A. the Dirac Hamiltonian in 2D Starting from the Time Dependent Dirac Equation: 휕 퐻 = 푖ℏ 훽 − 푐(푝⃗

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

Section 1: A. The Dirac Hamiltonian in 2D Starting from the time dependent Dirac equation: 휕 퐻 = 푖ℏ 훽 − 푐(푝⃗. 훾⃗) + 푚푐2퐼 푡 휕푡 4

Here m is the ‘rest-mass’, c the velocity, 푝⃗ = 푝푥푥̂ + 푝푦푦̂ is the momentum in 2D and 훾⃗ = 훾푥푥̂ + 퐼 0 훾 푦̂ + 훾 푧̂ are the Dirac gamma (γ) matrices. 훽 = ( ) where I is the 2 identity matrix. And 퐼 = 푦 푧 0 −퐼 4 퐼 0 ( ) 0 퐼 Keeping in mind the γ-matrices in the Dirac representation:

0 휎푥,푦,푧 훾푥,푦,푧 = ( ) −휎푥,푦,푧 0 0 1 0 −푖 1 0 With, 휎 = ( ), 휎 = ( ), 휎 = ( ) 푥 1 0 푦 푖 0 푧 0 −1 −푖휔푡 We will solve 퐻푡Ψ = 0 for Ψ = ϕe with ℏ휔 = 퐸 We get:

2 (퐸 + 푚푐 )퐼 −푐푝⃗. 휎⃗ 휙1 ( 2 ) ( ) = 0 푐푝⃗. 휎⃗ (−퐸 + 푚푐 )퐼 휙2

2 (퐸 + 푚푐 )퐼휙1 − 푐푝⃗. 휎⃗휙2 = 0 Or equivalently, { 2 (−퐸 + 푚푐 )퐼휙2 + 푐푝⃗. 휎⃗휙1 = 0 This can be rearranged into:

2 (−퐸 − 푚푐 )퐼휙1 + 푐푝⃗. 휎⃗휙2 = 0 { 2 (−퐸 + 푚푐 )퐼휙2 + 푐푝⃗. 휎⃗휙1 = 0 To finally get the equation in a slightly different form and a different basis:

2 (−퐸 + 푚푐 )퐼 푐푝⃗. 휎⃗ 휙2 ( 2 ) ( ) = 0 푐푝⃗. 휎⃗ (−퐸 − 푚푐 )퐼 휙1 From this we can define the time-independent Dirac Hamiltonian to be:

퐻 = 푐(푝⃗. 훼⃗) + 푚푐2훽 With the Dirac α matrices given by:

0 휎푥,푦,푧 훼푥,푦,푧 = ( ) 휎푥,푦,푧 0 Since we are interested in the problem in 2D, we can drop out z terms to get the following matrix Hamiltonian:

푚푐2 0 0 푐(푝푥 − 푖푝푦) 2 0 푚푐 푐(푝푥 + 푖푝푦) 0 퐻 = 0 푐(푝푥 − 푖푝푦) −푚푐2 0 푐(푝 + 푖푝 ) 0 2 ( 푥 푦 0 −푚푐 )

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

For a particle with spin-1/2, the wavefunction decomposed into a position dependent part and a spinor, takes the form:

푎1 푖 푎 푝⃗.푟⃗ 휓(푟⃗) = ( 2) 푒ℏ 푎3 푎4 So that we have the following eigenvalue problem that results from the Dirac equation 퐻휓= 퐸휓 can be written as:

2 0 푐(푝 − 푖푝 ) 푚푐 − 퐸 0 푥 푦 푎1 2 0 푚푐 − 퐸 푐(푝푥 + 푖푝푦) 0 푎2 (푎 ) = 0 0 푐(푝푥 − 푖푝푦) −푚푐2 − 퐸 0 3 푐(푝 + 푖푝 ) 0 2 푎4 ( 푥 푦 0 −푚푐 − 퐸) Eigenvalue It is easy to see that the eigenvalue problem decouples into two independent determinant equations so that we have:

