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====== Sub ======

###
We can subtract two response functions from each other. Note that the residue of a pole always needs to be positive. For $G = G_1 - G_2$ we start by setting $G=G_1$. For each pole in $G_2$ we look for the poles in $G_1$ at the same energy (a's). If this is present we subtract the residue of $G_2$ (b's) from $G$. If $G_2$ contains a pole at an energy that is not present in $G_1$ we subtract the residue from the pole with lower and higher energy such that the first moment locally is conserved.
###

===== Example =====

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For the case where the poles in $G_2$ are also present in $G_1$ we find
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==== Input ====
<code Quanty Example.Quanty>
a1 = {0, -1,-0.5, 0,   0.5,   1, 1.5}
b1 = {  0.2, 0.4, 0.2, 0.4, 0.6, 0.2}
G1 = ResponseFunction.New( {a1,b1,mu=0,type="ListOfPoles", name="G1"} )


a1 = {0, -1,-0.5, 0,   0.5,   1, 1.5}
b1 = {  0.1, 0.4, 0.1, 0.4, 0.3, 0.2}
G2 = ResponseFunction.New( {a1,b1,mu=0,type="ListOfPoles", name="G2"} )

G3 = G1-G2
print(G3)
</code>
with the output
==== Result ====
<file Quanty_Output>
{ { 0 , -1 , 0 , 1 } , 
  { 0.1 , 0.1 , 0.3 } ,
  name = G1 ,
  type = ListOfPoles ,
  mu = 0 }
</file>


===== Example =====

###
For the case where the poles in $G_2$ not present in $G_1$ we find
###

==== Input ====
<code Quanty Example.Quanty>
a1 = {0, -1,   0,   5,   6}
b1 = {  0.2, 0.2, 0.3, 0.3}
G1 = ResponseFunction.New( {a1,b1,mu=0,type="ListOfPoles", name="G1"} )


a1 = {0, -0.5, 5.75}
b1 = {    0.2,  0.2}
G2 = ResponseFunction.New( {a1,b1,mu=0,type="ListOfPoles", name="G2"} )

G3 = G1-G2
print(G3)
</code>
with the output
==== Result ====
<file Quanty_Output>
{ { 0 , -1 , 0 , 5 , 6 } , 
  { 0.1 , 0.1 , 0.25 , 0.15 } ,
  name = G1 ,
  mu = 0 ,
  type = ListOfPoles }
</file>

The pole in $G_2$ at energy $-0.5$ is equally divided between the poles at $-1$ and $0$ in $G_1$, such that each of these residues is reduced by $0.2/2=0.1$. The pole in $G_2$ at energy $5.75$ is divided between the poles at $5$ and $6$ in $G_1$. In oder to conserve the first moment we subtract $0.2 * (5.75 - 5) / (6-5) = 0.15$ from the residue at $6$ in $G_1$ and $0.2 * (6 - 5.75) / (6-5) = 0.05$ from the residue at $5$ in $G_1$.

===== Table of contents =====
{{indexmenu>../#2|tsort}}