# Berry's curvature and magnetic moment in TMDCs

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).
• +1. Welcome to our site, and thank you for asking this question here!!! We hope to see much more of you!! One problem might be that you have 3 questions here, not 1. This might be appropriate as 2 or 3 separate questions. Jul 17 '20 at 21:25
• I too am trying to master everything Berry in a study of the integer QHE! Rather than make a poor attempt at explaining what I know, I thought I would point you to a source that I found very useful: damtp.cam.ac.uk/user/tong/qhe/one.pdf. Section 1.5 is devoted to Berry phase. Also your mention of transition metal dichalcogenides caught my eye. 30 years ago I was studying charge density waves in them as a graduate student with STM. My avatar is an STM image of atoms + CDW in 1T-TiSe2. It is nice to see that they are still systems of scientific interest!
– CGS
Jul 17 '20 at 23:35

The Berry curvature is defined as:

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

where $$A_{\mu}(\mathbf{k})=\langle u_{\mathbf{k}}|i\partial_{\mu}u_{\mathbf{k}}\rangle$$ is the Berry connection, $$|u_{\mathbf{k}}\rangle$$ is a Bloch state, and $$\partial_\mu\equiv \frac{\partial}{\partial k_\mu}$$, 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:

$$\mathcal{I}|u_{\mathbf{k}}\rangle=e^{i\varphi_{\mathbf{k}}}|u_{\mathbf{k}}\rangle\Longrightarrow |u_{-\mathbf{k}}\rangle=e^{i\varphi_{\mathbf{k}}}|u_{\mathbf{k}}\rangle.\tag{2}$$

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

$$\begin{eqnarray} A_{\mu}(-\mathbf{k})&=&\langle u_{-\mathbf{k}}|i\partial_{\mu}u_{-\mathbf{k}}\rangle \tag{3}\\ &=& \langle u_{\mathbf{k}}|e^{-i\varphi_{\mathbf{k}}}i\partial_{\mu}\left(e^{i\varphi_{\mathbf{k}}}u_{\mathbf{k}}\right)\rangle \tag{4}\\ &=& \langle u_{\mathbf{k}}|e^{-i\varphi_{\mathbf{k}}}ie^{i\varphi_{\mathbf{k}}}\partial_{\mu}u_{\mathbf{k}}\rangle + \langle u_{\mathbf{k}}|e^{-i\varphi_{\mathbf{k}}}i^2e^{i\varphi_{\mathbf{k}}}u_{\mathbf{k}}\rangle\partial_{\mu}\varphi_{\mathbf{k}}\tag{4} \\ &=& \langle u_{\mathbf{k}}|i\partial_{\mu}u_{\mathbf{k}}\rangle -\partial_{\mu}\varphi_{\mathbf{k}} \tag{5}\\ &=&A_{\mu}(\mathbf{k})-\partial_{\mu}\varphi_{\mathbf{k}},\tag{6} \end{eqnarray}$$ 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, $$\mathcal{I}\Omega_{\mu\nu}(\mathbf{k})=\Omega_{\mu\nu}(-\mathbf{k}$$). If the system has inversion symmetry, then

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

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, $$\Omega_{\mu\nu}(\mathbf{k})=\Omega_{\mu\nu}(-\mathbf{k})$$.

Time reversal symmetry. You can use an analogous procedure (I encourage you to try) to prove that for a time reversal invariant system, $$\Omega_{\mu\nu}(\mathbf{k})=-\Omega_{\mu\nu}(-\mathbf{k})$$. 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.

