Chapter 2, Problem 2.54P

(a)

To determine

The elements of M matrix in terms of elements of S matrix, $R_l,T_l,R_r\text{and }{T}_r$ in terms of elements of M matrix.

Answer to Problem 2.54P

The elements of M matrix in terms of elements of S matrix is $\underset{\_}{\text{M}=\frac{1}{S_{12}}\left(\begin{array}{cc}-\det S& S_{22}\\ -S_{11}& 1\end{array}\right)}$, $R_l,T_l,R_r\text{and }{T}_r$ in terms of elements of M matrix  is $\underset{\_}{R_l={\left|-\frac{M_{21}}{M_{22}}\right|}^2,T_l={\left|\frac{\det M}{M_{22}}\right|}^2,R_r={\left|-\frac{M_{12}}{M_{22}}\right|}^2,T_r={\left|\frac{1}{M_{22}}\right|}^2}$.

Explanation of Solution

The element B can be expressed as follows,

    $\begin{array}{c}B=S_{11}A+S_{12}G\\ G=\frac{1}{S_{12}}\left(B-S_{11}A\right)\\ =M_{21}A+M_{22}B\left[\text{from matrix}\right]\\ \text{then}\\ M_{21}=-\frac{-S_{11}}{S_{12}}\\ M_{22}=\frac{1}{S_{12}}\end{array}$        (I)

The element F can be expressed as follows,

    $F=S_{21}A+S_{22}G$        (II)

Use the value of G from the equation (I) in (II).

    $\begin{array}{c}F=S_{21}A+S_{22}\frac{1}{S_{12}}\left(B-S_{11}A\right)\\ =-\frac{\left(S_{11}{S}_{22}-S_{12}{S}_{21}\right)}{S_{12}}A+\frac{S_{22}}{S_{12}}B\\ =M_{11}A+M_{12}B\\ M_{11}=-\frac{\det S}{S_{12}},M_{12}=\frac{S_{22}}{S_{12}}\end{array}$        (III)

From the equation (I) and (III), the matric M can be written as follows.

    $\text{M}=\frac{1}{S_{12}}\left(\begin{array}{cc}-\det S& S_{22}\\ -S_{11}& 1\end{array}\right)$        (IV)

Similarly define the element G and solve for $S_{12},S_{11}$.

    $\begin{array}{c}G=M_{21}A+M_{22}B\\ B=\frac{1}{M_{22}}\left(G-M_{21}A\right)\\ =S_{11}A+S_{12}G\\ S_{11}=-\frac{M_{21}}{M_{22}},S_{12}=\frac{1}{M_{22}}\end{array}$        (V)

Similarly define the element F and solve for $S_{21},S_{22}$.

    $F=M_{11}A+M_{12}B$        (VI)

Use the value of B from the equation (V) in (VI).

    $\begin{array}{c}F=M_{11}A+M_{12}\frac{1}{M_{22}}\left(G-M_{21}A\right)\\ =\frac{\left(M_{11}{M}_{22}-M_{12}{M}_{21}\right)}{M_{22}}A+\frac{M_{12}}{M_{22}}G\\ =S_{21}A+S_{22}G\\ S_{21}=\frac{\det M}{M_{22}},S_{22}=\frac{M_{12}}{M_{22}}\end{array}$        (VII)

From equation (V), and (VII).

    $\text{S}=\frac{1}{M_{22}}\left(\begin{array}{cc}-M_{21}& 1\\ -\det M& M_{12}\end{array}\right)$        (VIII)

From the equation (V), $R_l$ can be expressed as,

    $\begin{array}{c}{R}_l={\left|S_{11}\right|}^2\\ ={\left|-\frac{M_{21}}{M_{22}}\right|}^2\end{array}$        (IX)

From the equation (VII), $T_l$ can be expressed as,

    $\begin{array}{l}{T}_l={\left|S_{21}\right|}^2\\ ={\left|\frac{\det M}{M_{22}}\right|}^2\end{array}$        (X)

From the equation (V), $R_r$ can be expressed as,

    $\begin{array}{c}{R}_r={\left|S_{22}\right|}^2\\ ={\left|-\frac{M_{12}}{M_{22}}\right|}^2\end{array}$        (XI)

From the equation (VII), $T_r$ can be expressed as,

    $\begin{array}{c}{T}_r={\left|S_{12}\right|}^2\\ ={\left|\frac{1}{M_{22}}\right|}^2\end{array}$        (XII)

Conclusion:

Therefore, the elements of M matrix in terms of elements of S matrix is $\underset{\_}{\text{M}=\frac{1}{S_{12}}\left(\begin{array}{cc}-\det S& S_{22}\\ -S_{11}& 1\end{array}\right)}$, $R_l,T_l,R_r\text{and }{T}_r$ in terms of elements of M matrix  is $\underset{\_}{R_l={\left|-\frac{M_{21}}{M_{22}}\right|}^2,T_l={\left|\frac{\det M}{M_{22}}\right|}^2,R_r={\left|-\frac{M_{12}}{M_{22}}\right|}^2,T_r={\left|\frac{1}{M_{22}}\right|}^2}$.

