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.

  • 2
    $\begingroup$ It is first-order perturbation theory . Take the variation of $H_R|n\rangle = E_n|n\rangle$ and multiply on the left by $\langle m|$. $\endgroup$
    – mike stone
    Jul 4 at 0:03
  • $\begingroup$ 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! $\endgroup$
    – JoDraX
    Jul 4 at 1:20

1 Answer 1


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 $E_n\neq E_m$, $$\langle m|\nabla_R|n\rangle = \frac{\langle m|\big(\nabla_RH\big)|n\rangle}{E_m-E_n}$$

I thought it was a strict equality for any energy scale, not a perturbative one [...]

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: $$\frac{1}{1-x} = 1 + \color{red}{1}x + x^2 + \ldots \implies \frac{d}{dx}\left[\frac{1}{1-x}\right]\bigg|_{x=0} = \color{red}{1}$$

The power series may be an approximation, but the derivative is exact.


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