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For a charged particle in a uniform magnetic field, the symmetric gauge $\mathbf{A}=\mathbf{B}\times \mathbf{x}/2$ makes the displayed magnetic pseudomomentum a conserved quantity when no other position-dependent potential breaks translation symmetry. It differs from the kinetic momentum $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}=m\dot{\mathbf{x}}$. With $\mathbf{B}=B{\mathbf{e}}_3$, Hamilton's equations for $H=|\mathbf{p}-e\mathbf{A}{|}^2/(2m)$ give ${\dot{p}}_1=(eB/2){\dot{x}}_2$ and ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, proving conservation of $K_1=p_1-eB{x}_2/2$ and $K_2=p_2+eB{x}_1/2$. The charge factor is necessary unless it has already been absorbed into the field convention. In quantum mechanics the transverse components need not commute, even though each commutes with this Hamiltonian operator; magnetic translation algebra records that distinction.

Part (a) givesFortheir commutator is the scalarIt therefore commutes with both $M$ and $N$. Applying the stated Baker--Campbell--Hausdorff identity in both orders yields the magnetic translation algebra

Writing $\Phi =B\cdot (r\times r^{\prime })$, the translations commute exactly when their phase is one:Thus the cell flux must be an integral multiple of the magnetic flux quantum: