# Scattering States, Transmission and Reflection

We will now discuss scattering states in a one-dimensional potential. Scattering states are states that are not bound. Such states have energies larger than the potential at at least one of $x\to\pm\infty,$ and their energy spectrum forms a continuous band, rather than a discrete set as the bound states do. Unlike the bound case, the wave function does not have to vanish at infinity, though a particle cannot reflect from infinity, often giving a useful boundary condition. At any discontinuous changes in the potentials, the wave function must still be continuous and differentiable as for the bound states.

We are interested in obtaining the wave functions for these scattering states in order to discuss transmission and reflection of waves from one-dimensional potentials, and to find the transmission and reflection coefficients $T\!$ and $R,\!$ which give the probability that an incident wave will be transmitted and reflected, respectively.

## The Step Potential

As a first example, let us consider a one-dimensional potential step. The potential in this case is given by

$V(x) = \begin{cases} 0, & x < 0, \\ V_0, & x > 0. \end{cases}$

The corresponding Schrödinger equation is

$\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) = E \psi(x).$

Let us first consider states with energy $E>V_0.\!$ We will divide the one-dimensional space into two regions - region I, for which $x < 0,\!$ and region II, for which $x > 0.\!$ The Schrödinger equations for the two regions are

$- \frac{\hbar^2}{2m} \frac{d^2 \psi_{I}(x)}{dx^2} = E \psi_{I}(x)$

and

$\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V_0 \right) \psi_{II}(x) = E \psi_{II}(x),$

and thus the corresponding wave functions are

$\psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x}$

and

$\psi_{II}(x) = C e^{i k x} + D e^{-i k x}, \!$

where $k_0 = \sqrt{\frac{2mE}{\hbar^2}}$ and $k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}.$ We may consider the first term in $\psi_{I}(x),\,Ae^{ik_0x},$ to be an incident wave from the left, in which case the second term, $Be^{-ik_0x},$ is a reflected wave from the potential barrier and the first term in $\psi_{II}(x)\!$, $Ce^{ikx},\!$ is a transmitted wave. Similarly, we can think of the second term in $\psi_{II}(x),\,De^{-ikx},$ to be an incident wave from the right, in which case $Ce^{ikx}\!$ is now the reflected wave and $Be^{-ik_0x}\!$ is the transmitted wave. These interpretations of the various terms will become more obvious shortly.

The boundary conditions at $x=0\!$ require

$\psi_I(0) = \psi_{II}(0)\!$

and

$\left. \frac{d\psi_I(x)}{dx} \right|_{x=0} = \left. \frac{d\psi_{II}(x)}{dx} \right|_{x=0},$

which give us

$A + B = C + D \!$

and

$k_0\left(A-B\right) = k \left(C-D\right).$

If we assume that the wave is incident from the left, then we can set $D=0\!$. In this case, reflection occurs at the potential step, and there is transmission to the right. We then have

$A + B = C \!$

and

$\left(A-B\right) = \frac{k}{k_0} C.$

From the above equations, we obtain

$\frac{B}{A} = \frac{k_0-k}{k_0+k}$

and

$\frac{C}{A} = \frac{2k}{k_0+k}.$

To determine the probability of reflection and transmission, we must now find the current density,

$j = \frac{\hbar}{2m i} \left (\psi^* \frac{d\psi}{dx} - \frac{d\psi^*}{dx} \psi \right ),$

on each side of the barrier. Doing so, we obtain

$j = \begin{cases} \displaystyle \frac{\hbar k_0}{m} \left( \left|A\right|^2 - \left|B\right|^2 \right), & x < 0, \\ {} & {} \\ \displaystyle \frac{\hbar k}{m} \left|C\right|^2, & x > 0. \end{cases}$

We thus see more clearly that the terms that we earlier identified as incident, reflected, and transmitted waves are as we labeled them. We define the ratio of the reflected current density to the incident current density as the reflection coefficient, $R=\frac{\left |B\right |^2}{\left |A\right |^2},$ and the ratio of the transmitted current density to the incident current density as the transmission coefficient, $T=\frac{k}{k_0}\frac{\left |C\right |^2}{\left |A\right |^2}.$

The continuity equation for the current density in one dimension implies conservation of current density, and thus that

$R+T=1.\!$

If we determine the reflection and transmission coefficients for the problem at hand, we obtain

$R = \frac{\left|B\right|^2}{\left|A\right|^2} = \frac{\left( k_0 - k \right)^2}{\left(k_0 + k \right)^2}$

and

$T = \frac{k}{k_0} \frac{\left|C\right|^2}{\left|A\right|^2} = \frac{4k_0k}{\left(k_0 + k \right)^2}.$

We see that these expressions satisfy $R + T = 1,\!$ as expected.

