User:Dibyendum/sandbox

Definition
Consider a localized one dimensional potential barrier $$V(x)$$, subjected to a beam of quantum particles with energy $$E$$ .They incident on the potential barrier from left and right.The solution of Schrödinger's equation outside the potential barrier are plane waves and are given by:


 * $$\psi_L(x)= A e^{ikx} + B e^{-ikx}$$

for the region left to the potential barrier and


 * $$\psi_R(x)= C e^{ikx} + D e^{-ikx}$$

for the region right to the potential barrier, where $$k=\sqrt{2m E/\hbar^{2}}$$ is the wave vector while the terms with coefficients $$A$$ and $$D$$ represent the incoming waves whereas with the terms with coefficients $$B$$ and $$C$$ represent the outgoing waves.The outgoing waves are connected to the incoming waves by a linear combination described by the S-matrix
 * $$\begin{pmatrix}B \\ C \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix}\begin{pmatrix} A \\ D \end{pmatrix}\,$$.

Elements of this matrix completely characterize the scattering properties of the $$V(x)$$.
 * Let,
 * $$\Psi_{out}=\begin{pmatrix}B \\ C \end{pmatrix}$$,$$\Psi_{in}=\begin{pmatrix}A \\ D \end{pmatrix}$$,$$ S=\begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix}$$
 * then $$\Psi_{out}=S \Psi_{in}$$

Unitary property of S-matrix
The unitary property of S-matrix is directly related to conservation of probability current in quantum mechanics.The probability current $$J$$ of the wave function $$\psi(x)\,,$$ is defined as
 * $$ J = \frac{\hbar}{2mi}\left(\psi^* \frac{\partial \psi }{\partial x}- \psi \frac{\partial \psi^* }{\partial x} \right) $$.

The current density to the left of barrier $$J_L=\frac{\hbar k}{m}\left(|A|^2-|B|^2\right)$$,similarly the current density to the right of the barrier is $$J_R=\frac{\hbar k}{m}\left(|C|^2-|D|^2\right)$$. According to conservation of probability current density $$J_L=J_R$$.This implies the S-matrix is a unitary matrix
 * {| class="wikitable collapsible collapsed"

! Proof \begin{align} & J_L=J_R\\ & |A|^2-|B|^2=|C|^2-|D|^2\\ & |B|^2+|C|^2=|A|^2+|D|^2 \\ & \Psi_{out}^\dagger \Psi_{out}=\Psi_{in}^\dagger \Psi_{in}\\ & \Psi_{in}^\dagger S^\dagger S \Psi_{in}=\Psi_{in}^\dagger \Psi_{in}\\ & S^\dagger S=I\\ \end{align}$$
 * }

Time-reversal symmetry
If the potential $V(x)$ is real,then system posses time-reversal symmetry.Under this condition if $$\psi(x)$$ is a solution of Schrödinger's equation,then $$\psi^*(x)$$ is also a solution.The time-reversed solution is given by:
 * $$\psi^*_L(x)= A^* e^{-ikx} + B^* e^{ikx}$$

for the region left to the potential barrier and


 * $$\psi^*_R(x)= C^* e^{-ikx} + D^* e^{ikx}$$

for the region right to the potential barrier, where the terms with coefficient $B*$,$C*$represent incoming wave and terms with coefficient $A*$,$D*$ represent outgoing wave.So they are again related by the S-matrix :
 * $$\begin{pmatrix}A^* \\ D^* \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix}\begin{pmatrix} B^* \\ C^* \end{pmatrix}\,$$ or $$\Psi^*_{in}=S \Psi^*_{out}$$.

Now,the relations $$\Psi^*_{in}=S \Psi^*_{out}$$ and $$\Psi_{out}=S \Psi_{in}$$ together yield a condition $$S^*S=I$$. This condition in conjunction with the unitary relation implies that the S-matrix is symmetric as a result of time reversal symmetry:


 * $$S^T=S$$.

Transmission coefficient and Reflection coefficient
Transmission coefficient from the left of the potential barrier is $$T_L=\frac{|C|^2}{|A|^2}$$,when $$D=0$$.Thus $$T_L=|S_{21}|^2 $$.

Reflection Coefficient from the left of the potential barrier is $$R_L=\frac{|B|^2}{|A|^2}$$,when $$D=0$$.Thus $$R_L=|S_{11}|^2 $$. Similarly,Transmission Coefficient from the right of the potential barrier is $$T_R=\frac{|C|^2}{|D|^2}$$,when $$A=0$$.Thus $$T_R=|S_{22}|^2 $$.

Reflection Coefficient from the right of the potential barrier is $$R_R=\frac{|D|^2}{|D|^2}$$,when $$A=0$$.Thus $$R_R=|S_{12}|^2 $$.

The relation between transmission coefficien and reflection coefficient is: $$T_L+R_L=1$$ and $$T_R+R_R=1$$. This relation is the consequence of the unitary property of S-matrix.

Optical theorem in one dimension dimension
In the case of free particle $$V(x)=0$$.The S-matrix is then,$$ S=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$. Now whenever $V(x)$ is different from zero,there is A departure of S-matrix from the above form.This departure is measured by two complex functions of energy,r and t,which are defined by $$S_{11}=2ir$$ and $$S_{21}=1+2it$$.The relation between this two function is given by:

$$|r|^2+|t|^2=\operatorname{Im}(t)$$.

The analogue of this identity in three dimension is known as optical theorem.