Magnetic Translations

Magnetic translations are the correct translation operators for a charged particle moving in a magnetic field. They combine an ordinary spatial shift with a gauge-dependent phase so that the result is gauge covariant and commutes with the Hamiltonian when the magnetic field is uniform.

The surprise is that magnetic translations generally do not commute. Moving by $\mathbf a$ and then by $\mathbf b$ differs from moving by $\mathbf b$ and then by $\mathbf a$ by a phase proportional to the magnetic flux through the parallelogram spanned by $\mathbf a$ and $\mathbf b$.

That noncommuting translation algebra is a compact symmetry explanation of Landau-level degeneracy, magnetic flux quantization on a torus, and the magnetic-unit-cell structure behind lattice quantum Hall models.

Why Ordinary Translations Need Help

For a free particle, ordinary translations are generated by canonical momentum:

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

They act on a wavefunction as

$$(T(\mathbf a)\psi)(\mathbf r) = \psi(\mathbf r-\mathbf a).$$

In a vector potential, the Hamiltonian is

$$H = \frac{1}{2m} \left( \mathbf p-q\mathbf A(\mathbf r) \right)^2.$$

If $\mathbf A$ changes under a spatial shift, the ordinary translation operator need not commute with this particular gauge-fixed Hamiltonian even when the magnetic field itself is uniform. The physical field may be translation invariant while the chosen vector potential is not.

Magnetic translations repair this mismatch. They translate the wavefunction and include the phase needed to compare the vector potential at neighboring points.

Gauge-Covariant Translation

For a displacement $\mathbf a$, a gauge-covariant translation along the straight segment from $\mathbf r-\mathbf a$ to $\mathbf r$ has the form

$$(\mathsf T_B(\mathbf a)\psi)(\mathbf r) = \exp\left[ \frac{iq}{\hbar} \int_{\mathbf r-\mathbf a}^{\mathbf r} \mathbf A(\boldsymbol\ell)\cdot d\boldsymbol\ell \right] \psi(\mathbf r-\mathbf a).$$

Under a gauge transformation

$$\mathbf A' = \mathbf A+\nabla\chi, \qquad \psi' = \exp\left( \frac{iq\chi}{\hbar} \right)\psi,$$

the transformed magnetic translation satisfies

$$\mathsf T'_B(\mathbf a)\psi' = \exp\left( \frac{iq\chi}{\hbar} \right) \mathsf T_B(\mathbf a)\psi.$$

Thus magnetic translations are not gauge-invariant operators in isolation; they are gauge-covariant operations. They map gauge-related descriptions to gauge-related descriptions.

Uniform Magnetic Field

Now take a two-dimensional particle in a uniform field

$$\mathbf B = B\hat{\mathbf z}, \qquad B>0,$$

with Hamiltonian

$$H = \frac{\boldsymbol\pi^2}{2m}, \qquad \boldsymbol\pi = \mathbf p-q\mathbf A.$$

The kinetic momenta obey

$$[\pi_x,\pi_y] = i\hbar qB.$$

The ordinary kinetic momentum components therefore cannot both be used as commuting translation generators. Instead, use the guiding-center coordinates

$$X = x+\frac{\pi_y}{qB}, \qquad Y = y-\frac{\pi_x}{qB}.$$

They commute with the Hamiltonian:

$$[X,H]=[Y,H]=0,$$

but not with each other:

$$[X,Y] = -\frac{i\hbar}{qB}.$$

The noncommutativity of magnetic translations is the finite version of this guiding-center commutator.

Generator Form

With the conventions above, define the magnetic translation by

$$\mathsf T_B(\mathbf a) = \exp\left[ \frac{i}{\hbar} qB(a_xY-a_yX) \right].$$

It shifts the guiding center:

$$\mathsf T_B(\mathbf a)^\dagger X \mathsf T_B(\mathbf a) = X+a_x,$$

and

$$\mathsf T_B(\mathbf a)^\dagger Y \mathsf T_B(\mathbf a) = Y+a_y.$$

Because $X$ and $Y$ commute with $H$, the magnetic translation also commutes with the uniform-field Hamiltonian:

$$[\mathsf T_B(\mathbf a),H]=0.$$

This is the symmetry statement behind the fact that shifting a cyclotron orbit’s guiding center costs no energy in the ideal Landau problem.

