User:Dibyendum/sandbox

S-matrix in one dimension

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:

$\psi _{L}(x)=Ae^{ikx}+Be^{-ikx}$

for the region left to the potential barrier and

$\psi _{R}(x)=Ce^{ikx}+De^{-ikx}$

for the region right to the potential barrier,

where $k={\sqrt {2mE/\hbar ^{2}}}$ 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

Proof
${\begin{aligned}&J_{L}=J_{R}\\&|A|^{2}-|B|^{2}=|C|^{2}-|D|^{2}\\&|B|^{2}+|C|^{2}=|A|^{2}+|D|^{2}\\&\Psi _{out}^{\dagger }\Psi _{out}=\Psi _{in}^{\dagger }\Psi _{in}\\&\Psi _{in}^{\dagger }S^{\dagger }S\Psi _{in}=\Psi _{in}^{\dagger }\Psi _{in}\\&S^{\dagger }S=I\\\end{aligned}}$

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.