Spatial Symmetries

A translated wavepacket need not have the same energy as the original one: a trap selects a preferred position, while a free-particle Hamiltonian does not. Spatial symmetry asks which translations, rotations, inversions, or changes of inertial frame preserve the complete model, including its walls and external fields. A lattice illustrates an intermediate case: translations by one lattice period remain symmetries although arbitrary translations do not.

Helpful background. Continuous symmetries connects transformations to generators. Active and passive transformations fixes the direction of the coordinate and state changes.

Enter this chapter

Question	Read next
How does displacement lead to momentum?	Momentum Operator as Generator
Which potentials and boundaries conserve momentum?	Translation-invariant Hamiltonians
Why does changing inertial frame require a phase?	Galilean boosts
How do spatial rotations act on wavefunctions?	Rotations
How is inversion different from a rotation?	Parity
What labels survive a periodic potential?	Crystalline symmetry

Translating a wavepacket fixes the sign

On the full line, an active translation by $a$ is

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

If $\psi$ was centered at $x_0$, the translated function is largest near $x-a=x_0$, hence near $x_0+a$. This is a direct sign check. Expanding on differentiable vectors gives $P=-i\hbar\partial_x$, with the self-adjoint full-line momentum domain. The matching observable identity is $T^\dagger XT=X+aI$.

On an interval, half-line, or ring, the differential expression alone does not specify the operator. Boundary conditions determine whether the intended translation is available and whether its generator is self-adjoint. In particular, shifting a box wavefunction while leaving the walls fixed is not a symmetry of the box Hamiltonian.

Translation invariance is a property of the full model

A Hamiltonian is invariant under a translation when

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

for every allowed $\mathbf a$. For continuous translations, differentiating at the identity gives

$$[H,P_i]=0$$

for each invariant direction. Only those components are conserved. A system can be invariant along $y$ and $z$ while a potential varies along $x$.

For one particle,

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

and

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

Continuous translation symmetry therefore requires a constant potential on the modeled region. The nonzero derivative is the operator form of force:

$$\frac{d}{dt}\langle P\rangle = -\langle V'(X)\rangle.$$

The existence of the translation operator does not imply that it commutes with this Hamiltonian. Symmetry is a property of the full model, including its potential, boundaries, domain, and fixed external fields.

Total and mechanical momentum

Boundary conditions can break translation symmetry even when the differential expression for $H$ looks translation invariant. A particle in a box has a kinetic Hamiltonian in the interior, but the walls select locations. On a ring, continuous translations can survive because periodic boundary conditions are themselves translation invariant.

For many particles with pair interactions depending only on relative positions,

$$H = \sum_i\frac{\mathbf P_i^2}{2m_i} + \sum_{i<j}V(\mathbf X_i-\mathbf X_j),$$

simultaneous translation of every particle leaves the relative coordinates unchanged. The conserved generator is total momentum,

$$\mathbf P_{\mathrm{tot}} = \sum_i\mathbf P_i.$$

In electromagnetic backgrounds, distinguish canonical momentum from kinetic momentum. The Gauge, Phase, and Magnetic Geometry chapter explains why ordinary translations may have to be combined with gauge transformations, producing magnetic translations. The statement $\mathbf P=-i\hbar\boldsymbol\nabla$ identifies the canonical translation generator in a chosen representation; it is not automatically the mechanical momentum $m\dot{\mathbf X}$.

Boosts relate inertial frames

A Galilean boost relates inertial frames moving with constant relative velocity. In a passive convention,

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

and a particle’s momentum transforms as

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

Quantum wavefunctions require more than a coordinate substitution. They acquire a mass-dependent phase so that momentum and energy transform correctly. At fixed time, the boost generator can be written

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

For a free particle,

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

the explicit time dependence of $\mathbf K(t)$ cancels its commutator evolution:

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

Thus the boost generator is a constant of motion in the appropriate sense even though it depends explicitly on time.

The Galilei algebra contains

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

The mass appears as a central charge. This is tied to projective quantum representations and to the phase acquired under boosts. Galilean boosts are not low-speed Lorentz boosts with every relativistic feature retained; they belong to a different spacetime symmetry group and preserve absolute time.

Rotations and inversion have different generators

A proper rotation $\mathcal R$ acts on a spinless scalar wavefunction by $(U(\mathcal R)\psi)(\mathbf r)=\psi(\mathcal R^{-1}\mathbf r)$. Its infinitesimal generator is orbital angular momentum $\mathbf L=\mathbf X\times\mathbf P$. With spin present, the complete rotation generator is $\mathbf J=\mathbf L+\mathbf S$.

Full rotational invariance requires compatibility with every rotation, not just one axis. It conserves all components of $\mathbf J$, although those components cannot all have sharp values in a common basis. Continue to rotations and orbital angular momentum for the algebra and spherical harmonics.