Different sign conventions for active translations, charge $q$, or the orientation of $B$ change some displayed phases. The invariant content is always the same: the phase around a closed magnetic-translation loop is $q$ times the enclosed magnetic flux divided by $\hbar$.

Noncommuting Group Law

Let

$$A(\mathbf a,\mathbf b) = a_xb_y-a_yb_x$$

be the signed area of the parallelogram spanned by $\mathbf a$ and $\mathbf b$. Since $[X,Y]$ is central, the Baker-Campbell-Hausdorff formula gives

$$\mathsf T_B(\mathbf a) \mathsf T_B(\mathbf b) = \exp\left[ \frac{iqB}{2\hbar} A(\mathbf a,\mathbf b) \right] \mathsf T_B(\mathbf a+\mathbf b).$$

Swapping the two translations gives the opposite half-phase, so

$$\mathsf T_B(\mathbf a) \mathsf T_B(\mathbf b) = \exp\left[ \frac{iqB}{\hbar} A(\mathbf a,\mathbf b) \right] \mathsf T_B(\mathbf b) \mathsf T_B(\mathbf a).$$

Thus magnetic translations commute only when

$$\exp\left[ \frac{iqB}{\hbar} A(\mathbf a,\mathbf b) \right] = 1.$$

The exponent is the Aharonov–Bohm phase for the magnetic flux through the parallelogram:

$$\frac{q}{\hbar} \Phi_{\mathbf a,\mathbf b}, \qquad \Phi_{\mathbf a,\mathbf b} = B\,A(\mathbf a,\mathbf b).$$

The algebra is a projective representation of the ordinary translation group. The projective phase is not an arbitrary convention; it is magnetic flux.

Closed-Loop Interpretation

Perform four magnetic translations around a small parallelogram. Up to the orientation convention,

$$\mathsf T_B(-\mathbf b) \mathsf T_B(-\mathbf a) \mathsf T_B(\mathbf b) \mathsf T_B(\mathbf a) = \exp\left[ \frac{iqB}{\hbar} A(\mathbf a,\mathbf b) \right]I.$$

This says that returning to the same point need not return the same phase. The leftover phase is

$$\exp\left( \frac{iq}{\hbar} \oint \mathbf A\cdot d\mathbf r \right) = \exp\left( \frac{iq\Phi}{\hbar} \right),$$

the same holonomy that appears in the Aharonov–Bohm effect. Magnetic translations are therefore a local algebraic face of the same gauge geometry.

Landau-Level Degeneracy

The Landau Hamiltonian depends only on the kinetic momenta:

$$H = \frac{1}{2m} \left( \pi_x^2+\pi_y^2 \right).$$

The kinetic momenta create the cyclotron oscillator and determine the Landau-level energy. The guiding-center coordinates commute with that energy and label the degeneracy inside each Landau level.

Because

$$[X,Y] = -i\,\operatorname{sgn}(qB)\ell_B^2, \qquad \ell_B = \sqrt{ \frac{\hbar}{\lvert q\rvert B} },$$

the guiding-center plane is itself a noncommutative phase plane. A region of area $A$ supports roughly one independent guiding-center state per area $2\pi\ell_B^2$:

$$N_\Phi \approx \frac{A}{2\pi\ell_B^2} = \frac{\lvert q\rvert BA}{h}.$$

This is the degeneracy count derived in wave-mechanics language in Degeneracy of Landau Levels. Magnetic translations explain why the degeneracy is a symmetry structure, not an accidental feature of Landau gauge.

