# Berry's curvature and magnetic moment in TMDCs

Matter Modeling Asked by Carmen González on August 19, 2021

I am studying the transition metal dichalcogenides (TMDCs) and one of the applications that these materials have is their use in valleytronics. Valleytronics is related to the magnetic moment, Berry curvature, the spatial inversion symmetries, and the time-reversal symmetry.

According to the time-reversal symmetry, Berry’s curvature and magnetic moment are odd functions ($$mathbf{Omega(-k)}=-mathbf{Omega(k)}$$ and $$mathbf{m(-k)}=-mathbf{m(k)}$$).
According to the symmetry of spatial inversion, the functions are even ($$mathbf{Omega(-k)}=mathbf{Omega(k)}$$ and $$mathbf{m(-k)}=mathbf{m(k)}$$). Therefore, for valleytronics to exist, there does not have to be inversion symmetry, which occurs with a single-layer TMDCs.

• How can I demonstrate that the functions are odd according to the time reversal symmetry and even according to the spatial inversion symmetry?
• What is the physical interpretation of Berry’s curvature and Berry’s phase?
• H̶o̶w̶ ̶t̶o̶ ̶d̶e̶d̶u̶c̶e̶ ̶B̶e̶r̶r̶y̶’̶s̶ ̶e̶q̶u̶a̶t̶i̶o̶n̶s̶?̶ (Maybe a question for a new thread, since ProfM already answered 2 of the above and I answered the other).

The Berry curvature is defined as:

$$Omega_{munu}(mathbf{k})=partial_{mu}A_{nu}(mathbf{k})-partial_{nu}A_{mu}(mathbf{k}), tag{1}$$

where $$A_{mu}(mathbf{k})=langle u_{mathbf{k}}|ipartial_{mu}u_{mathbf{k}}rangle$$ is the Berry connection, $$|u_{mathbf{k}}rangle$$ is a Bloch state, and $$partial_muequiv frac{partial}{partial k_mu}$$, and $$mu,nu=x,y,z$$.

Invesion symmetry. Under inversion, $$mathbf{k}to-mathbf{k}$$, so that applying the inversion operation $$mathcal{I}$$ on a Bloch state gives $$mathcal{I}|u_{mathbf{k}}rangle=|u_{-mathbf{k}}rangle$$. If the system is invariant under inversion, then $$|u_{mathbf{k}}rangle$$ and $$|u_{-mathbf{k}}rangle$$ must be the same state up to a global phase, so that:

$$mathcal{I}|u_{mathbf{k}}rangle=e^{ivarphi_{mathbf{k}}}|u_{mathbf{k}}rangleLongrightarrow |u_{-mathbf{k}}rangle=e^{ivarphi_{mathbf{k}}}|u_{mathbf{k}}rangle.tag{2}$$

For the Berry connection, $$mathcal{I}A_{mu}(mathbf{k})=A_{mu}(-mathbf{k})$$. If the system has inversion symmetry, then

$$begin{eqnarray} A_{mu}(-mathbf{k})&=&langle u_{-mathbf{k}}|ipartial_{mu}u_{-mathbf{k}}rangle tag{3}\ &=& langle u_{mathbf{k}}|e^{-ivarphi_{mathbf{k}}}ipartial_{mu}left(e^{ivarphi_{mathbf{k}}}u_{mathbf{k}}right)rangle tag{4}\ &=& langle u_{mathbf{k}}|e^{-ivarphi_{mathbf{k}}}ie^{ivarphi_{mathbf{k}}}partial_{mu}u_{mathbf{k}}rangle + langle u_{mathbf{k}}|e^{-ivarphi_{mathbf{k}}}i^2e^{ivarphi_{mathbf{k}}}u_{mathbf{k}}ranglepartial_{mu}varphi_{mathbf{k}}tag{4} \ &=& langle u_{mathbf{k}}|ipartial_{mu}u_{mathbf{k}}rangle -partial_{mu}varphi_{mathbf{k}} tag{5}\ &=&A_{mu}(mathbf{k})-partial_{mu}varphi_{mathbf{k}},tag{6} end{eqnarray}$$ where in the second line I used the result for the Bloch state in a system with inversion symmetry, and in the third line the chain rule for differentiation. This result means that for a system that is invariant under inversion, then $$A_{mu}(mathbf{k})$$ and $$A_{mu}(-mathbf{k})$$ differ at most by a gauge transformation.

