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: $\mathbf{𝐅}=-\mathrm{\nabla }(-\mathit{\mu }\cdot \mathbf{𝐁})=\mathrm{\nabla }(\mathit{\mu }\cdot \mathbf{𝐁})$.

Using Hamiltonian formalism

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

Magnetic field in quantum mechanics

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

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

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

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

${\hat{H}}_{\mathbf{𝐀}=0}=\sum _{\mathbf{𝐦}}{\epsilon }_{\mathbf{𝐦}}|\mathbf{𝐦}⟩⟨\mathbf{𝐦}|+t\sum _{\mathbf{𝐦}}|\mathbf{𝐦}⟩⟨\mathbf{𝐦}+\mathit{\delta }|$

will be transformed into

${\hat{H}}_{\mathbf{𝐀}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{.}}=e^{ie\mathbf{𝐀}\cdot \widehat{\mathbf{𝐑}}/\hslash }{\hat{H}}_{\mathbf{𝐀}=0}{e}^{-ie\mathbf{𝐀}\cdot \widehat{\mathbf{𝐑}}/\hslash }=\sum _{\mathbf{𝐦}}{\epsilon }_{\mathbf{𝐦}}|\mathbf{𝐦}⟩⟨\mathbf{𝐦}|+t\sum _{\mathbf{𝐦}}|\mathbf{𝐦}⟩⟨\mathbf{𝐦}+\mathit{\delta }|=\sum _{\mathbf{𝐦}}{\epsilon }_{\mathbf{𝐦}}|\mathbf{𝐦}⟩⟨\mathbf{𝐦}|+t\sum _{\mathbf{𝐦}}{e}^{ie\mathbf{𝐀}\cdot \mathbf{𝐦}/\hslash }|\mathbf{𝐦}⟩⟨\mathbf{𝐦}+\mathit{\delta }|e^{-ie\mathbf{𝐀}\cdot (\mathbf{𝐦}+\mathit{\delta }/\hslash }$

${\hat{H}}_{\mathbf{𝐀}=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{.}}=\sum _{\mathbf{𝐦}}{\epsilon }_{\mathbf{𝐦}}|\mathbf{𝐦}⟩⟨\mathbf{𝐦}|+t\sum _{\mathbf{𝐦}}{e}^{-ie\mathbf{𝐀}\cdot \mathit{\delta }/\hslash }|\mathbf{𝐦}⟩⟨\mathbf{𝐦}+\mathit{\delta }|$

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:

$t_{\mathbf{𝐦}\mathbf{𝐧}}\to t_{\mathbf{𝐦}\mathbf{𝐧}}{e}^{-\frac{ie}{\hslash }\underset{{\mathbf{r}}_{\mathbf{𝐧}}}{\overset{{\mathbf{r}}_{\mathbf{𝐦}}}{\int }}d\mathbf{𝐫}\cdot \mathbf{𝐀}(\mathbf{𝐫},t)}$

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).