Torus and Flux Quantization

On a torus, translations around two fundamental cycles must be globally consistent. If the torus has side vectors $\mathbf L_1$ and $\mathbf L_2$, the two large magnetic translations obey

$$\mathsf T_B(\mathbf L_1) \mathsf T_B(\mathbf L_2) = \exp\left[ \frac{iq\Phi_{\mathrm{tot}}}{\hbar} \right] \mathsf T_B(\mathbf L_2) \mathsf T_B(\mathbf L_1),$$

where

$$\Phi_{\mathrm{tot}} = B\,A(\mathbf L_1,\mathbf L_2).$$

For ordinary periodic boundary conditions to be consistent, the phase must be unity:

$$\frac{q\Phi_{\mathrm{tot}}}{\hbar} = 2\pi N, \qquad N\in\mathbb Z.$$

Equivalently,

$$\Phi_{\mathrm{tot}} = N\frac{h}{q}$$

with orientation included in the sign of $q\Phi_{\mathrm{tot}}$. In magnitude, the condition is an integer number of single-particle flux quanta:

$$\frac{\lvert q\Phi_{\mathrm{tot}}\rvert}{h} \in \mathbb Z.$$

When this holds, each ideal Landau level on the torus has exactly that many orbital states.

Lattice Magnetic Translations

On a lattice with primitive vectors $\mathbf a_1,\mathbf a_2$, magnetic translations along the two lattice directions commute only if the flux through the primitive cell is an integer multiple of the flux quantum:

$$\exp\left[ \frac{iq\Phi_{\mathrm{cell}}}{\hbar} \right] = 1.$$

If the flux per cell is rational in flux-quantum units,

$$\frac{\Phi_{\mathrm{cell}}}{\Phi_0} = \frac{p}{Q}, \qquad \Phi_0=\frac{h}{\lvert q\rvert},$$

then an enlarged magnetic unit cell with $Q$ ordinary cells can restore commuting magnetic translations. This is the symmetry reason behind magnetic Bloch bands and the Hofstadter spectrum. The detailed lattice theory belongs to quantum matter; the essential point here is that magnetic flux changes the translation group itself.

Common Mistakes

• Saying a uniform magnetic field destroys translation symmetry. It destroys ordinary gauge-fixed translation symmetry, but magnetic translations remain.
• Using canonical momentum translations in a magnetic field and expecting them to commute with the Hamiltonian in every gauge.
• Forgetting that magnetic translations commute only up to a flux phase.
• Treating $k_y$ in Landau gauge as a universal physical momentum rather than one gauge-dependent way to label guiding centers.
• Ignoring flux quantization when imposing periodic boundary conditions in both directions.
• Losing the charge sign in the algebra. Many formulas change phase orientation when $q$ or $B$ is reversed.

Cross-Links

• Translations and Momentum
• Minimal Coupling in Wave Mechanics
• Gauge Transformations: First Encounter
• Particle in a Uniform Magnetic Field
• Landau Levels
• Landau Levels Revisited
• Peierls Phase Preview
• Degeneracy of Landau Levels
• Landau Gauge and Symmetric Gauge
• Aharonov–Bohm Effect
• Dirac Monopole Preview
• Heisenberg Group
• Landau-Level Formula Card
• Landau-Level System Reference
• From Phase Symmetry to Gauge Theory
• From Projective Representations to Anomalies Preview

References

• J. Zak, “Magnetic translation group,” Physical Review 134, A1602-A1606, 1964.
• E. Brown, “Bloch electrons in a uniform magnetic field,” Physical Review 133, A1038-A1044, 1964.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• R. E. Prange and S. M. Girvin, eds., The Quantum Hall Effect, 2nd ed., Springer, 1990.
• D. R. Hofstadter, “Energy levels and wave functions of Bloch electrons in rational and irrational magnetic fields,” Physical Review B 14, 2239-2249, 1976.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Derive the magnetic-translation commutator.

