Curl of Berry connection

Physics Asked by rt54 on May 8, 2021

If $|nrangle=|n( textbf{R}(t) ) rangle $ satisfies the equation $$H(textbf{R}(t))|n(textbf{R}(t)) rangle = E_{n}(textbf{R}(t))|n(textbf{R}(t))rangle$$ then the berry phase $gamma_{n}(t)$ for a closed path in the parameter space is given by: $$gamma_{n}(t) = ioint langle n|nabla_{textbf{R}}|nrangle dtextbf{R} = iiint(nabla_{textbf{R}} times langle n|nabla_{textbf{R}}|nrangle) dvec{S} = i iintsumlimits_{nneq m}frac{langle n|(nabla_{textbf{R}} H)|mrangle times langle m|(nabla_{textbf{R}}H)|nrangle}{(E_{m}-E_{n})²}dvec{S}$$

Im having trouble understanding the last equality. How does one get $$nabla_{textbf{R}} times langle n|nabla_{textbf{R}}|nrangle = sumlimits_{nneq m}frac{langle n|(nabla_{textbf{R}} H)|mrangle times langle m|(nabla_{textbf{R}}H)|nrangle}{(E_{m}-E_{n})²}~?$$

One Answer

This identity is extremely useful - it's much more convenient to differentiate the Hamiltonian as a function of its parameters than it is to differentiate the corresponding eigenstates - but it is not trivial, so you shouldn't feel bad for not seeing it immediately.

We start by assuming that the Hamiltonian $H$ and its eigenstates $|nrangle$ and eigenvalues $E_n$ depend on some set of parameters $mathbf R$. If this is the case, then

$$nabla_mathbf R(H|nrangle) = (nabla_mathbf R H)|nrangle + Hnabla_mathbf R|nrangle$$ $$ = nabla_mathbf R(E_n|nrangle) = (nabla_mathbf RE_n)|nrangle + E_nnabla_mathbf R |nrangle$$ and so

$$ (H-E_n) nabla_mathbf R|nrangle= big[nabla_mathbf R(H-E_n)big]|nrangle$$

Note that the operator $(H-E_n)$ annihilates any component of $nabla_mathbf R|nrangle$ which is proportional to $|nrangle$. To find the orthogonal component, we can apply the projection operator $P=sum_{mneq n} |mranglelangle m|$ to both sides, and then safely invert $(H-E_n)$ to obtain

$$nabla_mathbf R| nrangle =boldsymbolalpha(mathbf R)|nrangle + (H-E_n)^{-1}sum_{mneq n} |mranglelangle m| big[nabla_mathbf R(H - E_n)]|nrangle$$ $$ =boldsymbol alpha(mathbf R)|nrangle + sum_{mneq n} |mrangle frac{langle m|big[nabla_mathbf R(H-E_n)big] |nrangle}{E_m-E_n}$$

$$ = boldsymbolalpha(mathbf R) |nrangle + sum_{mneq n} |mrangle frac{langle m|(nabla_mathbf R H)|nrangle}{E_m-E_n}$$

for some (vector-valued!) $boldsymbolalpha(mathbf R)$. Finally, we note that

$$nabla_mathbf R times langle n|nabla_mathbf R |nrangle = big(nabla_mathbf R langle n|big) times big(nabla_mathbf R |nranglebig) + langle n| nabla_mathbf R times big(nabla_mathbf R |nranglebig)$$

$$=sum_{mneq n} frac{langle n|(nabla_mathbf R H)|mrangle times langle m | (nabla_mathbf R H)|nrangle}{(E_m-E_n)^2} + 0$$

I've skipped quite a large amount of algebra here, but that is an outline of the calculation.

Note also that we made several assumptions - most importantly that $|nrangle$ is non-degenerate, so $E_n neq E_m$ for all $nneq m$. If there is degeneracy, then we are going to have a problem, and indeed this is what happens when e.g. the band gap of some topological material closes.

As an example, the Chern number of a system is equal to the integral of the Berry curvature (summed over all occupied bands) and is proportional to the system's Hall conductivity. If we vary the parameters $mathbf R$ slowly, then the Berry curvature (and therefore the Chern number) should change continuously. However, since the Chern number can be shown to be an integer, the only way to change it continuously is to not change it at all, and so the Hall conductivity is topologically protected.

This goes out the window if, as we change the parameters $mathbf R$, we close the band gap, thereby introducing energy degeneracy. When the gap re-opens, it may have a different (but still topologically protected) Chern number and Hall conductivity; this is an example of a topological phase transition.

Correct answer by J. Murray on May 8, 2021

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