PT-symmetry leading to spin-degenerate band structures

Question

I am trying to understand the fundamental physics of a kind of problem. I will introduce a specific example and then list some of the broader qualms I hold. I have observed many band structures, be it spin-polarized or spin-degenerate. Let's take the case of CrI$_3$, a well-studied 2D van der Waals semiconducting magnet. The monolayer band structure is what one would expect - it has spin-polarized carriers at the band edges, i.e. the majority and minority carriers occupy different energy eigenstates. Moving onto the bilayer, however, the band structure is spin-degenerate. Many papers like this one [1] report PT-symmetry (Parity-Time) as the cause for the degeneracy: basically, the Hamiltonian commutes with the product of P$\cdot$T. My main questions, which is not just specific to CrI$_3$, are as follows:

1. Is there a rigorous way when one looks at new material, to say if PT-symmetry would lead to a spin degeneracy? Because from what I've read, and I could totally be wrong, but it looks like people do band structure calculations and then attributes such a degeneracy to PT-symmetry.

2. I do not hold a good theoretical physics footing, so I don't know much about PT symmetry with regard to specific systems, but I do know about the universal CPT symmetry that is preserved. Is one just a subset of the other?

3. Finally, I am having a hard time trying to understand how the system can break inversion symmetry and time-reversal symmetry but the product of these two symmetry operators is still a 'good' quantum number i.e. one that commutes with the Hamiltonian.

Monolayer CrI$_3$ with spin-polarized bandstructure:

Bilayer CrI$_3$, where spin carriers are degenerate:

PS: I am well aware that the answer might involve a thorough level of detail, so I'm very open to awarding bounty points for detailed answers.

Answer 1

Basically, your question is related to the relationship between symmetries and energy bands. In the following, I will give a summary you can refer to and give a course link to prove them.

• Time-reverasl symmetry ($\mathcal{T}$) implies:

$$E_{n,-\chi}(-\vec{k})=E_{n,\chi}(\vec{k}) \tag{1} $$

• Inversion or parity symmetry ($\mathcal{P}$) implies:

$$E_{n,\chi}(-\vec{k})=E_{n,\chi}(\vec{k}) \tag{2} $$

• Then in crystals that have inversion and time-reversal symmetry implies:

$$E_{n,-\chi}(\vec{k})=E_{n,\chi}(\vec{k}) \tag{3} $$

• Here $n$ is the energy quantum number, $\vec{k}$ the crystal momentum, and $\chi$ the spin quantum number. This expression $(3)$ should answer your question "$\mathcal{P}\mathcal{T}$-symmetry leading to spin-degenerate band structures"

• You may want to know how to derive these conclusions, please take a look at these slides. (All these conclusions are borrowed from that, in which all the derivations are very pedagogical if you have basic training in solid-state physics.)

• The paper you cited discussed the change of energy band when considering different symmetry breaking:

  • (a): $\mathcal{P}$ ($\checkmark$) and $\mathcal{T}$ ($\checkmark$)
  • (b): $\mathcal{P}$ ($\times$) and $\mathcal{T}$ ($\checkmark$)
  • (c): $\mathcal{P}$ ($\times$) and $\mathcal{T}$ ($\times$)

  and the material realization for case (c) with bilayer CrI$_3$. And the symmetry analysis of bilayer CrI$_3$ is also stated in the main manuscript (section: Symmetry of 2D magnetic insulator CrI$_3$):

The atomic crystal of CrI$_3$ exhibits inversion symmetry ($\mathcal{P}$) with two inversion centers, one inside the monolayer and the other in between neighboring monolayers. The AFM order reduces the crystal symmetry by removing the interlayer inversion center. Therefore, an AFM bilayer breaks the inversion symmetry, although an FM bilayer does not. Another important feature is that the AFM ordered phase preserves $\mathcal{P}\mathcal{T}$ symmetry while it breaks $\mathcal{P}$ and $\mathcal{T}$ independently; hence, no polarization exists in the ordered phase.

• Now I believe you are comfortable about what the author is talking about with (1), (2), and (3).

Disclaimer: The previous answer can not be accepted. Because it doesn't solve the main point: $\mathcal{P}$ and $\mathcal{T}$ are broken but the combined $\mathcal{PT}$ is conserved.

I have sent an email to the corresponding author of CrI3 and state the problem with him. I will list the main points below although I can't prove them.

#Second email:

1. TRS induces Kramers degeneracy, at k that respects TRS (like the Gamma point). This is because T is an anti-unitary operator. In other words, the Kramers theorem applies for any anti-unitary operator.
2. The combined symmetry PT is also anti-unitary, because P is unitary and T is anti-U. Therefore, PT leads to Kramers degeneracy, at k that respect PT. As you know, any k-points respect PT.

#First email:

I assume that you understand the Kramers degeneracy induced by T. Because PT is antisymmetric, PT leads to the same Kramers degeneracy at generic k point. Therefore, everywhere is doubly degenerate in the band structure.

This is irrelevant to whether P/T is the symmetry of system.

May it helps.