advanced30 min studyLesson 402 in Physics

Translation generators and momentum conservation

Derive the active translation sign and the precise conditions under which momentum is conserved.

Before you begin

On this page

The question for today

What you’ll understand

Derive the action of an active translation on a wavefunction.
Relate translation invariance to a conserved momentum distribution.
Distinguish continuous translations, lattice translations, and boundary restrictions.

Moving a state to the right makes its wavefunction depend on a shifted argument to the left. This elementary sign issue controls the definition of the momentum generator. Translation symmetry then explains a conservation law without solving the Schrödinger equation. The statement needs a specified system and domain: a particle in a fixed trap, a particle on a periodic ring, and an isolated collection of interacting particles have different translation symmetries.

Fix the active convention

Define an active displacement by $T(a)|x\rangle=|x+a\rangle$ , where $a$ is a length. The translated probability peak must move right when $a$ is positive. In position coordinates this requires $(T(a)\psi)(x)=\psi(x-a)$ : at the new peak position, the argument returns to the old peak. A passive shift of coordinate labels uses a different sign. Neither convention is wrong, but combining their formulas would move the peak the wrong way.

Expand a smooth wavefunction for a small displacement. Its first change is minus the displacement times its spatial derivative. Comparing that change with a unitary exponential identifies $p=-i\hbar\partial_x$ . Momentum has SI units kilograms metres per second, so $ap/\hbar$ is dimensionless. On the infinite line the operator needs its appropriate self-adjoint domain. On a finite interval, boundary conditions decide whether this expression actually generates allowed translations of the physical state space.

T(a)=e^{-iap/\hbar},\qquad (T(a)\psi)(x)=\psi(x-a),\qquad p=-i\hbar\frac{d}{dx}

This is the active translation convention on the line, for states in the generator domain. Here a is in metres and hbar has units joule seconds.↔ Scroll wide equations horizontally.

Invariance makes the generator conserved

For a time-independent Hamiltonian, translation invariance means $T(a)^\dagger HT(a)=H$ for every allowed displacement. Differentiating at zero yields a vanishing commutator with momentum, subject to domain compatibility. Conversely, a formal zero commutator on an inadequately chosen collection of functions is not sufficient to establish a full operator symmetry. Infinite walls are a useful warning: differentiating a wavefunction can violate the boundary condition even when the interior differential expressions appear compatible.

For $H=p^2/(2m)+V(x)$ with a differentiable potential, the momentum derivative is minus the mean potential gradient. A constant potential therefore conserves the mean momentum; a slope accelerates it. Full translation invariance preserves more than the mean: momentum spectral probabilities stay fixed under time evolution. An isolated many-particle system can have interactions depending on coordinate differences and still conserve total momentum, although the momentum of each interacting particle changes through exchanged forces.

[H,p]=i\hbar V^{\prime}(x),\qquad \frac{d\langle p\rangle}{dt}=\frac{i}{\hbar}\langle[H,p]\rangle=-\langle V^{\prime}(x)\rangle

The derivative formula assumes a time-independent p, suitable states and compatible operator domains. V is an energy and V prime is a force with the opposite sign.↔ Scroll wide equations horizontally.

Plane-wave phases and weaker spatial symmetries

A momentum eigenfunction acquires a phase under translation because it already has a definite spatial wave number. This does not mean that every localized packet merely gains a common phase. A packet contains several momenta; different phase factors translate its envelope. The magnitude of each momentum amplitude remains unchanged. This separation between phases and probabilities is the Fourier-space version of shifting a real-space probability distribution without changing its shape at that instant.

A periodic lattice preserves only discrete translations by its lattice spacing. Its eigenvalue is a phase defining crystal momentum modulo a reciprocal lattice vector; ordinary mechanical momentum need not be conserved for the particle alone. On a ring, periodic boundary conditions quantize the allowed wave numbers, while continuous rotations around the ring can still translate the state. Always ask which displacement group preserves both the Hamiltonian and the boundaries before assigning a conserved generator or claiming a degeneracy.

T(a)e^{ikx}=e^{-ika}e^{ikx},\qquad k_n=\frac{2\pi n}{L},\qquad p_n=\hbar k_n

The first identity uses wave number k in inverse metres. The final two expressions assume a ring of circumference L and an ordinary periodic scalar wavefunction, with integer n.↔ Scroll wide equations horizontally.

Work it through

A packet on a ring translated by a quarter turn

On a ring of circumference $L$ , prepare $(|n=0\rangle+|n=2\rangle)/\sqrt2$ . Translate it actively by $a=L/4$ . Find the two amplitudes and the probability of the original plus superposition.

The zero-momentum component gains phase $e^0=1$ . The second has wave number $4\pi/L$ , so its translation phase is $e^{-i(4\pi/L)(L/4)}=-1$ . The circumference cancels as the exponent requires.
The translated state is $(|0\rangle-|2\rangle)/\sqrt2$ . Both momentum probabilities remain one half; the operation changed only their relative phase.
Project onto the initial plus superposition. The overlap is $(1-1)/2=0$ , so its probability is zero. A later coherent comparison can detect a translation even though a momentum measurement cannot.

The result

The relative amplitude changes sign, and the original plus outcome becomes impossible under the ideal coherent measurement.

The living notebook

Explore the model

Interactive lab

Interactive lab · quantum

A two-momentum phase comparison

Encode the ring's two momentum labels as the lab's two basis labels. For independent paused comparisons Reset and set both controls to $(90^\circ,0^\circ)$ and then $(90^\circ,180^\circ)$ . Predict fixed half probabilities and Bloch x changing from +1 to −1. This is only the two-amplitude geometry; the lab has no position wavefunction, translation-distance control, or ring Hamiltonian.

Bloch vector / P(0)P(1)Phase latitude

Probability |0⟩: 71.1%
Probability |1⟩: 28.9%
Bloch (x, y, z): 0.78, 0.45, 0.42

Mixing angle θ65°Relative phase φ30°

t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forTranslation can preserve momentum probabilities while changing a measurable relative phase.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The force is minus the potential derivative, so Ehrenfest evolution gives minus F. The fixed linear potential breaks continuous translation invariance of the particle Hamiltonian.

Your turn

Try it yourself

For a ring state $(|0\rangle+|1\rangle)/\sqrt2$ , take $a=L/4$ . Find the relative phase and the probability of projection onto the initial state. Check that changing a to zero returns the correct limit.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The second component gains the phase with wave number two pi divided by the circumference. Take an overlap before squaring.

Compare with the solution

The phase is $e^{-i\pi/2}=-i$ , giving $(|0\rangle-i|1\rangle)/\sqrt2$ . Its overlap with the original state is $(1-i)/2$ , with squared magnitude $1/2$ . At zero displacement both factors are one, so the overlap is one and the probability is one, consistent with the identity translation.

Keep the idea

Take it with you

An active displacement to the right evaluates the original wavefunction at x minus the displacement.
Momentum conservation requires a translation symmetry of the Hamiltonian and its physical domain.
Discrete lattice symmetry conserves crystal momentum modulo reciprocal lattice vectors.

Read further

Original sources and deeper explanations to follow at your own pace.

Robert G. Littlejohn, Berkeley Physics 221: Spatial Degrees of Freedom, sections 3–6

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