# Momentum operator

In quantum mechanics, the momentum operator is the operator for measuring the momentum of particles . In the position representation , the momentum operator is given in one dimension by: ${\ displaystyle {\ hat {p}}}$

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

Here designated

• ${\ displaystyle \ mathrm {i}}$the imaginary unit
• ${\ displaystyle \ hbar}$the reduced Planck constant and
• ${\ displaystyle {\ frac {\ partial} {\ partial x}}}$the partial derivative in the direction of the location coordinate .${\ displaystyle x}$

With the Nabla operator one obtains the vector in three dimensions : ${\ displaystyle \ nabla}$

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

In quantum mechanics, the physical state of a particle is given mathematically by an associated vector of a Hilbert space . This state is thus described in Bra-Ket notation by the vector . The observables are represented by self-adjoint operators on . In particular, the momentum operator is the combination of the three observables such that ${\ displaystyle \ Psi \,}$ ${\ displaystyle {\ mathcal {H}}}$${\ displaystyle | \ Psi \ rangle}$${\ displaystyle {\ mathcal {H}}}$${\ displaystyle {\ hat {\ mathbf {p}}} = ({\ hat {p}} _ {1}, {\ hat {p}} _ {2}, {\ hat {p}} _ {3 })}$

${\ displaystyle E ({\ hat {p}} _ {j}) = \ langle \ Psi | {\ hat {p}} _ {j} \, | \ Psi \ rangle \, \ quad j = 1,2 , 3}$

is the mean ( expected value ) of the measurement results of the j th component of the momentum of the particle in the state . ${\ displaystyle \ Psi}$

## Definition and characteristics

${\ displaystyle [{\ hat {x}} _ {i}, {\ hat {p}} _ {j}] = \ mathrm {i} \, \ hbar \, \ delta _ {ij} \, \ quad [{\ hat {x}} _ {i}, {\ hat {x}} _ {j}] = 0 = [{\ hat {p}} _ {i}, {\ hat {p}} _ { j}] \, \ quad i, j \ in \ {1,2,3 \}}$
in analogy to the Poisson brackets of the Hamiltonian formulation
${\ displaystyle \ {x_ {i}, p_ {j} \} = \ delta _ {ij} \, \ quad \ {x_ {i}, x_ {j} \} = 0 = \ {p_ {i}, p_ {j} \} \ ,.}$
The factor is necessary for dimensional reasons , because place times momentum has the dimension of an angular momentum or an effect . The imaginary unit must appear because and are self-adjoint and their commutator therefore changes its sign with adjunction .${\ displaystyle \ hbar}$${\ displaystyle {\ rm {i}}}$${\ displaystyle {\ hat {x}} _ {i}}$${\ displaystyle {\ hat {p}} _ {j}}$
• From the canonical commutation relations it follows that the three components of the impulse can be measured together and that their spectrum (range of possible measured values ) consists of the entire space . The possible impulses are therefore not quantized, but rather continuous .${\ displaystyle \ mathbb {R} ^ {3}}$
• The location representation is defined by the spectral representation of the location operator. The Hilbert space is the space of the square integrable , complex functions of the spatial space every state is given by a spatial wave function. The position operators are the multiplication operators with the coordinate functions, i.e. H. the position operator acts on position wave functions by multiplying the wave function with the coordinate function :${\ displaystyle {\ mathcal {H}} = L ^ {2} (\ mathbb {R} ^ {3}; \ mathbb {C})}$ ${\ displaystyle \ mathbb {R} ^ {3};}$${\ displaystyle \ Psi}$ ${\ displaystyle \ psi (\ mathbf {x})}$${\ displaystyle {\ hat {\ mathbf {x}}} = ({\ hat {x}} _ {1}, {\ hat {x}} _ {2}, {\ hat {x}} _ {3 })}$${\ displaystyle {\ hat {x}} _ {i}}$${\ displaystyle x_ {i}}$
${\ displaystyle ({\ hat {x}} _ {i} \, \ psi) (\ mathbf {x}) = x_ {i} \, \ psi (\ mathbf {x}) \ ,.}$
The mathematical theorem of Stone and von Neumann then states that with a suitable choice of phases the momentum operator, which occurs in the canonical commutation relations, acts as a differential operator on spatial wave functions :
${\ displaystyle ({\ hat {p}} _ {j} \ psi) (\ mathbf {x}) = - {\ rm {i}} \, \ hbar \, \ left ({\ frac {\ partial} {\ partial x_ {j}}} \ psi \ right) (\ mathbf {x}) \ ,.}$
Its expected value is:
${\ displaystyle E ({\ hat {p}} _ {j}) = \ langle \ Psi | {\ hat {p}} _ {j} \, | \ Psi \ rangle = \ int {\ overline {\ psi (\ mathbf {x})}} \, \ left (- \ mathrm {i} \, \ hbar {\ frac {\ partial} {\ partial x_ {j}}} \ psi (\ mathbf {x}) \ right) \, \ mathrm {d} ^ {3} x \ ,.}$
• In the momentum representation, the momentum operator has a multiplicative effect on square integrable momentum wave functions :${\ displaystyle {\ tilde {\ psi}} (\ mathbf {p})}$
${\ displaystyle ({\ hat {p}} _ {j} \, {\ tilde {\ psi}}) (\ mathbf {p}) = p_ {j} \, {\ tilde {\ psi}} (\ mathbf {p})}$
and the position operator acts as a differential operator:
${\ displaystyle ({\ hat {x}} _ {i} \, {\ tilde {\ psi}}) (\ mathbf {p}) = \ mathrm {i} \, \ hbar \, \ left ({\ frac {\ partial} {\ partial p_ {i}}} {\ tilde {\ psi}} \ right) (\ mathbf {p}) \ ,.}$
${\ displaystyle {\ hat {x}} _ {i} = l_ {i} {\ frac {a_ {i} + a_ {i} ^ {\ dagger}} {\ sqrt {2}}} \, \ quad {\ hat {p}} _ {j} = {\ frac {\ hbar} {l_ {j}}} {\ frac {a_ {j} -a_ {j} ^ {\ dagger}} {{\ sqrt { 2}} \, \ mathrm {i}}} \ ,.}$
There are freely selectable lengths (greater than zero) and the creation and annihilation operators satisfy the canonical commutation relations:${\ displaystyle l_ {1}, l_ {2}, l_ {3}}$
${\ displaystyle [a_ {i}, a_ {j} ^ {\ dagger}] = \ delta _ {ij} \, \ quad [a_ {i}, a_ {j}] = 0 = [a_ {i} ^ {\ dagger}, a_ {j} ^ {\ dagger}] \, \ quad i, j \ in \ {1,2,3 \} \ ,.}$