(b)

To determine

Show that M matrix for the combination is the product of the two M matrix for each section separately.

Answer to Problem 2.54P

Showed that M matrix for the combination is the product of the two M matrix for each section separately.

Explanation of Solution

Consider the Figure below.

Consider the matrix.

    $\left(\begin{array}{l}F\\ G\end{array}\right)=M_2\left(\begin{array}{l}C\\ D\end{array}\right)$        (XIII)

Consider the matrix.

    $\left(\begin{array}{l}C\\ D\end{array}\right)=M_1\left(\begin{array}{l}A\\ B\end{array}\right)$        (XIV)

Use equation (XIV) in (XIII).

    $\begin{array}{c}\left(\begin{array}{l}F\\ G\end{array}\right)=M_2{M}_1\left(\begin{array}{l}A\\ B\end{array}\right)\\ =M\left(\begin{array}{l}A\\ B\end{array}\right)\end{array}$

Hence proved.

Conclusion:

Therefore, the M matrix for the combination is the product of the two M matrix for each section separately.

(c)

To determine

The matrix for the scattering from a single delta function potential at point a.

Answer to Problem 2.54P

The matrix for the scattering from a single delta function potential at point a is $\underset{\_}{\text{M}=\left(\begin{array}{cc}1+i\beta & i\beta e^{-2ika}\\ -i\beta e^{2ika}& 1-i\beta \end{array}\right)}$.

Explanation of Solution

The wave function for the scattering from a single delta function potential at point a is

    $\psi \left(x\right)=\left\{\begin{array}{l}A{e}^{ikx}+B{e}^{-ikx}\left(x<a\right)\\ F{e}^{ikx}+G{e}^{-ikx}\left(x>a\right)\end{array}\right\}$        (XV)

Using the continuity condition of the wave function, equation (XV) can be written as.

    $A{e}^{ika}+B{e}^{-ika}=F{e}^{ika}+G{e}^{-ika}$        (XVI)

Using the discontinuity condition of the derivative of wave function, equation (XV) can be written as.

    $\begin{array}{l}-ik\left(A{e}^{ika}-B{e}^{-ika}\right)=ik\left(F{e}^{ika}-G{e}^{-ika}\right)\\ -\frac{2m\alpha }{{\hslash }^2}\psi \left(a\right)=-\frac{2m\alpha }{{\hslash }^2}\left(A{e}^{ika}+B{e}^{-ika}\right)\end{array}$        (XVII)

Multiply the equation (XVI) with $e^{ika}$.

    $A{e}^{2ika}+B=F{e}^{2ika}+G$        (XVIII)

Multiply the equation (XVII) with $e^{ika}$.

    $A{e}^{2ika}-B+i\frac{2m\alpha }{{\hslash }^{2}k}\left(A{e}^{2ika}+B\right)=F{e}^{2ika}-G$        (XIX)

Add equation (XVIII) and (XIX), and solve for $M_{11},M_{12}$.

    $\begin{array}{c}2A{e}^{2ika}+i\frac{2m\alpha }{{\hslash }^{2}k}\left(A{e}^{2ika}+B\right)=2F{e}^{2ika}\\ F=\left(1+i\frac{m\alpha }{{\hslash }^{2}k}\right)A+i\frac{m\alpha }{{\hslash }^{2}k}{e}^{-2ika}B\\ =M_{11}A+M_{12}B\\ M_{11}=\left(1+i\beta \right),M_{12}=i\beta e^{-2ika}\end{array}$        (XX)

Where, $\beta =\frac{m\alpha }{{\hslash }^{2}k}$.

Subtract equation (XIX) from (XVIII), and solve for $M_{21},M_{22}$.

    $\begin{array}{c}2G=2B-2i\beta e^{2ika}A-2i\beta B\\ G=\left(1-i\beta \right)B-i\beta e^{2ika}A\\ =M_{21}A+M_{22}B\\ M_{21}=-i\beta e^{2ika},M_{22}=\left(1-i\beta \right)\end{array}$        (XXI)

From equation (XX), and (XXI), the matrix can be written as follows.

    $\text{M}=\left(\begin{array}{cc}1+i\beta & i\beta e^{-2ika}\\ -i\beta e^{2ika}& 1-i\beta \end{array}\right)$        (XXII)

Conclusion:

Therefore, the matrix for the scattering from a single delta function potential at point a is $\underset{\_}{\text{M}=\left(\begin{array}{cc}1+i\beta & i\beta e^{-2ika}\\ -i\beta e^{2ika}& 1-i\beta \end{array}\right)}$.

