Editing S-matrix (section)

==== Finite square barrier ====
The square ''barrier'' is similar to the square well with the difference that <math>V(x)=+V_0 > 0</math> for <math>|x|\le a</math>.

There are three different cases depending on the energy eigenvalue <math>E_k=\frac{\hbar^2 k^2}{2m}</math> of the plane waves (with wave numbers {{mvar|k}} resp. {{math|−''k''}}) far away from the barrier:
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| <math>E_k > V_0</math>: In this case <math>l = \sqrt{k^2-\frac{2mV_0}{\hbar^2}}</math> and the formulas for <math>S_{ij}</math> have the same form as is in the square well case, and the transmission is <math>T_k = |S_{21}|^2 = |S_{12}|^2 = \frac{1}{1+(\sin(2la))^2 \frac{(l^2 - k^2)^2}{4 k^2 l^2}}</math>

| <math>E_k = V_0</math>: In this case <math>\sqrt{k^2-\frac{2mV_0}{\hbar^2}} = 0</math> and the wave function <math>\psi(x)</math> has the property <math>\psi''(x)=0</math> inside the barrier and
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<math>S_{12}=S_{21}=\frac{\exp(-2ika)}{1-ika}</math>  and <math>S_{11} = S_{22} = \frac{-ika\cdot\exp(-2ika)}{1-ika}</math>
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The transmission is: <math>T_k=\frac{1}{1+k^2 a^2}</math>. This intermediate case is not singular, it's the limit (<math>l \to 0</math> resp. <math>\kappa \to 0</math>) from both sides.

| <math>E_k < V_0</math>:In this case <math>\sqrt{k^2-\frac{2mV_0}{\hbar^2}}</math> is an imaginary number. So the wave function inside the barrier has the components <math>e^{\kappa x}</math> and <math>e^{-\kappa x}</math> with <math>\kappa = \sqrt{\frac{2mV_0}{\hbar^2}-k^2}</math>. 
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The solution for the S-matrix is:<ref name="ucsd.edu">{{cite web |url=https://quantummechanics.ucsd.edu/ph130a/130_notes/node152.html |title=The Potential Barrier |access-date=1 November 2022 |website=quantummechanics.ucsd.edu}}</ref> <math>S_{12} = S_{21} = \frac{\exp(-2ika)}{\cosh(2\kappa a)-i\sinh(2\kappa a)\frac{k^2-{\kappa}^2}{2k\kappa}}</math> 
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and likewise: <math>S_{11}=-i\frac{k^2+\kappa^2}{2k\kappa}\sinh(2\kappa a)\cdot S_{12}</math> and also in this case <math>S_{22}=S_{11}</math>.
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The transmission is <math>T_k=|S_{21}|^2=|S_{12}|^2=\frac{1}{1+(\sinh(2\kappa a))^2\frac{(k^2+\kappa^2)^2}{4k^2\kappa^2}}</math>.
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