Now let us consider the case in which $0 < E < V_0.\!$ If we again assume that the wave is incident from the left, the wave functions become

$\psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x}$

and

$\psi_{II}(x) = C e^{-\kappa x}, \!$

where $\kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.$ Applying the same boundary conditions as before, we obtain

$\frac{B}{A} = \frac{k_0-i\kappa}{k_0+i\kappa}$

and

$\frac{C}{A} = \frac{2i\kappa}{k_0+i\kappa}.$

From the second wave function, $\psi_{II},\!$ we see that the transmitted wave decreases exponentially over a length scale given by $\frac{1}{\kappa}.$.

If we determine the current density in this case, we find that

$j = \begin{cases} \displaystyle \frac{\hbar k_0}{m} \left( \left|A\right|^2 - \left|B\right|^2 \right), & x < 0, \\ {} & {} \\ \displaystyle 0, & x > 0. \end{cases}$

We see that the wave is completely reflected; i.e., $R=1\!$ and $T=0.\!$

The reflected wave acquires a phase difference relative to the incident wave. To see this, we simply rewrite the wave function $\psi_I, \!$ as

\begin{align} \psi_I (x) &= e^{ik_0 x} + \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2 + \kappa^2} e^{-ik_0 x} \\ &= e^{ik_0x} + e^{i\theta} e^{-ik_0x}, \end{align}

where the phase difference $\theta\!$ of the reflected wave with respect to the incident wave is given by

$e^{i \theta} = \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2+\kappa^2},$

or

$\theta = \tan^{-1} \left( \frac{2\kappa k_0}{\kappa^2 - k_0^2} \right).$

## The Square Potential Barrier

For a square potential barrier, given by

$V(x) = \begin{cases} 0, & x < -a , \\ V_0, & -a < x < a, \\ 0, & x > a , \end{cases}$

we can write the general solution of the Schrödinger equation for $0 < E < V_0: \!$

$\psi(x) = \begin{cases} A e^{i k_0x} + B e^{-ik_0x}, & x < -a, \\ C e^{-\kappa x} + D e^{\kappa x}, & -a < x < a, \\ F e^{i k_0x} + G e^{-ik_0x}, & x > a, \end{cases}$

where $k_0 = \frac{\sqrt{2mE}}{\hbar} \!$ and $\kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}.\!$

The boundary conditions are the same as before; the boundary conditions at $x = -a, \!$ give us

$A e^{-ik_0 a} + B e^{ik_0a} = C e^{\kappa a} + D e^{-\kappa a },$
$A e^{-ik_0 a} - B e^{ik_0a} = \frac{i\kappa}{k_0} \left( C e^{\kappa a} - D e^{-\kappa a } \right).$

Similarly, the boundary conditions at $x = a \!$ give us

$C e^{-\kappa a} + D e^{\kappa a} = F e^{i k_0 a} + G e^{- i k_0 a} ,$
$C e^{-\kappa a} - D e^{\kappa a} = - \frac{ik_0}{\kappa} \left( F e^{i k_0 a} - G e^{- i k_0 a} \right).$

For the convenience, let us express the coefficients of these linear homogeneous relations in terms of matrices:

$\begin{bmatrix} A \\ B \end{bmatrix} = \frac{1}{2} \begin{bmatrix} \left( 1 + \frac{i\kappa}{k_0} \right) e^{\kappa a + i k_0 a} && \left( 1 - \frac{i\kappa}{k_0} \right)e^{\kappa a - i k_0 a} \\ \left( 1 - \frac{i\kappa}{k_0} \right) e^{-\kappa a + i k_0 a} && \left( 1 + \frac{i\kappa}{k_0} \right)e^{-\kappa a - i k_0 a} \end{bmatrix} \begin{bmatrix} C \\ D \end{bmatrix}$
$\begin{bmatrix} C \\ D \end{bmatrix} = \frac{1}{2} \begin{bmatrix} \left( 1 - \frac{ik_0}{\kappa} \right)e^{\kappa a + i k_0 a} && \left( 1 + \frac{ik_0}{\kappa} \right)e^{-\kappa a + i k_0 a} \\ \left( 1 + \frac{ik_0}{\kappa} \right)e^{\kappa a - i k_0 a} && \left( 1 - \frac{ik_0}{\kappa} \right)e^{-\kappa a - i k_0 a} \end{bmatrix} \begin{bmatrix} F \\ G \end{bmatrix}$

