Translation-Invariant Hamiltonians

A Hamiltonian is translation invariant when translating the system does not change the Hamiltonian. Momentum conservation follows only under that condition. The existence of a momentum operator is not enough.

For translations by $a$ in one dimension,

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

The Hamiltonian is invariant under these translations when

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

for every real displacement $a$, with the Hamiltonian domain preserved. Equivalently,

$$[H,T(a)]=0$$

for every $a$. For continuous translations, differentiation on a suitable common domain gives

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

Thus translation invariance is the symmetry reason momentum can be conserved.

From Finite Translations to the Commutator

For smooth test states on the full line, differentiate the invariance condition

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

at $a=0$. Since

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

one has

$$\begin{aligned} T(a)HT(a)^\dagger &= \left(I-\frac{iaP}{\hbar}\right) H \left(I+\frac{iaP}{\hbar}\right) +O(a^2) \\ &= H-\frac{ia}{\hbar}[P,H]+O(a^2). \end{aligned}$$

For this to equal $H$ to first order for all $a$,

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

or equivalently

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

The same argument in three dimensions gives one condition for each translation direction:

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

If the Hamiltonian is invariant only along one direction, only the corresponding momentum component is conserved.

Momentum Conservation

For a time-independent momentum operator, the expectation-value equation gives

$$\frac{d}{dt}\langle P\rangle = \frac{i}{\hbar}\langle[H,P]\rangle.$$

If

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

then

$$\frac{d}{dt}\langle P\rangle=0$$

for every state in the appropriate domain. More strongly, the full momentum measurement distribution is preserved under time evolution when the spectral projectors of $P$ commute with $H$.

This is the quantum-mechanical Noether pattern:

$$\text{translation symmetry} \quad \Longrightarrow \quad \text{momentum conservation}.$$

The general commutator logic is developed in Commutators and Conservation Laws and summarized in Noether Theorem in Quantum Mechanics.

Single Particle in a Potential

On the full line, consider

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

The kinetic term is translation invariant because it is built from $P$, which commutes with itself. The potential term is the test.

Using

$$[V(X),P]=i\hbar V'(X),$$

one finds

$$[H,P] = i\hbar V'(X).$$

Therefore momentum is conserved only when

$$V'(X)=0$$

on the full line, assuming $V$ is differentiable. That means the potential is constant. A free particle is the special case $V=0$.

The same statement appears from finite translations. With the active convention,

$$T(a)XT(a)^\dagger=X-a,$$

so

$$T(a)V(X)T(a)^\dagger = V(X-a).$$

Invariance for every $a$ requires

$$V(X-a)=V(X)$$

for every allowed $a$. On the full line, that forces $V$ to be constant.

Free and Constant-Potential Examples

For the free particle,

$$H=\frac{P^2}{2m},$$

so

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

Momentum eigenstates diagonalize the Hamiltonian:

$$H\lvert p\rangle = \frac{p^2}{2m}\lvert p\rangle.$$

A constant potential

$$H=\frac{P^2}{2m}+V_0$$

does not change the symmetry. It shifts all energies by $V_0$ but leaves momentum conserved.

By contrast, a harmonic oscillator potential,

$$V(X)=\frac12m\omega^2X^2,$$

selects an origin. It is not translation invariant, and momentum is not conserved.

Force as Broken Translation Symmetry

For a uniform force in one dimension,

$$V(X)=-FX.$$

Then

$$V'(X)=-F,$$

so

$$[H,P]=-i\hbar F.$$

The expectation-value equation gives

$$\frac{d}{dt}\langle P\rangle = F.$$

This is the Ehrenfest form of Newton’s law. Here $T(a)HT(a)^\dagger=H+FaI$, so translations preserve the force law and change the energy only by a constant. They do not satisfy the strict $H$-invariance condition used above, and $P$ is not conserved. Instead, the explicitly time-dependent observable $P-FtI$ is conserved. This example distinguishes exact operator invariance from invariance of the dynamics up to an overall time-dependent phase.

Boundary Conditions Matter

Translation invariance is a statement about the full problem, including boundary conditions.

On the full line, arbitrary real translations are allowed. On a ring or a periodic box, translations are allowed modulo the period, and momentum becomes quantized. For a ring of circumference $L$,

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

On a hard-wall interval, arbitrary translations do not preserve the boundaries. The Hamiltonian may contain $P^2/(2m)$ locally, but the boxed system is not continuously translation invariant. Momentum is not generally a conserved observable for a particle in a hard-wall box.

This is why boundary conditions are part of the symmetry data. They can preserve, reduce, or destroy translation symmetry.

Periodic Potentials

A periodic potential satisfies

$$V(x+R)=V(x)$$

for lattice translations $R=na$, not for arbitrary real translations. The Hamiltonian is invariant under the discrete translation operators

$$T(na), \qquad n\in\mathbb Z,$$

but not under all $T(a')$ with arbitrary $a'$.

