Timeline for Is my attempt to prove that Berry's phase is quantized in inversion symmetric systems true? Do I violate gauge invariance?
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14 events
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| Nov 30, 2017 at 9:40 | comment | added | L.K. | Let us continue this discussion in chat. | |
| Nov 30, 2017 at 9:02 | comment | added | David Bar Moshe | cont. The computation in the question was performed with the original state first, then with the time reversed state, and consequence is obtained by equating the Zak phases of the initial state and of the time reversed state. | |
| Nov 30, 2017 at 9:02 | comment | added | David Bar Moshe | @L.K. It's no bother; I am afraid that I misled you in my previous comment. In our case we indeed have: $|u(\mathbf{k})\rangle = \mathcal{T} |u(\mathbf{k})\rangle = e^{i\phi{\mathbf{k}}} |u(-\mathbf{k})\rangle $, because we are talking about a time reversal invariant system. The Zak-Berry phase in the question is cyclic, i.e. we always start from a given state say $|u(\mathbf{k})\rangle$ and we finish at the same state. | |
| Nov 29, 2017 at 17:08 | comment | added | L.K. | I am sorry to bother you so much. But there should be phase due to TR is because of the reason that it will matter the path we take to go back to the time reversed state. It is will be unity if we come back to initial state $e^{i\phi(\pi)} = e^{i\phi(-\pi)}$. | |
| Nov 29, 2017 at 16:03 | comment | added | David Bar Moshe | @L.K. If I understood you correctly; I meant that the time reversed version of $|u(\mathbf{k})\rangle$ is $e^{i\phi{(\mathbf{k}})} |u(-\mathbf{k})\rangle$, (the states are not equal), and indeed the time reversal operation acts on the states by reversing the momentum sign and adding a phase factor. | |
| Nov 29, 2017 at 15:49 | comment | added | L.K. | Many thanks!! If I understand correctly. The $| u(\mathbf{k}) \rangle$ states are time reversal upto a phase factor, of the form written above(iff $| u(\mathbf{k}) \rangle$ is time reversed one)? | |
| Nov 29, 2017 at 15:09 | comment | added | David Bar Moshe | Cont. The point is that if the winding number of this phase is not zero, it cannot be removed by a gauge transformation. | |
| Nov 29, 2017 at 15:07 | comment | added | David Bar Moshe | @L.K. We know that the time conjugation operation reverses the momentum, therefore maps state indexed by the lattice momentum $\mathbf{k}$ to a state indexed by $-\mathbf{k}$. But since a multiplicative phase does not change the state, the most general form of the time reversed state can be of the form $e^{i\phi({\mathbf{k}})} |u(-\mathbf{k})\rangle$. The exact form of $\phi{(\mathbf{k})}$ depends on the Hamiltonian, whose specific form may necessitate such a phase in order to be time reversal invariant. | |
| Nov 29, 2017 at 14:17 | comment | added | L.K. | Excuse me for one query: $\mathcal{T} | u(\mathbf{k}) \rangle = e^{i\phi(\mathbf{k})} | u(\mathbf{-k}) \rangle$ .How do we get this? I seem to be lost, I saw the reference also. Forgive me for my ignorance. | |
| Sep 1, 2017 at 17:26 | history | edited | David Bar Moshe | CC BY-SA 3.0 |
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| Sep 1, 2017 at 17:15 | history | edited | David Bar Moshe | CC BY-SA 3.0 |
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| Aug 31, 2017 at 14:47 | history | edited | David Bar Moshe | CC BY-SA 3.0 |
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| Aug 31, 2017 at 14:05 | history | edited | David Bar Moshe | CC BY-SA 3.0 |
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| Aug 31, 2017 at 11:32 | history | answered | David Bar Moshe | CC BY-SA 3.0 |