If we combine these two equations, we have

$\begin{bmatrix} A \\ B \end{bmatrix} =\begin{bmatrix} \left( \cosh{2\kappa a}+ \frac{i\varepsilon}{2} \sinh{2\kappa a} \right) e^{2i k_0 a} && -\frac{i\eta}{2} \sinh{2\kappa a} \\ \frac{i\eta}{2} \sinh{2\kappa a} && \left( \cosh{2\kappa a} - \frac{i\varepsilon}{2} \sinh{2\kappa a} \right) e^{- 2i k_0 a} \end{bmatrix} \begin{bmatrix} F \\ G \end{bmatrix}$

where $\varepsilon = \frac{\kappa}{k_0} - \frac{k_0}{\kappa} \!$ and $\eta = \frac{\kappa}{k_0} + \frac{k_0}{\kappa}. \!$

Note that $\eta^2 - \varepsilon^2 = 4. \!$

Let us now assume that there is only an incident wave coming from the left; i.e., $G=0.\!$ By similar arguments as before, we may identify the transmission coefficient as

$T=\frac{|F|^2}{|A|^2}=\frac{4}{4\cosh^2{2\kappa a}+\varepsilon^2\sinh^2{2\kappa a}}$

and the reflection coefficient as

$R=\frac{|B|^2}{|A|^2}=\frac{\eta^2\sinh^2{2\kappa a}}{4\cosh^2{2\kappa a}+\varepsilon^2\sinh^2{2\kappa a}}.$

## Finite Asymmetric Square Well

We now consider an asymmetric square well potential, given by

$V(x) = \begin{cases} V_1, & x < -a , \\ 0, & -a < x < a, \\ V_2, & x > a. \end{cases}$

In this case, the wave functions are given by

$\psi(x) = \begin{cases} A_{1}e^{ik_{1}x}+A_{2}e^{-ik_{2}x}, & x < -a, \\ B_{1}e^{ik_{2}x}+B_{2}e^{-ik_{2}x}, & -a < x < a, \\ C_{1}e^{ik_{3}x}+C_{2}e^{-ik_{3}x}, & x > a, \end{cases}$

where $k_{1}=\frac{\sqrt{2m(E-V_{1})}}{\hbar},\,k_{2}=\frac{\sqrt{2mE}}{\hbar},$ and $k_{3}=\frac{\sqrt{2m(E-V_{2})}}{\hbar}.$

Applying the boundary conditions at $x=-a,\!$ we obtain

$A_{1}e^{-ik_{1}a}+A_{2}e^{ik_{1}a}=B_{1}e^{-ik_{2}a}+B_{2}e^{ik_{2}a}$

and

$ik_{1}A_{1}e^{-ik_{1}a}-ik_{1}A_{2}e^{ik_{1}a}=ik_{2}B_{1}e^{-ik_{2}a}-ik_{2}B_{2}e^{ik_{2}a},$

while those at $x=a\!$ give us

$B_{1}e^{ik_{2}a}+B_{2}e^{-ik_{2}a}=C_{1}e^{ik_{3}a}+C_{2}e^{ik_{3}a}$

and

$ik_{2}B_{1}e^{ik_{2}a}-ik_{2}B_{2}e^{-ik_{2}a}=ik_{3}C_{1}e^{ik_{3}a}+C_{2}e^{-ik_{3}a}.$

