Editing S-matrix (section)

== In one-dimensional quantum mechanics ==
A simple prototype in which the ''S''-matrix is 2-dimensional is considered first, for the purposes of illustration. In it, particles with sharp energy {{math|''E''}} scatter from a localized potential {{math|''V''}} according to the rules of 1-dimensional quantum mechanics. Already this simple model displays some features of more general cases, but is easier to handle.

Each energy {{math|''E''}} yields a matrix {{math|1=''S'' = ''S''(''E'')}} that depends on {{math|''V''}}. Thus, the total ''S''-matrix could, figuratively speaking, be visualized, in a suitable basis, as a "continuous matrix" with every element zero except for {{math|2 × 2}}-blocks along the diagonal for a given {{math|''V''}}.

=== Definition ===
Consider a localized one dimensional [[potential barrier]] {{math|''V''(''x'')}}, subjected to a beam of quantum particles with energy {{math|''E''}}. These particles are incident on the potential barrier from left to right.

The solutions of the [[Schrödinger equation]] outside the potential barrier are [[plane waves]] given by
<math display="block">\psi_{\rm L}(x)= A e^{ikx} + B e^{-ikx}</math>
for the region to the left of the potential barrier, and
<math display="block">\psi_{\rm R}(x)= C e^{ikx} + D e^{-ikx}</math>
for the region to the right to the potential barrier, where
<math display="block">k=\sqrt{2m E/\hbar^{2}}</math>
is the [[wave vector]]. The time dependence is not needed in our overview and is hence omitted. The term with coefficient {{math|''A''}} represents the incoming wave, whereas term with coefficient {{math|''C''}} represents the outgoing wave. {{math|''B''}} stands for the reflecting wave. Since we set the incoming wave moving in the positive direction (coming from the left), {{math|''D''}} is zero and can be omitted.

The "scattering amplitude", i.e., the transition overlap of the outgoing waves with the incoming waves is a linear relation defining the ''S''-matrix,
<math display="block">\begin{pmatrix} B \\ C \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix} \begin{pmatrix} A \\ D \end{pmatrix}.</math>

The above relation can be written as
<math display="block">\Psi_{\rm out}=S \Psi_{\rm in}</math>
where
<math display="block">\Psi_{\rm out}=\begin{pmatrix}B \\ C \end{pmatrix}, \quad \Psi_{\rm in}=\begin{pmatrix}A \\ D \end{pmatrix}, \qquad S=\begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix}.</math>
The elements of {{math|''S''}} completely characterize the scattering properties of the potential barrier {{math|''V''(''x'')}}.

=== Unitary property ===
The unitary property of the ''S''-matrix is directly related to the conservation of the [[probability current]] in [[quantum mechanics]].

The probability current density {{mvar|J}} of the [[wave function]] {{math|''ψ''(''x'')}} is defined as
<math display="block"> J = \frac{\hbar}{2mi}\left(\psi^* \frac{\partial \psi }{\partial x}- \psi \frac{\partial \psi^* }{\partial x} \right) .</math>
The probability current density <math>J_{\rm L}(x)</math> of <math>\psi_{\rm L}(x)</math> to the left of the barrier is
<math display="block">J_{\rm L}(x)=\frac{\hbar k}{m}\left(|A|^2-|B|^2\right),</math>
while the probability current density <math>J_{\rm R}(x)</math> of <math>\psi_{\rm R}(x)</math> to the right of the barrier is
<math display="block">J_{\rm R}(x)=\frac{\hbar k}{m}\left(|C|^2-|D|^2\right).</math>

For conservation of the probability current, {{math|1=''J''<sub>L</sub> = ''J''<sub>R</sub>}}. This implies the ''S''-matrix is a [[unitary matrix]].
{{Cleanup|proof|reason=Proofs should be in prose|date=January 2025}}
{{math proof|proof=<math display="block">
\begin{align}
 & J_{\rm L} = J_{\rm R} \\
 & |A|^2-|B|^2=|C|^2-|D|^2 \\
 & |B|^2+|C|^2=|A|^2+|D|^2 \\
 & \Psi_\text{out}^\dagger \Psi_\text{out}=\Psi_\text{in}^\dagger \Psi_\text{in} \\
 & \Psi_\text{in}^\dagger S^\dagger S \Psi_\text{in}=\Psi_\text{in}^\dagger \Psi_\text{in} \\
 & S^\dagger S = I\\
\end{align}</math>}}

=== Time-reversal symmetry ===
If the potential {{math|''V''(''x'')}} is real, then the system possesses [[time-reversal symmetry]]. Under this condition, if {{math| ''ψ''(''x'')}} is a solution of the Schrödinger equation, then {{math| ''ψ''*(''x'')}} is also a solution.

The time-reversed solution is given by
<math display="block">\psi^*_{\rm L}(x)= A^* e^{-ikx} + B^* e^{ikx}</math>
for the region to the left to the potential barrier, and
<math display="block">\psi^*_{\rm R}(x)= C^* e^{-ikx} + D^* e^{ikx}</math>
for the region to the right to the potential barrier,
where the terms with coefficient {{math|''B''*}}, {{math|''C''*}} represent incoming wave, and terms with coefficient {{math|''A''*}}, {{math|''D''*}} represent outgoing wave.

