Momentum Operator as Generator

The momentum operator is the self-adjoint generator of continuous spatial translations. In position representation on the full line, this statement gives the familiar differential expression

$$P=-i\hbar\frac{d}{dx}.$$

The point of the derivation is not to memorize a derivative. It is to understand why momentum, translations, plane waves, and the canonical commutator are the same structure viewed from different sides.

This page uses the active convention: translating the state to the right by $a$ moves a wavepacket centered at $x_0$ to one centered at $x_0+a$.

Helpful background. One-parameter unitary groups and Generators explain how a continuous transformation determines a self-adjoint operator. Position–Momentum Representations fixes the Fourier convention used for momentum eigenstates.

Translation on Wavefunctions

For a particle on the full real line, the Hilbert space is modeled by $L^2(\mathbb R)$. The active translation operator is defined by

$$(T(a)\psi)(x) = \psi(x-a).$$

This preserves normalization:

$$\begin{aligned} \int_{-\infty}^{\infty} |(T(a)\psi)(x)|^2\,dx &= \int_{-\infty}^{\infty} |\psi(x-a)|^2\,dx \\ &= \int_{-\infty}^{\infty} |\psi(y)|^2\,dy. \end{aligned}$$

Since $T(-a)$ is its inverse, norm preservation makes $T(a)$ unitary. The operators also form a representation of the additive translation group:

$$T(a)T(b)=T(a+b), \qquad T(0)=I.$$

The sign in $\psi(x-a)$ is not arbitrary. A bump originally at $x=0$ becomes a bump at $x=a$, because the translated wavefunction has its old value when $x-a=0$.

Infinitesimal Derivation

For a smooth rapidly decreasing wavefunction (or a vector in $D(P^2)$ for a norm estimate),

$$\psi(x-a) = \psi(x)-a\frac{d\psi}{dx}(x)+O(a^2).$$

On the other hand, a continuous unitary group generated by $P$ has the form

$$T(a) = \exp\!\left(-\frac{iaP}{\hbar}\right),$$

so infinitesimally

$$T(a)\psi = \psi-\frac{ia}{\hbar}P\psi+O(a^2).$$

Comparing the first-order terms,

$$-\frac{i}{\hbar}P\psi = -\frac{d\psi}{dx},$$

and therefore

$$P\psi = -i\hbar\frac{d\psi}{dx}.$$

More generally, for $\psi$ in the generator domain $D(P)$, the derivative exists as a limit in the $L^2$ norm:

$$P\psi = i\hbar\lim_{a\to0}\frac{T(a)\psi-\psi}{a} = -i\hbar\psi'.$$

Here $\psi'$ is the weak derivative; the full-line domain is specified below. This is the translation instance of the general formula in Generators.

Translation of the Position Operator

The same convention gives a clean operator statement:

$$T(a)^\dagger X T(a) = X+a.$$

To check it, compute on a test wavefunction:

$$\begin{aligned} (T(a)^\dagger X T(a)\psi)(x) &= (T(-a)X T(a)\psi)(x) \\ &= (X T(a)\psi)(x+a) \\ &= (x+a)\psi(x). \end{aligned}$$

Thus a translated state has its position expectation value shifted by $a$:

$$\langle X\rangle_{T(a)\psi} = \langle X\rangle_\psi+a.$$

Canonical Commutator

On the common invariant core of smooth rapidly decreasing Schwartz functions, expand the conjugation formula for small $a$. The following operator expansions are understood through their action on such functions:

$$T(a)^\dagger X T(a) = \left(I+\frac{iaP}{\hbar}\right) X \left(I-\frac{iaP}{\hbar}\right) +O(a^2).$$

Therefore

$$T(a)^\dagger X T(a) = X+\frac{ia}{\hbar}[P,X]+O(a^2).$$

Since the left side must equal $X+a$, one obtains

$$\frac{i}{\hbar}[P,X]=1,$$

or

$$[X,P]=i\hbar I.$$

The canonical commutation relation is therefore the infinitesimal form of the fact that momentum translates position.

