Translations and Momentum

Momentum is the generator of spatial translations. This statement is one of the cleanest examples of how a one-parameter unitary group becomes an observable in quantum mechanics.

In one dimension, a translation by $a$ is represented by

$$T(a)=e^{-iaP/\hbar}.$$

The operator $P$ is the momentum.

Active Translation of a Wavefunction

With the active convention, translating a wavefunction to the right by $a$ gives

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

For small $a$,

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

The infinitesimal unitary form is

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

Comparing the two expressions gives the position-space momentum operator:

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

Translation of Position

The translated position operator satisfies

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

This formula says that after translating the state to the right by $a$, the expectation value of position shifts by $a$:

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

The sign convention is consistent with $(T(a)\psi)(x)=\psi(x-a)$.

Canonical Commutator

Using the infinitesimal form,

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

Equating this with $X+a$ gives

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

Thus the canonical commutation relation is the infinitesimal statement that momentum generates translations.

Momentum Eigenstates

A momentum eigenstate satisfies

$$P\lvert p\rangle=p\lvert p\rangle.$$

Under translations,

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

In position representation, the generalized eigenfunctions are plane waves:

$$\langle x|p\rangle \propto e^{ipx/\hbar}.$$

The proportionality depends on normalization convention.

Translation-Invariant Hamiltonians

A Hamiltonian is translation invariant when

$$T(a)HT^\dagger(a)=H$$

for all $a$. Infinitesimally this is

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

For

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

translation invariance requires $V(X)$ to be constant on the translated region. The free particle is the basic example.

Higher Dimensions

In three dimensions,

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

with

$$P_j=-i\hbar\frac{\partial}{\partial x_j}.$$

The components commute:

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

reflecting the abelian nature of ordinary spatial translations.

Common Mistakes

• Reversing the sign in $(T(a)\psi)(x)=\psi(x-a)$.
• Confusing active translation of the state with passive coordinate relabeling.
• Treating plane waves as normalizable states on the full line.
• Forgetting that translation symmetry is broken by position-dependent potentials.
• Ignoring boundary conditions; translations on a ring, interval, or lattice require modified domains or discrete translation operators.

Cross-Links

• Spatial Symmetries
• One-Parameter Unitary Groups
• Generators
• Momentum Operator as Generator
• Translation-Invariant Hamiltonians
• Crystalline Symmetry Preview
• Rotations Preview
• Galilean Boosts
• Magnetic Translations
• Commutators and Conservation Laws
• Momentum Operator
• Canonical Commutation Relations
• Heisenberg Group
• Position and Momentum Representations
• Free Particle

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.

Exercises

1. Starting from $(T(a)\psi)(x)=\psi(x-a)$, derive $P=-i\hbar\,d/dx$.

Solution

Expand both expressions for small $a$:

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

and

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

Equating the first-order terms gives

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

so $P=-i\hbar\,d/dx$.

2. Show that $T(a)\lvert p\rangle=e^{-iap/\hbar}\lvert p\rangle$ if $P\lvert p\rangle=p\lvert p\rangle$.

Solution

Use the power-series definition of the exponential:

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