he only way two bands can cross without opening a ...

Think of the two bands as two “songs” playing in the same room. A gap opens only if the two songs can mix. Symmetry tells us whether mixing is allowed. If the two wavefunctions transform differently under the crystal symmetries, the Hamiltonian has no allowed matrix element connecting them, so they pass through each other instead of repelling.

1. What “D4h point group” means

The point group $D_{4h}$ is a tetragonal symmetry group. It contains operations such as fourfold rotation $C_4$, twofold rotations, mirror operations, and inversion $i$. In a crystal with this symmetry, the electronic Hamiltonian must be unchanged when these symmetry operations are applied. Standard $D_{4h}$ character tables classify states into symmetry types such as $A_{1g}, A_{2g}, B_{1g}, B_{2g}, E_g$, and their odd-parity $u$ partners. The subscripts $g/u$ refer to even/odd behavior under inversion. (Newcastle University Staff)

For band theory, the important object is often not the full crystal point group, but the little group of the particular momentum $\mathbf{k}$: the subgroup of symmetries that leaves that $\mathbf{k}$-point or $\mathbf{k}$-line unchanged. IRVSP assigns irreducible representations of those little groups to each band.

2. What an irreducible representation is

An irreducible representation, or irrep, is the smallest symmetry “behavior type” a wavefunction can have. When a symmetry operation acts on a wavefunction, the wavefunction may stay the same, change sign, rotate into a partner state, or pick up a complex phase. The collection of those transformation rules is the representation.

For example, under a mirror symmetry, one state might be mirror-even:

$$M\psi_1 = +\psi_1$$

while another might be mirror-odd:

$$M\psi_2 = -\psi_2$$

Those two states belong to different symmetry sectors. They are orthogonal in the group-theory sense. More generally, characters of different irreps are orthogonal, and Schur’s lemma says that a symmetry-respecting operator has zero matrix element between inequivalent irreps. (Xie Chen)

That is the mathematical version of “the bands are invisible to each other.”

3. Why two bands normally avoid crossing

Near a possible crossing, you can model two bands by a $2 \times 2$ Hamiltonian:

$$H(k)= \begin{pmatrix} E_1(k) & V(k) \\ V^*(k) & E_2(k) \end{pmatrix}.$$

Here $E_1(k)$ and $E_2(k)$ are the uncoupled band energies, and $V(k)$ is the coupling or hybridization between the two wavefunctions.

If $V(k)\neq 0$, the two eigenvalues become

$$E_\pm(k)=\frac{E_1+E_2}{2} \pm \sqrt{\left(\frac{E_1-E_2}{2}\right)^2+|V|^2}.$$

At the point where $E_1=E_2$, the splitting is $2|V|$. That splitting is the gap. This is called an avoided crossing.

So the key question is: does symmetry force $V=0$?

If yes, the Hamiltonian becomes diagonal in the two-band subspace:

$$H(k)= \begin{pmatrix} E_1(k) & 0 \\ 0 & E_2(k) \end{pmatrix}.$$

Then when $E_1(k)=E_2(k)$, the bands really cross.

4. Why different irreps force $V=0$

Suppose the two states are $\psi_a$ and $\psi_b$. The coupling is

$$V=\langle \psi_a | H | \psi_b\rangle.$$

Because the Hamiltonian respects the crystal symmetry, applying any symmetry operation $G$ must not change the value of this matrix element.

Now imagine a simple case where the two states have different mirror eigenvalues:

$$M\psi_a=+\psi_a, \qquad M\psi_b=-\psi_b.$$

Then

$$\langle \psi_a|H|\psi_b\rangle = \langle M\psi_a|H|M\psi_b\rangle = (+1)(-1)\langle\psi_a|H|\psi_b\rangle = -\langle\psi_a|H|\psi_b\rangle.$$

The only number equal to its own negative is zero:

$$V=0.$$

That is the selection rule. The two states cannot hybridize because doing so would violate the symmetry.

The same logic applies to $C_4$ rotation eigenvalues, inversion parity, mirror eigenvalues, or double-group spinor eigenvalues. If two bands have incompatible symmetry labels, the off-diagonal coupling is forbidden.

5. What DT4 and DT5 mean

The labels DT4 and DT5 are IRVSP/Bilbao-style labels for irreducible representations. The exact meaning depends on the space group, the k-point or k-line, and whether spin-orbit coupling is included. IRVSP is specifically designed to compute irreducible representations of electronic states from VASP wavefunctions and symmetry operations, and it supports spin-orbit-coupled calculations using double groups. (arXiv)

So when the paper says one band is DT4 and another is DT5, it is not just naming them. It is saying:

Band 1 and Band 2 transform differently under the relevant symmetry operations.

Therefore, they belong to different symmetry blocks of the Hamiltonian. A symmetric Hamiltonian cannot contain a term that mixes those two blocks. In matrix language:

$$\langle DT4|H|DT5\rangle = 0$$

as long as the protecting symmetry remains present.

6. Why this protects the crossing

If the DT4 band and DT5 band approach each other in energy, there are two possibilities.

If they belonged to the same irrep, the Hamiltonian would usually allow a coupling:

$$\langle DT4|H|DT4\rangle \neq 0,$$

and an avoided crossing would appear.

But because they are different irreps,

$$DT4 \neq DT5,$$

the hybridization matrix element is forbidden. So the bands can cross cleanly. That crossing is called symmetry-protected.

This is what the paper means by saying the two bands are “invisible” to each other. They are not literally noninteracting. They exist in the same crystal and feel the same periodic potential. But symmetry prevents the particular mixing term that would open a gap.

7. Why the crossing can still disappear

The protection is conditional. It survives only while the relevant symmetry survives.

A gap can open if:

1. the crystal is distorted so the protecting symmetry is broken;
2. strain lowers the point group;
3. magnetic order breaks time-reversal symmetry;
4. spin-orbit coupling changes the relevant irreps, depending on the case;
5. disorder or surface effects break the symmetry locally;
6. the crossing moves away from the high-symmetry line or plane where the little-group classification applies.

This is why papers often say “protected along the $\Gamma$-Z line” or “protected by $C_{4v}$, $C_{3v}$, mirror, inversion, etc.” The protection is not everywhere in the Brillouin zone. It is tied to a symmetry-preserving path or plane.

8. Why this can produce a Dirac point

If the system has spin-orbit coupling, time-reversal symmetry, and inversion symmetry, bands are often Kramers-degenerate. Each “band” may actually be a twofold-degenerate pair. If two such doublets cross, the crossing is fourfold-degenerate. That is a Dirac point.

So the logic is:

$$\text{two different irreps} \Rightarrow \text{no hybridization} \Rightarrow \text{protected crossing} \Rightarrow \text{Dirac point if the crossing is fourfold-degenerate}.$$

In a related IRVSP-analyzed example, the authors explicitly report a crossing between bands with different irreducible representations and state that this symmetry difference protects the crossing as a Dirac point. (APS Harvest)

9. The intuitive picture

A band crossing is like two roads crossing at an intersection.

If the two roads are allowed to connect, cars can turn from one road to the other. In band language, the wavefunctions hybridize, and a gap opens.

If symmetry puts a wall between the roads, cars cannot turn. The roads pass through each other. In band language, the two wavefunctions belong to different irreps, the coupling matrix element is zero, and the crossing remains.

So the statement

“one band is DT4 and the other is DT5”

really means:

the two bands live in different symmetry sectors of the crystal Hamiltonian; because the Hamiltonian preserves the symmetry, it cannot mix those sectors, so the crossing is protected and no gap opens.