Lets say we have a tunelling problem in the picture, where $W_p$ is a finite potential step:

enter image description here

If particle is comming from the left a general solutions to the Schrödinger equations for sepparate intervals I, II and II are:

\begin{align} \text{I:}& & \psi_1 &= \overbrace{A e^{i\mathcal L x}}^{\psi_{in}} + \overbrace{Be^{-i \mathcal L x}}^{\psi_{re}}& \mathcal L &= \sqrt{\tfrac{2mW}{\hbar^2}}\\ \text{II:}& & \psi_2 &= C e^{\mathcal K x} + De^{-\mathcal K x}& \mathcal K &= \sqrt{-\tfrac{2m(W-W_p)}{\hbar^2}}\\ \text{III:}& & \psi_3 &= \underbrace{E e^{i \mathcal L x}}_{\psi_{tr}}& &\\ \end{align}

Where $\psi_{in}$ is an incomming wave, $\psi_{re}$ is a reflected wave and $\psi_{tr}$ is transmitted wave. I used the boundary conditions and got a system of 4 equations:

\begin{align} {\tiny\text{boundary}}&{\tiny\text{conditions at x=0:}} & {\tiny\text{boundary conditions}}&{\tiny\text{at x=d:}}\\ A + B &= C + D & Ce^{\mathcal K d} + De^{-\mathcal K d} &= E e^{i \mathcal L d}\\ i \mathcal L A - i \mathcal L B &= \mathcal KC - \mathcal K D & \mathcal K C e^{\mathcal K d} - \mathcal K D e^{-\mathcal K d}&= i \mathcal L E e^{i \mathcal L d} \end{align}

So now i decided to calculate coefficient of transmission $T$:

\begin{align} T &= \dfrac{|j_{tr}|}{|j_{in}|} \!=\! \Bigg|\dfrac{\dfrac{\hbar }{2mi}\! \left( \dfrac{d\overline{\psi}_{tr}}{dx}\, \psi_{tr} - \dfrac{d \psi_{tr}}{dx}\, \overline{\psi}_{tr} \right)}{\dfrac{\hbar}{2mi} \!\left( \dfrac{d\overline{\psi}_{in}}{dx}\, \psi_{in} - \dfrac{d\psi_{in}}{dx}\, \overline{\psi}_{in} \right) }\Bigg| \!=\! \Bigg|\dfrac{\frac{d}{dx}\big(\overbrace{Ee^{-i\mathcal L x}}^{\text{konjug.}}\big) Ee^{i\mathcal L x} - \frac{d}{dx} \left( Ee^{i\mathcal L x}\right)\! \overbrace{Ee^{-i\mathcal L x}}^{\text{konjug.}}}{ \frac{d}{dx}\big(\underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}\big) Ae^{i\mathcal L x} - \frac{d}{dx} \left( Ae^{i\mathcal L x}\right)\! \underbrace{Ae^{-i\mathcal L x}}_{\text{konjug.}}}\Bigg|\! = \nonumber\\ &=\Bigg|\dfrac{-i\mathcal L Ee^{-i\mathcal L x} E e^{i \mathcal L x} - i\mathcal L E e^{i \mathcal L x} Ee^{-i \mathcal L x}}{-i \mathcal L A e^{-i\mathcal L x} Ae^{i \mathcal L x} - i \mathcal L A e^{i \mathcal L x}Ae^{-i \mathcal L x} }\Bigg|=\Bigg|\dfrac{-i\mathcal L E^2 - i\mathcal L E^2}{-i \mathcal L A^2 - i \mathcal L A^2}\Bigg|=\Bigg|\dfrac{-2 i \mathcal L E^2}{-2i\mathcal L A^2}\Bigg| = \frac{|E|^2}{|A|^2} \end{align}

It accured to me that if out of 4 system equations i can get amplitude ratio $E/A$, i can calculate $T$ quite easy. Could anyone show me how do i get this ratio?

  • 1
    $\begingroup$ Is there anything that prevents you from just eliminating $B$, $C$ and $D$ from the four equations you listed? $\endgroup$
    – Slaviks
    Apr 22, 2013 at 12:58
  • $\begingroup$ I don't know. Why would i just eliminate those? $\endgroup$
    – 71GA
    Apr 25, 2013 at 5:26
  • $\begingroup$ The algebra here is quite tricky the first time you meet it, You can find it all worked through in wiki, I think, and also here: users.physics.ox.ac.uk/~Steane/teaching/waves_on_barrier.pdf $\endgroup$ Dec 6, 2018 at 13:16

1 Answer 1


Strictly speaking, you have 4 equations and 5 unknowns. However, given that the coefficient A is applied to the incoming wave-function, you could arbitrarily set it equal to 1 (because it represents 100% of the wave) and solve the system of equations for E. Then $T=E$. This is how the problem is handled in most cases. Alternatively, if you absolutely cannot set $A=1$, then try assuming A is a given and solve the 4 equations for B, C, D, and E in terms of A. Then, again, perform $T=E/A$.

In theory, the ratio for any A will be the same as for A=1.

(I checked, it is, the A divides out in the end).


You can easily solve for B,C,D, and E using matrices, where your four system equations are:

$$\begin{pmatrix}-1 & 1 & 1 & 0 \\ i \mathcal L & \mathcal K & -\mathcal K & 0 \\ 0 & e^{\mathcal Kd} & e^{-\mathcal Kd} & -e^{i\mathcal Ld} \\ 0 & \mathcal Ke^{\mathcal Kd} & -\mathcal Ke^{-\mathcal Kd} & -i\mathcal Le^{i\mathcal Ld}\end{pmatrix} \begin{pmatrix}B \\ C \\ D \\ E\end{pmatrix}=\begin{pmatrix}A \\ i\mathcal LA \\ 0 \\ 0\end{pmatrix} $$

Optionally, $A=1$. But if you invert the matrix and solve for E, you should get:

$$E={4iA\mathcal K\mathcal L\over\mathcal K^2 e^{i\mathcal Ld-\mathcal Kd}-\mathcal K^2 e^{d\mathcal K+id\mathcal L}+2i\mathcal L\mathcal Ke^{id\mathcal L-d\mathcal K}+2i\mathcal L\mathcal Ke^{id\mathcal L+d\mathcal K}-\mathcal L^2 e^{id\mathcal L-d\mathcal K}+\mathcal L^2 e^{id\mathcal L+d\mathcal K}}$$

And, of course, A=1

  • $\begingroup$ This is the first time i came across a matrix solving of 4 system equation. I understand how you wrote down a a matrix form, but i would need some explaination on, how you got the equation for $E$ in the end. I mean do i have to find an inverse matrix? Please be descriptive. $\endgroup$
    – 71GA
    Apr 25, 2013 at 5:28
  • $\begingroup$ yes, as I stated, you have to find an inverse matrix, multiply it by the RHS and that gives you (B,C,D,E) $\endgroup$
    – Jim
    Apr 25, 2013 at 17:03

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