Scattering States, Transmission and Reflection: Difference between revisions

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{{Quantum Mechanics A}}
{{Quantum Mechanics A}}
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 <math>x\to\pm\infty,</math> 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 <math>T\!</math> and <math>R,\!</math> which give the probability that an incident wave will be transmitted and reflected, respectively.


==The Step Potential==
==The Step Potential==


Let's consider one dimensional potential step with an energy <math> E > V_0 \!</math>. That is, we have a potential
As a first example, let us consider a one-dimensional potential step. The potential in this case is given by
:<math> V(x) =  
:<math> V(x) =  
\begin{cases}
\begin{cases}
Line 11: Line 15:
</math>
</math>


The [[Schrödinger equation]] is  
The corresponding Schrödinger equation is  
:<math>
:<math>
\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) = E \psi(x).
\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) = E \psi(x).
</math>
</math>


If we divide the region I and the region II for each <math> x < 0 \!</math> and <math> x > 0 \! </math>, the Schrödinger equations for each region are
Let us first consider states with energy <math>E>V_0.\!</math>  We will divide the one-dimensional space into two regions - region I, for which <math> x < 0,\! </math> and region II, for which <math> x > 0.\!</math> The Schrödinger equations for the two regions are
 
:<math>
:<math>
- \frac{\hbar^2}{2m} \frac{d^2 \psi_{I}(x)}{dx^2} = E \psi_{I}(x),
- \frac{\hbar^2}{2m} \frac{d^2 \psi_{I}(x)}{dx^2} = E \psi_{I}(x)
</math>
</math>
and


:<math>
:<math>
\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V_0 \right) \psi_{II}(x) = E \psi_{II}(x).
\left( - \frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V_0 \right) \psi_{II}(x) = E \psi_{II}(x),
</math>
</math>


The general wave functions for each region are
and thus the corresponding wave functions are
:<math> \psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x}, </math>
:<math> \psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x} </math>
 
and
 
:<math> \psi_{II}(x) = C e^{i k x} + D e^{-i k x}, \!</math>
:<math> \psi_{II}(x) = C e^{i k x} + D e^{-i k x}, \!</math>
where
:<math> k_0 = \sqrt{\frac{2mE}{\hbar^2}} \mbox{    and    } k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}. </math>


The boundary conditions at <math> x=0 \!</math> require
where <math> k_0 = \sqrt{\frac{2mE}{\hbar^2}} </math> and <math>k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}.</math>  We may consider the first term in <math>\psi_{I}(x),\,Ae^{ik_0x},</math> to be an incident wave from the left, in which case the second term, <math>Be^{-ik_0x},</math> is a reflected wave from the potential barrier and the first term in <math>\psi_{II}(x)\!</math>, <math>Ce^{ikx},\!</math> is a transmitted wave.  Similarly, we can think of the second term in <math>\psi_{II}(x),\,De^{-ikx},</math> to be an incident wave from the right, in which case <math>Ce^{ikx}\!</math> is now the reflected wave and <math>Be^{-ik_0x}\!</math> is the transmitted wave.  These interpretations of the various terms will become more obvious shortly.
:<math> \psi_I(0) = \psi_{II}(0) \mbox{      and     } \left. \frac{d\psi_I(x)}{dx} \right|_{x=0} = \left. \frac{d\psi_{II}(x)}{dx} \right|_{x=0} </math>
 
and we have
The boundary conditions at <math> x=0\! </math> require
:<math> A + B = C + D, \!</math>
:<math> \psi_I(0) = \psi_{II}(0)\!</math>
 
and
 
:<math>\left. \frac{d\psi_I(x)}{dx} \right|_{x=0} = \left. \frac{d\psi_{II}(x)}{dx} \right|_{x=0}, </math>
 
which give us
 
:<math> A + B = C + D \!</math>
 
and
 
:<math> k_0\left(A-B\right) = k \left(C-D\right). </math>
:<math> k_0\left(A-B\right) = k \left(C-D\right). </math>


If we assume the waves incident from the left to the right, we can set <math> D = 0 \! </math>. In this case, reflection occurs at the potential step, and there is transmission to the right. We then have
If we assume that the wave is incident from the left, then we can set <math>D=0\!</math>. In this case, reflection occurs at the potential step, and there is transmission to the right. We then have
:<math> A + B = C, \!</math>
:<math> A + B = C \!</math>
 
and
 
:<math> \left(A-B\right) = \frac{k}{k_0} C. </math>
:<math> \left(A-B\right) = \frac{k}{k_0} C. </math>


