Let us denote the potentials for Scatters 1 and 2 by 
and
, respectively. These two potentials are related to each
other by reflection about the point L/2 so that 
.
This linear mapping may be verified by noting that under this transform
the point -a maps to 
and the point 0 maps to
, and from Figure 4, which are indeed
the corresponding points in the two potentials.
If 
is the scattering solution for , by symmetry we
expect that 
will be a solution for the
scattering problem associated with the potential . This is
verified by inserting 
directly into the TISE for the
potential ,
Algebraically, this means that if we start with the pictured in
4c,
where ?(x) indicates the unspecified form for the solution within
the scattering region, and then form 
, we
find a solution to the
TISE for Scatterer 2 of the form
This algebraic relationship may be summarized in a very intuitive way by stating that we may flip around any Feynman diagram, as indicated in Figure 5, and produce a valid solution to the TISE.
Figure 5: Reflection symmetry expressed in Feynman Diagrams
The connection between the quantum amplitudes comes about when we
inspect the result of the flipping process. The result in Figure
5 shows a perfect, properly normalized scattering
wave function, with a unit wave approaching Scatterer 2 from the left
and no returning current from the side opposite the source. We may
therefore read the quantum amplitudes for left-incident reflection
and transmission from Scatter 2 from the new diagram as 
and 
,
respectively. However, from Figure 4c we see that
these quantities were originally defined as 
and 
.
Therefore we conclude
that the quantum amplitudes for reflection and transmission when approaching mirror image potentials from opposite directions are exactly equal, which is consistent with what we expected on physical grounds.