2 푚푐 − 퐸 푐(푝푥 − 푖푝푦) | 2 | = 0 푐(푝푥 + 푖푝푦) −푚푐 − 퐸 Or

± 2 4 2 2 2 2 4 2 2 퐸 = ±√푚 푐 + 푐 (푝푥 + 푝푦) = ±√푚 푐 + 푐 푝

In terms of the wavevector k:

± 2 4 2 2 2 퐸 = ±√푚 푐 + (ℏ푐) (푘푥 + 푘푦)

2 Let us define Δ ≡ 푚푐 and try to plot E(px,py) for different characteristic values of Δ while keeping c constant. For simplicity let us use the following parameters: Δ = 0meV, 10meV and 100meV for Fig. 1(a-c) respectively, and ℏ푐 = 6푒푉Å.

FIG 1. (a-c) Three Dirac fermion energy-momentum spectra with Δ = 0meV, 10meV and 100meV respectively and ℏ푐 = 6푒푉Å.

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

The solid-state physicist’s take on this result: While in high-energy physics the Dirac equation first described the behavior of free electrons and their antiparticle the positron that has an opposite energy, in a solid, the Dirac equation can describe a semiconductor with an energy gap equal to 2Δ and a band structure that is non-parabolic away from the band gap:

Since 퐸(푝) = ±√푚2푐4 + 푐2푝2

퐸±(0) = ±푚2푐4

Therefore, the energy gap, 퐸+(0) − 퐸−(0) = 2푚2푐4 = 2Δ Limiting case 1:

Also notice the limiting case, for 푚2푐4 ≪ 푐2푝2, 퐸 → 푐푝. The bands are linear at large p. Limiting case 2:

푝2 푝2 When 푚2푐4 ≫ 푐2푝2, 퐸(푝) = ±푚2푐4√1 + = ±푚2푐4 (1 + ). The bands are parabolic at 푚2푐2 2푚2푐2 small p. In assignment 1, you will also show that the ‘effective mass’ of Dirac particles is energy dependent. Limiting case 3: Using ℏ푐푘 = 푐푝

푑퐸 ℏ푐2푘 푣푓 = = ℏ푑푘 √푚2푐4 + (ℏ푐푘)2

When 푘 → ∞

ℏ푐2 푣푓 = → 푐 푚2푐4 √ + (ℏ푐)2 푘2

Eigenvector Let us now proceed and extract the eigenvectors resulting from the Dirac eigenvalue problem.

2 ± 0 푐(푝 − 푖푝 ) 푚푐 − 퐸 0 푥 푦 푎1 2 ± 0 푚푐 − 퐸 푐(푝푥 + 푖푝푦) 0 푎2 (푎 ) = 0 0 푐(푝푥 − 푖푝푦) −푚푐2 − 퐸± 0 3 푐(푝 + 푖푝 ) 0 2 ± 푎4 ( 푥 푦 0 −푚푐 − 퐸 ) The problem decoupled into two independent 2x2 eigenvector equations so that for each energy we get two independent 2x1 eigen-spinors:

2 ± (푚푐 − 퐸 )푎1 + 푐(푝푥 − 푖푝푦)푎4 = 0

2 ± (푚푐 − 퐸 )푎2 + 푐(푝푥 + 푖푝푦)푎3 = 0

2 ± 푐(푝푥 − 푖푝푦)푎2 − (푚푐 + 퐸 )푎3 = 0

2 ± 푐(푝푥 + 푖푝푦)푎1 − (푚푐 + 퐸 )푎4 = 0

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

Or equivalently:

푐(푝푥 + 푖푝푦) 푐(푝푥 − 푖푝푦) 푎 = 푎 푎푛푑 푎 = 푎 4 (푚푐2 + 퐸±) 1 3 (푚푐2 − 퐸±) 2 So that we have two possible eigenvector solutions given by:

1 0 0 1 휙1 = 0 and 휙2 = 푐(푝푥−푖푝푦) 푐(푝푥+푖푝푦) (푚푐2−퐸±) 2 ± ((푚푐 +퐸 )) ( 0 ) Normalized;

