Magnetic translation

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.

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

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

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

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

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

${\displaystyle \nabla \times \left[{\bf {A}}({\bf {r+a}})-{\bf {A}}({\bf {r}})\right]={\bf {B}}({\bf {r+a}})-{\bf {B}}({\bf {r}})=0.}$

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

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

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

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

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

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

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

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

Thus the failure of the commutativity of two translations is captured by the flux ${\displaystyle \Phi =Bab}$ 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 ${\displaystyle \Phi _{0}={\frac {2\pi \hbar }{q}},}$

i.e., ${\displaystyle \Phi =n\Phi _{0}}$ ${\displaystyle [M({\bf {a}}),M({\bf {b}})]=0}$.

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