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 $x_0+a$.

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:

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

Thus $T(a)$ is 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 wavefunction,

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

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

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

This is the position-space action of the momentum generator on wavefunctions for which the derivative expression is meaningful.

Generator as a Derivative of Translations

Equivalently, the generator can be recovered from the unitary family itself:

$$P\psi = i\hbar \left. \frac{d}{da}T(a)\psi \right|_{a=0}.$$

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

$$\left. \frac{d}{da} \psi(x-a) \right|_{a=0} = -\frac{d\psi}{dx}(x),$$

so again

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

This is the translation analogue of the general generator 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

Expand the conjugation formula for small $a$:

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

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

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

with solutions

$$\phi_p(x) = C e^{ipx/\hbar}.$$

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, a standard self-adjoint momentum operator acts on wavefunctions that are sufficiently regular and whose derivative is square-integrable. A rigorous page would state this in Sobolev-space language.

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 formal derivative may still be useful in calculations, but there is no full continuous translation symmetry of the boxed system. 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 $e^{ika}$, 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 is the canonical page for 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 the translated direction, or when the system has the relevant translation symmetry. The detailed Hamiltonian test is Translation-Invariant Hamiltonians; the conceptual link is developed in Commutators and Conservation Laws.

Common Mistakes

• Reversing the sign between active translations and the exponential.
• Forgetting that $P=-i\hbar d/dx$ is a representation-dependent expression, not the abstract definition.
• Treating the formal derivative as self-adjoint without specifying a domain.
• Assuming hard-wall boxes have continuous translation symmetry.
• Treating plane waves as normalizable states on the full line.
• Confusing canonical momentum with kinetic momentum in electromagnetic fields.
• Concluding that momentum is conserved before checking whether the Hamiltonian is translation invariant.

Cross-Links

• Translations and Momentum
• Translation-Invariant Hamiltonians
• Generators
• One-Parameter Unitary Groups
• Commutators and Conservation Laws
• Canonical Commutation Relations
• Position and Momentum Representations
• Heisenberg Group
• Unbounded Operators
• Translation Operator
• Momentum Operator
• Free Particle
• Magnetic Translations

References

• H. Weyl, The Theory of Groups and Quantum Mechanics, Dover, 1950.
• 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.
• 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.

Exercises

1. Derive the differential momentum operator from the active translation convention.

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.

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.