From the above equations, we can get
From the above equations, we obtain
:<math> \frac{B}{A} = \frac{k_0-k}{k_0+k} \mbox{  and }
:<math> \frac{B}{A} = \frac{k_0-k}{k_0+k}</math>
\frac{C}{A} = \frac{2k}{k_0+k}. </math>
 
and
 
:<math>\frac{C}{A} = \frac{2k}{k_0+k}. </math>
 
To determine the probability of reflection and transmission, we must now find the current density,
 
:<math>j = \frac{\hbar}{2m i} \left (\psi^* \frac{d\psi}{dx} - \frac{d\psi^*}{dx} \psi \right ),</math>


The current density which is defined by
on each side of the barrier.  Doing so, we obtain
:<math> j = \frac{\hbar}{2m i} \left[ \psi^* \frac{d\psi}{dx} - \frac{d\psi^*}{dx} \psi \right] </math>
can be written by
:<math> j =  
:<math> j =  
\begin{cases}
\begin{cases}
  \displaystyle \frac{\hbar k_0}{m} \left( \left|A\right|^2 - \left|B\right|^2 \right) & \left( x < 0 \right), \\
  \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 & \left( x < 0 \right).
  \displaystyle \frac{\hbar k}{m} \left|C\right|^2, & x > 0.
\end{cases}
\end{cases}
</math>
</math>


The continuity of waves and the current density implies the relation
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'', <math>R=\frac{\left |B\right |^2}{\left |A\right |^2},</math> and the ratio of the transmitted current density to the incident current density as the ''transmission coefficient'', <math>T=\frac{k}{k_0}\frac{\left |C\right |^2}{\left |A\right |^2}.</math>
 
The continuity equation for the current density in one dimension implies conservation of current density, and thus that
:<math>R+T=1.\!</math>
 
If we determine the reflection and transmission coefficients for the problem at hand, we obtain
 
:<math>
:<math>
\frac{\left|B\right|^2}{\left|A\right|^2} + \frac{k}{k_0} \frac{\left|C\right|^2}{\left|A\right|^2} = 1.
  R = \frac{\left|B\right|^2}{\left|A\right|^2} = \frac{\left( k_0 - k \right)^2}{\left(k_0 + k \right)^2}
</math>
</math>


The first is called the ''reflection coefficient'' and the second term is called the ''transmission coefficient'' which are defined by
and
:<math>
  R = \frac{\left|B\right|^2}{\left|A\right|^2} = \frac{\left( k_0 - k \right)^2}{\left(k_0 + k \right)^2},
</math>


:<math>
:<math>
Line 71: Line 99:
</math>
</math>


Thus, we ensure that <math> R + T = 1 \! </math>.
We see that these expressions satisfy <math>R + T = 1,\!</math> as expected.


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


Now, let's consider <math> 0 < E < V_0 \! </math> case. In this case, <math> k_0 = \frac{\sqrt{2mE}}{\hbar} \! </math> is still real, but <math> k = \frac{\sqrt{2m\left(E-V_0\right)}}{\hbar} \!</math> is imaginary. If we define <math> \kappa \! </math> as a real value following as
:<math> \psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x} </math>
:<math> \kappa = \frac{\sqrt{2m\left(V_0-E\right)}}{\hbar},
 
</math>
and
then we have
 
:<math> k = i\kappa . \!</math>
:<math> \psi_{II}(x) = C e^{-\kappa x}, \!</math>
 
where <math>\kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.</math> Applying the same boundary conditions as before, we obtain
 
:<math> \frac{B}{A} = \frac{k_0-i\kappa}{k_0+i\kappa}</math>
 
and
 
:<math>\frac{C}{A} = \frac{2i\kappa}{k_0+i\kappa}. </math>


Therefore, we have
From the second wave function, <math>\psi_{II},\!</math> we see that the transmitted wave decreases exponentially over a length scale given by <math> \frac{1}{\kappa}.</math>.
:<math>
\psi_I = e^{ik_0x} + \frac{k_0-i\kappa}{k_0+i\kappa} e^{-ik_0 x} ,
</math>
:<math>
\psi_{II} = \frac{2k_0}{k_0 + i\kappa} e^{-\kappa x}.  
</math>