They are again related by the ''S''-matrix,
<math display="block">\begin{pmatrix}A^* \\ D^* \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix}\begin{pmatrix} B^* \\ C^* \end{pmatrix}\,</math>
that is,<br />
<math display="block">\Psi^*_{\rm in}=S \Psi^*_{\rm out}.</math>
Now, the relations
<math display="block">\Psi^*_{\rm in} = S \Psi^*_{\rm out}, \quad \Psi_{\rm out}=S \Psi_{\rm in}</math>
together yield a condition
<math display="block">S^*S=I</math>
This condition, in conjunction with the unitarity relation, implies that the ''S''-matrix is symmetric, as a result of time reversal symmetry,
<math display="block">S^T=S.</math>

By combining the symmetry and the unitarity, the S-matrix can be expressed in the form:
<math display="block">\begin{pmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{pmatrix} = \begin{pmatrix} e^{i\varphi} e^{i\delta} \cdot r & e^{i\varphi} \sqrt{1-r^2}  \\ e^{i\varphi}\sqrt{1-r^2} & -e^{i\varphi} e^{-i\delta} \cdot r  \end{pmatrix} = e^{i\varphi} \begin{pmatrix} e^{i\delta} \cdot r & \sqrt{1-r^2} \\ \sqrt{1-r^2} & -e^{-i\delta} \cdot r  \end{pmatrix}</math> 
with <math>\delta,\varphi \in [0;2\pi]</math> and <math>r\in [0;1]</math>. So the S-matrix is determined by three real parameters.

==== Transfer matrix ====
The ''transfer matrix'' <math>M</math> relates the plane waves <math>C e^{ikx}</math> and <math>D e^{-ikx}</math> on the ''right'' side of scattering potential to the plane waves <math>A e^{ikx}</math> and <math>B e^{-ikx}</math> on the ''left'' side:<ref name="cas.cz">{{cite web |url=https://gemma.ujf.cas.cz/~krejcirik/AAMP13/slides/Mostafazadeh.pdf |website=gemma.ujf.cas.cz |title=Transfer Matrix Formulation of Scattering Theory in Arbitrary Dimensions |access-date=29 October 2022}}</ref>  

<math display="block">\begin{pmatrix}C \\ D \end{pmatrix} = \begin{pmatrix} M_{11} & M_{12} \\ M_{21} & M_{22} \end{pmatrix}\begin{pmatrix} A \\ B \end{pmatrix}</math> and its components can be derived from the components of the S-matrix via:<ref name="ucr.edu">{{cite web |url=https://intra.ece.ucr.edu/~korotkov/courses/EE201/EE201-slides/EE201-Lec-6.pdf |title=EE201/MSE207 Lecture 6 |website=intra.ece.ucr.edu |access-date=29 October 2022}}</ref> <math>M_{11}=1/S_{12}^*= 1/S_{21} ^* {,}\  M_{22}= M_{11}^*</math> and <math>M_{12}=-S_{11}^*/S_{12}^* = S_{22}/S_{12} {,}\ M_{21} = M_{12}^*</math>, whereby time-reversal symmetry is assumed.

In the case of time-reversal symmetry, the transfer matrix <math>\mathbf{M}</math> can be expressed by three real parameters:

<math display="block">M = \frac{1}{\sqrt{1-r^2}} \begin{pmatrix} e^{i\varphi} & -r\cdot e^{-i\delta} \\ -r\cdot e^{i\delta} & e^{-i\varphi} \end{pmatrix}</math> 
with <math>\delta,\varphi \in [0;2\pi]</math> and <math>r\in [0;1]</math> (in case {{math|1=''r'' = 1}} there would be no connection between the left and the right side)

==== 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>

==== 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:
{{unordered list
| <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
{{pb}}
<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>
{{pb}}
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>. 
{{pb}}
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> 
{{pb}}
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>.
{{pb}}
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>.
}}

=== Transmission coefficient and reflection coefficient ===
The [[transmission coefficient]] from the left of the potential barrier is, when {{math|1=''D'' = 0}},
<math display="block">T_{\rm L}=\frac{|C|^2}{|A|^2} = |S_{21}|^2. </math>

The [[reflection coefficient]] from the left of the potential barrier is, when {{math|1=''D'' = 0}},
<math display="block">R_{\rm L}=\frac{|B|^2}{|A|^2}=|S_{11}|^2.</math>

Similarly, the transmission coefficient from the right of the potential barrier is, when {{math|1=''A'' = 0}},
<math display="block">T_{\rm R}=\frac{|B|^2}{|D|^2}=|S_{12}|^2.</math>

The reflection coefficient from the right of the potential barrier is, when {{math|1=''A'' = 0}},
<math display="block">R_{\rm R}=\frac{|C|^2}{|D|^2}=|S_{22}|^2.</math>

The relations between the transmission and reflection coefficients are
<math display="block">T_{\rm L}+R_{\rm L}=1</math>
and
<math display="block">T_{\rm R}+R_{\rm R}=1.</math>
This identity is a consequence of the unitarity property of the ''S''-matrix.

With time-reversal symmetry, the S-matrix is symmetric and hence <math>T_{\rm L}=|S_{21}|^2=|S_{12}|^2=T_{\rm R}</math> and <math>R_{\rm L} = R_{\rm R}</math>.

=== Optical theorem in one dimension ===
In the case of [[free particle]]s {{math|1=''V''(''x'') = 0}}, the ''S''-matrix is<ref>{{harvnb|Merzbacher|1961}} Ch 6. A more common convention, utilized below, is to have the ''S''-matrix go to the identity in the free particle case.</ref>
<math display="block"> S=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.</math>
Whenever {{math|''V''(''x'')}} is different from zero, however, there is a departure of the ''S''-matrix from the above form, to
<math display="block"> S = \begin{pmatrix} 2ir & 1+2it \\ 1+2it &2ir^* \frac{1+2it}{1-2it^*} \end{pmatrix}.</math>
This departure is parameterized by two [[complex functions]] of energy, {{math|''r''}} and {{math|''t''}}.
From unitarity there also follows a relationship between these two functions,
<math display="block">|r|^2+|t|^2 = \operatorname{Im}(t).</math>

The analogue of this identity in three dimensions is known as the [[optical theorem]].