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I would like to determine the transmission coefficient $\mathcal{T}$ for a particle beam $$\Psi(x,t) = A_o e^{ikx}e^\frac{-iE_ot}{\hbar}$$ with energy $$E = \frac{\hbar^2k^2}{2m}$$ incident upon a negative linear potential (see diagram; I apologize for its crudeness, I made it myself) defined by:

$$V(x) = \begin{cases} \quad 0 &\text{for $x<0$}\\ -|V_o|(x/L) &\text{for $0<x<L$}\\ -|V_o| &\text{for $x>L$} \end{cases}$$

The transmission coefficient $\mathcal{T}$ is found by making use of probability density current $J$ as:

$$\mathcal{T} = \frac{|J_{trans}|}{|J_{inc}|}$$

where $|J_{trans}|$ and $|J_{inc}|$ are the transmitted and incident probability currents. Probability current $J$ can be obtained by:

$$J = \frac{i\hbar}{2m}\{\Psi(x,t) \frac{\partial}{\partial x}(\Psi^*(x,t))-\Psi^*(x,t)\frac{\partial}{\partial x}(\Psi(x,t))\}$$

So for the case of a single step-down potential (green line in diagram) we can solve the TISE in each region and employ boundary conditions to resolve coefficients and dependence on $k_1$ and $k_2$, which represent either incident/reflected beams, or transmitted beams respectively. Let the location of the single step be at $x=0$ instead of the $L/2$ shown in the diagram.

$$\psi(x) = \begin{cases} A_oe^{ik_1x} + Be^{-ik_1x}&\text{for $x<0$}\\ Ce^{ik_2x}+De^{-ik_2x} &\text{for $x>0$} \end{cases}$$ where $$(k_2)^2 = \frac{2m}{\hbar^2}(E+|V_o|)$$

$A_o$ represents the incident beam, $B$ represents the reflected beam, and $C$ represents the transmitted beam, and $D=0$. Imposing boundary conditions, we require the wavefunction $\psi$ and its derivative $\frac{d\psi}{dx}$ to be continuous at $x=0$, which leads to:

$$A_o + B = C$$ and $$ik_1(A_o-B) = ik_2C$$

The above conditions allow us to determine coefficients $B$ and $C$ in terms of our known $A_o$: $$B = \frac{A_o(k_1-k_2)}{k_1+k_2}$$ $$C = \frac{A_o2k_1}{k_1+k_2}$$

Noticing that the temporal part of $\Psi(x,t)$ goes to zero in the equation for $J$, we use $\psi(x) = A_o e^{ikx}$ to find the probability current of the incident beam, $J_{inc}$

$$J_{inc} = \frac{i\hbar}{2m}\{A_o e^{ik_1x}(- A_oik e^{-ik_1x})-A_o e^{-ik_1x} A_oik e^{ik_1x}\}$$

$$ = \frac{-2i^2|A_o|^2\hbar k_1}{2m}$$

$$J_{inc} = |A_o|^2 \frac{\hbar k_1}{m}$$

Consequently, $J_{refl}$ and $J_{trans}$ are given similarly by:

$$J_{refl} = -|B|^2\frac{\hbar k_1}{m}$$ $$J_{trans} = |C|^2\frac{\hbar k_2}{m} = (\frac{A_o2k_1}{k_1+k_2})^2 \frac{\hbar k_2}{m}$$

Finally, the transmission coefficient $\mathcal{T}$ can be found by $$\mathcal{T} = \frac{|J_{trans}|}{|J_{inc}|} = \frac{(\frac{A_o2k_1}{k_1+k_2})^2k_2}{|A_o|^2 k_1} = \frac{4k_1k_2}{(k_1+k_2)^2}$$

I apologize for the length up to this point, but now I wonder if I can estimate a transmission coefficient $\mathcal{T}$ for a linear potential (with $E>0$) like that in the diagram by using multiple step potentials along length $L$. The single step potential is not a good approximation because there is no $L$ dependence. So now if I examine the potential given by the red line in the diagram, I obtain

$$V(x) = \begin{cases} \quad 0 &\text{for $x<0$}\\ -|V_o|/3 &\text{for $0<x<L/3$}\\ -2|V_o|/3 &\text{for $L/3<x<2L/3$}\\ -|V_o| &\text{for $2L/3<x<L$, $x>L$}\\ \end{cases}$$

Now if I impose boundary conditions on $\psi$ and $\frac{d\psi}{dx}$ at each of the step down functions, I am essentially repeating what is shown above, but three times, with three different transmission beams, and seeing how the transmission coefficient changes between each region will let me find an expression for the transmission coefficient itself.

Is there an easier way to do this? Obviously if I modeled an infinite number of step potentials by breaking up $L$ and $V_o$ into infinite parts, this would give an ideal approximation for $\mathcal{T}$. Is there a way to do this mathematically/numerically, even graphically?


2 Answers 2


Have you considered finding an analytical solution?

On the interval $0\leq x\leq L$ the SE is:

$$-\frac{\hbar^2}{2m}\psi''-|V_0|\frac{x}{L}\psi=E\psi$$ $$\implies\psi''+\frac{2m}{\hbar^2}\Big[E+|V_0|\frac{x}{L}\Big]\psi=0$$ This type of DE has exact solutions, involving Airy functions $A_i$ and $B_i$ but as the latter tends to infinity, assume $c_2=0$.

Then with the boundary condition $\psi(0)=A_0$ it might be possible to find $\psi$ in $x=L$ and thus the transmission coefficient.

  • $\begingroup$ Could you elaborate on how I would use the boundary conditions for the method you've described? From my understanding, the transmission coefficient will require the ratio of incident and transmitted probability currents. So say I have the analytic solution for $\psi$ involving $A_i$, would I then calculate $J$ using the bounds imposed at $x=0,L$? It just isn't clear how I would obtain $\mathcal{T}$ from what you state in the last sentence $\endgroup$
    – bleuofblue
    Commented Nov 9, 2016 at 23:13

If you want to do it exactly, you need Airy functions for the wave solutions in the linear potential region and find the reflection and transmission coefficients by matching the wave functions and derivatives at the junctions. An approximate method would be the Wentzel-Kramers-Brillouin (WKB) approximation.

See https://en.wikipedia.org/wiki/WKB_approximation

  • $\begingroup$ I am interested in trying the WKB approximation, but am unsure how it would be applied for my question. I looked at the section on schrodinger eqn. How could I begin for my problem? $\endgroup$
    – bleuofblue
    Commented Nov 9, 2016 at 5:00
  • $\begingroup$ Also, is the method similar to what @Gert described in his answer? $\endgroup$
    – bleuofblue
    Commented Nov 9, 2016 at 5:00
  • $\begingroup$ @bleuofblue - I looked at the answer of Gert. He also mentions the Airy functions which are orthogonal solutions of the Schroedinger equation for the case of a linear potential just like sin and cos (or exponentials) are for constant potential. These solutions would be exact. The WKB method is an approximation for slowly varying potentials described in QM textbooks. $\endgroup$
    – freecharly
    Commented Nov 10, 2016 at 0:00

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