We may express these in matrix form as

$\begin{bmatrix} A_{1} \\ A_{2} \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} \left (1+\frac{k_2}{k_1}\right )e^{i(k_1-k_2)a} && \left (1-\frac{k_2}{k_1}\right )e^{i(k_1+k_2)a} \\ \left (1-\frac{k_2}{k_1}\right )e^{-i(k_1+k_2)a} && \left (1+\frac{k_2}{k_1}\right )e^{-i(k_1-k_2)a} \end{bmatrix} \begin{bmatrix} B_{1} \\ B_{2} \end{bmatrix}$

and

$\begin{bmatrix} B_{1} \\ B_{2} \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} \left (1+\frac{k_3}{k_2}\right )e^{i(k_2-k_3)a} && \left (1-\frac{k_3}{k_2}\right )e^{i(k_2+k_3)a} \\ \left (1-\frac{k_3}{k_2}\right )e^{-i(k_2+k_3)a} && \left (1+\frac{k_3}{k_2}\right )e^{-i(k_2-k_3)a} \end{bmatrix} \begin{bmatrix} C_{1} \\ C_{2} \end{bmatrix}.$

If we combine these, we obtain

$\begin{bmatrix} A_{1} \\ A_{2} \end{bmatrix} =\tfrac{1}{2} \begin{bmatrix} \left (1+\frac{k_3}{k_1}\right )e^{i(k_1-k_3)a} && \left (1-\frac{k_3}{k_1}\right )e^{i(k_1+k_3)a} \\ \left (1-\frac{k_3}{k_1}\right )e^{-i(k_1+k_3)a} && \left (1+\frac{k_3}{k_1}\right )e^{-i(k_1-k_3)a} \end{bmatrix} \begin{bmatrix} C_{1} \\ C_{2} \end{bmatrix}.$

Let us once again assume that there is only an incident wave from the left, so that $C_{2}=0.\!$ Using the same arguments as before, we may identify the transmission coefficient as

$T=\frac{k_3|C_1|^2}{k_1|A_1|^2}=\frac{4k_3}{k_1}\frac{k_1^2}{(k_1+k_3)^2}=\frac{4k_1k_3}{(k_1+k_3)^2}$

and the reflection coefficient as

$R=\frac{|A_2|^2}{|A_1|^2}=\frac{(k_1-k_3)^2}{(k_1+k_3)^2}.$

## The Dirac Delta Function Potential

We now consider scattering from a Dirac delta function potential, $V(x)=V_0\delta(x).\!$ For $x\neq 0,\!$ the Schrödinger equation is just that for a free particle,

$\frac{d^2 \psi(x)}{dx^2}=-\frac{2mE}{\hbar^2}\psi(x) = -k^{2}\psi,$

where

$k = \sqrt{\frac{2mE}{\hbar^2}}.$

The general solution for $x<0\!$ is

$\psi_{1}(x)=Ae^{ikx}+Be^{-ikx}, \!$

while that for $x>0 \!$ is

$\psi_{2}(x)=Fe^{ikx}+Ge^{-ikx}.\!$

As in the other cases, the wave function must be continuous at $x=0,\!$ so

$F+G=A+B.\!$

The derivative of the wave function, however, is not continuous, as noted when we studied the bound states. The discontinuity of the derivative is given by

$\left[\frac{d \psi_2(0)}{dx}-\frac{d \psi_1(0)}{dx}\right]-\frac{2mV_0}{\hbar^2}\psi(0)=0,$

which yields the relation,

$ik(F-G-A+B) = -\frac{2mV_0}{\hbar^2}(A+B),$

or

$F-G = A(1+2i\beta)-B(1-2i\beta), \!$

where

$\beta=\frac{mV_{0}}{\hbar^{2}k}.\!$

We once again assume that the incoming particles are coming from the left, so that $G=0.\!$ Consideration of the current densities on each side of the potential tells us that the reflection coefficient $R=\frac{|B|^2}{|A|^2}$ and that the transmission coefficient is $T=\frac{|F|^2}{|A|^2}.$ We thus wish to solve for $B\!$ and $F\!$ in terms of $A;\!$ doing so, we obtain

$B=\frac{i\beta}{1-i\beta}A$

and

$F=\frac{1}{1-i\beta}A.$

The reflection coefficient is thus

$R=\frac{\beta^{2}}{1+\beta^{2}},$

and the transmission coefficient is

$T=\frac{1}{1+\beta^{2}}.$

In terms of the energy, these become

$R=\frac{1}{1+(2\hbar^{2}E/mV_{0}^{2})}$ and $T=\frac{1}{1+(mV_{0}^{2}/2\hbar^{2}E)}.$