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In Nakahara 10.6.2 the case of system with fast $r$ and slow $R$ degrees (might be more than one of each) of freedom is discussed. The Hamiltonian is -

$$H=\frac{p^2}{2m}+\frac{P^2}{2M} + V(r;R)$$ Where $p(P)$ is the momentum associated with $r(R)$.

In describing the Schrodinger equation for the entire system, the total wavefunction of both dofs $\Psi(r,R)=\Phi(R)\left|R\right\rangle$ where $\Phi(R)$ is the slow dof wavefunction, $\left|R \right\rangle$ is the fast dofs wavefunction for fixed values of $R$.

In the derivation Nakahara apears to use this equality -

$$\left\langle R|\nabla^2_R|R \right\rangle=\left\langle R|\nabla_R|R \right\rangle)^2$$

However, I do not understand how it is reached. When starting with the LHS, and using integration by parts, I find a minus sign which I cannot wave off.

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I do not have a copy of Nakahara to hand, but I don't think it is an equality, but rather an approximation. Your notation is a bit confusing as $$ \Psi(r,R)= \Phi(R)|R\rangle $$ does not make sense as written. The LHS is a function and $|R\rangle$ is a vector in the Hilbert space, so the two sides are quite different things. I think what is meant is that we can factor the combined wavefunction for the slow nucleaus and the fast electron as $$ \Psi(r,R)= \psi(r,R)\Phi(R) $$ where the wavefunction for the fast moving electron is $$ \langle r|R\rangle= \psi(r,R). $$ Here $R$ is a parameter. Berry connection is $$ i {\bf A}(R)= \langle R| \nabla_R |R\rangle=\int \psi^*(r,R)\nabla_R \psi(r,R) d^3{\bf r}. $$ We want an equation for the $\Phi(R)$ wavefunction to include the gauge covariant derivative $$ (\nabla_R+i{\bf A}(R))^2. $$ However, when we plug the factored wavefunction into the full Shroedinger equation, we get $\langle R|\nabla^2_R|R\rangle$ instead of the $(iA)^2$. We therefore insert a complete set of states between the two $\nabla_R$'s but it is pecisely at this point that we make the adiabatic approximation by only keeping the single intermediate state $|R\rangle \langle R|$.

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  • $\begingroup$ The notation is indeed confusing but is not one (and so is accurate), I edited my question to explain it. Can you elaborate on the approximation you made? why is a consequence of the adiabatic approximation? $\endgroup$ – proton Feb 22 at 17:37
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    $\begingroup$ When you make a small change in $R$ the $\psi(r,R)$ ony changes a little. By far its largest overlap is still with the unchanged $\psi(r,R)$. The essence of the adiabatic approx is that we ignore the overlap with all states orthogonal to the original $\psi(r,R)$. $\endgroup$ – mike stone Feb 22 at 17:40

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