Consequences:

• ordinary continuous momentum is not generally conserved;
• the discrete translation operator can be diagonalized;
• with the active convention here, the eigenvalue of $T(a)$ may be written $e^{-ika}$;
• the equivalent wavefunction phase is $\psi_k(x+a)=e^{ika}\psi_k(x)$;
• $k$ is crystal momentum or quasimomentum, defined modulo reciprocal lattice shifts.

This is the symmetry preview of Bloch theory. The detailed band-structure machinery belongs to quantum-matter pages; here the important point is that continuous translation symmetry has been reduced to a discrete subgroup.

Many-Body Total Momentum

For $N$ particles on the line, the total momentum is

$$P_{\mathrm{tot}} = \sum_{\alpha=1}^N P_\alpha.$$

It generates simultaneous translation of all particles:

$$X_\alpha \longmapsto X_\alpha+a \qquad \text{for every }\alpha.$$

Consider a Hamiltonian of the form

$$H = \sum_{\alpha=1}^N \frac{P_\alpha^2}{2m_\alpha} + \sum_{\alpha<\beta} W_{\alpha\beta}(X_\alpha-X_\beta).$$

The interaction depends only on relative positions, so a common shift of all positions leaves it unchanged:

$$X_\alpha-X_\beta \longmapsto (X_\alpha+a)-(X_\beta+a) = X_\alpha-X_\beta.$$

Therefore

$$[H,P_{\mathrm{tot}}]=0.$$

Individual particle momenta need not be conserved, because interactions exchange momentum between particles. The conserved quantity is the total momentum of the closed translation-invariant system.

If an external potential is added,

$$\sum_\alpha V_{\mathrm{ext}}(X_\alpha),$$

then translation invariance usually fails unless the external potential is constant or has a residual discrete symmetry.

Magnetic and Gauge-Coupled Systems

In electromagnetic backgrounds, ordinary translation invariance can be obscured by the vector potential. The physical magnetic field may be spatially uniform while a particular gauge choice for $\mathbf A$ is not invariant under ordinary translations.

In such cases the correct symmetry may be a magnetic translation: an ordinary spatial shift accompanied by a compensating gauge phase. Ordinary canonical momentum, kinetic momentum, and magnetic translation generators should not be conflated. See Magnetic Translations for the modified translation operators.

Exercises

1. Let

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

Use $[V(X),P]=i\hbar V'(X)$ to find the condition for momentum conservation.

Solution

Since $[P^2,P]=0$,

$$[H,P]=[V(X),P]=i\hbar V'(X).$$

Momentum is conserved as an operator statement when

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

so the condition is

$$V'(X)=0.$$

On the full line, this means $V$ is constant.

2. For $V(X)=-FX$, compute $d\langle P\rangle/dt$.

Solution

Here

$$V'(X)=-F,$$

so

$$[H,P]=i\hbar V'(X)=-i\hbar F.$$

The expectation-value equation gives

$$\frac{d}{dt}\langle P\rangle = \frac{i}{\hbar}\langle[H,P]\rangle = \frac{i}{\hbar}(-i\hbar F) = F.$$

3. Why does a periodic potential conserve crystal momentum rather than ordinary continuous momentum?

Solution

A periodic potential satisfies $V(x+a)=V(x)$ for a lattice spacing $a$, but not generally for arbitrary displacements. Therefore the Hamiltonian commutes with the discrete translation operator $T(a)$, not with all continuous translations $T(a')$. With the active convention, a translation eigenstate can be written $T(a)|\psi_k\rangle=e^{-ika}|\psi_k\rangle$, equivalently $\psi_k(x+a)=e^{ika}\psi_k(x)$. The label $k$ is defined modulo reciprocal lattice shifts, so it is crystal momentum or quasimomentum, not ordinary continuous momentum.

4. Show why pair interactions depending only on relative positions conserve total momentum.

For two particles, take

$$H = \frac{P_1^2}{2m_1} + \frac{P_2^2}{2m_2} + W(X_1-X_2), \qquad P_{\mathrm{tot}}=P_1+P_2.$$

Solution

The kinetic terms commute with $P_{\mathrm{tot}}$. For the interaction, use

$$[W(X_1-X_2),P_1] = i\hbar W'(X_1-X_2),$$

and

$$[W(X_1-X_2),P_2] = -i\hbar W'(X_1-X_2).$$

Adding them gives

$$[W(X_1-X_2),P_1+P_2]=0.$$

Therefore

$$[H,P_{\mathrm{tot}}]=0.$$

The interaction can exchange momentum between particles, but it conserves their total momentum.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• 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.
• E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of Atomic Spectra, Academic Press, 1959.