c
c###################################################################
c
      subroutine heexcen(n, l, mult, sumen, eexc, kermsg)
c
c     this subroutine passes the excitation energy for excited states
c     of  helium taken from tables a.4 and a.5 given by janev et al.
c     (see doc=h-he-plasma.)
c
c     the input subroutine parameters are
c
c    n = principal quantum number of the excited electronic state
c
c    l = orbital angular momentum quantum number of the exited state
c
c    mult = the spin multiplicity (2s+1) of the state
c
c    sumen =  indicates choice of excitation enegy to be returned.
c             if sumen=0  the excitation energy for the specific
c              state (quantified by n, l amd mult) is returned
c             if sumen=1  the excitation energy taken as an average
c             over angular momentum and toatal spin is returned
c
c     the output subroutine parameters are
c
c    eexc = excitation energy
c
c    kermsg = blank if no errors
c
c     written by j. j. smith , iaea atomic and molecular data unit
c
c------------------------------------------------------------------------
c
      character*(*) kermsg
      integer n, l, mult, sumen, lp, multp
      dimension enl(4,4,2), en(7)
      data (((enl(i,j,k),i=1,4),j=1,4),k=1,2) /32*0.0/
c
c----   n=1 state
c
c     data enl(1,1,1)
c    *    / 24.588 /,
c
c----   n=2 excited states
c
c    1   enl(2,1,1),enl(2,1,2),enl(2,2,1),enl(2,2,2)
c    *    / 20.614, 19.818, 21.217, 20.963 /,
c
c----   n=3 excited states
c
c    2   enl(3,1,1),enl(3,1,2),enl(3,2,1),enl(3,2,2),enl(3,3,1),
c    *   enl(3,3,2)
c    *    / 22.919, 22.717, 23.086, 23.006, 23.073, 23.072 /,
c
c----   n=4 excited states
c
c    3   enl(4,1,1),enl(4,1,2),enl(4,2,1),enl(4,2,2),enl(4,3,1),
c    *   enl(4,3,2),enl(4,4,1),enl(4,4,2)
c    *    / 23.672, 23.529, 23.741, 23.706, 23.735, 23.735, 23.736,
c    *     23.736 /
c
c-----   excitation energies summed over angular momentum and
c-----   total spin
c
      enl(1,1,1)=24.588
c----   n=2 excited states
      enl(2,1,1)=20.614
      enl(2,1,2)=19.818
      enl(2,2,1)=21.217
      enl(2,2,2)=20.963
c----   n=3 excited states
      enl(3,1,1)=22.919
      enl(3,1,2)=22.717
      enl(3,2,1)=23.086
      enl(3,2,2)=23.006
      enl(3,3,1)=23.073
      enl(3,3,2)=23.072
c----   n=4 excited states
      enl(4,1,1)=23.672
      enl(4,1,2)=23.529
      enl(4,2,1)=23.741
      enl(4,2,2)=23.706
      enl(4,3,1)=23.735
      enl(4,3,2)=23.735
      enl(4,4,1)=23.736
      enl(4,4,2)=23.736
      data (en(k),k=1,2) /2*0.0/
      data en(3),en(4),en(5),en(6),en(7)
     *    / 22.9799 , 23.699, 24.07, 24.30, 24.71/
c
      kermsg =' '
      if (sumen .eq. 1 .or. n .gt. 4) then
        if (n .ge. 8) then
          eexc =enl(1,1,1) - 13.58/(n*n)
          return
        else
           eexc =en(n)
           if (eexc .eq. 0.0) kermsg =
     *        'excitation energy for n value not in table in heexcen'
        endif
      else
           lp = l+1
           if (mult .eq. 1) then
             multp = 1
           else if (mult .eq. 3) then
               multp = 2
             else
                 kermsg = 'invalid spin multipclity as input in heexcen'
             endif
           eexc = enl(n,lp,multp)
           if (eexc .eq. 0.0) kermsg =
     *        'excitation energy not in table in heexcen'
      endif
      return
      end
c##################################################################
c
c###################################################################
c
      subroutine heionen(n, l, mult, sumen, eion, kermsg)
c
c     this subroutine passes the ionization energy for excited states
c     of  helium taken from tables a.4 and a.5 given by janev et al.
c     (see doc=h-he-plasma.)
