Momentum operator

In

. For the case of one particle in one spatial dimension, the definition is:

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

where ħ is , x is the spatial coordinate, and a partial derivative (denoted by ${\displaystyle \partial /\partial x}$) is used instead of a total derivative (d/dx) since the wave function is also a function of time. The "hat" indicates an operator. The "application" of the operator on a differentiable wave function is as follows:

In a basis of

.

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

of a single free particle, where p is interpreted as momentum in the x-direction and E is the particle energy. The first order partial derivative with respect to space is

This suggests the operator equivalence

so the momentum of the particle and the value that is measured when a particle is in a plane wave state is the of the above operator.

Since the partial derivative is a

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:

and the gradient is where e_x, e_y, and e_z are the unit vectors for the three spatial dimensions, hence

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]

where ∇ is the .

In one spatial dimension, this becomes[3]

This is the expression for the

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

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

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

(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:

The

.

Fourier transform

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

so the tilde represents the Fourier transform, in converting from coordinate space to momentum space. It then holds that that is, the momentum acting in coordinate space corresponds to spatial frequency,

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

leading to further useful relations, where δ stands for .

Derivation from infinitesimal translations

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

that becomes

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:

so for infinitesimal values of ε:

As it is known from

]

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

thus

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

4-momentum operator

Inserting the 3d momentum operator above and the

):

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

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

.

The

:

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

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.

See also

• Mathematical descriptions of the electromagnetic field
• Translation operator (quantum mechanics)
• Relativistic wave equations
• Pauli–Lubanski pseudovector

References