Magnetic Translations

Magnetic translations move the guiding center of a charged particle in a uniform magnetic field without changing its cyclotron energy. They combine a spatial shift with a phase. We first consider a spinless particle of nonzero charge on the entire plane, with no electric potential or confining boundary; either can break the translation symmetry.

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.

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

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

Kinetic momentum is not conserved: the Lorentz force rotates it during cyclotron motion. 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.

Explicit action in symmetric gauge

Define the conserved momentum-like generator

$$\mathbf K=(-qBY,qBX) =\boldsymbol\pi+q\mathbf B\times\mathbf r, \qquad \mathsf T_B(\mathbf a)=e^{-i\mathbf a\cdot\mathbf K/\hbar}.$$

Its commutators are $[K_i,\pi_j]=0$ and $[K_x,K_y]=-i\hbar qB$. In symmetric gauge, $\mathbf A=\tfrac12\mathbf B\times\mathbf r$, we have $\mathbf K=\mathbf p+q\mathbf A$. The translation acts as

$$(\mathsf T_B(\mathbf a)\psi)(\mathbf r) =\exp\!\left[\frac{iqB}{2\hbar}(a_xy-a_yx)\right] \psi(\mathbf r-\mathbf a).$$

The phase gradient compensates the change in $\mathbf A$ under the shift, so $\boldsymbol\pi\mathsf T_B=\mathsf T_B\boldsymbol\pi$ and consequently $[H,\mathsf T_B]=0$.

In another gauge, $\mathbf A'=\mathbf A+\nabla\chi$, use

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

The coordinate-space phase gains $q[\chi(\mathbf r)-\chi(\mathbf r-\mathbf a)]/\hbar$. Thus the symmetry has a gauge-covariant description.

Parallel transport is a different operation

A straight Wilson-line transport is

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

This also transforms covariantly, but covariance alone does not imply a symmetry. In symmetric gauge its phase is $-qB(a_xy-a_yx)/(2\hbar)$, the opposite of the magnetic-translation phase above. Its infinitesimal generator is $\boldsymbol\pi$, which does not commute with $H$. Use $W$ to compare phases along a path and $\mathsf T_B$ to translate a Landau orbit without changing its energy.

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.

Order and the Closed-Loop Phase

The algebraic group commutator follows directly from the multiplication law:

$$\mathsf T_B(\mathbf a)\mathsf T_B(\mathbf b) \mathsf T_B(-\mathbf a)\mathsf T_B(-\mathbf b) =e^{iqB A(\mathbf a,\mathbf b)/\hbar}I.$$

Operators act from right to left. Chronologically moving by $\mathbf a$, then $\mathbf b$, then $-\mathbf a$, then $-\mathbf b$ instead gives

$$\mathsf T_B(-\mathbf b)\mathsf T_B(-\mathbf a) \mathsf T_B(\mathbf b)\mathsf T_B(\mathbf a) =e^{-iqB A(\mathbf a,\mathbf b)/\hbar}I.$$

Both formulas express the same noncommuting symmetry algebra. A Wilson transport around the geometrical loop has phase $e^{+iq\Phi/\hbar}$ with its chosen path orientation; it is a different operation. Keeping the operator order and the transport definition explicit avoids assigning the wrong sign by an appeal to flux alone.

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

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

Nonzero flux requires gauge-compatible boundary conditions, rather than one globally periodic vector potential and scalar wavefunction. Write the two boundary identifications as $\mathsf T_B(\mathbf L_i)\psi=e^{i\theta_i}\psi$. Applying them in either order must agree, so 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:

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

For nonzero flux, each ideal Landau level on the torus then has exactly $|N|$ orbital states. A finite sample with edges instead has only an approximate bulk count; edge energies can disperse.

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,

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

with $p$ and $Q$ coprime, then an enlarged magnetic unit cell with $Q$ ordinary cells can restore commuting magnetic translations. This assumes the scalar lattice potential also respects the corresponding Bravais translations. This is the symmetry reason behind magnetic Bloch bands and the Hofstadter spectrum. The Peierls phase realizes this algebra in a lattice hopping model.

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.

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. Show why gauge covariance does not by itself guarantee a translation symmetry. Compute $[\pi_x,H]$ and $[\pi_y,H]$, then compare them with the conserved generators $K_x=\pi_x-qBy$ and $K_y=\pi_y+qBx$.

Solution

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

$$[\pi_x,H]=\frac{i\hbar qB}{m}\pi_y, \qquad [\pi_y,H]=-\frac{i\hbar qB}{m}\pi_x.$$

Also $[y,H]=i\hbar\pi_y/m$ and $[x,H]=i\hbar\pi_x/m$. Hence the two terms in $[K_x,H]$ cancel, as do those in $[K_y,H]$.

The straight Wilson-line shift has infinitesimal generator $\boldsymbol\pi$ and therefore is not a symmetry of this Hamiltonian. The magnetic translation generated by $\mathbf K$ is. Both operations have a consistent transformation law under a change of gauge.

References

• E. Brown, “Bloch electrons in a uniform magnetic field,” Physical Review 133, A1038-A1044, 1964.
• D. R. Hofstadter, “Energy levels and wave functions of Bloch electrons in rational and irrational magnetic fields,” Physical Review B 14, 2239–2249, 1976.
• 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.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• J. Zak, “Magnetic translation group,” Physical Review 134, A1602-A1606, 1964.