6
$\begingroup$

I'm trying to understand the reasoning behind Peierls substitution. The final result seems to be simply replacing the hopping elements

$$t_{ij} \to t_{ij} e^{i \frac{q}{\hbar} \int_i^j \vec{A} \cdot d \vec{r}}$$

The line integral in the exponent is taken along the straight line connecting the two sites. Aside from having the shortest euclidean distance, I do not see what is so special about this path? From my understanding, the difference of the line integrals along two different paths is proportional to the magnetic flux enclosed by the two paths.

$$\oint \vec{A} \cdot d\vec{r} = \iint \vec{B} \cdot d\vec{S}$$

So it seems to me the choice of path is actually quite important.

$\endgroup$
4
  • $\begingroup$ Does this answer your question: Peierls substitution vs minimal coupling ? @YuanFang $\endgroup$
    – Roger V.
    Commented Oct 11 at 10:15
  • $\begingroup$ I want to know how to choose the integration path of the vector potential. $\endgroup$
    – Yuan Fang
    Commented Oct 11 at 10:25
  • $\begingroup$ It's just a choice adopted to make the hamiltonian gauge invariant. The tight-binding model with hopping terms is just an approximation after all, so making the straight-line path approximation is just another one. $\endgroup$
    – mike stone
    Commented Oct 11 at 11:31
  • $\begingroup$ @YuanFang I expanded my answer to say more about the path. $\endgroup$
    – Roger V.
    Commented Oct 11 at 11:50

1 Answer 1

0
$\begingroup$

The question seems as nearly a duplicate of Peierls substitution vs minimal coupling.

Indeed, the point of Peierls substitution is introducing magnetic field into a tight-binding Hamiltonian or even a band-Hamiltonian (with non-quadratic dependence on momentum or in an effective mass approximation.)

In either case, the problem contains two elements:

  • We know how to include magnetic field (via vector potential) into a Hamiltonian of a free particle, but both tight-binding approximation and effective mass approximation are approximations, and the electrons in solid state are not free electrons, but quasiparticles, obtained, e.g., by trancating other bands (and neglecting a bunch of interactions.) There is no guarantee that we can treat them as if they were free, even if we expanded the dispersion relation to the second order in the wave vector and assumed spherical symmetry.
  • Another problem is that adding vector potential to the real electronic Hamiltonian destroys its translational symmetry and Bloch theorem, on which all the solid state physics (including the tight-binding approximation) is based on.

Thus, Peierls substitution is a non-trivial statement that, even for tight-binding and effective mass approximations we can treat the vector potential, as if we were dealing with free particles, forgetting about the lattice background... provided that the magnetic field is weak (negligible flux through a unit cell.)

Gauge invariance
The integration path for vector potential is subject to gauge invariance, just as for free particles. What really matters is the sum of phases accumulated along any closed path, which is the flux through the area enclosed by the path (in lattice problems one typically assures that there is the same flux through every unit cell.)

As an example, consider the tight-binding/hopping Hamiltonian description of the Aharonov-Bohm interferometer below. It could be described by a Hamiltonian: $$ H=\sum_{k,\nu=L,R}E_{k\nu}c_{k\nu}^\dagger c_{k\nu} + \sum_{i=1,2}E_i d_i^\dagger d_i + \sum_{k,\nu=L,R, i=1, 2}V_{\nu i}\left[c_{k\nu}^\dagger d_i + h.c.\right] $$ where the hopping matrix elements have phases: $$ V_{\nu i}=|V_{\nu i}|e^{i\phi_{\nu i}}. $$ Calculating the properties of this Hamiltonian - e.g., the transmission from left to right - would involve matrix element products like $$ V_{L1}^*V_{R1}V_{R2}^*V_{L2} = |V_{L1}^*V_{R1}V_{R2}^*V_{L2}|e^{i(-\phi_{L1}+\phi_{R1}-\phi_{R2}+\phi_{L2})}=|V_{L1}^*V_{R1}V_{R2}^*V_{L2}|e^{i\Phi}, $$ which describe the path around the interferometer. Only the phase $\Phi$ will appear in the results for the physically measurable properties, which means that phases of the matrix elements can be varied at will, as long as they sum up to the correct quantity: $$ \Phi = -\phi_{L1}+\phi_{R1}-\phi_{R2}+\phi_{L2}. $$ E.g., a possible choice is setting $\phi_{L1}=-\Phi, \phi_{R1}=\phi_{R2}=\phi_{L2}=0.$

enter image description here
(image source)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.