# Solving scattering problem including spin flip

## Homework Statement

The Hamiltonian is given below:
$H=\frac {\mathbf p^2}{2m} -\frac {\partial^2_z}{2m} +V(z) +\gamma V'(z)(\hat{\mathbf z} \times \mathbf p)\cdot \vec{\sigma}$
Where this term $\gamma V'(z)(\hat{\mathbf z} \times \mathbf p)\cdot \vec{\sigma}$ represents the spin orbit coupling term.
$\vec P$ is the two dimensional momentum in the x-y plane.
$V(z)$ is the potential step.
$\sigma$ is a vector of pauli matrices.
$\gamma$ is a material dependent parameter which describes the strength of SOC(spin orbit coupling) at the interface.

because of the translational invariance in the x-y plane the eigenfunctions read:
$\psi_{\vec p,z}=e^{i(\vec p \cdot \vec{\rho})}\varphi_k(z)$
$\rho=(x,y)$ is the in plane coordinate and $\varphi_k(z)$ are spinor scattering states in z direction.
In the complete set of scattering states we distinguish two orthogonal subsets of eigenfunctions:
(i) the states $\vec{\varphi}$ incoming from the left.
(ii) the states $\varphi^{\leftarrow}$ incoming from the right.
Away from the interface the wave functions corresponding to the energy $\varepsilon=\frac{(p^2+k^2)}{2m}$ have the following form:

${\vec{\varphi_{k\sigma}}=\begin{cases} (e^{ikz} +\hat{r_k}e^{-ikz}){\chi}_{\sigma} & if z<0 \\ \hat{t_k}e^{ik'z}{\chi}_{\sigma} & if z>0 \\ \end{cases}}$

${\leftarrow\\{\varphi_{k\sigma}}=\begin{cases} (e^{ik'z} +\hat{r'_k}e^{ik'z}){\chi}_{\sigma} & if z>0 \\ \hat{t'_k}e^{-ikz}{\chi}_{\sigma} & if z<0 \\ \end{cases}}$

$\hat{r_k}$ and$\hat{r'_k}$ are the 2*2 matrix reflection coefficients.
$\hat{t_k}$and$\hat{t'_k}$ are the matrix reflection coefficients.
$\chi_{\sigma}$ with $\sigma=\pm$ are the basis vectors spanning the spinor subspace.
We choose $\chi_{\sigma}$ to be the eigenfunctions of $\sigma_{x}$.
The matrix scattering coefficients can be represented as follows:

$\hat{r_k}=r_{0}\sigma_0 +{\vec r} \cdot {\vec{\sigma}}$

$\hat{r'_k}=r'_{0}\sigma_0 +{\vec r'} \cdot {\vec{\sigma}}$

$\hat{t_k}=t_0\sigma_0 +{\vec t} \cdot {\vec{\sigma}}$

$\hat{t'_k}=t_0\sigma_0 +{\vec t'} \cdot {\vec{\sigma}}$

$\sigma_{0}$ is a 2*2 unit matrix $\vec r,\vec r',\vec t,\vec t'$ describe the spin dependent (spin flip) part of scattering at the interface.

The problem we want to solve is to have a "closed analytical expressions for the scattering coefficients".
We will choose the potential to be $V(z)=V\Theta(z)$ which implies $V'(z)=V\delta(z)$

We want to prove that by solving explicitly the scattering problem for the hamiltonian H with $V(z)=V\Theta(z)$ we find:

$t_0=\frac{2k}{k+k'}*\frac{1}{1+\frac{(\upsilon^2sp)^2}{(k+k')^2}}$
$t'_0=\frac{2k'}{k+k'}*\frac{1}{1+\frac{(\upsilon^2sp)^2}{(k+k')^2}}$
$r_0=-1+t_0=\frac{(k-k')-\frac{(\upsilon^2 sp)^2}{k+k'}}{k+k'}*\frac{1}{1+\frac{(\upsilon^2sp)^2}{(k+k')^2}}$
$r'_0=-1+t'_0=\frac{(k'-k)-\frac{(\upsilon^2 sp)^2}{k+k'}}{k+k'}*\frac{1}{1+\frac{(\upsilon^2sp)^2}{(k+k')^2}}$

$\vec t=\vec r=i(\hat z \times \vec p)\frac{s\upsilon^2}{k+k'}t_0$

$\vec t'=\vec r'=i(\hat z \times \vec p)\frac{s\upsilon^2}{k+k'}t'_0$

## Homework Equations

$H[\vec{\varphi_{k\sigma}}(z)]=(\frac {\mathbf p^2}{2m} -\frac {\partial^2_z}{2m} +V(z) +\gamma V'(z)(\hat{\mathbf z} \times \mathbf p)\cdot \vec{\sigma})[\vec{\varphi_{k\sigma}}(z)]=(\frac{\mathbf p^2}{2m}+\frac{\mathbf k^2}{2m})[\vec{\varphi_{k\sigma}}(z)]$
$V(z)=V\Theta(z)$ and$V(z)=V\delta(z)$
$\sigma_x= \begin{pmatrix} 0 &1\\ 1 &0 \end{pmatrix}$
$\sigma_y= \begin{pmatrix} 0 & -i\\ i &0 \end{pmatrix}$
$\sigma_z= \begin{pmatrix} 1 & 0\\ 0 &-1 \end{pmatrix}$