From the second wave function, <math> \psi_{II} \! </math>, we know that the transmitted waves decrease exponentially with relaxation length <math> \frac{1}{\kappa} \!</math>.
If we determine the current density in this case, we find that


If we calculate the reflection coefficient and the transmission coefficient in this case,
:<math> j =  
:<math> R = \left|\frac{j_{ref}}{j_{inc}}\right| = \left| \frac{k_0 - i\kappa}{k_0 + i\kappa} \right|^2 = 1,  
\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}
</math>
</math>
:<math> T = \left|\frac{j_{tr}}{j_{inc}}\right| = 0, </math>
because <math> j_{tr} = 0 \! </math>. That is, the incident waves are totally reflected.


The reflected waves have a phase difference from the incident waves. If we rewrite the wave function <math> \psi_I \! </math>,
We see that the wave is completely reflected; i.e., <math>R=1\!</math> and <math>T=0.\!</math>
 
The reflected wave acquires a phase difference relative to the incident wave. To see this, we simply rewrite the wave function <math> \psi_I, \! </math> as
:<math>  
:<math>  
\begin{align}
\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} \\
\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} \\
&= e^{ik_0x} + e^{i\theta} e^{-ik_0x},
&= 2 e^{i \frac{\theta}{2}} \cos \left( k_0 x - \frac{\theta}{2} \right),
\end{align}
\end{align}
</math>
</math>
where the phase difference of the reflected waves with respect to the incident waves defined by
where the phase difference <math>\theta\!</math> of the reflected wave with respect to the incident wave is given by
:<math> e^{i \theta} = \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2+\kappa^2}.</math>
:<math> e^{i \theta} = \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2+\kappa^2},</math>
Therefore,
or
:<math> \theta = \tan^{-1} \left( \frac{2\kappa k_0}{\kappa^2 - k_0^2} \right). </math>
:<math> \theta = \tan^{-1} \left( \frac{2\kappa k_0}{\kappa^2 - k_0^2} \right). </math>


==The Square Potential Barrier==
==The Square Potential Barrier==


For the square potential barrier with
For a square potential barrier, given by
:<math>  
:<math>  
V(x) =
V(x) =
\begin{cases}
\begin{cases}
0, & x < -a ,  \\
0, & x < -a ,  \\
V_0, & -a < x < 0, \\
V_0, & -a < x < a, \\
0, & a < a ,
0, & x > a ,
\end{cases}
\end{cases}
</math>
</math>
we can write the general solution of the Schrödinger equation for <math> 0 < E < V_0 \! </math> :
we can write the general solution of the Schrödinger equation for <math> 0 < E < V_0: \! </math>
:<math> \psi(x) =  
:<math> \psi(x) =  
\begin{cases}
\begin{cases}
  A e^{i k_0x} + B e^{-ik_0x} \equiv \psi_{I}(x) & \left( x < -a \right), \\
  A e^{i k_0x} + B e^{-ik_0x}, & x < -a, \\
  C e^{-\kappa x} + D e^{\kappa x} \equiv \psi_{II}(x) & \left( -a < x < 0 \right), \\
  C e^{-\kappa x} + D e^{\kappa x}, & -a < x < a, \\
  F e^{i k_0x} + G e^{-ik_0x} \equiv \psi_{III}(x) & \left( a < x \right),
  F e^{i k_0x} + G e^{-ik_0x}, & x > a,
\end{cases}
\end{cases}
</math>
</math>
where <math> k_0 = \frac{\sqrt{2mE}}{\hbar} \!</math> and <math> \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}\!</math>.
where <math> k_0 = \frac{\sqrt{2mE}}{\hbar} \!</math> and <math> \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar}.\!</math>