1 0 (푚푐2+퐸±)2 0 (푚푐2+퐸±)2 1 휙 = √ and 휙 = √ 푐(푝푥−푖푝푦) 1 (푚푐2+퐸±)2+(푐푝)2 0 2 (푚푐2−퐸±)2+(푐푝)2 푐(푝푥+푖푝푦) (푚푐2−퐸±) 2 ± ((푚푐 +퐸 )) ( 0 ) We have 2 solutions for each energy eigenvalue. Section 1: B. Weyl fermions Going back to Klein-Gordon Hamiltonian (in 3D) we have:

ℏ2휕2 [− + ℏ2∇2] Ψ = 푚푐2Ψ 푐2휕푡2 In a seminal paper in 1929 Hermann Weyl noticed a special solution to the Dirac equation when m=0. In this case, the Hamiltonian above reduced to:

ℏ2휕2 [− + ℏ2∇2] Ψ = 0 푐2휕푡2

So that if Ψ = 휓e−푖휔푡, we have,

[퐸2 + 푐2ℏ2∇2]Ψ = 0

2 Keeping in mind the properties of the Pauli matrices, in particular that σi =I, we can write this as:

퐸Ψ = ±푖푐ℏ∇⃗⃗⃗. 휎⃗ Or, Using the Weyl representation for the 훾 matrices, one can decouple the system to get two that would have the form.

±푐(푝⃗. 휎⃗)휓 = 퐸휓 We’re going to call these two equations the Weyl equations.

Let’s work out the Weyl problem in the 3D case. Starting from the 3D Weyl equation (퐻푊), and assuming the wavefunction is a planewave that multiplies a spinor we get: − 퐸 − 푐푝푧 −푐푝 푎1 ( + ) ( ) = 0 −푐푝 퐸 + 푐푝푧 푎2

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

− 퐸 + 푐푝푧 푐푝 푎3 ( + ) ( ) = 0 푐푝 퐸 − 푐푝푧 푎4 A small parenthesis: if we were two consolidate these two equations in a single 4x4 system we get the Weyl system: − 0 0 퐸 − 푐푝푧 −푐푝 푎1 + 0 0 −푐푝 퐸 + 푐푝푧 푎2 ( − ) ( ) = 0 퐸 + 푐푝푧 푐푝 0 0 푎3 + 푎 푐푝 퐸 − 푐푝푧 0 0 4 The problem remains decoupled of course, but we get a sense of why Weyl introduced a different basis with different α matrices (the Weyl representation), in which: 0 0 1 0 훽 = (0 0 0 1) 1 0 0 0 0 1 0 0 If we start from the conventional Dirac equation 퐻 = 푐(푝⃗. 훼⃗) + 푚푐2훽

Using the 훽 matrix given above, we recover the 4x4 Weyl system. Let us know go back to the 2x2 system and solve it: − 퐸 − 푐푝푧 −푐푝 푎1 ( + ) ( ) = 0 −푐푝 퐸 + 푐푝푧 푎2 − 퐸 + 푐푝푧 푐푝 푎3 ( + ) ( ) = 0 푐푝 퐸 − 푐푝푧 푎4 Eigenvalue The eigenvalue is simple to obtain in both cases:

2 2 2 2 So that, 퐸 − 푐 (푝푥 + 푖푝푦)(푝푥 − 푖푝푦) − 푐 푝푧 = 0 Or,

± 2 2 2 퐸 = ±푐√푝푥 + 푝푦 + 푝푧 = ±푐푝

We get a 3D massless Dirac cone with a linear dispersion in all momentum directions Eigenvector For the first system of equations:

퐸± − 푐푝 −푐푝− 푎 푧 1 ( + ± ) (푎 ) = 0 −푐푝 퐸 + 푐푝푧 2 We get:

± − (퐸 − 푐푝푧)푎1 − 푐푝 푎2 = 0 푝− ± 휓1 = (±푝 − 푝푧) 1

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

For the second system we get: − 퐸 + 푐푝푧 푐푝 푎3 ( + ) ( ) = 0 푐푝 퐸 − 푐푝푧 푎4 ± − (퐸 + 푐푝푧)푎3 + 푐푝 푎4 = 0 푝− ± − 휓2 = ( ±푝 + 푝푧) 1 It is convenient to work in 2D in order to get a sense of the implications of the Weyl treatment. Let us then set pz=0 without any loss of generality. We get the following normalized solutions: 푝− ± 1 ± 휓 = ( 푝 ) 1 2 √ 1 푝− ± 1 ∓ 휓 = ( 푝 ) 2 2 √ 1

Spin expectation value and helicity The handedness of the solutions is set by their helicity. The helicity can be computed by looking at the ⃗⃗ Σ. 푝⃗⁄ ⃗⃗ ⃗⃗ expectation value of the operator |푝| with Σ given by: Σ = 휎푥푥̂ + 휎푦푦̂. Before we examine helicity and its meaning, let us simply compute the spin (x, y) expectation values for each one of the E+ branch For the (1) eigenvector: 푝− 1 푝+ 0 1 〈휎 〉 = 〈휓+휎 휓+〉 = ( 1) ( ) ( 푝 ) 푥 1 푥 1 2 푝 1 0 1 풑+ + 풑− 풑 〈흈 〉 = = 풙 풙 ퟐ풑 풑

푝− 1 푝+ 0 −푖 〈휎 〉 = 〈휓+휎 휓+〉 = ( 1) ( ) ( 푝 ) 푦 1 푦 1 2 푝 푖 0 1 + − −풊풑 + 풊풑 풑풚 〈흈 〉 = = 풚 ퟐ풑 풑

For the (2) eigenvector: 푝− 1 푝+ 0 1 − 〈휎 〉 = 〈휓+휎 휓+〉 = (− 1) ( ) ( 푝 ) 푥 2 푥 2 2 푝 1 0 1

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf

풑+ + 풑− 풑 〈흈 〉 = − = − 풙 풙 ퟐ풑 풑

푝+ 푝− 1 − 0 −푖 − 〈휎 〉 = 〈휓+휎 휓+〉 = ( ) ( ) ( 푝 ) 푦 2 푦 2 2 푝 푖 0 1 1

− + −풊풑 + 풊풑 풑풚 〈흈 〉 = = − 풚 ퟐ풑 풑

We can now take a look at the helicity of the spinor in each respective case. The helicity is the sum of the spin expectation values projected onto the momenta directions. It is thus a measure of the spin ‘handed-ness’ with respect to the momentum.

Σ⃗⃗. 푝⃗ 휎푥푝푥 + 휎푦푝푦 〈 〉 = 〈 〉 |푝| |푝|

+ For 휓1 : Σ⃗⃗. 푝⃗ 푝2 푝2 〈 〉 = 푥 + 푥 = +1 |푝| |푝|2 |푝|2

+ For 휓2 :

2 2 Σ⃗⃗. 푝⃗ 푝 푝푦 〈 〉 = − 푥 − = −1 |푝| |푝|2 |푝|2

+ + Thus, helicity defines the ‘winding’ direction of the spin expectation value. It is clear the 휓1 and 휓2 + wind in opposite directions, hence the term right-handed (R) solution to refer to 휓1 and left-handed (L) + solution to refer to 휓2 . This is illustrated in FIG. 3

FIG 3. Fermi surface contours showing spin angular momentum direction for the L- and R- handed solutions. Degeneracy and absence of spin-momentum locking Note that the solution ends up being a superposition of two massless Dirac cones (or Weyl cones) of opposite helicity, meaning the overall helicity of the fermions is zero. A recent major accomplishment in condensed-matter physics demonstrated that in certain materials having inversion asymmetry the two Weyl pairs can be separated in k-space yielding real helical Weyl fermions at different k-points. One

PHYS 90507. Topology and Dirac fermions in condensed matter Assaf can also think of the surface state of topological insulators as 2D Weyl fermions that are separated in real space.