Scattering States, Transmission and Reflection

The step potential

Let's consider one dimensional potential step with an 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 \!} . That is, we have a potential

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 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). }

If we divide the region I and the region II for each 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 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 each region 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), }

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). }

The general wave functions for each region 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}, }

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}} \mbox{ and } k = \sqrt{\frac{2m(E-V_0)}{\hbar^2}}. }

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

and we 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 + D, \!}

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 the waves incident from the left to the right, 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, \!}

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 can get

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

The current density which is defined 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 j = \frac{\hbar}{2m i} \left[ \psi^* \frac{d\psi}{dx} - \frac{d\psi^*}{dx} \psi \right] }

can be written 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 j = \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}{m} \left|C\right|^2 & \left( x < 0 \right). \end{cases} }

The continuity of waves and the current density implies the relation

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{\left|B\right|^2}{\left|A\right|^2} + \frac{k}{k_0} \frac{\left|C\right|^2}{\left|A\right|^2} = 1. }

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

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}. }

Thus, we ensure 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 \! } .

Now, let's consider 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 \! } case. In this 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 k_0 = \frac{\sqrt{2mE}}{\hbar} \! } is still real, but 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 = \frac{\sqrt{2m\left(E-V_0\right)}}{\hbar} \!} is imaginary. If we define 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 \! } as a real value following as

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\left(V_0-E\right)}}{\hbar}, }

then we 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 k = i\kappa . \!}

Therefore, we 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 \psi_I = e^{ik_0x} + \frac{k_0-i\kappa}{k_0+i\kappa} e^{-ik_0 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 \psi_{II} = \frac{2k_0}{k_0 + i\kappa} e^{-\kappa x}. }

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 know that the transmitted waves decrease exponentially with relaxation length 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 calculate the reflection coefficient and the transmission coefficient in this 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 R = \left|\frac{j_{ref}}{j_{inc}}\right| = \left| \frac{k_0 - i\kappa}{k_0 + i\kappa} \right|^2 = 1, }

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 = \left|\frac{j_{tr}}{j_{inc}}\right| = 0, }

because 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_{tr} = 0 \! } . 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 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 \! } ,

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 \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} \\ &= 2 e^{i \frac{\theta}{2}} \cos \left( k_0 x - \frac{\theta}{2} \right), \end{align} }

where the phase difference of the reflected waves with respect to the incident waves defined 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 e^{i \theta} = \frac{k_0^2 - \kappa^2 - 2i\kappa k_0}{k_0^2+\kappa^2}.}

Therefore,

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

The square potential barrier

For the square potential barrier with

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 < -a , \\ V_0, & -a < x < 0, \\ 0, & a < a , \end{cases} }

we can write the general solution of the Schrödinger equation for 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 \! }  :

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(x) = \begin{cases} A e^{i k_0x} + B e^{-ik_0x} \equiv \psi_{I}(x) & \left( x < -a \right), \\ C e^{-\kappa x} + D e^{\kappa x} \equiv \psi_{II}(x) & \left( -a < x < 0 \right), \\ F e^{i k_0x} + G e^{-ik_0x} \equiv \psi_{III}(x) & \left( a < x \right), \end{cases} }

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

Whit 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 = -a \! } , we 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 e^{-ik_0 a} + B e^{ik_0a} = C e^{\kappa a} + D e^{-\kappa a }, }

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

Also, there are another 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 = a \! } and it requires

$Ce^{-\kappa a}+De^{\kappa a}=Fe^{ik_{0}a}+Ge^{-ik_{0}a},$

$Ce^{-\kappa a}-De^{\kappa a}=-{\frac {ik_{0}}{\kappa }}\left(Fe^{ik_{0}a}-Ge^{-ik_{0}a}\right).$