Whit the boundary conditions at <math> x = -a \! </math>, we have
The boundary conditions are the same as before; the boundary conditions at <math> x = -a, \! </math> give us
:<math>
:<math>
A e^{-ik_0 a} + B e^{ik_0a} = C e^{\kappa a} + D e^{-\kappa a }, </math>
A e^{-ik_0 a} + B e^{ik_0a} = C e^{\kappa a} + D e^{-\kappa a }, </math>
Line 136: Line 170:
A e^{-ik_0 a} - B e^{ik_0a} = \frac{i\kappa}{k_0} \left( C e^{\kappa a} - D e^{-\kappa a } \right). </math>
A e^{-ik_0 a} - B e^{ik_0a} = \frac{i\kappa}{k_0} \left( C e^{\kappa a} - D e^{-\kappa a } \right). </math>


Also, there are another boundary conditions at <math> x = a \! </math> and it requires
Similarly, the boundary conditions at <math> x = a \! </math> give us
:<math>
:<math>
C e^{-\kappa a} + D e^{\kappa a} = F e^{i k_0 a} + G e^{- i k_0 a} , </math>
C e^{-\kappa a} + D e^{\kappa a} = F e^{i k_0 a} + G e^{- i k_0 a} , </math>
Line 142: Line 176:
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). </math>
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). </math>


For the convenience, let's express the coefficients of these linear homogeneous relations in terms of matrices:
For the convenience, let us express the coefficients of these linear homogeneous relations in terms of matrices:
:<math>
:<math>
\left( \begin{align} A \\ B \end{align}\right)
\begin{bmatrix}
= \frac{1}{2} \left(\begin{align}
A \\
\left( 1 + \frac{i\kappa}{k_0} \right) e^{\kappa a + i k_0 a} \\
B
\left( 1 - \frac{i\kappa}{k_0} \right) e^{\kappa a - i k_0 a}  
\end{bmatrix}
\end{align} \right.
= \frac{1}{2}
\left. \begin{align}
\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}
\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{align} \right)
\end{bmatrix}
\left( \begin{align} C \\ D \end{align} \right)
\begin{bmatrix}
C \\
D
\end{bmatrix}
</math>
</math>


:<math>
:<math>
\left( \begin{align} C \\ D \end{align} \right)
\begin{bmatrix}
= \frac{1}{2} \left( \begin{align}
C \\
\left( 1 - \frac{ik_0}{\kappa} \right) e^{\kappa a + i k_0 a} \\
D
\left( 1 + \frac{ik_0}{\kappa} \right) e^{-\kappa a + i k_0 a}  
\end{bmatrix}
\end{align} \right.
= \frac{1}{2}
\left. \begin{align}
\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}
\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{align} \right)
\end{bmatrix}
\left( \begin{align} F \\ G \end{align} \right)
\begin{bmatrix}
F \\
G
\end{bmatrix}
</math>
</math>


If we combine these two equations, we have  
If we combine these two equations, we have  
:<math>
:<math>
\left( \begin{align} A \\ B \end{align} \right)
\begin{bmatrix}
= \left( \begin{align}
A \\
\left( \cosh 2\kappa a + \frac{i\varepsilon}{2} \sinh 2\kappa a \right) e^{2i k_0 a} \\
B
-\frac{i\eta}{2} \sinh 2 \kappa a  
\end{bmatrix}
\end{align} \right.
=\begin{bmatrix}
\left. \begin{align}
\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 \\
\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}
\left( \cosh 2 \kappa a - \frac{i\varepsilon}{2} \sinh 2 \kappa a \right) e^{- 2i k_0 a}
\end{bmatrix}
\end{align} \right)
\begin{bmatrix}
\left( \begin{align} F \\ G \end{align} \right)
F \\
G
\end{bmatrix}
</math>
</math>
where <math> \varepsilon = \frac{\kappa}{k_0} - \frac{k_0}{\kappa} \!</math> and <math> \eta = \frac{\kappa}{k_0} + \frac{k_0}{\kappa} \! </math>.


Note that <math> \eta^2 - \varepsilon^2 = 4 \! </math>.
where <math> \varepsilon = \frac{\kappa}{k_0} - \frac{k_0}{\kappa} \!</math> and <math> \eta = \frac{\kappa}{k_0} + \frac{k_0}{\kappa}. \! </math>
 
Note that <math> \eta^2 - \varepsilon^2 = 4. \! </math>
 
Let us now assume that there is only an incident wave coming from the left; i.e., <math>G=0.\!</math>  By similar arguments as before, we may identify the transmission coefficient as
 
