Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|improve this question
Is there anything that prevents you from just eliminating $B$, $C$ and $D$ from the four equations you listed? –  Slaviks Apr 22 '13 at 12:58
I don't know. Why would i just eliminate those? –  71GA Apr 25 '13 at 5:26

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
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. –  71GA Apr 25 '13 at 5:28
yes, as I stated, you have to find an inverse matrix, multiply it by the RHS and that gives you (B,C,D,E) –  Jimself Apr 25 '13 at 17:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.