In three dimensions, inversion $\mathbf r\mapsto-\mathbf r$ has determinant $-1$ and is not a proper rotation. Its standard orbital representative $P$ satisfies $P^2=I$. Position and momentum reverse sign, but angular momentum, an axial vector, does not. Since inversion is discrete, it has no continuous infinitesimal parameter whose derivative gives a parity generator. If the Hamiltonian and its domain preserve inversion, energy eigenstates can be chosen with parity $\pm1$; parity develops the resulting matrix-element rules.

A lattice preserves discrete translations

A nonconstant periodic potential generally preserves translations only by vectors in its translation lattice:

$$V(\mathbf r+\mathbf R)=V(\mathbf r), \qquad \mathbf R\in\Lambda.$$

The symmetry condition is

$$[H,T(\mathbf R)]=0 \qquad \text{for every }\mathbf R\in\Lambda.$$

This does not imply $[H,\mathbf P]=0$. Continuous momentum conservation is generally lost, while the commuting lattice translations still supply simultaneous labels. Their eigenstates satisfy

$$T(\mathbf R)|\psi_{\mathbf k}\rangle = e^{-i\mathbf k\cdot\mathbf R} |\psi_{\mathbf k}\rangle.$$

Reciprocal-lattice vectors $\mathbf G$ obey

$$e^{i\mathbf G\cdot\mathbf R}=1,$$

so

$$\mathbf k\sim\mathbf k+\mathbf G.$$

Crystal momentum is therefore a label modulo reciprocal-lattice vectors, not ordinary momentum with an unrestricted unique value. A crystal also retains a finite point group of rotations, reflections, or inversions compatible with the lattice. The combination of translations and point operations forms a space group, with additional nonsymmorphic possibilities beyond this preview.

Exercises

1. On the full line, let $V(x+a_0)=V(x)$ for one nonzero period $a_0$. Explain why $T(a_0)$ can commute with $H=P^2/(2m)+V(X)$ even though $P$ need not commute with $H$.

Solution

Translation by the lattice period sends the potential to itself:

$$\begin{aligned} T(a_0)V(X)T^\dagger(a_0) &= V(X-a_0), \\ &= V(X). \end{aligned}$$

The negative shift follows from $TXT^\dagger=X-a_0I$. The kinetic term is also invariant, so $[H,T(a_0)]=0$.

Continuous invariance would require the same statement for every real $a$. Differentiating that stronger condition would give $[H,P]=0$. A nonconstant periodic potential obeys

$$[H,P]=i\hbar V'(X)\ne0$$

in general. Thus discrete lattice translation symmetry survives while continuous translation symmetry and ordinary momentum conservation do not.

2. Let $H=P^2/(2m)+V(X)$ with $V(-x)=V(x)$. If $|e\rangle$ and $|o\rangle$ have even and odd parity, respectively, determine whether $\langle e|X|e\rangle$, $\langle o|X|o\rangle$, and $\langle e|X|o\rangle$ are forbidden by parity.

Solution

Position is parity odd:

$$\Pi X\Pi^{-1}=-X.$$

For a parity eigenstate $|\pi\rangle$,

$$\begin{aligned} \langle\pi|X|\pi\rangle &= \langle\pi|\Pi^{-1}(\Pi X\Pi^{-1})\Pi|\pi\rangle \\ &=-\langle\pi|X|\pi\rangle. \end{aligned}$$

Therefore both diagonal matrix elements vanish. Between opposite-parity states, the two state parities and the odd operator multiply to an even overall factor, so $\langle e|X|o\rangle$ is allowed by parity. It may still vanish for another reason.

3. A lattice translation eigenstate satisfies $T(\mathbf R)|\psi_{\mathbf k}\rangle=e^{-i\mathbf k\cdot\mathbf R}|\psi_{\mathbf k}\rangle$. Show why $\mathbf k$ and $\mathbf k+\mathbf G$ label the same translation eigenvalues when $\mathbf G$ is reciprocal.

Solution

By definition of a reciprocal-lattice vector,

$$e^{i\mathbf G\cdot\mathbf R}=1$$

for every lattice vector $\mathbf R$. Therefore

$$\begin{aligned} e^{-i(\mathbf k+\mathbf G)\cdot\mathbf R} &= e^{-i\mathbf k\cdot\mathbf R} e^{-i\mathbf G\cdot\mathbf R} \\ &= e^{-i\mathbf k\cdot\mathbf R}. \end{aligned}$$

All lattice translations act with the same phases, so the two labels represent the same character of the discrete translation group. Crystal momentum is defined modulo reciprocal-lattice vectors.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• V. Bargmann, “On unitary ray representations of continuous groups,” Annals of Mathematics 59, 1–46 (1954).
• 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.
• H. Weyl, The Theory of Groups and Quantum Mechanics, translated by H. P. Robertson, Dover, 1950.