## Why is the momentum operator a differential operator in position representation?

According to Noether's theorem , every continuous symmetry of the effect has a conservation quantity . Conversely, every conserved quantity implies the existence of an (at least infinitesimal) symmetry of the effect. For example, the momentum is obtained if and only if the effect is translation-invariant. In Hamilton's formulation, the conserved quantity generates the symmetry transformation in phase space through its Poisson bracket, the momentum generates shifts.

Applied to a wave function , every shift by the shifted function results , which has the value at every point that had on the archetype , ${\ displaystyle \ psi}$${\ displaystyle a}$${\ displaystyle (T_ {a} \, \ psi)}$${\ displaystyle x}$${\ displaystyle \ psi}$${\ displaystyle xa}$

${\ displaystyle (T_ {a} \, \ psi) (x) = \ psi (xa) = \ sum _ {n = 0} ^ {\ infty} {{\ frac {1} {n!}} \ left (-a {\ frac {\ partial} {\ partial x}} \ right) ^ {n}} \ psi = \ exp \ left (-a {\ frac {\ partial} {\ partial x}} \ right) \ psi}$(i.e.: via Taylor series to a formal exponential function).

The infinitesimal generator of this one-parameter family of displacements defines the momentum up to a factor , that is, the momentum fulfills by definition ${\ displaystyle - \ mathrm {i} / \ hbar}$${\ displaystyle {\ hat {p}} _ {x}}$

${\ displaystyle T_ {a} \, \ psi = \ exp \ left (-a {\ frac {\ partial} {\ partial x}} \ right) \ psi = \ exp {\ left (- {\ rm {i }} \, a {\ frac {{\ hat {p}} _ {x}} {\ hbar}} \ right)} \, \ psi \ ,.}$

The factor occurs for dimensional reasons, because the product of momentum and location has the dimension of an angular momentum or an effect. The imaginary unit is required because is a unitary operator and the momentum should be self-adjoint . One conducts the equation ${\ displaystyle \ hbar}$ ${\ displaystyle \ mathrm {i}}$${\ displaystyle T_ {a}}$

${\ displaystyle \ left (\ exp {\ left (- {\ rm {i}} \, {\ frac {{\ hat {p}} _ {j} \, a ^ {j}} {\ hbar}} \ right)} \, \ psi \ right) (x) = \ psi (xa)}$

after at from, the momentum operator results as a derivative according to the location, ${\ displaystyle a ^ {j}}$${\ displaystyle a = 0}$

${\ displaystyle ({\ hat {p}} _ {j} \, \ psi) (x) = \ left. \ mathrm {i} \, \ hbar \, {\ frac {\ partial} {\ partial a ^ {j}}} \ right | _ {a = 0} \ psi (xa) = - \ mathrm {i} \, \ hbar {\ frac {\ partial} {\ partial x ^ {j}}} \ psi ( x) \ ,.}$

The fact that the momentum operator assumes this form in spatial space can be read from Noether's theorem as follows , even without knowing the associated unitary operator : One first reconstructs the associated Lagrange density from the Schrödinger equation and then explicitly determines that in the case of an infinitesimal shift of the wave function received expected value. ${\ displaystyle T_ {a}}$