# What is the physical meaning of an electronic system evolving adiabatically through a closed path?

I am trying to understand Physics behind the Weyl Fermion in Condensed Matter Systems.

Electrons show Weyl fermionic behaviour in the vicinity of so called 'Diabolical Points' in the band structure. If I understand it correctly, these are accidental touchings between successive energy bands of a system. The paper I am referring to (Scientific Reports 5, Article number: 7816 (2015)) states that "Diabolical Points were made prominent by Berry (4,5), who showed that a system accrues a phase when it evolves adiabatically through a closed path in parameter space enclosing the DP: the Berry phase, or more precisely, a topological Berry phase (6)."

As someone looking at this problem from the point of view of experimental condensed matter physics, I find it difficult to understand the physical picture associated with the statement "system evolves adiabatically through a closed path in the parameter space". Is it the Hamiltonian of a system that is evolving with time? If so, what are the physical properties of the system that are changing? In fact, if I am given a compound which is said to contain such Weyl fermion like electronic excitations, then what does it mean to say this system is evolving adiabatically in time? What exactly is changing? And what does it mean to say that the system has acquired a phase? How does band structure and other characteristics change when the system has acquired this phase?

Please forgive my naivete, but I am finding it very difficult to unite the quantum mechanical and condensed matter picture in my head.

Thanks in advance!

• I am having difficulties establishing connections between quantum mechanics and the so-called "physical picture" - what does it mean for a wavefunction to "gain a phase"? What happens to an electron when its wavefunction gains a phase? – Gamora Aug 19 '15 at 13:39
• To "gain a phase" in QM means quite the same thing that it usually means. When any wave "gain a phase", it means that the time reference of the wave is shifted. And such phase shift can usually be detected by using some interferometry (for optics) or by looking in the Fourier domain (or for electrons, in the momentum distribution). – dolun Aug 21 '15 at 8:52
• So, the time reference of the electron wave has shifted by the magnitude of the Berry phase it gained on physically translating across the lattice in such a way that it encircled a Diabolical point? Am I seeing this correctly? And would this show up in bulk transport measurements? What signs would such a phenomenon show in the resistivity, for instance? – Gamora Aug 22 '15 at 6:51
• Yes you could say something like that. In order to calculate transport properties like current, one has to sum up over all the $\vert\textbf{k}\rangle$ states. Topological properties of the band structure then may introduce some additional terms in the continuity equation I guess, i.e. : $$\partial_t\rho+\mathbf{\nabla}\cdot\textbf{j}=\text{stuff}$$ Moreover, to the point of view of the resistivity, the Ohm law may fail to describe transport properties. – dolun Aug 24 '15 at 12:45
• In the case of Weyl fermions, since they are chiral excitations, maybe some anisotropy in the conductivity could also be measured. – dolun Aug 24 '15 at 12:51

## 1 Answer

I'm not sure if I can help you on the part concerning the Weyl fermions. But your question seems to deal rather with what is a geometrical phase.

Parallel transport and geometrical phase

Maybe the more intuitive thing to do first is to draw a parallel between geometrical phase and parallel transport. As shown in the image of this wikipedia article, considere a sphere $S$ with a vector $\textbf{e}$ lying tangentially to its surface. In the usual spherical coordinate system $\{\textbf{u}_r,\textbf{u}_\theta,\textbf{u}_\phi\}$ let say that initially : $$\textbf{e}=-\textbf{u}_\theta$$ which correspond to a blue arrow on the image.

Now let say that we perform a parallel transport of this vector along a path $C$ which lies on the contour an eigth of the sphere (as shown on the image), which is a closed path. Here parallel means that the vector conserves its direction along the motion path.

What you see is that the directions of the vector before and after transport is different (compare blue and red arrows). More precisely the vector has transformed following : $$\textbf{e}=-\textbf{u}_\theta\;\rightarrow\;\textbf{e}'=\textbf{u}_\phi\quad\text{i.e.}\quad\textbf{e}'=e^{-\frac{\mathrm{i}\pi}{2}}\textbf{e}$$ which correspond to $-\pi/2$ rotation. The reason why it is $\pi/2$ (let forget about the signe) is that the contour $C$ embodies a $4\pi/8=\pi/2$ solid angle. This $\pi/2$ is what we call geometrical phase since it only depends on the choice of the path $C$ and not on the time taken to perform the transport along $C$. More precisely, this phase reflects the topology of $S$, namely here its curvature.

Quantum mechanics and Berry phase

Now there is quantum analogue of such phase which is called the Berry phase. The principle is basically the same but here instead of a vector $\textbf{e}$ you have an initial quantum eigenstate $\vert\phi_\ell\rangle$ of an hamiltonian $\mathcal{H}(\mathbf{\Gamma}(t))$.

Here $\mathbf{\Gamma}(t)$ are some slow time-dependant coupling parameters. The vector $\mathbf{\Gamma}$ lives in a parameter space $\mathcal{M}$ which will play the role of the sphere $S$. The reason why $\mathbf{\Gamma}$ is said to be slow is that it guaranties the adiabaticity of the time evolution of the quantum state, namely, that : $$\forall t>0,\,\vert\Psi(t)\rangle\sim\vert\phi_\ell(\mathbf{\Gamma}(t))\rangle$$ Such condition is analogous to kipping the direction of the vector $\textbf{e}$ constant when performing parallel transport.

If $\mathbf{\Gamma}$ is a $T$-periodic function of time, this means that one can do some closed path $C: \mathbf{\Gamma}(0)\rightarrow\mathbf{\Gamma}(T)$ in the parameter space $\mathcal{M}$.

By doing so, one can print a Berry phase $\varphi_B$ on the quantum state : $$\forall t>0,\,\vert\Psi(t)\rangle\sim e^{\mathrm{i}\varphi_B}\vert\phi_\ell(\mathbf{\Gamma}(0))\rangle$$ with the interesting feature that $\varphi_B$ does not depend on $T$. For details about the calculations you can check this PSE question.