Suppose we have normalized states $| n(\vec{R})\rangle$ indexed by continuous variable $\vec{R}$. Then fixing our choice of gauge and ignoring dynamic phase, the phase difference between two states is the Berry's Phase:

$$\tag{1} \langle n(\vec{R}_0)| n(\vec{R}_0 +\Delta \vec{r} ) \rangle ~=~ e^{i\gamma}$$

where, if $C$ is some curve that goes between $\vec{R}_0$ and $\vec{R}_0 + \Delta \vec{r}$, $$\tag{2} \gamma~=~i\int_C \langle n(\vec{R}) | \nabla_{\vec{R}} | n(\vec{R})\rangle \cdot d\vec{R}$$

If $\Delta \vec{r}$ is small, then $$\tag{3} \gamma \approx i\langle n(\vec{R}_0) | \nabla_{\vec{R}} | n(\vec{R}) \rangle \Big|_{\vec{R}=\vec{R}_0} \cdot \Delta \vec{r}$$

However, we can directly calculate this as well:

$$\langle n(\vec{R}_0)| n(\vec{R}_0 +\Delta \vec{r} ) \rangle ~\approx~ \langle n(\vec{R}_0) | n(\vec{R}_0) \rangle + \langle n(\vec{R}_0) | \nabla_{\vec{R}}| n(\vec{R}) \rangle \Big|_{\vec{R}=\vec{R}_0} \cdot \Delta \vec{r}$$

$$\tag{4} \approx 1 + \langle n(\vec{R}_0) | \nabla_{\vec{R}} | n(\vec{R}) \rangle \Big|_{\vec{R}=\vec{R}_0} \cdot \Delta \vec{r} ~\approx~ \text{exp}\left[ \langle n(\vec{R}_0) | \nabla_{\vec{R}} | n(\vec{R}) \rangle \Big|_{\vec{R}=\vec{R}_0} \cdot \Delta \vec{r} \right]$$

and therefore

$$\tag{5} \langle n(\vec{R}_0)| n(\vec{R}_0 +\Delta \vec{r} ) \rangle ~\approx~ e^{-i\gamma}$$

There's a minus sign now! What am I doing wrong here?

• I suspect there is a mistake in your Taylor expansion, but I wasn't able to track it down. It strikes me as odd that you start with only $\tau$ and $\Delta\tau$ but take a $t$ derivative. – Flavin Nov 7 '13 at 14:43
• @Flavin I don't see anything wrong with the Taylor expansion: $f(a+h) \approx f(a)+h f'(a)$ – ChickenGod Nov 7 '13 at 21:28

Comments to the question (v2): OP's second equation

$$\tag{2} \gamma~=~i\int_{0}^{t_f} \langle \psi_n(t) | \frac{d}{dt} | \psi_n(t)\rangle dt.$$

is not correct. It should read

$$\tag{A} \gamma_n(t_f) ~=~i\int_{0}^{t_f} \! dt \langle n, R(t)| \frac{d}{dt} | n,R(t)\rangle,$$

Here

$$\tag{B} |\psi_n(t_f) \rangle~=~e^{i\gamma_n(t_f)}\exp\left[-\frac{i}{\hbar}\int_{0}^{t_f} \! dt ~E_n(R(t)) \right] | n,R(t_f)\rangle,$$

cf. Wikipedia. In particular, the rhs. of OP's second eq. (2) [multiplied by $\hbar$] reads

$$\tag{C} i\hbar\int_{0}^{t_f} \! dt \langle\psi_n(t)| \frac{d}{dt} | \psi_n(t)\rangle ~=~ i\int_{0}^{t_f} \! dt \langle\psi_n(t)| \hat{H} | \psi_n(t)\rangle ~=~\int_{0}^{t_f} \! dt ~E_n(R(t)).$$

Comments to the update (v3): In the revision (v3), the Berry phase $\gamma$ in eq. (1) has the wrong sign.

• Maybe I'm being really stupid right now, but I fail to see the difference between your equation (A) and my equation (1) if you set $\tau=0$, $\Delta \tau = t_f$, and symbolically change $\psi$ to $R$. – ChickenGod Nov 7 '13 at 21:32
• I updated the answer. – Qmechanic Nov 7 '13 at 21:53
• Sorry, I guess the way I posed my question was confusing. I didn't intend $t$ to be time, but rather just some parameter to index the states. I revised my question for clarity. – ChickenGod Nov 8 '13 at 3:17
• In the revision (v3), the Berry phase $\gamma$ in eq. (1) has the wrong sign. – Qmechanic Nov 8 '13 at 19:36
• Your previous comment was very cryptic. At first glance, my eq. (1) looks like it agrees with your eq. (B). However, after some deep thought, I think I understand. Thanks! – ChickenGod Nov 9 '13 at 1:33