• +10. Another great one by ProfM! Jul 18 '20 at 9:53
• @CarmenGonzález $\mu$ and $\nu$ are indices (a bit like the $i$ and $j$ for a matrix $A_{ij}$ but Greek indices are often used when working in spacetime while Latin indices are often used when there's only spatial indices (search "general relativity" in this article to see this). I have added the definition of the derivative symbol to the answer. $R$ is often a nuclear coordinate. As for $\varphi$, it's a phase: remember that $|e^{i\theta}|=1$ so states can have a phase (see Bloch sphere). Jul 18 '20 at 21:58
• @CarmenGonzález I am no expert in the magnetic moment, but in a finite system (e.g. atom) it is given by $\mathbf{m}\propto\int d\mathbf{r}\mathbf{r}\times\mathbf{j}(\mathbf{r})$, so I imagine that you need to use how $\mathbf{r}$ and $\mathbf{j}$ individually transform under inversion and time reversal. Hopefully someone more knowledgeable on that will provide a detailed answer. Jul 19 '20 at 8:09
• @CarmenGonzález I also see that you have materials questions on Phys.SE that have got 0 upvotes and 0 answers in 5+ months: (1) physics.stackexchange.com/q/530539/134583, (2) physics.stackexchange.com/q/530511/134583, (3) physics.stackexchange.com/q/530341/134583, (4) physics.stackexchange.com/q/439184/134583. Why not copy and paste them here where everyone is nicer and more welcoming? We need more questions to convince the Stack Exchange company to keep our site live: it costs them money to store our data Jul 19 '20 at 16:56
• @CarmenGonzález and NikeDattani, Berry phase-like quantities can be defined with respect to any varying parameter, in the context of a Brillouin zone, the parameter that varies is the wave vector $\mathbf{k}$, so indeed $\mu\in\{x,y,z\}$. I realize this may cause confusion with general relativity notation, but I used it rather than $(i,j)$ because "$i$" already refers to $\sqrt{-1}$. Happy to change notation if you think this is confusing. Jul 19 '20 at 18:24

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

If the system is time-reversally invariant:

$$T|u_k\rangle=e^{i\varphi_{\mathbf{k}}}|u_{\mathbf{k}}\rangle\Rightarrow |u_{-\mathbf{k}}^{*}\rangle=e^{i\varphi_{\mathbf{k}}}|u_{\mathbf{k}}\rangle\tag{2}$$

The time reversal symmetry operator applied to Berry's curvature

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

$$\Omega_{\mu\nu}(\mathbf{k})=-\Omega_{\mu\nu}(-\mathbf{k})\tag{7}$$

• +1. For all the effort that you've shown us that you've put, into answering your own question after the help of ProfM. It might take time for one of us to be able to double-check all your work. By the way, I've edited one of your "equation arrays" so that the equal signs line up properly, can you please do the same for the other equation array, and add \tag{1}, \tag{2}, etc. to label the equations like I did for ProfM's answer? Jul 20 '20 at 2:43
• +1, great effort! The derivation is almost correct, but not quite. For example, under time reversal the Berry connection transforms as $TA_{\mu}(\mathbf{k})=A_{\mu}(-\mathbf{k})$. Given that you have started so well, I would still recommend that you try to complete the proof yourself, and this online resource explicitly goes through the steps so it can guide you: www-personal.umich.edu/~sunkai/teaching/Fall_2012/… Jul 20 '20 at 7:50
• Also, writing the Berry curvature and connection with subindices indicates the vector or tensor components, so the symbols should not be in bold face. There is often some confusion on this point because although for $n$ parameters the Berry connection is a $n$-component vector and the Berry curvature a $n\times n$ second-rank tensor, if you have 3 parameters (like in this example), then the Berry curvature can be written as a pseudo-vector, so you can work with it as if it were a "vector". This would be an interesting topic for a new question, given our need for questions in the beta... Jul 20 '20 at 7:57
• In my notation it should still be an $i$ rather than a $-i$ in the Berry connection (whether it is plus or minus $i$ is a convention, and both definitions are used). Then you would get $A_{\mu}(-\mathbf{k})=A_{\mu}(\mathbf{k})-\partial_{\mu}\varphi(\mathbf{k})$. For the Berry curvature, it turns out that the application of time reversal symmetry gives $T\Omega_{\mu\nu}(\mathbf{k})=-\Omega_{\mu\nu}(-\mathbf{k})$. Are you happy to have a go or would you like me to provide the solution? Jul 21 '20 at 12:48
• @CarmenGonzález, I have now added the solution for the Berry curvature, explicitly working in the position representation to keep track of the complex conjugates more easily. Jul 22 '20 at 9:01