Using

$$[X,Y]=-\frac{i\hbar}{qB},$$

and

$$\mathsf T_B(\mathbf a) = \exp\left[ \frac{i}{\hbar} qB(a_xY-a_yX) \right],$$

show that

$$\mathsf T_B(\mathbf a) \mathsf T_B(\mathbf b) = \exp\left[ \frac{iqB}{\hbar} A(\mathbf a,\mathbf b) \right] \mathsf T_B(\mathbf b) \mathsf T_B(\mathbf a).$$

Solution

Let

$$G(\mathbf a) = qB(a_xY-a_yX).$$

Then

$$[G(\mathbf a),G(\mathbf b)] = -i\hbar qB \left( a_xb_y-a_yb_x \right).$$

Because this commutator is a scalar, Baker-Campbell-Hausdorff gives

$$\mathsf T_B(\mathbf a)\mathsf T_B(\mathbf b) = \exp\left[ \frac{iqB}{2\hbar}A(\mathbf a,\mathbf b) \right] \mathsf T_B(\mathbf a+\mathbf b).$$

Swapping $\mathbf a$ and $\mathbf b$ flips the sign of $A$. Taking the ratio gives

$$\mathsf T_B(\mathbf a) \mathsf T_B(\mathbf b) = \exp\left[ \frac{iqB}{\hbar} A(\mathbf a,\mathbf b) \right] \mathsf T_B(\mathbf b) \mathsf T_B(\mathbf a).$$

2. Find when two magnetic lattice translations commute.

Let the flux through a lattice unit cell be $\Phi_{\mathrm{cell}}$. What condition makes magnetic translations along the two primitive lattice vectors commute?

Solution

They commute when the flux phase is unity:

$$\exp\left( \frac{iq\Phi_{\mathrm{cell}}}{\hbar} \right) = 1.$$

Therefore

$$\frac{q\Phi_{\mathrm{cell}}}{\hbar} = 2\pi n, \qquad n\in\mathbb Z.$$

Equivalently,

$$\frac{\lvert q\Phi_{\mathrm{cell}}\rvert}{h} \in \mathbb Z.$$

3. Connect the algebra to Landau degeneracy.

Use the guiding-center commutator to explain why an area $A$ supports roughly $A/(2\pi\ell_B^2)$ states in one Landau level.

Solution

The guiding-center coordinates obey

$$[X,Y] = -i\,\operatorname{sgn}(qB)\ell_B^2.$$

This is analogous to a canonical phase plane with effective Planck area $2\pi\ell_B^2$. Therefore a region of ordinary area $A$ contains roughly

$$N_\Phi = \frac{A}{2\pi\ell_B^2}$$

independent guiding-center states. Since the Landau Hamiltonian depends on the cyclotron variables rather than on $X,Y$, these states have the same Landau-level energy in the ideal problem.

4. Check gauge covariance of the Wilson-line translation.

Show that the straight-line expression for $\mathsf T_B(\mathbf a)$ transforms covariantly under $\mathbf A\mapsto\mathbf A+\nabla\chi$ and $\psi\mapsto e^{iq\chi/\hbar}\psi$.

Solution

The line integral changes by an endpoint term:

$$\int_{\mathbf r-\mathbf a}^{\mathbf r} \left( \mathbf A+\nabla\chi \right)\cdot d\boldsymbol\ell = \int_{\mathbf r-\mathbf a}^{\mathbf r} \mathbf A\cdot d\boldsymbol\ell + \chi(\mathbf r)-\chi(\mathbf r-\mathbf a).$$

Acting on the transformed wavefunction gives an additional factor

$$\exp\left[ \frac{iq\chi(\mathbf r-\mathbf a)}{\hbar} \right]$$

from $\psi'(\mathbf r-\mathbf a)$. Multiplying the two endpoint factors leaves

$$\exp\left[ \frac{iq\chi(\mathbf r)}{\hbar} \right],$$

which is exactly the transformed phase at the final point. Hence

$$\mathsf T'_B(\mathbf a)\psi' = e^{iq\chi/\hbar} \mathsf T_B(\mathbf a)\psi.$$