0
$\begingroup$

Do you know any way besides Conjugate to normalize the following symmetric function and calculate "ParSym"? I realized that Conjugate does not allow calculations.

phi1[n1_,l1_,m1_,Z_,r1_,theta1_,
phi1_]:=(((2*Z)/n1)^3*(n1-l1-1)!/(2*n1*((n1+l1)!)))^(1/2)*
Exp[-((Z*r1)/n1)]*((2*Zr1/n1)^l1)*
LaguerreL[n1-l1-1,2*l1+1,2*Z*r1/n1]*
SphericalHarmonicY[l1,m1,theta1,phi1]

phi2[n2_,l2_,m2_,Z_,r2_,theta2_,
phi2_]:=(((2*Z)/n2)^3*(n2-l2-1)!/(2*n2*((n2+l2)!)))^(1/2)*
Exp[-((Z*r2)/n2)]*((2*Zr2/n2)^l2)*
LaguerreL[n2-l2-1,2*l2+1,2*Z*r2/n2]*
SphericalHarmonicY[l2,m2,theta2,phi2]

PsiSym[r1_,r2_,theta1_,phi1_,theta2_,phi2_,Z_]:=
1/Sqrt[2]*((phi1[1,0,0,Z,r1,theta1,phi1]*
phi2[2,1,1,Z,r2,theta2,phi2])+(phi2[1,0,0,Z,r2,theta2,phi2]*phi1[2,1,1,Z,r1,theta1,phi1]))

NormSym[Z_]:=
NIntegrate[
PsiSym[r1,r2,theta1,phi1,theta2,phi2,Z]*
Conjugate[PsiSym[r1,r2,theta1,phi1,theta2,phi2,Z]]*r1^2*
r2^2*Sin[theta1]*Sin[theta2],{r1,0,Infinity},{r2,0,
Infinity},{theta1,0,Pi},{theta2,0,Pi},{phi1,0,
2*Pi},{phi2,0,2*Pi}]

ParSym[Z_]:=-NIntegrate[
PsiSym[r1,r2,theta1,phi1,theta2,phi2,Z]*
Conjugate[PsiSym[r1,r2,theta1,phi1,theta2,phi2,Z]]*
Log[PsiSym[r1,r2,theta1,phi1,theta2,phi2,Z]*
Conjugate[PsiSym[r1,r2,theta1,phi1,theta2,phi2,Z]]]*r1^2*
r2^2*Sin[theta1]*Sin[theta2],{r1,0,Infinity},{r2,0,
Infinity},{theta1,0,Pi},{theta2,0,Pi},{phi1,0,
2*Pi},{phi2,0,2*Pi}]

Table[{Z,NormSym[Z],ParSym[Z]},{Z,2,30}]
$\endgroup$
  • $\begingroup$ Your PsiSymC is not defined. Please try your own code before posting. $\endgroup$ – Roman Jul 3 '19 at 8:23
  • $\begingroup$ I think the only thing that needs to be complex-conjugated in your expressions is the phase of the spherical harmonics. Therefore, replacing phi with -phi will do the trick. $\endgroup$ – Roman Jul 3 '19 at 8:26
  • $\begingroup$ Maybe define PsiSymC[r1_, r2_, theta1_, phi1_, theta2_, phi2_, Z_] := ComplexExpand[Conjugate[PsiSym[r1, r2, theta1, phi1, theta2, phi2, Z]]] under the assumption that all variables are real. $\endgroup$ – Roman Jul 3 '19 at 10:37
  • $\begingroup$ Hi. I was wrong to name functions and variables like phi1 and phi2. That was the mistake, Conjugate works fine. An apology. $\endgroup$ – Arturo García Flores Jul 4 '19 at 4:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.