[![enter image description here](https://i.stack.imgur.com/AxJcG.jpg)](https://i.stack.imgur.com/AxJcG.jpg)

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 $$\mathcal{T} = \frac{|J\_{trans}|}{|J\_{inc}|} = \frac{(\frac{A\_o2k\_1}{k\_1+k\_2})^2k\_2}{|A\_o|^2 k\_1} = \frac{4k\_1k\_2}{(k\_1+k\_2)^2}$$

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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?