<math>T=\frac{|F|^2}{|A|^2}=\frac{4}{4\cosh^2{2\kappa a}+\varepsilon^2\sinh^2{2\kappa a}}</math>
 
and the reflection coefficient as
 
<math>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}}.</math>


==The Dirac Delta Function Potential==
==Finite Asymmetric Square Well==
Consider the previous [[The Dirac Delta function potential|Dirac delta function potential]], but this time we have [[Scattering states|scattering states]] with <math> E>0 </math>. For <math> x<0 </math> the Schrödinger equation reads
 
We now consider an asymmetric square well potential, given by
:<math>  
V(x) =
\begin{cases}
V_1, & x < -a ,  \\
0, & -a < x < a, \\
V_2, & x > a.
\end{cases}
</math>


:<math>\frac{d^2 \psi(x)}{dx^2}=-\frac{2mE}{\hbar^2}\psi(x) = -k^{2}\psi</math>
In this case, the wave functions are given by
:<math> \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}
</math>


where
where
<math>k_{1}=\frac{\sqrt{2m(E-V_{1})}}{\hbar},\,k_{2}=\frac{\sqrt{2mE}}{\hbar},</math> and <math>k_{3}=\frac{\sqrt{2m(E-V_{2})}}{\hbar}.</math>
Applying the boundary conditions at <math>x=-a,\!</math> we obtain
<math>A_{1}e^{-ik_{1}a}+A_{2}e^{ik_{1}a}=B_{1}e^{-ik_{2}a}+B_{2}e^{ik_{2}a}</math>
and
<math>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},</math>


:<math> k = \sqrt{\frac{2mE}{\hbar^2}} </math>.
while those at <math>x=a\!</math> give us


The general solution is
<math>B_{1}e^{ik_{2}a}+B_{2}e^{-ik_{2}a}=C_{1}e^{ik_{3}a}+C_{2}e^{ik_{3}a}</math>


:<math> \psi_{1}(x)=Ae^{ikx}+Be^{-ikx} \!</math>.
and


Similarly, for <math> x>0 \!</math>,
<math>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}.</math>


:<math> \psi_{2}(x)=Fe^{ikx}+Ge^{-ikx} \!</math>.
We may express these in matrix form as


The continuity of <math> \psi(x)\!</math> at <math>x=0\!</math> requires that
<math>
\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}
</math>


:<math> F+G=A+B \!</math>
and


And the other boundary condition, the discontinuity of <math> \frac{d\psi(x)}{dx} \!</math> at <math> x = 0 \!</math>, can be obtained by integrating the Schrödinger equation from <math> -\epsilon \!</math> to <math>\epsilon \!</math> and then letting <math> \epsilon \rightarrow 0 \!</math>
<math>
\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}.
</math>


Integrating the whole equation across the potential gives
If we combine these, we obtain


:<math>
<math>
\int_{-\epsilon}^{\epsilon} \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{+\epsilon} (-V_0\delta(x))\psi(x)dx=\int_{-\epsilon}^{\epsilon} E \psi(x)dx
\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}.
</math>
</math>
in the limit <math>\epsilon \rightarrow 0 \!</math>, we have


:<math>\frac{-\hbar^2}{2m}\left[\frac{d \psi_2(0)}{dx}-\frac{d \psi_1(0)}{dx}\right]+V_0\psi(0)=0</math>
Let us once again assume that there is only an incident wave from the left, so that <math>C_{2}=0.\!</math>  Using the same arguments as before, we may identify the transmission coefficient as
 
<math>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}</math>


which yields the relation:
and the reflection coefficient as


:<math> ik(F-G-A+B) = -\frac{2mV_0}{\hbar^2}(A+B) </math>.
<math>R=\frac{|A_2|^2}{|A_1|^2}=\frac{(k_1-k_3)^2}{(k_1+k_3)^2}.</math>


or, more compactly,
==The Dirac Delta Function Potential==
We now consider scattering from a [[The Dirac Delta Function Potential|Dirac delta function potential]], <math>V(x)=V_0\delta(x).\!</math>  For <math> x\neq 0,\!</math> the [[Schrödinger Equation|Schrödinger equation]] is just that for a free particle,


