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Emilio Pisanty
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Is my attempt to prove that Berry's phase is quantized in inversion symmetric systems true?Do Do I violate gauge invariance?

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}\begin{align*} \gamma & =\oint_\mathrm{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_\mathrm{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$$$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$$

then we have \begin{align*} U_{p}H(\mathbf{k})U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(-\mathbf{k})U_{p}U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(\mathbf{k})U_{p}U(-\mathbf{k}) & =E(-\mathbf{k})U_{p}U(-\mathbf{k}) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-\mathbf{k})=e^{i\phi(\mathbf{k})}U(\mathbf{k})$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(\mathbf{k}) & =i(U_{p}U(-\mathbf{k})e^{-i\phi(\mathbf{k})})^{\dagger}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U_{P}U(-\mathbf{k})\\ & =i(U(-\mathbf{k}))^{\dagger}e^{i\phi(\mathbf{k})}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U(-\mathbf{k})\\ & =-\mathcal{A}(-\mathbf{k})+\nabla_{\mathbf{k}}\phi(\mathbf{k}) \end{align*}

and the Berry's phase is for a 1d $BZ$$\mathrm{BZ}$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspissionsuspicion arises from the last step. The Berry phase should be gauge independent, but here it looks like gauge term determines its value.

Is my attempt to prove that Berry's phase is quantized in inversion symmetric systems true?Do I violate gauge invariance?

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(\mathbf{k})U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(-\mathbf{k})U_{p}U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(\mathbf{k})U_{p}U(-\mathbf{k}) & =E(-\mathbf{k})U_{p}U(-\mathbf{k}) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-\mathbf{k})=e^{i\phi(\mathbf{k})}U(\mathbf{k})$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(\mathbf{k}) & =i(U_{p}U(-\mathbf{k})e^{-i\phi(\mathbf{k})})^{\dagger}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U_{P}U(-\mathbf{k})\\ & =i(U(-\mathbf{k}))^{\dagger}e^{i\phi(\mathbf{k})}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U(-\mathbf{k})\\ & =-\mathcal{A}(-\mathbf{k})+\nabla_{\mathbf{k}}\phi(\mathbf{k}) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

Is my attempt to prove that Berry's phase is quantized in inversion symmetric systems true? Do I violate gauge invariance?

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_\mathrm{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_\mathrm{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$$

then we have \begin{align*} U_{p}H(\mathbf{k})U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(-\mathbf{k})U_{p}U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(\mathbf{k})U_{p}U(-\mathbf{k}) & =E(-\mathbf{k})U_{p}U(-\mathbf{k}) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-\mathbf{k})=e^{i\phi(\mathbf{k})}U(\mathbf{k})$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(\mathbf{k}) & =i(U_{p}U(-\mathbf{k})e^{-i\phi(\mathbf{k})})^{\dagger}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U_{P}U(-\mathbf{k})\\ & =i(U(-\mathbf{k}))^{\dagger}e^{i\phi(\mathbf{k})}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U(-\mathbf{k})\\ & =-\mathcal{A}(-\mathbf{k})+\nabla_{\mathbf{k}}\phi(\mathbf{k}) \end{align*}

and the Berry's phase is for a 1d $\mathrm{BZ}$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspicion arises from the last step. The Berry phase should be gauge independent, but here it looks like gauge term determines its value.

