Galilean Boosts

Galilean boosts relate inertial frames moving at constant relative velocity in nonrelativistic mechanics. Quantum mechanically, they are not just coordinate substitutions. A wavefunction also acquires a mass-dependent phase. That phase is the simplest physical doorway to the projective representation of the Galilei group and to mass as a central charge.

This page uses a passive convention: a boost by velocity $\mathbf v$ means describing the same state in a frame moving with velocity $\mathbf v$ relative to the original frame. In that convention,

$$\mathbf x' = \mathbf x-\mathbf v t, \qquad \mathbf p' = \mathbf p-m\mathbf v.$$

Active conventions differ by signs. The important habit is to state the convention before interpreting the boost generator.

Classical Transformation

For a nonrelativistic particle of mass $m$, a Galilean boost to a frame moving with constant velocity $\mathbf v$ uses

$$t'=t, \qquad \mathbf x' = \mathbf x-\mathbf v t.$$

Velocities and momenta transform as

$$\dot{\mathbf x}' = \dot{\mathbf x}-\mathbf v, \qquad \mathbf p' = \mathbf p-m\mathbf v.$$

For a free particle,

$$E = \frac{\mathbf p^2}{2m}$$

becomes

$$\begin{aligned} E' &= \frac{ (\mathbf p-m\mathbf v)^2 }{2m}\\ &= E-\mathbf v\cdot\mathbf p + \frac12m\mathbf v^2. \end{aligned}$$

The quantum transformation must reproduce these momentum and energy shifts while preserving probabilities.

Wavefunction Transformation

For a free spinless particle, the boosted wavefunction is

$$\psi'(\mathbf x,t) = \exp \left[ -\frac{i}{\hbar} \left( m\mathbf v\cdot\mathbf x + \frac12m\mathbf v^2t \right) \right] \psi(\mathbf x+\mathbf v t,t).$$

Here $\mathbf x$ is the coordinate in the boosted frame after relabeling. The shift $\mathbf x+\mathbf v t$ samples the original wavefunction at the corresponding old-frame point, while the exponential supplies the mass-dependent phase.

For a plane wave,

$$\psi(\mathbf x,t) = \exp \left[ \frac{i}{\hbar} \left( \mathbf p\cdot\mathbf x -Et \right) \right],$$

the transformed wave has

$$\mathbf p' = \mathbf p-m\mathbf v, \qquad E' = E-\mathbf v\cdot\mathbf p + \frac12m\mathbf v^2.$$

Thus the phase is not optional. Without it, the momentum and energy of the transformed plane wave would be wrong.

Boost Generator

At time $t$, the standard passive boost unitary can be written

$$U(\mathbf v,t) = \exp \left( -\frac{i}{\hbar} \mathbf v\cdot\mathbf K(t) \right),$$

with boost generator

$$\mathbf K(t) = m\mathbf X-t\mathbf P.$$

This convention gives the operator transformations

$$U(\mathbf v,t)^\dagger \mathbf P U(\mathbf v,t) = \mathbf P-m\mathbf v,$$

and

$$U(\mathbf v,t)^\dagger \mathbf X U(\mathbf v,t) = \mathbf X-\mathbf v t.$$

These are the quantum versions of the passive Galilean transformation.

For the free Hamiltonian

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

the boost generator is conserved in the Heisenberg sense:

$$\frac{d\mathbf K}{dt} = \frac{i}{\hbar}[H,\mathbf K] + \frac{\partial\mathbf K}{\partial t} = \mathbf 0.$$

That conservation expresses uniform center-of-mass motion.

Relation to the Schrödinger Equation

The free Schrödinger equation is

$$i\hbar \frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2\psi.$$

If $\psi$ solves this equation, then the boosted $\psi'$ above also solves it. The phase is precisely what cancels the extra terms produced by differentiating $\psi(\mathbf x+\mathbf v t,t)$.

One way to remember the phase is to demand that the plane-wave phase

$$\frac{1}{\hbar} \left( \mathbf p\cdot\mathbf x-Et \right)$$

transform into the same form with $\mathbf p'$ and $E'$, up to the coordinate substitution. That requirement gives the factor

$$\exp \left[ -\frac{i}{\hbar} \left( m\mathbf v\cdot\mathbf x + \frac12m\mathbf v^2t \right) \right].$$

Galilei Algebra Preview

Let $\mathbf P$ generate translations, $\mathbf J$ generate rotations, $H$ generate time translations, and $\mathbf K$ generate boosts. For a spinless free particle, the boost-momentum commutator is

$$[K_i,P_j] = i\hbar m\,\delta_{ij}.$$

The boost-Hamiltonian commutator is

$$[K_i,H] = i\hbar P_i.$$

Rotations act on boosts as vectors:

$$[J_i,K_j] = i\hbar\epsilon_{ijk}K_k.$$

The first commutator is the one to remember. The right-hand side is not zero; it is proportional to the mass. Since $mI$ commutes with all observables, mass appears as a central element in the quantum representation of the Galilei algebra.

Mass as a Central Charge

In classical Galilean mechanics, mass is a parameter in the equations of motion. In quantum mechanics, it also labels the projective representation of the Galilei group.

At $t=0$,

$$U(\mathbf v) = \exp \left( -\frac{i}{\hbar} m\mathbf v\cdot\mathbf X \right), \qquad T(\mathbf a) = \exp \left( -\frac{i}{\hbar} \mathbf a\cdot\mathbf P \right).$$

Because $[\mathbf X,\mathbf P]\ne0$, boosts and translations commute only up to a phase:

$$U(\mathbf v)T(\mathbf a) = \exp \left( -\frac{i}{\hbar} m\mathbf v\cdot\mathbf a \right) T(\mathbf a)U(\mathbf v).$$

The phase does not change a single ray, but it cannot be ignored in the group composition law. This is the same projective-representation logic introduced in Projective Representations. The mass $m$ labels the central extension.

