# Strange manipulation of Hamiltonian operator and gradient

I'm reading M. V. Berry's Quantal Phase Factors Accompanying Adiabatic Changes and came across an unfamiliar identity in eq. (8), namely $$\langle m | \nabla _Rn \rangle = \frac{\langle m | \nabla_R \hat{H} | n \rangle}{E_n - E_m}$$ provided $$m\neq n$$, where the gradient is taken over the space of slowly time-varying parameters of the Hamiltonian (as is assumed in the adiabatic approximation).

I'm just unsure of where the Hamiltonian comes from in this identity. I would greatly appreciate any reference explaining or proving this identity.

• It is first-order perturbation theory . Take the variation of $H_R|n\rangle = E_n|n\rangle$ and multiply on the left by $\langle m|$. Jul 4 at 0:03
• That makes perfect sense. I thought it was a strict equality for any energy scale, not a perturbative one, but that makes perfect sense, thank you! Jul 4 at 1:20

If $$H|n\rangle = E_n|n\rangle$$ for each $$R$$ (with the dependence on $$R$$ suppressed for readability), then
$$\nabla_R \langle m|H|n\rangle = \nabla_RE_n\delta_{nm} = E_n(R)\bigg[\nabla_R \langle m|\bigg]|n\rangle + E_m\langle m|\bigg[\nabla_R|n\rangle\bigg] + \langle m|\big(\nabla_RH\big)| n\rangle$$
Noting that $$\nabla_R\langle m|n\rangle = 0 \implies \bigg[\nabla_R\langle m|\bigg]|n\rangle = - \langle m|\bigg[\nabla_R |n\rangle\bigg]$$, it follows that for $$E_n\neq E_m$$, $$\langle m|\nabla_R|n\rangle = \frac{\langle m|\big(\nabla_RH\big)|n\rangle}{E_m-E_n}$$
This is not an approximate result. It can be obtained by expanding $$|n\rangle$$ in a perturbation series about $$R=R_0$$ and then reading off the first term, but that's the same as finding the derivative of a function by computing its power series and reading off the coefficient of the first term: $$\frac{1}{1-x} = 1 + \color{red}{1}x + x^2 + \ldots \implies \frac{d}{dx}\left[\frac{1}{1-x}\right]\bigg|_{x=0} = \color{red}{1}$$