The phase difference acquired over the course of a closed loop which results from the geometrical properties of the parameter space of the Hamiltonian.
Questions tagged [berry-pancharatnam-phase]
213 questions
25
votes
4 answers
Crystal momentum and the vector potential
I noticed that the Aharonov–Bohm effect describes a phase factor given by $e^{\frac{i}{\hbar}\int_{\partial\gamma}q A_\mu dx^\mu}$. I also recognize that electrons in a periodic potential gain a phase factor given by…
Alex Eftimiades
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Where does the Berry phase of $\pi$ come from in a topological insulator?
The Berry connection and the Berry phase should be related. Now for a topological insulator (TI) (or to be more precise, for a quantum spin hall state, but I think the Chern parities are calculated in the same fashion for a 3D TI). I can follow the…
Mike
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17
votes
2 answers
How is Berry phase connected with chiral anomaly?
Recently I've read in one article about very strange way to describe chiral anomaly on quasiclassical level (i.e., on the level of Boltzmann equation and distribution function).
Starting from Weyl hamiltonian
$$
H= \sigma \cdot \mathbf p,
$$
which…
Name YYY
- 9,071
16
votes
1 answer
Fermi statistics and Berry phase
When the positions of two fermions are exchanged adiabatically in three-dimensional space, we know that the wave function gains a factor of $-1$. Is this related to Berry's phase?
leongz
- 4,174
15
votes
2 answers
Adiabatic theorem and Berry phase
As far as I can check, the adiabatic theorem in quantum mechanics can be proven exactly when there is no crossing between (pseudo-)time-evolved energy levels. To be a little bit more explicit, one describes a system using the Hamiltonian…
FraSchelle
- 11,033
15
votes
1 answer
Calculating the Berry curvature in case of degenerate levels (Non abelian Berry curvature): issue
The Berry phase accumulated on a path can be described by a matrix when we look at adiabatic time evolution with a Hamiltonian with degenerate energy levels.
The Berry phase matrix is given by
$$
\gamma_{mn}= \int_\mathcal{C} \left\langle m(R)…
13
votes
1 answer
Physical Interpretation of Relationship Between Hall Conductivity and Berry Curvature?
Why is the Hall conductivity in a 2D material
$$\tag{1} \sigma_{xy}=\frac{e^2}{2\pi h} \int dk_x dk_y F_{xy}(k)$$
where the integral is taken over the Brillouin Zone and $F_{xy}(k)$ is the Berry curvature of the filled bands?
What is the physical…
ChickenGod
- 2,255
13
votes
1 answer
Aharonov-Bohm effect as a geometric phase-Adiabatic transfer not needed?
In his 1984 paper,
Michael Berry proved that the Aharonov-Bohm effect is the same as a geometric phase. He did this by transferring a box containing charged particles around a solenoid. However, he mentions between equations 33 and 34 that the box…
TheQuantumMan
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12
votes
4 answers
Book recommendations - Topological Insulators for dummies
Is there a pedagogical explanation of what is a topological insulator for those that do not even know what the Berry phase is but have a basic understanding of quantum mechanics and solid state physics?
Zythos
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11
votes
1 answer
Berry phase and Wilson loop
According to the definition, the Wilson loop is
\begin{equation}
W[\mathcal{C}] =\operatorname{Tr}\left[\mathcal{P} \exp\left\{i\oint _{\mathcal{C}} A_{\mu } dx^{\mu }\right\}\right]
\end{equation}
where $\mathcal{P} $ is the path ordering, $A_{\mu…
Fang Lyu
- 111
11
votes
1 answer
Why is Berry connection a connection?
The Berry connection, following the derivation of the Berry phase for a non degenerate system, is
$\mathcal{A}_{k}(\lambda) = i \langle n|\frac{\partial}{\partial \lambda^{k}}|n\rangle$
This result is gauge-dependent, as seen making an analogy with…
vbarcelo
- 373
11
votes
1 answer
A question on the Chern number and the winding number?
Let $\mid \psi(x,y) \rangle$ be a normalized wavefunction living in a $d$-dimensional Hilbert space and depend on two real parameters $(x,y)$ that belong to a closed surface (e.g., $S^2, T^2$, ...). The Chern number of $\mid \psi(x,y) \rangle$ then…
Kai Li
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10
votes
2 answers
Berry curvature concentration around nodal points
It is well-known that in TI-symmetric semi-metals the Berry curvature on the Brillouin torus vanishes away from the nodal points (eg. [XCN10, III.B] [Van18, p. 105]).
But even for non-TI-symmetric semi-metals, the Berry curvature turns out, in…
Urs Schreiber
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votes
1 answer
Why is the Berry curvature odd under time reversal?
The question is: Why is the Berry curvature, defined as
$$\mathcal{F}=-\mathrm i\, \epsilon_{ij}\, \left\langle\partial_{ki}u_{n}(k)\mid \partial_{kj}u_{n}(k) \right\rangle ,$$
odd if I apply time reversal? From my understanding, the time reversal…
Suppenkasper
- 838
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votes
1 answer
Topological insulators - Surface states have a phase?
When I look at the circle of the Dirac cone around the Dirac point of, let's say, $Bi_2Se_3$, then the electron winds around and it is true that it goes from momentum $-k$ and spin-up to $+k$ and spin-down. Now how can I use this fact to show that…
Mike
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