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 $𝐀(𝐫)$, where $𝐁=\mathrm{\nabla }\times 𝐀(𝐫)$:

$H=\frac{[𝐩-q𝐀(𝐫){]}^2}{2m}+V(𝐫)$.

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

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

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

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

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

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

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

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

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

$[H,M(𝐚)]=0\text{ and }[H,M(𝐛)]=0$.

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

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

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

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