How to put magnetic field into tight-binding Hamiltonian

Magnetic field in classical physics

Using Newton second law

• Lorentz force on charge particle: $\mathbf {F} =q\mathbf {v} \times \mathbf {B}$.
• Force on magnetic moment: $\mathbf {F} =-\nabla (-{\boldsymbol {\mu }}\cdot \mathbf {B} )=\nabla ({\boldsymbol {\mu }}\cdot \mathbf {B} )$.

Using Hamiltonian formalism

• Free particle: $H=\mathbf {p} ^{2}/2m$.
• Particle in magnetic field (replace kinetic with canonical momentum): $H=(\mathbf {p} -q\mathbf {A} )^{2}/2m-{\boldsymbol {\mu }}\cdot \mathbf {B}$.

Magnetic field in quantum mechanics

• Convert c-numbers into operators in the classical Hamiltonian: ${\hat {H}}=({\hat {\mathbf {p} }}-q{\hat {\mathbf {A} }})^{2}/2m-g\mu _{B}{\hat {\boldsymbol {\sigma }}}\cdot \mathbf {B}$, where ${\hat {\boldsymbol {\sigma }}}=({\hat {\sigma }}_{x},{\hat {\sigma }}_{y},{\hat {\sigma }}_{z})$ is the vector of Pauli matrices, $g$ is the gyromagentic ratio and $\mu _{B}$ is the Bohr magenton.

• Assuming $\mathbf {A} =\mathrm {const.}$, we see that ${\hat {\mathbf {p} }}-e\mathbf {A} =e^{ie\mathbf {A} \cdot {\hat {\mathbf {R} }}/\hbar }{\hat {\mathbf {p} }}e^{-ie\mathbf {A} \cdot {\hat {\mathbf {R} }}/\hbar }$, which suggests that vector potential can equivalently be introduced into the Hamiltonian using ${\hat {H}}\mapsto e^{ie\mathbf {A} \cdot {\hat {\mathbf {R} }}/\hbar }{\hat {H}}e^{-ie\mathbf {A} \cdot {\hat {\mathbf {R} }}/\hbar }$. This is derivation can be used only as a hint of heuristic value since $\mathbf {A} =\mathrm {const.}$ means that $\mathbf {B} =\nabla \times \mathbf {A} \equiv 0$.

Heuristic version of the Pierls substitution to introduce magnetic field into tight-binding Hamiltonian

• Continuing with the trivial case $\mathbf {A} =\mathrm {const.}$, we see that tight-binding Hamiltonian:

${\hat {H}}_{\mathbf {A} =0}=\sum _{\mathbf {m} }\varepsilon _{\mathbf {m} }|\mathbf {m} \rangle \langle \mathbf {m} |+t\sum _{\mathbf {m} }|\mathbf {m} \rangle \langle \mathbf {m} +{\boldsymbol {\delta }}|$

will be transformed into

${\hat {H}}_{\mathbf {A} =\mathrm {const.} }=e^{ie\mathbf {A} \cdot {\hat {\mathbf {R} }}/\hbar }{\hat {H}}_{\mathbf {A} =0}e^{-ie\mathbf {A} \cdot {\hat {\mathbf {R} }}/\hbar }=\sum _{\mathbf {m} }\varepsilon _{\mathbf {m} }|\mathbf {m} \rangle \langle \mathbf {m} |+t\sum _{\mathbf {m} }|\mathbf {m} \rangle \langle \mathbf {m} +{\boldsymbol {\delta }}|=\sum _{\mathbf {m} }\varepsilon _{\mathbf {m} }|\mathbf {m} \rangle \langle \mathbf {m} |+t\sum _{\mathbf {m} }e^{ie\mathbf {A} \cdot \mathbf {m} /\hbar }|\mathbf {m} \rangle \langle \mathbf {m} +{\boldsymbol {\delta }}|e^{-ie\mathbf {A} \cdot (\mathbf {m} +{\boldsymbol {\delta }}/\hbar }$

${\hat {H}}_{\mathbf {A} =\mathrm {const.} }=\sum _{\mathbf {m} }\varepsilon _{\mathbf {m} }|\mathbf {m} \rangle \langle \mathbf {m} |+t\sum _{\mathbf {m} }e^{-ie\mathbf {A} \cdot {\boldsymbol {\delta }}/\hbar }|\mathbf {m} \rangle \langle \mathbf {m} +{\boldsymbol {\delta }}|$

by the introduction of constant vector potential.

Exact version of the Pierls substitution to introduce magnetic field into tight-binding Hamiltonian

The correct result for the tight-binding hopping modified by the so-called Peierls substation looks similar to heuristic expressions above, except that one has to integrate $d\mathbf {r} \cdot \mathbf {A} (\mathbf {r} ,t)$ along the line joining the two sites:

$t_{\mathbf {mn} }\rightarrow t_{\mathbf {mn} }e^{-{\frac {ie}{\hbar }}\int \limits _{\mathbf {r_{n}} }^{\mathbf {r_{m}} }d\mathbf {r} \cdot \mathbf {A} (\mathbf {r} ,t)}$

since vector potential is never homogeneous for non-zero magnetic field.

Example of magnetic field in $3\times 3$ tight-binding Hamiltonian

• Assuming a homogeneous magnetic field orthogonal to the plane of 2D lattice, $\mathbf {B} =B\mathbf {e} _{z}$, and gauge in which $\mathbf {A} (\mathbf {r} )={\frac {1}{2}}\mathbf {B} \times \mathbf {r} =(0,Bx,0)$, the $9\times 9$ matrix of the tight-binding Hamiltonian is shown below explicitly:

Matrix elements of the tight-binding Hamiltonian defined on $3\times 3$ lattice of sites (with single orbital per site) which is placed in orthogonal magnetic field $\mathbf {B} =B\mathbf {e} _{z}$.

References

• J. G. Analytis, S. J. Blundell, and A. Ardavan, Landau levels, molecular orbitals, and the Hofstadter butterfly in finite systems, Am. J. Phys. 72, 5 (2004).