Three Dimensions

For translations by a vector $\mathbf a$,

$$T(\mathbf a) = \exp\!\left( -\frac{i}{\hbar}\mathbf a\cdot\mathbf P \right).$$

On scalar wavefunctions,

$$(T(\mathbf a)\psi)(\mathbf r) = \psi(\mathbf r-\mathbf a).$$

The components of momentum act as

$$P_j = -i\hbar\frac{\partial}{\partial x_j}, \qquad j=1,2,3.$$

Ordinary spatial translations commute with each other, so their generators commute:

$$[P_i,P_j]=0.$$

The position commutators are

$$[X_i,P_j]=i\hbar\delta_{ij}I.$$

Momentum Eigenstates

If

$$P\lvert p\rangle=p\lvert p\rangle,$$

then translations act by a phase:

$$T(a)\lvert p\rangle = e^{-iap/\hbar}\lvert p\rangle.$$

In position representation, the eigenvalue equation becomes

$$-i\hbar\frac{d}{dx}\phi_p(x) = p\,\phi_p(x),$$

The solutions are plane waves. With $\langle p|p'\rangle=\delta(p-p')$ and the standard Fourier phase convention, choose

$$\phi_p(x) = \langle x|p\rangle = \frac{e^{ipx/\hbar}}{\sqrt{2\pi\hbar}}.$$

On the full line these plane waves are generalized eigenfunctions, not normalizable vectors in $L^2(\mathbb R)$. They are basis distributions used to build normalizable wave packets.

Domains and Boundary Conditions

The formula

$$P=-i\hbar\frac{d}{dx}$$

is a differential expression. An operator also needs a domain. On the full line its self-adjoint domain is $H^1(\mathbb R)$: square-integrable functions whose weak first derivative is also square-integrable. “Weak derivative” allows the derivative to be defined by integration against smooth test functions; it need not exist pointwise everywhere.

Boundary conditions can change the story.

On a ring of circumference $L$, periodic boundary conditions preserve continuous translations around the ring. Momentum eigenfunctions satisfy

$$\psi(x+L)=\psi(x),$$

so

$$e^{ipL/\hbar}=1.$$

Therefore

$$p_n = \frac{2\pi\hbar n}{L}, \qquad n\in\mathbb Z.$$

On a finite interval with hard walls, ordinary continuous translations do not preserve the interval and its boundary conditions. The derivative with zero boundary values at both endpoints is symmetric but not self-adjoint as a momentum operator. Its self-adjoint extensions instead identify the two endpoints with a phase, $\psi(L)=e^{i\chi}\psi(0)$, describing a ring-like geometry rather than two hard walls. The hard-wall kinetic Hamiltonian can still be self-adjoint even though this momentum operator is not. Momentum is then not a conserved generator in the same way it is on the line or ring.

On a lattice, only discrete translations may remain. The translation operator for one lattice spacing can have eigenvalues $e^{-ika}$ in the active convention, but there need not be a self-adjoint generator for arbitrary continuous displacements. The label $k$ is then crystal momentum or quasimomentum, defined modulo reciprocal lattice vectors.

Canonical Versus Kinetic Momentum

The generator above is the canonical momentum associated with ordinary position translations. In electromagnetic backgrounds, the kinetic or mechanical momentum is often

$$\boldsymbol\Pi = \mathbf P-q\mathbf A(\mathbf R).$$

This is the momentum related to velocity in the minimally coupled Hamiltonian, but it is not the same as the generator of ordinary translations when $\mathbf A$ is present. Magnetic fields also make spatial translation symmetry subtler; Magnetic Translations explains the extra gauge phase in that setting.

