0
$\begingroup$

I'm learning about Berry phase for a presentation and am working through Berry's original paper.

I can't quite make the connection between equation 4

$$ \dot{\gamma}(t) = i \langle n(\vec{R}) |\nabla_R |n(\vec{R})\rangle \cdot \dot{\vec{R}}(t) $$

where and equation 6

$$ {\gamma}(C) = i \oint_C \langle n(\vec{R}) |\nabla_R |n(\vec{R})\rangle \cdot d{\vec{R}} $$

Equation 4 is derived from postulating that another phase (other than from Hamiltonian evolution) exists, and if so then that phase will obey that form. He then essentially jumps to equation 6.

I assume there is some calculus relation or detail from the adiabatic approximation that I am not seeing, any ideas?

Improve this question
$\endgroup$
1
  • $\begingroup$ Please read this and take appropriate actions (also for all other questions). $\endgroup$
    – Tobias Fünke
    Commented Apr 30 at 5:41

1 Answer 1

Reset to default
2
$\begingroup$

Equation $(6)$ is just obtained by integrating $(4)$ along a closed loop $C$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I believe I had some errors in my original post for equation 6, and have now been fixed. Is this still true? I don't see how the time derivatives are delt with. $\endgroup$
    – Hurricane
    Commented Apr 30 at 0:59
  • 2
    $\begingroup$ @Hurricane Yes, it's still true. Do you remember how to compute a line integral? You first define a parameterized curve $\vec R(t)$ such that $\vec R(0)$ is the starting point and $\vec R(T)$ is the ending point (in this case, they are the same point, since the curve is closed). Then $\int \vec F(\vec R) \cdot \mathrm d\vec R \equiv \int_0^T \vec F\big(\vec R(t)\big) \cdot \dot{\vec R}(t) \mathrm dt$. $\endgroup$
    – J. Murray
    Commented Apr 30 at 1:02

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.