## The Attempt at a Solution

for z<0
[/B] $\vec{\varphi_{k+}(z)}=e^{ikz}$ $\chi_+$+ $[\begin{pmatrix} r_0 & 0 \\ 0 & r_0 \end{pmatrix}+r_x\begin{pmatrix} 0 & 1\\ 1& 0 \end{pmatrix}+r_y\begin{pmatrix} 0 & -i\\ i& 0 \end{pmatrix}]\chi_+e^{-ikz}$
$=\frac{1}{\sqrt 2} \begin{pmatrix} e^{ikz}\\ e^{ikz} \end{pmatrix} +[ \begin{pmatrix} r_0 & 0 \\ 0 & r_0 \end{pmatrix} + \begin{pmatrix} 0 & {r_x-ir_y}\\ {r_x+iry} & 0 \end{pmatrix} ]\begin{pmatrix} {\frac{1}{\sqrt2}} \\ {\frac{1}{\sqrt2}} \end{pmatrix}e^{-ikz}$

$=\frac{1}{\sqrt 2} \begin{pmatrix} {e^{ikz}+(r_0+r_x-ir_y)e^{-ikz} } \\ {e^{ikz}+(r_0+r_x+ir_y)e^{-ikz} }\end{pmatrix}$
for z>0
$\vec{\varphi_{k+}(z)}=$ $[\begin{pmatrix} t_0 & 0 \\ 0 & t_0 \end{pmatrix}+(t_x\vec i+ t_y\vec j)\cdot (\sigma_x\vec i +\sigma_y\vec y) ]e^{ik'z}\chi_{\sigma}$
$=\frac{1}{\sqrt 2}\begin{pmatrix} t_0+t_x-it_y\\ t_0+t_x+it_y\end{pmatrix}e^{ik'z}$
for z<0
$\vec{\varphi_k-(z)}(=\frac{1}{\sqrt 2} \begin{pmatrix} e^{ikz}\\ -e^{ikz} \end{pmatrix} +[ \begin{pmatrix} {r_0} & {r_x-ir_y} \\ {r_x+ir_y} & {r_0} \end{pmatrix}]\chi_- e^{-ikz}$
$=\frac{1}{\sqrt 2}\begin{pmatrix} {e^{ikz}+e^{-ikz}(r_0-(r_x-ir_y))}\\ {-e^{ikz}+e^{-ikz}(-r_0+(r_x+ir_y))} \end{pmatrix}$
for z>0
$\vec{\varphi_{k-}(z)}=\begin{pmatrix} t_0 &{t_x-ity}\\{t_x+t_y}&{t_0} \end{pmatrix}\begin{pmatrix} {\frac{1}{\sqrt 2}}\\{-\frac{1}{\sqrt 2}}\end{pmatrix}e^{ik'z}$
$=\frac{1}{\sqrt 2} \begin{pmatrix}{t_0-(t_x-it_y)}\\{-t_0+(t_x+it_y)}\end{pmatrix} e^{ik'z}$

As the wave function must be continuous, we can write:
$\lim_{z \rightarrow +0^-}\vec{\varphi_{k+}(z)}=\lim_{z \rightarrow +0^+}\vec{\varphi_{k+}(z)}=\vec{\varphi_{k\sigma}(0)}$

$\lim_{z \rightarrow 0^-}[e^{ikz}+(r_0+r_x-ir_y)e^{-ikz}]=\lim_{z \rightarrow 0^+}[(t_0+t_x-it_y)e^{ik'z}]=\vec{\varphi_{k+}(0)}$
$\rightarrow 1+r_o+r_x-ir_y=t_0+t_x-it_y=\vec{\varphi_{k+}(0)}$ ................................... (1)
$\lim_{z \rightarrow 0^-}[e^{ikz}+(r_0+r_x+ir_y)e^{-ikz}]=\lim_{z \rightarrow 0^+}[(t_0+t_x+it_y)e^{ik'z}]=\vec{\varphi_{k+}(0)}$
$\rightarrow 1+r_0+r_x+ir_y=t_0+t_x+it_y=\varphi_{k+}(0)$ ................................................ (2)
now for$\chi_-$
$\lim_{z \rightarrow 0^-}[e^{ikz}+(r_0-(r_x-ir_y))e^{-ikz}]=\lim_{z \rightarrow 0^+}[(t_0+t_x-it_y)e^{ik'z}]=\vec{\varphi_{k-}(0)}$
$\rightarrow1+r_0-r_x+ir_y=t_0-t_x+it_y=\varphi_{k-}(0)$ ..................................................... (3)

