# Solving scattering problem including spin flip

In summary, the conversation discusses the Hamiltonian and eigenfunctions for a spin orbit coupling problem. The Hamiltonian includes terms for momentum, potential, and spin orbit coupling. Two orthogonal subsets of eigenfunctions are distinguished for states incoming from the left and right. The problem aims to find closed analytical expressions for the scattering coefficients, which can be represented as matrices. The potential is chosen to be a step function, and the wave functions must be continuous at the interface. The solution involves finding relationships between the reflection and transmission coefficients, and the spin dependent part of the scattering is described by the Pauli matrices.

## 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?

Hi
In fact I'm working on a paper.In this paper I am 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).

#### Attachments

• 1706.04797 (1).pdf
2.6 MB · Views: 291

## 1. What is the scattering problem and why is it important in physics?

The scattering problem refers to the study of how particles interact and change direction when they encounter each other. It is important in physics because it allows us to understand and predict the behavior of particles, which is crucial in fields such as nuclear and particle physics.

## 2. What is spin flip in the context of scattering?

Spin flip refers to the phenomenon where the spin of a particle changes during a scattering event. This can occur when the particle interacts with another particle or with an external force, causing its spin to change direction.

## 3. How is spin flip incorporated into the solution of the scattering problem?

In order to solve the scattering problem including spin flip, the equations of motion for the particles must be modified to take into account the changing spin orientations. This can be done through the use of quantum mechanical calculations and techniques such as spinor formalism.

## 4. What are the factors that can influence spin flip in a scattering event?

There are several factors that can influence spin flip in a scattering event, including the type of particles involved, the energy of the particles, and the strength of the interaction between them. In addition, external forces or magnetic fields can also play a role in inducing spin flip.

## 5. Are there any real-world applications for solving scattering problems with spin flip?

Yes, there are several real-world applications for understanding and solving scattering problems with spin flip. These include the development of new materials for use in electronic devices, the study of magnetic and superconducting materials, and the design of more efficient particle accelerators and nuclear reactors.

Replies
9
Views
1K
Replies
1
Views
1K
Replies
58
Views
5K
Replies
7
Views
1K
Replies
1
Views
839
Replies
14
Views
1K
Replies
4
Views
2K
Replies
3
Views
1K