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The scattering matrix. The theory of scattering generalizes in a pretty obvious way to arbitrary localized potentials to the left (Region I),$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$so$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{ikx}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{ikx}}\mathbf{,}\mathbf{鈥�}\mathbf{where}\mathbf{}\mathbf{k}\mathbf{鈮�}\frac{\sqrt{\mathbf{2}\mathbf{mE}}}{\mathbf{鈩�}}$

To the right (Region III),$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is again zero, so$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Fe}}^{\mathbf{ikx}}\mathbf{+}{\mathbf{Ge}}^{\mathbf{-}\mathbf{ikx}}$In between (Region II), of course, I can't tell you what is until you specify the potential, but because the Schr枚dinger equation is a linear, second-order differential equation, the general solution has got to be of the form

$\mathbf{蠄}$where $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}$and $\mathbf{}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}$are two linearly independent particular solutions. 48 There will be four boundary conditions (two joining Regions I and II, and two joining Regions II and III). Two of these can be used to eliminate C and D, and the other two can be "solved" for B and F in terms of \(A\) and G

$\mathbf{B}\mathbf{=}{\mathbf{S}}_{\mathbf{11}}\mathbf{A}\mathbf{+}{\mathbf{S}}_{\mathbf{12}}\mathbf{G}\mathbf{.}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}\mathbf{F}\mathbf{=}{\mathbf{S}}_{\mathbf{21}}\mathbf{A}\mathbf{+}{\mathbf{S}}_{\mathbf{22}}\mathbf{G}$

The four coefficients${\mathit{S}}_{\mathbf{i}\mathbf{j}}$which depend on k (and hence on E), constitute a$\mathbf{2}\mathbf{脳}\mathbf{2}\mathbf{}\mathbf{}$matrix s called the scattering matrix (or S-matrix, for short). The S-matrix tells you the outgoing amplitudes (B and F) in terms of the incoming amplitudes (A and G):

$\left(\begin{array}{c}B\\ F\end{array}\right)\mathbf{=}\left(\begin{array}{cc}{S}_{11}& S_{21}\\ S_{21}& S_{22}\end{array}\right)\left(\begin{array}{c}A\\ G\end{array}\right)$

In the typical case of scattering from the left, $\mathbf{G}\mathbf{=}\mathbf{0}$so the reflection and transmission coefficients are

${\mathbf{R}}_{\mathbf{l}}\mathbf{=}\frac{\mathbf{|}\mathbf{B}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{A}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}}_{\mathbf{G}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{11}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}{\mathbf{T}}_{\mathbf{I}}\mathbf{=}\frac{\mathbf{|}\mathbf{F}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{A}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}}_{\mathbf{G}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{2}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}$

For scattering from the right, and

${\mathbf{R}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{|}\mathbf{F}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{G}{\mathbf{|}}^{\mathbf{2}}}{\overline{)\mathbf{}}}_{\mathbf{A}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{22}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}{\mathbf{T}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{|}\mathbf{B}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{G}{\mathbf{|}}^{\mathbf{2}}}{\overline{)\mathbf{}}}_{\mathbf{A}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{12}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}$

(a) Construct the S-matrix for scattering from a delta-function well (Equation 2.114). (b) Construct the S-matrix for the finite square well (Equation 2.145). Hint: This requires no new work, if you carefully exploit the symmetry of the problem.

Step by step solution

01

Define the Schrödinger equation

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

02

Step 2: Construct the S-matrix

(a)

Given equation from 2.133 :

$F+G=A+B$ 鈥︹�︹�︹�︹��.. (1)

Again from the equation 2.135 :

$$\begin{gathered} \mathrm{F}-\mathrm{G}=(1+2\mathrm{颈尾})\mathrm{A}-(1-2\mathrm{颈尾})\mathrm{B} \\ \mathrm{尾}=\mathrm{尘伪}/{鈩�}^2\mathrm{k}鈥�.\left(2\right) \end{gathered}$$

Subtracting equations 1 and 2 as:

$$\begin{gathered} 2G=-2\mathrm{颈尾}A+2\left(1-i尾\right)B \\ B=\frac{1}{1-\mathrm{颈尾}}\left(\mathrm{颈尾A}+\mathrm{G}\right) \end{gathered}$$

Then multiply equation 1 by$(1-2i尾)$ as:

role="math" localid="1658292860542" $$\begin{gathered} 2\left(1-\mathrm{颈尾}\right)\mathrm{F}-2\mathrm{颈尾G}=2\mathrm{A} \\ \mathrm{F}=\frac{1}{1-\mathrm{颈尾}}\left(\mathrm{A}+\mathrm{颈尾G}\right) \\ \mathrm{S}=\frac{1}{1-\mathrm{颈尾}}\left[\begin{array}{cc}\mathrm{颈尾}& 1\\ 1& \mathrm{颈尾}\end{array}\right] \end{gathered}$$

Hence, the S matrix is $\frac{1}{1-i尾}\left[\begin{array}{cc}i尾& 1\\ 1& i尾\end{array}\right]$.

03

Form the matrix.

(b)

Given infinite square well:

$$\begin{gathered} \mathrm{S}=\frac{1}{1-\mathrm{颈尾}}\left[\begin{array}{cc}\mathrm{颈尾}& 1\\ 1& \mathrm{颈尾}\end{array}\right] \\ {\mathrm{S}}_{21}=\frac{{\mathrm{e}}^{-2\mathrm{ika}}}{\cos 2\mathrm{la}-\mathrm{i}\frac{\left({\mathrm{k}}^2+{\mathrm{l}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}} \\ {S}_{11}=\frac{\mathrm{i}\frac{\left({\mathrm{l}}^2+{\mathrm{k}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}{e}^{-2ika}}{\cos 2\mathrm{la}-\mathrm{i}\frac{\left({\mathrm{k}}^2+{\mathrm{l}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}}\begin{array}{}\end{array} \\ \mathrm{S}=\frac{{\mathrm{e}}^{-2\mathrm{ika}}}{\cos 2\mathrm{la}-\mathrm{i}\frac{\left({\mathrm{k}}^2+{\mathrm{l}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}}\left[\begin{array}{cc}\mathrm{i}\frac{\left({\mathrm{l}}^2+{\mathrm{k}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}& 1\\ 1& \mathrm{i}\frac{\left({\mathrm{l}}^2+{\mathrm{k}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}\end{array}\right] \end{gathered}$$

Thus, the S-matrix for the finite square well$\mathrm{S}=\frac{{\mathrm{e}}^{-2\mathrm{ika}}}{\cos 2\mathrm{la}-\mathrm{i}\frac{\left({\mathrm{k}}^2+{\mathrm{l}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}}\left[\begin{array}{cc}\mathrm{i}\frac{\left({\mathrm{l}}^2+{\mathrm{k}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}& 1\\ 1& \mathrm{i}\frac{\left({\mathrm{l}}^2+{\mathrm{k}}^2\right)}{2\mathrm{kl}}\sin 2\mathrm{la}\end{array}\right]$.

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