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

$\left({\begin{aligned}A\\B\end{aligned}}\right)={\frac {1}{2}}\left({\begin{aligned}\left(1+{\frac {i\kappa }{k_{0}}}\right)e^{\kappa a+ik_{0}a}\\\left(1-{\frac {i\kappa }{k_{0}}}\right)e^{\kappa a-ik_{0}a}\end{aligned}}\right.\left.{\begin{aligned}\left(1-{\frac {i\kappa }{k_{0}}}\right)e^{-\kappa a+ik_{0}a}\\\left(1+{\frac {i\kappa }{k_{0}}}\right)e^{-\kappa a-ik_{0}a}\end{aligned}}\right)\left({\begin{aligned}C\\D\end{aligned}}\right)$

$\left({\begin{aligned}C\\D\end{aligned}}\right)={\frac {1}{2}}\left({\begin{aligned}\left(1-{\frac {ik_{0}}{\kappa }}\right)e^{\kappa a+ik_{0}a}\\\left(1+{\frac {ik_{0}}{\kappa }}\right)e^{-\kappa a+ik_{0}a}\end{aligned}}\right.\left.{\begin{aligned}\left(1+{\frac {ik_{0}}{\kappa }}\right)e^{\kappa a-ik_{0}a}\\\left(1-{\frac {ik_{0}}{\kappa }}\right)e^{-\kappa a-ik_{0}a}\end{aligned}}\right)\left({\begin{aligned}F\\G\end{aligned}}\right)$

If we combine these two equations, we have

$\left({\begin{aligned}A\\B\end{aligned}}\right)=\left({\begin{aligned}\left(\cosh 2\kappa a+{\frac {i\varepsilon }{2}}\sinh 2\kappa a\right)e^{2ik_{0}a}\\-{\frac {i\eta }{2}}\sinh 2\kappa a\end{aligned}}\right.\left.{\begin{aligned}{\frac {i\eta }{2}}\sinh 2\kappa a\\\left(\cosh 2\kappa a-{\frac {i\varepsilon }{2}}\sinh 2\kappa a\right)e^{-2ik_{0}a}\end{aligned}}\right)\left({\begin{aligned}F\\G\end{aligned}}\right)$

where $\varepsilon ={\frac {\kappa }{k_{0}}}-{\frac {k_{0}}{\kappa }}\!$ and $\eta ={\frac {\kappa }{k_{0}}}+{\frac {k_{0}}{\kappa }}\!$.

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

The Dirac delta function potential

Consider the previous Dirac delta function potential, but this time we have scattering states with $E>0$. For $x<0$ the Schrödinger equation reads

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

where

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

The general solution is

$\psi _{1}(x)=Ae^{ikx}+Be^{-ikx}\!$.

Similarly, for $x>0\!$,

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

The continuity of $\psi (x)\!$ at $x=0\!$ requires that

$F+G=A+B\!$

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

Integrating the whole equation across the potential gives

$\int _{-\epsilon }^{\epsilon }{\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}dx+\int _{-\epsilon }^{+\epsilon }(-V_{0}\delta (x))\psi (x)dx=\int _{-\epsilon }^{\epsilon }E\psi (x)dx$

in the limit $\epsilon \rightarrow 0\!$, we have

${\frac {-\hbar ^{2}}{2m}}\left[{\frac {d\psi _{2}(0)}{dx}}-{\frac {d\psi _{1}(0)}{dx}}\right]+V_{0}\psi (0)=0$

which yields the relation:

$ik(F-G-A+B)=-{\frac {2mV_{0}}{\hbar ^{2}}}(A+B)$.

or, more compactly,

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

where

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

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:

$G=0\!$ (for scattering from the left).

$A\!$ is then the amplitude of the incident wave, $B\!$ is the amplitude of the reflected wave, and $f\!$ is the amplitude of the transmitted wave. Solving the equations of boundary conditions, we find

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

and

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

Now, the reflection coefficient:

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

meanwhile, the transmission coefficient:

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

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

$R+T=1\!$.

Notice that $R\!$ and $T\!$ are functions of $\beta \!$, and hence of $E\!$:

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

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

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.