:<math> F-G = A(1+2i\beta)-B(1-2i\beta), \!</math>
:<math>\frac{d^2 \psi(x)}{dx^2}=-\frac{2mE}{\hbar^2}\psi(x) = -k^{2}\psi,</math>


where
where


:<math> \beta= \frac{mV_{0}}{\hbar^{2}k} \!</math>.
:<math> k = \sqrt{\frac{2mE}{\hbar^2}}. </math>


In atypical scattering experiment particles are fired in from one direction-let's say, from the left. In that case the amplitude of the wave coming in from the right will be zero:
The general solution for <math>x<0\!</math> is


:<math> G=0 \!</math> (for scattering from the left).
:<math> \psi_{1}(x)=Ae^{ikx}+Be^{-ikx}, \!</math>


<math>A\!</math> is then the amplitude of the '''incident wave''', <math>B\!</math> is the amplitude of the '''reflected wave''', and <math>f\!</math> is the amplitude of the '''transmitted wave'''. Solving the equations of boundary conditions, we find
while that for <math> x>0 \!</math> is


:<math> B= \frac{i\beta}{1-i\beta}A,</math>.
:<math> \psi_{2}(x)=Fe^{ikx}+Ge^{-ikx}.\!</math>


and
As in the other cases, the wave function must be continuous at <math>x=0,\!</math> so
 
:<math> F+G=A+B.\!</math>
 
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
 
:<math>\left[\frac{d \psi_2(0)}{dx}-\frac{d \psi_1(0)}{dx}\right]-\frac{2mV_0}{\hbar^2}\psi(0)=0,</math>
 
which yields the relation,
 
:<math> ik(F-G-A+B) = -\frac{2mV_0}{\hbar^2}(A+B),</math>
 
or
 
:<math> F-G = A(1+2i\beta)-B(1-2i\beta), \!</math>
 
where


:<math> F= \frac{1}{1-i\beta}A,</math>.
:<math>\beta=\frac{mV_{0}}{\hbar^{2}k}.\!</math>


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


:<math> R= \frac{|B|^{2}}{|A|^{2}}=\frac{\beta^{2}}{1+\beta^{2}},</math>
:<math>B=\frac{i\beta}{1-i\beta}A</math>


meanwhile, the transmission coefficient:
and


:<math> T= \frac{|F|^{2}}{|A|^{2}}=\frac{1}{1+\beta^{2}}</math>.
:<math>F=\frac{1}{1-i\beta}A.</math>


Of course, the sum of these two coefficients should be 1, and it is:
The reflection coefficient is thus


:<math> R+T=1\!</math>.
:<math>R=\frac{\beta^{2}}{1+\beta^{2}},</math>


Notice that <math> R\!</math> and <math> T\!</math> are functions of <math>\beta\!</math>, and hence of <math>E\!</math>:
and the transmission coefficient is


:<math> R=\frac{1}{1+(2\hbar^{2}E/mV_{0}^{2})} </math>, 
:<math>T=\frac{1}{1+\beta^{2}}.</math>
:<math> T=\frac{1}{1+(mV_{0}^{2}/2\hbar^{2}E)}</math>.


In terms of the energy, these become


These results for R and T are considering the conditions of the potential such as the parameter k is the same in both regions. Is a good exercise to compare this result with the transmission and reflection coefficients in the step potential.
<math> R=\frac{1}{1+(2\hbar^{2}E/mV_{0}^{2})}</math> and <math> T=\frac{1}{1+(mV_{0}^{2}/2\hbar^{2}E)}.</math>

Latest revision as of 15:44, 2 August 2013

Quantum Mechanics A
SchrodEq.png
Schrödinger Equation
The most fundamental equation of quantum mechanics; given a Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , it describes how a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.
Basic Concepts and Theory of Motion
UV Catastrophe (Black-Body Radiation)
Photoelectric Effect
Stability of Matter
Double Slit Experiment
Stern-Gerlach Experiment
The Principle of Complementarity
The Correspondence Principle
The Philosophy of Quantum Theory
Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
Some Consequences of the Uncertainty Principle
Linear Vector Spaces and Operators
Commutation Relations and Simultaneous Eigenvalues
The Schrödinger Equation in Dirac Notation
Transformations of Operators and Symmetry
Time Evolution of Expectation Values and Ehrenfest's Theorem
One-Dimensional Bound States
Oscillation Theorem
The Dirac Delta Function Potential
Scattering States, Transmission and Reflection
Motion in a Periodic Potential
Summary of One-Dimensional Systems
Harmonic Oscillator Spectrum and Eigenstates
Analytical Method for Solving the Simple Harmonic Oscillator
Coherent States
Charged Particles in an Electromagnetic Field
WKB Approximation
The Heisenberg Picture: Equations of Motion for Operators
The Interaction Picture
The Virial Theorem
Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x) = \begin{cases} 0, & x < 0, \\ V_0, & x > 0. \end{cases} }

