0
$\begingroup$

The relative phase of two quantum states $|\psi_1\rangle,|\psi_2\rangle$ can be written, $$\gamma_{12}=-\text{arg}\langle\psi_1|\psi_2\rangle=-\text{arg}\big[|\langle\psi_1|\psi_2\rangle|e^{-i\gamma_{12}}\big].$$ If these two states are very close to each other then, $$e^{-i\Delta\gamma}=\frac{\langle\psi(R)|\psi(R+dR)\rangle}{|\langle\psi(R)|\psi(R+dR)\rangle|}.$$ From this one is supposed to be able to find, $$\Delta\gamma=i\langle\psi(R)|\nabla_R|\psi(R)\rangle dR.$$ This can be found in some power points when searching for Berry phase as relative phase and in the book A short course on topological insulators - Asbóth, link: http://optics.szfki.kfki.hu/~asboth/topins_course/book.pdf

See page 25 of the book. I get something slightly different.

Attempt at solution:

Assuming the phase between two neighbouring states is small we can expand the left side, $$e^{-i\Delta\gamma}=1-i\Delta\gamma.$$ We can also expand the ket, $$|\psi(R+dR)\rangle=|\psi(R)\rangle +\nabla_R|\psi(r)\rangle dR.$$ Together then I get, $$1-i\Delta\gamma=\frac{\langle\psi(R)|\psi(R)\rangle +\langle\psi(R)|\nabla_R|\psi(r)\rangle dR}{|\langle\psi(R)|\psi(R)\rangle +\langle\psi(R)|\nabla_R|\psi(r)\rangle dR|}.$$ It is only possible to get the correct solution if the denominator can be removed (equals 1).

Why is $|\langle\psi(R)|\psi(R)\rangle +\langle\psi(R)|\nabla_R|\psi(r)\rangle dR|=1$ to first order in $dR$?

$\endgroup$
  • $\begingroup$ Use the fact that $|\psi(R)\rangle$ and $|\psi(R + dR)\rangle$ have to be normalized, i.e. $\langle \psi(R) | \psi(R) \rangle = \langle \psi(R+ dR) | \psi(R + dR) \rangle = 1$. $\endgroup$ – Dominic Else Dec 4 '17 at 19:12
  • $\begingroup$ Could you elaborate? Obviously $|\langle\psi(R)|\psi(R)\rangle +\langle\psi(R)|\nabla_R|\psi(r)\rangle dR|=|1+\langle\psi(R)|\nabla_R|\psi(r)\rangle dR|$. But where can I make use of the second normalization condition? It does not appear anywhere. $\endgroup$ – Jens Roderus Dec 4 '17 at 19:18
  • $\begingroup$ Well, if you do the Taylor expansion of the second normalization condition in terms of $dR$ then you will find a term involving $\langle \psi(R) | \nabla_R | \psi(r) \rangle$, which also appears in your denominator. $\endgroup$ – Dominic Else Dec 4 '17 at 19:26
  • $\begingroup$ Yes, sure but it does not help. $1=\langle\psi(R+dR)|\psi(R+dR)\rangle $ gives, $0=\langle \psi|\nabla \psi\rangle dR+ \langle \nabla\psi|\psi\rangle dR$. $\endgroup$ – Jens Roderus Dec 4 '17 at 19:34
  • $\begingroup$ Expand the denominator also to first order in dR and you get the exact same quantity as on the RHS of that equation (plus 1). $\endgroup$ – Dominic Else Dec 4 '17 at 19:45
1
$\begingroup$

$\langle\psi(R)|\nabla_R|\psi(r)\rangle dR $ is purely imaginary because, $$0=\nabla_R\langle\psi(R)|\psi(R)\rangle=\langle\psi(R)|\nabla_R\psi(R)\rangle+\langle\nabla_R\psi(R)|\psi(R)\rangle=$$ $$=\langle\nabla_R\psi(R)|\psi(R)\rangle+\langle\nabla_R\psi(R)|\psi(R)\rangle^*.$$ The absolute value of $a+ib$ is $|a+ib|=\sqrt{a^2-b^2}$. Therfore, $$|\langle\psi(R)|\psi(R)\rangle +\langle\psi(R)|\nabla_R|\psi(r)\rangle dR|=\sqrt{1+O(dR^2)}=1.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.