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 $dR$?

  • $\begingroup$ 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

$\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, $$|\langle\psi(R)|\psi(R)\rangle +\langle\psi(R)|\nabla_R|\psi(r)\rangle dR|=\sqrt{1+O(dR^2)}=1.$$


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