Editing S-matrix (section)

==== Finite square well ====
The one-dimensional, non-relativistic problem with time-reversal symmetry of a particle with mass m that approaches a (static) [[finite potential well|finite square ''well'']], has the potential function {{math|''V''}} with
<math display="block">V(x) = \begin{cases}
 -V_0 & \text{for}~~ |x| \le a ~~ (V_0 > 0) \quad\text{and}\\[1ex]
 0 & \text{for}~~ |x|>a
\end{cases}</math>
The scattering can be solved by decomposing the [[wave packet]] of the free particle into plane waves <math>A_k\exp(ikx)</math> with wave numbers <math>k>0</math> for a plane wave coming (faraway) from the left side or likewise <math>D_k\exp(-ikx)</math> (faraway) from the right side.

The S-matrix for the plane wave with wave number {{mvar|k}} has the solution:<ref name="ucr.edu"/>
<math display="block">S_{12}=S_{21}=\frac{\exp(-2ika)}{\cos(2la)-i\sin(2la)\frac{l^2+k^2}{2kl}}</math>
and <math>S_{11}=S_{12}\cdot i\sin(2la)\frac{l^2-k^2}{2kl}</math>  ; hence <math>e^{i\delta}=\pm i</math> and therefore  <math>-e^{-i\delta}=e^{i\delta}</math> and <math>S_{22}=S_{11}</math> in this case.

Whereby <math>l = \sqrt{k^2+\frac{2mV_0}{\hbar^2}}</math> is the (increased) wave number of the plane wave inside the square well, as the energy [[eigenvalue]] <math>E_k</math> associated with the plane wave has to stay constant: <math>E_k = \frac{\hbar^2 k^2}{2m}=\frac{\hbar^2 l^2}{2m}-V_0</math>

The transmission is <math>T_k = |S_{21}|^2=|S_{12}|^2=\frac{1}{(\cos(2la))^2+(\sin(2la))^2\frac{(l^2+k^2)^2}{4k^2l^2}}=\frac{1}{1+(\sin(2la))^2\frac{(l^2-k^2)^2}{4 k^2 l^2}}</math>

In the case of <math>\sin(2la)=0</math> then <math>\cos(2la)=\pm 1</math> and therefore <math>S_{11} = S_{22} = 0</math> and <math>|S_{21}| = |S_{12}| = 1</math> i.e. a plane wave with wave number k passes the well without reflection if <math>k^2+\frac{2mV_0}{\hbar^2}=\frac{n^2 \pi^2}{4a^2}</math> for a <math>n\in\mathbb{N}</math>