \frac{d}{dx}=1-\frac{i}{\hslash }\epsilon \left(-i\hslash \frac{d}{dx}\right)}$$

As it is known from classical mechanics, the momentum is the generator of translation, so the relation between translation and momentum operators is: $${\displaystyle T(\epsilon )=1-\frac{i}{\hslash }\epsilon \hat{p}}$$ thus $${\displaystyle \hat{p}=-i\hslash \frac{d}{dx}.}$$

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: $${\displaystyle {\hat{P}}_{\mu }=\left(\frac{1}{c}\hat{E},-\hat{\mathbf{p}}\right)=i\hslash \left(\frac{1}{c}\frac{\mathrm{\partial }}{\mathrm{\partial }t},\mathrm{\nabla }\right)=i\hslash {\mathrm{\partial }}_{\mu }}$$ 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: $${\displaystyle {\gamma }^{\mu }{\hat{P}}_{\mu }=i\hslash {\gamma }^{\mu }{\mathrm{\partial }}_{\mu }=\hat{P}=i\hslash \mathrm{\partial }/}$$

If the signature was (− + + +), the operator would be $${\displaystyle {\hat{P}}_{\mu }=\left(-\frac{1}{c}\hat{E},\hat{\mathbf{p}}\right)=-i\hslash \left(\frac{1}{c}\frac{\mathrm{\partial }}{\mathrm{\partial }t},\mathrm{\nabla }\right)=-i\hslash {\mathrm{\partial }}_{\mu }}$$ instead.