Potential scattering

We now consider a piecewise constant potential $V(x)$ with the corresponding wave function given by

$$V(x)=\left\{ \begin{array}{cc} V_1, & \\ V_2, & \\ V_3, & \\ ... ... ...a_{N+1}e^{ik_{N+1}x}+b_{N+1}e^{-ik_{N+1}x}, & x_N<x< \infty \end{array} \right.$$ (119)

$k_j=\sqrt{({2m}/{\hbar^2})\left(E-V_j\right)}$.

We first consider the case $E>V_1,V_{N+1}$ such that $k_1$ and $k_{N+1}$ are real wave vectors and $\psi(x)$ describes running waves outside the `scattering region' $[x_1,x_N]$. Our aim now is the following: we would like to determine solutions of the Schrödinger equation, i.e. wave functions $\psi(x)$, under the scattering condition $b_{N+1}=0$, i.e. we seek solutions that have only waves $a_{N+1}e^{ik_{N+1}x}$ traveling to the right (away from the scattering zone) on the right side of the potential. On the left side, we have the wave $a_1e^{ik_1x}+b_1e^{-ik_1x}$, i.e. a superposition of a right-going (incoming) and a left-going (outgoing) wave. We would like to know how much of an incoming wave gets reflected on the left side (coefficient $b_1$) and how much gets transmitted on the right side ($a_{N+1}$).

Step 1: we demand that $\psi(x)$ and its derivative $\psi'(x)$ are continuous at $x=x_1$. This gives two equations

$$a_1e^{ik_1x_1}+b_1e^{-ik_1x_1} = a_2e^{ik_2x_1}+b_2e^{-ik_2x_1}$$

$$a_1e^{ik_1x_1}-b_1e^{-ik_1x_1} = (k_2/k_1)(a_2e^{ik_2x_1}-b_2e^{-ik_2x_1})$$ (120)

or

$$a_1 = \frac{1}{2}\left(\frac{k_2}{k_1}+1\right)e^{i(k_2-k_1)x_1}a_2 + \frac{1}{2}\left(1-\frac{k_2}{k_1}\right)e^{-i(k_2+k_1)x_1}b_2$$

$$b_1 = \frac{1}{2}\left(1-\frac{k_2}{k_1}\right)e^{i(k_2+k_1)x_1}a_2 + \frac{1}{2}\left(1+\frac{k_2}{k_1}\right)e^{-i(k_2-k_1)x_1}b_2$$ (121)

which can be written in a matrix form

$${\bf u}_1=T^1{\bf u}_2,\quad {\bf u}_i= \left( \begin{array}{c} a_i \\ b_i \end{array} \right),\quad i=1,2,$$ (122)

with

$$T^1=\frac{1}{2k_1}\left( \begin{array}{cc} (k_1+k_2)e^{i(k_2-k_1)... ...\ (k_1-k_2)e^{i(k_2+k_1)x_1} & (k_1+k_2)e^{-i(k_2-k_1)x_1} \end{array} \right).$$ (123)

Step 2: In completely the same manner, we obtain the transfer matrix $T^2$ at the `slice' $x=x_2$ and

$${\bf u}_2=T^2{\bf u}_3 \leadsto {\bf u}_1=T^1{\bf u}_2 = T^1 T^2{\bf u}_3.$$ (124)

Doing this for all the slices $x_1,...,x_N$, we obtain the complete transfer matrix $M$ that connects the wave function on the left side of the potential with the one on the right side,

$${\bf u}_1=M{\bf u}_{N+1},\quad M=T^1T^2...T^N.$$ (125)

Step 3: We use the continuity equation

$$\partial_t\rho(x,t)+\partial_x j(x,t) =$$ 0 (126)

$$\rho(x,t) := \psi(x,t)\psi^*(x,t)$$

$$j(x,t) := -\frac{i\hbar}{2m} \left[ \psi(x,t)^*\partial_x\psi(x,t)- \psi(x,t)\partial_x\psi^*(x,t)\right]$$

$$\psi(x,t) = \psi(x)e^{-iEt/\hbar}.$$ (127)

The current density is time-independent and can be written as

$$j(x) := -\frac{i\hbar}{2m} \left[ \psi(x)^*\partial_x\psi(x)- \psi(x)\partial_x\psi^*(x)\right]= \frac{\hbar}{m}{\rm Im} \left[ \psi^*(x)\psi'(x)\right].$$ (128)

The current density on the right and left side of the potential is

$$j(x>x_N) = \frac{\hbar}{m}{\rm Im} (ik_{N+1} \vert a_{N+1}\vert^2) =\vert a_{N+1}\vert^2 \frac{\hbar k_{N+1}}{m}$$

$$j(x<x_1) = \frac{\hbar}{m}{\rm Im} \left[(a_1^*e^{-ik_1x}+b_1^*e^{ik_1x})ik_1 (a_1e^{ik_1x}-b_1e^{-ik_1x})\right]$$

$$= \frac{\hbar k_1}{m}[\vert a_1\vert^2 - \vert b_1\vert^2].$$ (129)

The current density $j(x>x_N)$ describes a particle flow to the right of the potential, `out-flowing' to $x\to \infty$. On the other hand, the current density $j(x<x_1)$ is the difference of an in-flowing positive current density and an out-flowing negative current density. The former describes an incoming particle, the latter a particle that is reflected back from the potential and returning back to $x=-\infty$.

Step 4: We define the transmission coefficient $T$ as the ratio of the right out-flowing current density to the left in-flowing current density,

$$T:=\frac{k_{N+1}}{k_1}\left\vert\frac{a_{N+1}}{a_1}\right\vert^2.$$ (130)

From

$$\left( \begin{array}{c} a_1 \\ b_1 \end{array}\right)= \left... ... \left( \begin{array}{c} a_{N+1} \\ b_{N+1} \end{array}\right)$$ (131)

and the scattering condition $b_{N+1}=0$ it follows

$$T=\frac{k_{N+1}}{k_1}\frac{1}{\vert M_{11}\vert^2}.$$ (132)

To calculate the transmission coefficient through a piecewise constant one-dimensional potential, it is therefore sufficient to know the total transfer matrix $M$. The fact that $M=T^1T^2...T^N$ is just the product of the individual two-by two transfer matrices makes it a very convenient tool for computations.

Figure: Reflection and Transmission

Step 5: In completely the same manner, we define the reflection coefficient $R$ as the ratio of the out-flowing current density on the left and the in-flowing (reflected) current density on the left, i.e.

$$R:=\left\vert\frac{b_1}{a_1}\right\vert^2=\left\vert\frac{M_{21}}{M_{11}}\right\vert^2.$$ (133)

The last equality is left as an exercise.

For eigenstates of energy $E$ we have from the continuity equation

$$\partial_t\rho(x,t)\equiv\partial_t\rho(x)=0\leadsto \partial_x j(x)=0\leadsto j(x)= const.$$ (134)

Using Eq. (2.52) and the definition of $T$ and $R$, this leads to

$$T+R=1.$$ (135)