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physshyp
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The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(k)U(k) & =E(k)U_{p}U(k)\\ H(-k)U_{p}U(k) & =E(k)U_{p}U(k)\\ H(k)U_{p}U(-k) & =E(-k)U_{p}U(-k) \end{align*}\begin{align*} U_{p}H(\mathbf{k})U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(-\mathbf{k})U_{p}U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(\mathbf{k})U_{p}U(-\mathbf{k}) & =E(-\mathbf{k})U_{p}U(-\mathbf{k}) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-k)=U(k)$$U_{p}U(-\mathbf{k})=e^{i\phi(\mathbf{k})}U(\mathbf{k})$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =i(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}\begin{align*} \mathcal{A}(\mathbf{k}) & =i(U_{p}U(-\mathbf{k})e^{-i\phi(\mathbf{k})})^{\dagger}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U_{P}U(-\mathbf{k})\\ & =i(U(-\mathbf{k}))^{\dagger}e^{i\phi(\mathbf{k})}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U(-\mathbf{k})\\ & =-\mathcal{A}(-\mathbf{k})+\nabla_{\mathbf{k}}\phi(\mathbf{k}) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(k)U(k) & =E(k)U_{p}U(k)\\ H(-k)U_{p}U(k) & =E(k)U_{p}U(k)\\ H(k)U_{p}U(-k) & =E(-k)U_{p}U(-k) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-k)=U(k)$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =i(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(\mathbf{k})U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(-\mathbf{k})U_{p}U(\mathbf{k}) & =E(\mathbf{k})U_{p}U(\mathbf{k})\\ H(\mathbf{k})U_{p}U(-\mathbf{k}) & =E(-\mathbf{k})U_{p}U(-\mathbf{k}) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-\mathbf{k})=e^{i\phi(\mathbf{k})}U(\mathbf{k})$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(\mathbf{k}) & =i(U_{p}U(-\mathbf{k})e^{-i\phi(\mathbf{k})})^{\dagger}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U_{P}U(-\mathbf{k})\\ & =i(U(-\mathbf{k}))^{\dagger}e^{i\phi(\mathbf{k})}\nabla_{\mathbf{k}}e^{-i\phi(\mathbf{k})}U(-\mathbf{k})\\ & =-\mathcal{A}(-\mathbf{k})+\nabla_{\mathbf{k}}\phi(\mathbf{k}) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

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physshyp
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The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(k)U(k) & =E(k)U_{p}U(k)\\ H(-k)U_{p}U(k) & =E(k)U_{p}U(k)\\ H(k)U_{p}U(-k) & =E(-k)U_{p}U(-k) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-k)=U(k)$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =i(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(k)U(k) & =E(k)U_{p}U(k)\\ H(-k)U_{p}U(k) & =E(k)U_{p}U(k)\\ H(k)U_{p}U(-k) & =E(-k)U_{p}U(-k) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-k)=U(k)$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

The Berry's or Zak's phase is given as

\begin{align*} \gamma & =\oint_{BZ}d\mathbf{k}\mathcal{\mathcal{\mathcal{A}}}(\mathbf{k})\ \ \mbox{mod }2\pi\\ & =i\oint_{BZ}d\mathbf{k}\langle u_{n}(\mathbf{k})|\nabla_{\mathbf{k}}u_{n}(\mathbf{k})\rangle\ \ \mbox{mod }2\pi\\ \end{align*}

which can take any value between $[0,2\pi]$.

Now we want to learn the implications of the symmetries on this Berry's phase first assume that our system has only inversion symmetry such that the hamiltonian obeys

$U_{p}H(\mathbf{k})=H(-\mathbf{k})U_{p}$

then we have \begin{align*} U_{p}H(k)U(k) & =E(k)U_{p}U(k)\\ H(-k)U_{p}U(k) & =E(k)U_{p}U(k)\\ H(k)U_{p}U(-k) & =E(-k)U_{p}U(-k) \end{align*}

such that $E(k)=E(-k)$ and $U_{p}U(-k)=U(k)$ where $U_{k}$ is the vector representation of $|u(k)\rangle$. Now the Berry connection is

\begin{align*} \mathcal{A}(k) & =i(U_{p}U(-k))^{\dagger}\nabla_{k}U_{P}U(-k)\\ & =i(U(-k))^{\dagger}\nabla_{k}U(-k)\\ & =-\mathcal{A}(-k) \end{align*}

and the Berry's phase is for a 1d $BZ$ for simplicity

\begin{align} \gamma & =\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)\\ & =-\intop_{-\pi}^{\pi}dk\,[\mathcal{A}(-k)+\nabla_{k}\phi(k)]\\ & =\left.\phi(k)\right|_{-\pi}^{\pi}-\underbrace{\intop_{-\pi}^{\pi}dk\,\mathcal{A}(k)}_{\gamma}\label{eq:-18}\\ & =2\pi n-\gamma\label{eq:-17} \end{align}

thus we have \begin{align*} \gamma=\pi n \end{align*}

since $\gamma$ is well defined only up to $\mbox{mod }2\pi$ it can be $0$ or $\pi$ while obeying the last condition. My suspission arises from the last step. Berry phase should be gauge independent but here it looks like gauge term determines its value.

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