Physics:Magnetic translation

In quantum mechanics, the action of symmetries on physical states are represented by either linear or, more generally, projective representations. For a particle moving in a crystal without a magnetic field, spatial translations are represented linearly and the corresponding translation operators commute with one another and with the Hamiltonian. However, in the presence of a magnetic field, even when the magnetic field configuration is translationally invariant, wave functions fail to transform linearly under translation. Instead, they are represented projectively, acquiring position-dependent phase factors. The resulting operators are known as magnetic translation operator .

Magnetic Symmetry Operator

In this section, we will start with the discussion of the more well-known magnetic translation operator and generalize it to other spatial symmetries such as rotations.

Magnetic Translation Operator

To be more specific, consider the Hamiltonian of a quantum particle (with charge $q$ and mass $m$) in a magnetic field explicitly depends on the magnetic vector potential $\mathbf{𝐀}(\mathbf{𝐫})$, where $\mathbf{𝐁}=\mathrm{\nabla }\times \mathbf{𝐀}(\mathbf{𝐫})$:

$H=\frac{[\mathbf{𝐩}-q\mathbf{𝐀}(\mathbf{𝐫}){]}^2}{2m}+V(\mathbf{𝐫})$.

For a uniform magnetic field $\mathbf{𝐁}(\mathbf{𝐫})=B\hat{z}$, the ordinary translation operator $T(\mathbf{𝐚})=e^{-i\mathbf{𝐩}\cdot \mathbf{𝐚}/\hslash }$ does not commute with the Hamiltonian even when $V(\mathbf{𝐫}\mathbf{+}\mathbf{𝐚})=V(\mathbf{𝐫})$ because

$T(\mathbf{𝐚}{)}^{-1}[\mathbf{𝐩}-q\mathbf{𝐀}(\mathbf{𝐫})]T(\mathbf{𝐚})=\mathbf{𝐩}-q\mathbf{𝐀}(\mathbf{𝐫}\mathbf{+}\mathbf{𝐚})=\mathbf{𝐩}-q\mathbf{𝐀}(\mathbf{𝐫})-q[\mathbf{𝐀}(\mathbf{𝐫}\mathbf{+}\mathbf{𝐚})-\mathbf{𝐀}(\mathbf{𝐫})]=\mathbf{𝐩}-q\mathbf{𝐀}(\mathbf{𝐫})-q\mathrm{\nabla }{\chi }_{\mathbf{𝐚}}(\mathbf{𝐫}).$

In the third equality, one writes $\mathbf{𝐀}(\mathbf{𝐫}\mathbf{+}\mathbf{𝐚})-\mathbf{𝐀}(\mathbf{𝐫})=\mathrm{\nabla }{\chi }_{\mathbf{𝐚}}(\mathbf{𝐫})$ using the fact that

$\mathrm{\nabla }\times [\mathbf{𝐀}(\mathbf{𝐫}\mathbf{+}\mathbf{𝐚})-\mathbf{𝐀}(\mathbf{𝐫})]=\mathbf{𝐁}(\mathbf{𝐫}\mathbf{+}\mathbf{𝐚})-\mathbf{𝐁}(\mathbf{𝐫})=0.$

For a uniform magnetic field, this has a solution (which can be identified through vector calculus identities):

${\chi }_{\mathbf{𝐚}}(\mathbf{𝐫})=\underset{\mathbf{𝐫}}{\overset{\mathbf{𝐫}\mathbf{+}\mathbf{𝐚}}{\int }}\mathbf{𝐀}\cdot d\mathit{\ell }+\mathbf{𝐁}\times \mathbf{𝐚}\cdot \mathbf{𝐫}+\text{Const.}$

Here, the line integral should be taken along a straight line. The failure of $T(\mathbf{𝐚})$ to commute with the Hamiltonian is due to the $-q\mathrm{\nabla }{\chi }_{\mathbf{𝐚}}(\mathbf{𝐫})$ term. One could remedy this by multiplying $T(\mathbf{𝐚})$ by an appropriate phase factor:

$M(\mathbf{𝐚})=e^{-\frac{i}{\hslash }\mathbf{𝐩}\cdot \mathbf{𝐚}}{e}^{\frac{iq{\chi }_{\mathbf{𝐚}}(\mathbf{𝐫})}{\hslash }}=e^{\frac{iq{\chi }_{\mathbf{𝐚}}(\mathbf{𝐫}\mathbf{-}\mathbf{𝐚})}{\hslash }}{e}^{-\frac{i}{\hslash }\mathbf{𝐩}\cdot \mathbf{𝐚}}.$

$M(\mathbf{𝐚})$ then commutes with the Hamiltonian for any $\mathbf{𝐚}$. In particular, the translation along $\mathbf{𝐚}$ and $\mathbf{𝐛}$ satisfy

$[H,M(\mathbf{𝐚})]=0\text{ and }[H,M(\mathbf{𝐛})]=0$.

But they fail to commute with each other in general; instead, they satisfy

$M(\mathbf{𝐚})M(\mathbf{𝐛})=e^{\frac{iq}{\hslash }\mathbf{𝐁}\cdot \mathbf{𝐚}\times \mathbf{𝐛}}M(\mathbf{𝐛})M(\mathbf{𝐚})=e^{\frac{iq}{2\hslash }\mathbf{𝐁}\cdot \mathbf{𝐚}\times \mathbf{𝐛}}M(\mathbf{𝐚}+\mathbf{𝐛}).$

Thus the failure of the commutativity of two translations is captured by the flux $\mathrm{\Phi }=Bab$ through the rectangle enclosed by the two translations. In particular, the two magnetic translations commute whenever this flux is an integer multiple of the flux quantum ${\mathrm{\Phi }}_0=\frac{2\pi \hslash }{q},$

i.e., $\mathrm{\Phi }=n{\mathrm{\Phi }}_0$ $[M(\mathbf{𝐚}),M(\mathbf{𝐛})]=0$.

The magnetic translation operators $M(\mathbf{𝐚})$ and $M(\mathbf{𝐛})$ now form a set of commuting symmetry operators.