Momentum operator

In position representation, momentum is minus i hbar times spatial differentiation.

The generators of spacetime translations in a quantum field theory. With the mostly-minus Minkowski metric, $P\cdot a=H{a}^0-\mathbf{P}\cdot \mathbf{a}$. A momentum eigenstate satisfies $P^{\mu }|p⟩=p^{\mu }|p⟩$, where $p^{\mu }$ is its four-momentum. After subtracting the vacuum energy, a translation-invariant vacuum state has eigenvalue zero.

The spatial translation operator is the unitary generated by the momentum operator. With the convention $U(\mathbf{y})=e^{-i\mathbf{P}\cdot \mathbf{y}}$, a momentum eigenstate of eigenvalue $\mathbf{p}$ acquires the phase $e^{-i\mathbf{p}\cdot \mathbf{y}}$, while $U(\mathbf{y})\varphi (\mathbf{x})U(\mathbf{y}{)}^{-1}=\varphi (\mathbf{x}+\mathbf{y})$.

In one dimension the momentum operator acts on a position-space wavefunction as $p=-i\hslash {\partial }_x$, and $p^2=-{\hslash }^2{\partial }_x^2$.

In position representation, $x$ acts by multiplication and $p_x=-i\hslash {\partial }_x$, so

Products of noncommuting linear operators depend on their order. For position and momentum, $xp-px=i\hslash$; substituting classical commuting expressions into a product can therefore lose an additive quantum term. For example, expansion of the oscillator product $a^{†}a$ includes a commutator contribution $-1/2$, leading to the correct zero-point energy in $H=\hslash \omega (a^{†}a+1/2)$.

On a common invariant domain for the operator products, let the position operator and momentum operator obey $[\hat{x},\hat{p}]=i\hslash I$. Then $\hat{x}\hat{p}=\hat{p}\hat{x}+i\hslash I$, so both displayed expressions equal $(\hat{p}\hat{x}{)}^2+i\hslash \hat{p}\hat{x}$. In position representation both act on a smooth function $f$ as $-{\hslash }^2(x^2{f}^{\prime \prime }+2x{f}^{\prime })$. This particular equality does not imply that arbitrary classical reorderings remain equivalent in quantum mechanics.

These bounded-unitary relations encode a pair of canonical commutation relations without multiplying unbounded position and momentum operators. In a fixed convention on $L^2(\mathbb{R})$, $W(x,y)f(t)=e^{ixy+2iyt}f(t+x)$ and $a\wedge b=x_1{y}_2-y_1{x}_2$. Strong continuity is part of a regular Weyl representation. Different nonzero choices of the central normalization rescale the canonical coordinates.

Every irreducible strongly continuous representation of the Weyl relations with fixed nonzero central character is unitarily equivalent to the translation-modulation representation on $L^2(\mathbb{R})$. The Gaussian projection in a Weyl representation supplies a cyclic vector whose Weyl-orbit inner products agree with those of a Gaussian in the standard representation. Matching those orbit vectors gives the unitary intertwiner, with no appeal to a classification assertion in place of the proof.

The integrated Gaussian in any nonzero regular representation of the Weyl relations is an orthogonal projection and obeys $PW(a)P=e^{-|a{|}^2/2}P$. Gaussian integration with the Weyl phase proves both identities. It cannot be zero: conjugating it by all Weyl operators would otherwise make the Fourier transform of every Gaussian-weighted matrix coefficient vanish, including its nonzero value at the identity. In an irreducible representation its range has dimension one, and the orbit of a unit vector there gives a dense cyclic space with universal Gaussian inner products.