Momentum Conservation Needs Hamiltonian Symmetry

The existence of a momentum operator does not by itself imply that momentum is conserved. Conservation requires translation invariance of the Hamiltonian:

$$[H,P]=0.$$

For

$$H=\frac{P^2}{2m}+V(X),$$

this holds only when $V(X)$ is constant along that direction, with a translation-invariant domain. Discrete lattice symmetry alone does not imply $[H,P]=0$. The detailed Hamiltonian test is Translation-Invariant Hamiltonians; the conceptual link is developed in Commutators and Conservation Laws.

A coordinate on a ring is not a global Cartesian position

Although ring momentum is well-defined, the coordinate $x$ on $[0,L)$ jumps at the chosen cut. Multiplication by $x$ does not preserve the periodic momentum domain. Consequently $[X,P]=i\hbar I$ cannot be used as a global operator identity on that domain. Periodic functions such as $e^{2\pi iX/L}$ give the well-defined position observables around the ring.

Exercises

1. Derive the differential momentum operator from the active translation convention for a smooth rapidly decreasing wavefunction.

Use

$$(T(a)\psi)(x)=\psi(x-a)$$

and

$$T(a)=I-\frac{iaP}{\hbar}+O(a^2).$$

Solution

Expand the translated wavefunction:

$$\psi(x-a) = \psi(x)-a\psi'(x)+O(a^2).$$

Compare with

$$T(a)\psi = \psi-\frac{ia}{\hbar}P\psi+O(a^2).$$

The first-order terms give

$$-\frac{i}{\hbar}P\psi = -\psi',$$

so

$$P\psi=-i\hbar\psi'.$$

2. Derive $[X,P]=i\hbar I$ from the translated position operator on the Schwartz core.

Solution

The convention gives

$$T(a)^\dagger X T(a)=X+a.$$

Using $T(a)=I-iaP/\hbar+O(a^2)$,

$$T(a)^\dagger X T(a) = X+\frac{ia}{\hbar}[P,X]+O(a^2).$$

Equating first-order terms gives

$$\frac{i}{\hbar}[P,X]=1.$$

Thus

$$[X,P]=i\hbar I.$$

3. Quantize momentum on a ring.

For a ring of circumference $L$, impose $\psi(x+L)=\psi(x)$ on $\psi(x)=e^{ipx/\hbar}$. Find the allowed $p$ values.

Solution

Periodicity requires

$$e^{ip(x+L)/\hbar}=e^{ipx/\hbar}.$$

Therefore

$$e^{ipL/\hbar}=1,$$

so

$$\frac{pL}{\hbar}=2\pi n, \qquad n\in\mathbb Z.$$

The allowed momenta are

$$p_n=\frac{2\pi\hbar n}{L}.$$

4. Why is a hard-wall box not continuously translation invariant?

Solution

In a hard-wall box, wavefunctions must obey boundary conditions at fixed endpoints. Translating a wavefunction by an arbitrary small amount generally moves its support and boundary values relative to those fixed endpoints. The translated function need not satisfy the same boundary conditions. Therefore arbitrary continuous translations are not symmetries of the boxed system, even though derivative operators still appear in the Hamiltonian.

5. A normalized wave packet on the full line has finite mean $x_0$ and position variance $s^2$. After active translation by $a$, find its mean and variance.

Solution

The translated density is $|\psi(x-a)|^2$. With $y=x-a$,

$$\langle X\rangle_{T(a)\psi} =\int_{\mathbb R}(y+a)|\psi(y)|^2\,dy =x_0+a.$$

The centered coordinate satisfies $x-(x_0+a)=y-x_0$, so

$$\operatorname{Var}_{T(a)\psi}(X) =\int_{\mathbb R}(y-x_0)^2|\psi(y)|^2\,dy =s^2.$$

Translation moves the packet without changing its shape or position uncertainty.

References

• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• H. Weyl, The Theory of Groups and Quantum Mechanics, Dover, 1950 reprint of the 1931 English translation.