For many particles, the central charge is the total mass. If

$$\mathbf K = \sum_\alpha m_\alpha\mathbf X_\alpha -t\sum_\alpha\mathbf P_\alpha,$$

then

$$[K_i,P_j^{\mathrm{tot}}] = i\hbar M\,\delta_{ij}, \qquad M = \sum_\alpha m_\alpha.$$

This is why center-of-mass separation and total mass are structurally tied to Galilean symmetry.

Relation to Lorentz Boosts

Galilean boosts are nonrelativistic spacetime symmetries. They keep time absolute:

$$t'=t.$$

Lorentz boosts replace this with spacetime mixing between $t$ and $\mathbf x$. Relativistic quantum theory also changes the status of mass, spin, antiparticles, and particle number. The nonrelativistic Galilei group is therefore not a small notation variant of the Lorentz group; it is a different symmetry group, recovered as a low-velocity limit under appropriate assumptions.

For the bridge language, see Symmetries. Detailed relativistic representation theory belongs beyond this page.

Common Mistakes

• Omitting the mass-dependent phase in the wavefunction transformation.
• Mixing active and passive sign conventions in the same calculation.
• Thinking boosts commute with translations exactly on Hilbert-space vectors.
• Forgetting that the projective phase is physically meaningful in the representation law even though global phase is unobservable for one state.
• Treating Galilean boosts as the same as Lorentz boosts with $c$ omitted.
• Using the single-particle mass formula without replacing $m$ by total mass for a many-body center-of-mass boost.

Cross-Links

• Translations and Momentum
• Generators
• Commutators and Conservation Laws
• Projective Representations
• From Projective Representations to Anomalies Preview
• Unitary Symmetries
• Free Particle
• Representations
• Symmetries

References

• V. Bargmann, “On unitary ray representations of continuous groups,” Annals of Mathematics 59, 1-46, 1954.
• J.-M. Lévy-Leblond, “Galilei Group and Galilean Invariance,” in Group Theory and Applications, Vol. II, Academic Press, 1972.
• 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.
• H. Bacry and J.-M. Lévy-Leblond, “Possible kinematics,” Journal of Mathematical Physics 9, 1605-1614, 1968.

Exercises

1. Check the plane-wave momentum shift.

Apply the boost wavefunction formula to

$$\psi(\mathbf x,t) = \exp \left[ \frac{i}{\hbar} \left( \mathbf p\cdot\mathbf x-Et \right) \right].$$

Show that the new momentum is $\mathbf p-m\mathbf v$.

Solution

The boosted wavefunction is

$$\psi'(\mathbf x,t) = \exp \left[ -\frac{i}{\hbar} \left( m\mathbf v\cdot\mathbf x + \frac12m\mathbf v^2t \right) \right] \exp \left[ \frac{i}{\hbar} \left( \mathbf p\cdot(\mathbf x+\mathbf v t) -Et \right) \right].$$

The coefficient of $\mathbf x$ in the total phase is

$$\mathbf p-m\mathbf v.$$

Thus the transformed momentum is

$$\mathbf p' = \mathbf p-m\mathbf v.$$

2. Derive the boost-momentum commutator.

Using $K_i=mX_i-tP_i$ and $[X_i,P_j]=i\hbar\delta_{ij}$, compute $[K_i,P_j]$.

Solution

Since $[P_i,P_j]=0$,

$$\begin{aligned} [K_i,P_j] &= [mX_i-tP_i,P_j]\\ &= m[X_i,P_j] -t[P_i,P_j]\\ &= i\hbar m\,\delta_{ij}. \end{aligned}$$

3. Show that the boost generator is conserved for a free particle.

Let $H=\mathbf P^2/(2m)$ and $\mathbf K=m\mathbf X-t\mathbf P$. Show that $d\mathbf K/dt=\mathbf0$.

Solution

Use the Heisenberg-picture derivative:

$$\frac{dK_i}{dt} = \frac{i}{\hbar}[H,K_i] + \frac{\partial K_i}{\partial t}.$$

First,

$$[H,mX_i] = \frac{1}{2} [P_jP_j,X_i] = -i\hbar P_i,$$

and $[H,-tP_i]=0$. Therefore

$$\frac{i}{\hbar}[H,K_i] = P_i.$$

Also,

$$\frac{\partial K_i}{\partial t} = -P_i.$$

The two terms cancel, so $dK_i/dt=0$.

4. Find the projective phase.

At $t=0$, use the Baker-Campbell-Hausdorff formula to show that

$$U(\mathbf v)T(\mathbf a) = e^{-im\mathbf v\cdot\mathbf a/\hbar} T(\mathbf a)U(\mathbf v).$$

Solution

Let

$$A = -\frac{i}{\hbar} m\mathbf v\cdot\mathbf X, \qquad B = -\frac{i}{\hbar} \mathbf a\cdot\mathbf P.$$

Their commutator is the scalar

$$[A,B] = -\frac{i}{\hbar} m\mathbf v\cdot\mathbf a.$$

Since this commutes with both $A$ and $B$,

$$e^Ae^B = e^Be^Ae^{[A,B]}.$$

Thus

$$U(\mathbf v)T(\mathbf a) = e^{-im\mathbf v\cdot\mathbf a/\hbar} T(\mathbf a)U(\mathbf v).$$