We are now ready to look at the Berry curvature. Under inversion, $$mathcal{I}Omega_{munu}(mathbf{k})=Omega_{munu}(-mathbf{k}$$). If the system has inversion symmetry, then

$$begin{eqnarray} Omega_{munu}(-mathbf{k})&=&partial_{mu}A_{nu}(-mathbf{k})-partial_{nu}A_{mu}(-mathbf{k}) tag{7}\ &=&partial_{mu}left(A_{nu}(mathbf{k})-partial_{nu}varphi_{mathbf{k}}right)-partial_{nu}left(A_{mu}(mathbf{k})-partial_{mu}varphi_{mathbf{k}}right) tag{8}\ &=&partial_{mu}A_{nu}(mathbf{k})-partial_{nu}A_{mu}(mathbf{k})-partial_{mu}partial_{nu}varphi_{mathbf{k}}+partial_{nu}partial_{mu}varphi_{mathbf{k}} tag{9}\ &=&partial_{mu}A_{nu}(mathbf{k})-partial_{nu}A_{mu}(mathbf{k}) tag{10}\ &=&Omega_{munu}(mathbf{k})tag{11}, end{eqnarray}$$

where in the second line I used the result for the Berry connection in a system with inversion symmetry. This proves that for a system with inversion symmetry, $$Omega_{munu}(mathbf{k})=Omega_{munu}(-mathbf{k})$$.

Time reversal symmetry. You can use an analogous procedure (I encourage you to try) to prove that for a time reversal invariant system, $$Omega_{munu}(mathbf{k})=-Omega_{munu}(-mathbf{k})$$. All you need to know is how the time reversal operator acts on a Bloch state, $$mathcal{T}|u_{mathbf{k}}rangle=|u_{mathbf{-k}}^{ast}rangle$$, and the rest of the proof proceeds in the same way.

Physical interpretation. Berry phase-like quantities look at the evolution of Bloch states at neighbouring $$mathbf{k}$$-points in the Brillouin zone. As an example, the Berry connection is looking at the overlap between a state $$|u_{mathbf{k}}rangle$$ and a state infinitessimally away from it, $$partial_{mu}|u_{mathbf{k}}rangle$$. As such, they are useful for calculation properties that depend on the structure of the Block states across the Brillouin zone. A well-known example is the calculation of topological invariants of materials, which measure the "twists" that the electronic wave function has when crossing the Brillouin zone. I am not familiar with applications in valleytronics, so will leave that for someone more knowledgeable.

Futher reading. An excellent book to learn about Berry phase-like quantities and applications (modern theory of polarization, topological materials, etc.) is David Vanderbilt's book.

Correct answer by ProfM on August 19, 2021

Resolution for the time reversal symmetry:

I need to demonstrate: $$Omega(-mathbf{k})=-Omega(mathbf{k})$$ (Berry's curvature is a odd function under time reversal symmetry)

Berry's curvature: $$Omega_{munu}(mathbf{k})=partial_{mu}A_{nu}(mathbf{k})-partial_{nu}A_{mu}(mathbf{k})tag{1}$$

If the system is time-reversally invariant:

$$T|u_krangle=e^{ivarphi_{mathbf{k}}}|u_{mathbf{k}}rangleRightarrow |u_{-mathbf{k}}^{*}rangle=e^{ivarphi_{mathbf{k}}}|u_{mathbf{k}}rangletag{2}$$

The time reversal symmetry operator applied to Berry's curvature

begin{align} TOmega_{munu}(mathbf{k})&=langlepartial_{mu}Tu_{mathbf{k}}|ipartial_{nu}Tu_{mathbf{k}}rangle-langlepartial_{nu}Tu_{mathbf{k}}|ipartial_{mu}Tu_{mathbf{k}}rangletag{5} \ &=iint dmathbf{r}partial_{mu}Tu^{ast}_{mathbf{k}}partial_{nu}Tu_{mathbf{k}}-iint dmathbf{r}partial_{nu}Tu^{ast}_{mathbf{k}}partial_{mu}Tu_{mathbf{k}} \ &=iint dmathbf{r}partial_{mu}u_{-mathbf{k}}partial_{nu}u^{ast}_{-mathbf{k}}-iint dmathbf{r}partial_{nu}u_{-mathbf{k}}partial_{mu}u^{ast}_{-mathbf{k}}\ &=Omega_{numu}(-mathbf{k})\ &=-Omega_{munu}(-mathbf{k}), end{align} where I have used the position representation. Also, the Berry curvature is fully gauge invariant, so time reversal symmetry implies $$TOmega_{munu}(mathbf{k})=Omega_{munu}(mathbf{k})$$. Putting together the two expressions for $$TOmega_{munu}(mathbf{k})$$ gives:

$$Omega_{munu}(mathbf{k})=-Omega_{munu}(-mathbf{k})tag{7}$$

Answered by Carmen González on August 19, 2021