$\lim_{z \rightarrow 0^-}[-e^{ikz}+(-r_0+(r_x-ir_y))e^{-ikz}]=\lim_{z \rightarrow 0^+}[(-t_0+t_x-it_y)e^{ik'z}]=\vec{\varphi_{k-}(0)}$
$\rightarrow -1-r_0+r_x+ir_y=-t_0+t_x+it_y$ ............................................................................ (4)

So we from (1),(2),(3) and (4) we can deduce that :
$r_x=t_x\\ r_y=t_y\\ 1+r_0=t_0$
$\vec r=r_x\vec i+ r_y\vec j=t_x\vec i +t_y\vec j=\vec t$

So we can write now $\vec{\varphi_{k+}}(0)= \begin{pmatrix}{ t_0+t_x-it_y}\\{t_0+t_x+it_y} \end{pmatrix}$
$\vec{\varphi_{k-}}(0)= \begin{pmatrix}{ t_0-t_x+it_y}\\{-t_0+t_x+it_y} \end{pmatrix}$

$\hat{z} \times \vec p=p_x\vec j-p_y\vec i$

$(p_x\vec j -p_y\vec i) \cdot(\sigma_x\vec i+\sigma_y\vec j)=p_x\sigma_y-p_y\sigma_x=\begin{pmatrix} 0 & -{p_xi}\\{p_xi} & 0 \end{pmatrix}-\begin{pmatrix} 0 & {p_y}\\{p_y}& 0 \end{pmatrix}=\begin{pmatrix} 0 & -{p_y+p_xi}\\{-p_y+p_xi}& 0 \end{pmatrix}$

I went through this approach to solve out the scattering coefficients:

$(\frac {-\nabla^2}{2m} -\frac {\partial^2_z}{2m})\int_{-\varepsilon}^{\varepsilon}\vec{\varphi_{k\sigma}(z)} \, dz +\int_{-\varepsilon}^{\varepsilon}V\Theta(z)\vec{\varphi_{k\sigma}(z)} \, dz +\gamma \int_{-\varepsilon}^{\varepsilon}V\delta(z)(\hat{\mathbf z} \times \mathbf p)\cdot \vec{\sigma}\vec{\varphi_{k\sigma}(z)} \, dz =(\frac{p^2}{2m}+\frac{k^2}{2m})\int_{-\varepsilon}^{\varepsilon} \vec{\varphi_{k\sigma}(z)} \, dz$
As the wave functions are continuous we can say that:
$\int_{-\varepsilon}^{\varepsilon}V\Theta(z) \vec{\varphi_{k\sigma}(z)} \, dz=0$
and $(\frac{p^2}{2m}+\frac{k^2}{2m})\int_{-\varepsilon}^{\varepsilon} \vec{\varphi_{k\sigma}(z)} \, dz=0$
it follows that $-[\frac{\partial}{\partial z}_{z>0}\vec{\varphi_{k\sigma}}(0)-\frac{\partial}{\partial z}_{z<0}\vec{\varphi_{k\sigma}}(0)]=$ $-\gamma V(\hat z \times \vec p)\cdot \vec{\sigma}\vec{\varphi_{k\sigma}(0)}$

By solving the last equation for both cases:$\chi_{+}$ and$\chi_{-}$

I got these relations:
for $\chi_+$
$2m\gamma Vp_yt_y+2m\gamma V P_x(t_0+t_x)=-(t_0+t_x)(k'+k)+2k$
$t_y(k'+k)=2m\gamma V[-p_y(t_0+t_x)+p_xt_y]$

$(t_0+t_x)(k'+k)-2k=2m\gamma V[p_yt_y+p_x(t_0+t_x)]$
$-t_y(k'+k)=2m\gamma V[-p_y(t_0+t_x)+p_xty]$

for$\chi_-$

$2m\gamma V[p_yt_y+p_x(-t_0+t_x)]=-[(t_0-t_x)(k'+k)-2k]$
$t_y(k'+k)=2m\gamma V[p_y(-t_0+t_x)-p_xt_y]$

$(-t_0+tx)(k'+k)=2m\gamma V[p_yt_y+p_x(t_0+t_x)]$
$-p_y2m\gamma V(t_0 +t_x)+2m\gamma Vp_xt_y=-ty(k'+k)$

The problem is that the equations I get make really contradictory results, which don't lead to the scattering coefficients that I must derive.
Is my approach wrong?
what went wrong?

Related Advanced Physics Homework Help News on Phys.org
Hi
In fact I'm working on a paper.In this paper Im trying to derive the scattering coefficients of a scattering problem.
So I will upload the paper, I think the statement of my problem will be more clear if you check section 2(the model).