The corresponding Schrödinger equation is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E>V_0.\!} We will divide the one-dimensional space into two regions - region I, for which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x < 0,\! } and region II, for which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x > 0.\!} The Schrödinger equations for the two regions are

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle - \frac{\hbar^2}{2m} \frac{d^2 \psi_{I}(x)}{dx^2} = E \psi_{I}(x) }

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x} }

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{II}(x) = C e^{i k x} + D e^{-i k x}, \!}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_0 = \sqrt{\frac{2mE}{\hbar^2}} } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}.} We may consider the first term in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{I}(x),\,Ae^{ik_0x},} to be an incident wave from the left, in which case the second term, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Be^{-ik_0x},} is a reflected wave from the potential barrier and the first term in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{II}(x)\!} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Ce^{ikx},\!} is a transmitted wave. Similarly, we can think of the second term in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{II}(x),\,De^{-ikx},} to be an incident wave from the right, in which case Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Ce^{ikx}\!} is now the reflected wave and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Be^{-ik_0x}\!} is the transmitted wave. These interpretations of the various terms will become more obvious shortly.

The boundary conditions at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0\! } require

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_I(0) = \psi_{II}(0)\!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. \frac{d\psi_I(x)}{dx} \right|_{x=0} = \left. \frac{d\psi_{II}(x)}{dx} \right|_{x=0}, }

which give us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A + B = C + D \!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D=0\!} . In this case, reflection occurs at the potential step, and there is transmission to the right. We then have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A + B = C \!}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(A-B\right) = \frac{k}{k_0} C. }

From the above equations, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{B}{A} = \frac{k_0-k}{k_0+k}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{C}{A} = \frac{2k}{k_0+k}. }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R+T=1.\!}

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R = \frac{\left|B\right|^2}{\left|A\right|^2} = \frac{\left( k_0 - k \right)^2}{\left(k_0 + k \right)^2} }

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R + T = 1,\!} as expected.

Now let us consider the case in which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < E < V_0.\!} If we again assume that the wave is incident from the left, the wave functions become

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{I}(x) = A e^{i k_0x} + B e^{-ik_0x} }

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{II}(x) = C e^{-\kappa x}, \!}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.} Applying the same boundary conditions as before, we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{B}{A} = \frac{k_0-i\kappa}{k_0+i\kappa}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{C}{A} = \frac{2i\kappa}{k_0+i\kappa}. }

From the second wave function, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{II},\!} we see that the transmitted wave decreases exponentially over a length scale given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\kappa}.} .

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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., and

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

where the phase difference of the reflected wave with respect to the incident wave is given by

or

The Square Potential Barrier

For a square potential barrier, given by

we can write the general solution of the Schrödinger equation for

where and

The boundary conditions are the same as before; the boundary conditions at give us

Similarly, the boundary conditions at give us

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

If we combine these two equations, we have

where and

Note that

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

and the reflection coefficient as

Finite Asymmetric Square Well

We now consider an asymmetric square well potential, given by

In this case, the wave functions are given by

where and

Applying the boundary conditions at we obtain

and

while those at give us

and

We may express these in matrix form as

and

If we combine these, we obtain

Let us once again assume that there is only an incident wave from the left, so that Using the same arguments as before, we may identify the transmission coefficient as

and the reflection coefficient as

The Dirac Delta Function Potential

We now consider scattering from a Dirac delta function potential, For the Schrödinger equation is just that for a free particle,

where

The general solution for is

while that for is

As in the other cases, the wave function must be continuous at so

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

which yields the relation,

or

where

We once again assume that the incoming particles are coming from the left, so that Consideration of the current densities on each side of the potential tells us that the reflection coefficient and that the transmission coefficient is We thus wish to solve for and in terms of doing so, we obtain

and

The reflection coefficient is thus

and the transmission coefficient is

In terms of the energy, these become

and

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