# Symmetries and band structure

I am currently trying to understand the relation between the symmetries of a crystal and its band structure. However, I am having a hard time figuring out how symmetries protect degeneracies in the band structure.

For instance, some literatures suggest that mirror symmetries could protect nodal line in nodal-line semimetals, while rotational symmetries (alongside inversion and time reversal symmetries) could protect Dirac nodes in Dirac semimetals, etc.

Does anyone have any documents or papers that show how these symmetries can protect band degeneracies? I am quite new to this field, so perhaps something not mathematically too complicated would be very helpful.

Thank you!

## 1 Answer

The answer to that might be very long and you want to avoid math in here so I will try my best to explain how I understand it but with a little math at the end.

Let’s start from the general connection between symmetry and band structures, in specific Translation symmetry which is responsible for the existence of energy bands in crystal structures, this symmetry allows for the formation of periodicity, where the arrangement of atoms repeats regularly in space. Such periodicity leads to the concept of a Brillouin zone and the formation of energy bands in the band structure of materials. In other words, the Hamiltonian of the system is invariant under translations, meaning that the energy of the system doesn’t change when the crystal is translated by a lattice vector. Thus, translation symmetry gives rise to distinct energy bands separated by energy gaps. On the other hand, point symmetries such as rotation, reflection and inversion may determine the nature of those energy bands.

Now regarding degeneracy, what does that mean? we said symmetries in a crystal lattice led to degeneracy, which means that multiple states (eigenfunctions) have the same energy (eigenvalue). When a crystal undergoes a symmetry operation, its structure remains the same, and the energy associated with the formation of this structure from isolated atoms is the same for any equivalent configuration. This translates to a single eigenvalue for a set of eigenfunctions corresponding to the various equivalent positions, which is what we mean by degeneracy.

Now you can imagine As the amount of symmetry in a system decreases, the degree of degeneracy also decreases. When symmetries are reduced or broken, the constraints that lead to degeneracy are lifted, and the energy levels associated with different configurations may no longer be the same. This reduction in symmetry results in the splitting of energy levels and a decrease in degeneracy.

To understand this connection between degeneracy and symmetry, let's consider a 1D harmonic oscillator, The Schrödinger equation corresponding to it is: $$(\frac{d^2\psi}{dx^2})+(\frac{2m}{\hslash^2})[E-(kx^2/2)]\psi = 0,\tag{1}$$

In which $$k$$ is Hooke's law constant. If we then use $$\alpha=2mE/\hslash^2$$ and $$\beta^2=mk/\hslash^2$$, the previous equation will become: $$(\frac{d^2\psi}{dx^2}) + (\alpha - \beta^2x)\psi = 0,\tag{2}$$

The solutions to this equation are the Hermite functions $$\psi_n = exp(\frac{-\alpha x^2}{2})H_n(x)$$, in which the eigenvalues are $$E_n = (n + \frac{1}{2})hv,\tag{3}$$

As you can see here for $$n=0,1,2..$$ there is a distinct solution $$\psi_n(x)$$ for each value of $$n$$ and therefore the system is nondegenerate. Now let's consider the three-dimensional Schrödinger equation of the general form $$(\frac{-\hslash^2}{2m})\nabla^2\psi(r)+V(r)\psi(r)=E\psi(r),\tag{4}$$

When applied to an oscillator with unequal constants $$k_x$$, $$k_y$$, $$k_z$$ along the three coordinate axes, it can be solved by the separation of variables, resulting in three ordinary differential equations with a similar form to the first equation we have. Then, the eigenvalues will be: $$E_{n_x,n_y,n_z}=[(n_x+\frac{1}{2})hv_x+(n_y+\frac{1}{2})hv_y+(n_z+\frac{1}{2})hv_z],\tag{5}$$

In the special case of an isotropic oscillator, the constants are all equal and eqaul to $$k$$, so the energy levels become $$E=(n_x+n_y+n_z+\frac{3}{2})hv$$ or $$(n+\frac{3}{2})hv$$, you can see that the total energy now depends only on $$n$$, and for a given value of $$n$$, there are $$(n+1)$$ choices for $$n_x$$; $$n_x=0,1,2,..,n$$, then for $$n_x=0$$, there are $$(n+1)$$ choices for $$n_y$$, for $$n_x=1$$, there are $$n$$ choices for $$n_y$$, for $$n_x=n$$, there is one choice for $$n_y$$. In all in all, the degree of degenracy can be expressed as $$\frac{1}{2}[(n+1)+1](n+1)=\frac{1}{2}(n+1)(n+2),\tag{6}$$

Such degeneracy is a result of the symmetry. For your reference and if you want to go deeper into that you can refer to Crystal Symmetry, Group Theory, and Band Structure Calculations by Allen Nussbaum.

• This statement is not correct in all cases : "Hamiltonian of the system is invariant under translations" Commented Jan 11 at 8:53
• @M06-2x you are right, would you like to provide some of those cases in an answer, that would be great. Commented Jan 11 at 11:03