(d)

To determine

The matrix for scattering from the double delta function, and the transmission coefficient for this potential.

Answer to Problem 2.54P

The matrix for scattering from the double delta function is $\underset{\_}{\text{M}=\left(\begin{array}{cc}1+2i\beta +{\beta }^2\left(e^{-4ika}-1\right)& 2i\beta \left(\cos 2ka-\beta \sin 2ka\right)\\ -2i\beta \left(\cos 2ka-\beta \sin 2ka\right)& 1-2i\beta +{\beta }^2\left(e^{4ika}-1\right)\end{array}\right)}$, and the transmission coefficient for this potential is $\underset{\_}{T=\frac{1}{1+4{\beta }^2{\left(\cos 2ka-\beta \sin 2ka\right)}^2}}$.

Explanation of Solution

The matrix $M_1$ and $M_2$ can be expressed as.

    $\begin{array}{l}{M}_2=\left(\begin{array}{cc}1+i\beta & i\beta e^{-2ika}\\ -i\beta e^{2ika}& 1-i\beta \end{array}\right)\\ M_1=\left(\begin{array}{cc}1+i\beta & i\beta e^{2ika}\\ -i\beta e^{-2ika}& 1-i\beta \end{array}\right)\end{array}$        (XXIII)

From the result of part (b), the matrix for scattering from the double delta function can be expressed as.

    $\begin{array}{c}M=M_1{M}_2\\ =\left(\begin{array}{cc}1+i\beta & i\beta e^{-2ika}\\ -i\beta e^{2ika}& 1-i\beta \end{array}\right)\left(\begin{array}{cc}1+i\beta & i\beta e^{2ika}\\ -i\beta e^{-2ika}& 1-i\beta \end{array}\right)\\ =\left(\begin{array}{cc}1+2i\beta +{\beta }^2\left(e^{-4ika}-1\right)& 2i\beta \left(\cos 2ka-\beta \sin 2ka\right)\\ -2i\beta \left(\cos 2ka-\beta \sin 2ka\right)& 1-2i\beta +{\beta }^2\left(e^{4ika}-1\right)\end{array}\right)\end{array}$        (XXIV)

The transmission coefficient can be calculated using the equation (XXIV).

    $\begin{array}{c}T=T_l\\ =T_r\\ =\frac{1}{{\left|M_{22}\right|}^2}\\ =\frac{1}{\left[1+2i\beta +{\beta }^2\left(e^{-4ika}-1\right)\right]\left[1-2i\beta +{\beta }^2\left(e^{4ika}-1\right)\right]}\\ T^{-1}=1-2i\beta +{\beta }^2{e}^{4ika}-{\beta }^2+2i\beta +4{\beta }^2+2i{\beta }^3{e}^{4ika}-2i{\beta }^3+{\beta }^2{e}^{-4ika}\\ -{\beta }^2-2i{\beta }^3{e}^{-4ika}+2i{\beta }^3+{\beta }^4\left(1-e^{-4ika}-e^{4ika}+1\right)\end{array}$        (XXV)

Simplify the equation (XXV).

    $\begin{array}{c}{T}^{-1}=1+2{\beta }^2+{\beta }^2\left(e^{-4ika}+e^{4ika}\right)-2i{\beta }^3\left(e^{-4ika}-e^{4ika}\right)\\ +2{\beta }^4-{\beta }^4\left(e^{-4ika}+e^{4ika}\right)\\ =1+2{\beta }^2+2{\beta }^2\cos 4ka+2i{\beta }^{3}2i\sin 4ka+2{\beta }^4-2{\beta }^4\cos 4ka\\ =1+2{\beta }^2\left(1+\cos 4ka\right)-4{\beta }^3\sin 4ka+2{\beta }^4\left(1-\cos 4ka\right)\\ =1+4{\beta }^2{\cos }^{2}2ka-8{\beta }^3\sin 2ka-4{\beta }^4{\sin }^{2}2ka\\ =1+4{\beta }^2{\left(\cos 2ka-\beta \sin 2ka\right)}^2\end{array}$

Conclusion:

Therefore, The matrix for scattering from the double delta function is $\underset{\_}{\text{M}=\left(\begin{array}{cc}1+2i\beta +{\beta }^2\left(e^{-4ika}-1\right)& 2i\beta \left(\cos 2ka-\beta \sin 2ka\right)\\ -2i\beta \left(\cos 2ka-\beta \sin 2ka\right)& 1-2i\beta +{\beta }^2\left(e^{4ika}-1\right)\end{array}\right)}$, and the transmission coefficient for this potential is $\underset{\_}{T=\frac{1}{1+4{\beta }^2{\left(\cos 2ka-\beta \sin 2ka\right)}^2}}$.