content_image/402690.jpg content_image/402687.jpg content_image/402686.jpg content_image/402689.jpg content_image/402688.jpg content_image/402685.jpg
| r − r $^{′}$|  → 2 R ; this is the manifestation of the usual break- down of the quasiclassical approximation in the vicinity of the classical turning point. ( j ) ( j ) 

(iii) To fix the relative phases of G + and G − we choose a given circular trajectory and consider the evolu- tion of G ( r , r $^{′}$; ϵ ) as r moves along the circle. Namely, as r passes the turning point at | r − r $^{′}$| = 2 R , G ( j ) + ( r , r $^{′}$; ϵ ) should transform into G ( j ) − ( r , r $^{′}$; ϵ ). In the vicinity of the turning point, | r − r $^{′}$| = 2 R (1 + z ), z ≪ 1, we have ϑ$_{±}$ ≈ π ∓ √ − 2 z . Thus, ϑ$_{±}$ can be viewed as two branches of the same analytical function, and ϑ$_{+}$ → ϑ$_{−}$ when z makes a circle around z = 0 in the complex plane. Since ϑ$_{±}$ + sin ϑ$_{±}$ ≈ π ∓ ( − 2 z ) 3 / $^{2}$/ 6, in order to get a decaying exponential in the classically forbidden region | r − r $^{′}$| > 2 R (the positive semiaxis of z ), the circle must be counterclockwise. Thus, 

<img src='content_image/402690.jpg'>

precisely as in Eq. (A14). 

(iv) To match the phases of G ( j ) − and G ( j +1 + ), we con- sider r → r ′ and require that G ( j ) − ( r , r $^{′}$; ϵ )+ G ( j +1) + ( r , r $^{′}$; ϵ ) satisfies the homogeneous Dirac equation. Indeed, the δ -function term is produced by G (0) + component, see re- mark (i). At r → r ′ the Green’s function satisfying the homogeneous Dirac equation can be can be constructed in the B → 0 limit. Let us introduce ¯ G$_{0}$ ( r − r $^{′}$; ϵ ), the Green’s function with analytical properties opposite to those of G$_{0}$ : 

<img src='content_image/402687.jpg'>

It also satisfies Eq. (A12), as seen from the antisymme- try of the Dirac operator with respect to the simultaneous change ϵ → − ϵ , i → − i , γ$_{ϵ}$ → − γ$_{ϵ}$ . Thus, G$_{0}$ ( r − r $^{′}$; ϵ ) − ¯ G$_{0}$ ( r − r $^{′}$; ϵ ) represents the sought solution, corresponding to the flux of particles focusing at the point r $^{′}$. Hence, the correspondence is G ( j ) − ( r → r $^{′}$; ϵ ) ∼ − ¯ G$_{0}$ ( r − r $^{′}$; ϵ ) (converging wave), G ( j +1) + ( r r $^{′}$; ϵ ) ∼ G$_{0}$ ( r − r $^{′}$; ϵ ) (out- going wave), which produces precisely the combination of signs as in Eq. (A14). 

Remarks (iii) and (iv) can be restated in very simple terms. Upon every half-rotation sin ϑ changes sign, so the square root produces a factor of e iπ/ $^{2}$. In addition, upon a full rotation ψ$_{n}$ acquires the Berry phase e $^{iπ}$. Hence, the sum over j in Eqs. (A13), (A14) reduces to a simple geometric progression, 

<img src='content_image/402686.jpg'>

which determines the poles corresponding to πϵR/ ℏ v be- ing an integer multiple of 2 π , i. e., Bohr-Sommerfeld quantization rule. It gives the exact expression for the Landau levels, | ϵ$_{n}$/v | = √ 2 n | e ℏ B/c | . 

## 3. Two-phonon Raman matrix element 

Here we consider the matrix element M ( q ; ϕ$_{in}$, ϕ$_{out}$ ) for the transition from the initial state, corresponding to the incident photon with frequency ω$_{in}$ polarized at the angle ϕ$_{in}$ to the x axis, and no phonons, to the fi- nal state, corresponding to the scatterd photon with fre- quency ω$_{in}$ − 2 ω$_{q}$ polarized at the angle ϕ$_{out}$ to the x axis, and two phonons with momenta q , − q and frequencies ω$_{q}$ ( q is measured from the K or K ′ point, and for the phonons in the two valleys the frequencies ω ( K ) q = ω ( K $^{′}$) − q due to the time-reversal symmetry; we denote ω ( K ) q ≡ ω$_{q}$ and omit the valley index). Then, for unpolarized excita- tion and detection, the frequency-resolved intensity I (Ω) is given by 

<img src='content_image/402689.jpg'>

The matrix element is given by the loop of four Green’s function with electron-photon and electron-phonon ver- tices (here we write it in the coordinate representation) $^{29}$: 

<img src='content_image/402688.jpg'>

We will assume γ$_{ω}$$_{in}$$_{/}$$_{2}$R/v ≫ 1 (non-quantizing field), then the probability of a half- or full rotation is exponen- tially small, and only the G (0) + contribution to the Green’s functions remains. The integral over ϵ is dominated by small | ϵ | ∼ γ$_{ω}$$_{in}$$_{/}$$_{2}$ , so the energy arguments of the Green’s functions can be assumed to have definite signs. Thus, the imaginary part of the exponent can be written as 

<img src='content_image/402685.jpg'>

where the action S is that given by Eq. (A6), with the “+” subscript omitted and the energy argument explic- itly introduced. 