c
c     the input subroutine parameters are
c
c    n = principal quantum number of the excited electronic state
c
c    l = orbital angular momentum quantum number of the exited state
c
c    mult = the spin multiplicity (2s+1) of the state
c
c    sumen =  indicates choice of ionization energy to be returned.
c             if sumen=0  the ionization energy for the specific
c              state (quantified by n, l amd mult) is returned
c             if sumen=1  the ionization energy taken as an average
c             over angular momentum and toatal spin is returned
c
c     the output subroutine parameters are
c
c    eion = ionization energy
c
c    kermsg = blank if no errors
c
c     written by j. j. smith , iaea atomic and molecular data unit
c
c------------------------------------------------------------------------
c
      character*(*) kermsg
      integer n, l, mult, sumen, lp, multp
      dimension enl(4,4,2), en(7)
      data (((enl(i,j,k),i=1,4),j=1,4),k=1,2) /32*0.0/
c
c     data enl(1,1,1)
c    *    / 24.588 /,
c
c----   n=2 excited states
c
c    1  enl(2,1,1),enl(2,1,2),enl(2,2,1),enl(2,2,2)
c    *    / 3.973, 4.769, 3.371, 3.625 /,
c
c----   n=3 excited states
c
c    2   enl(3,1,1),enl(3,1,2),enl(3,2,1),enl(3,2,2),enl(3,3,1),
c    *   enl(3,3,2)
c    *    / 1.699, 1.871, 1.502, 1.582, 1.515, 1.515 /,
c
c----   n=4 excited states
c
c    3   enl(4,1,1),enl(4,1,2),enl(4,2,1),enl(4,2,2),enl(4,3,1),
c    *   enl(4,3,2),enl(4,4,1),enl(4,4,2)
c    *    / 0.9155, 0.9951, 0.8470, 0.8812, 0.8529, 0.8530, 0.8520,
c    *      0.8520 /
c
c-----   ionization energies summed over angular momentum and
c-----   total spin
c
      data (en(k),k=1,2) /2*0.0/
      data en(3),en(4),en(5),en(6),en(7)
     *    / 1.609 , 0.8811, 0.52, 0.37, 0.29/
c
      enl(1,1,1)=24.588
c----   n=2 excited states
      enl(2,1,1)=3.973
      enl(2,1,2)=4.769
      enl(2,2,1)=3.371
      enl(2,2,2)=3.625
c----   n=3 excited states
      enl(3,1,1)=1.699
      enl(3,1,2)=1.871
      enl(3,2,1)=1.502
      enl(3,2,2)=1.582
      enl(3,3,1)=1.515
      enl(3,3,2)=1.515
c----   n=4 excited states
      enl(4,1,1)=0.9155
      enl(4,1,2)=0.9951
      enl(4,2,1)=0.8470
      enl(4,2,2)=0.8812
      enl(4,3,1)=0.8529
      enl(4,3,2)=0.8530
      enl(4,4,1)=0.8520
      enl(4,4,2)=0.8520
      kermsg =' '
      if (sumen .eq. 1) then
        if (n .ge. 8) then
          eion =13.58/(n*n)
          return
        else
           eion =en(n)
           if (eion .eq. 0.0) kermsg =
     *        'ionization energy for n value not in table in heionen'
        endif
      else
           lp = l+1
           if (mult .eq. 1) then
             multp = 1
           else if (mult .eq. 3) then
               multp = 2
             else
                 kermsg = 'invalid spin multipclity as input in heionen'
             endif
           eion = enl(n,lp,multp)
           if (eion .eq. 0.0) kermsg =
     *        'ionization energy not in table in heionen'
      endif
      return
      end
c###################################################################
c
c###################################################################
c
      subroutine jeexc3(pe, pcf, kncf, pxs, kermsg)
c
c     this is a subroutine to calculate a cross section
c     in cm[2] as function of electron energy in ev for dipole
c     forbidden transitions with spin change.
c     for details see doc=h-he-plasma , used for reactions 2.3.3
c     spin changing transitions only
c
c     pcf is the coefficient data array, where
c
c     pcf(1)  =   coefficient a
c
c     pcf(2)  =   coefficient b
c
c     pcf(3)  =   n, the principal quantum number of the final state
c
c     pcf(4)  =   l, the angular momentum of the the final state
c
c    - warning- .
c
c        the coefficient array pcf is updated by this routine to
c     include energy independent constants. these coefficients can be
c     used in subsequent calls for the same entry. the coefficeients
c     added are:
c
c     pcf(5) = threshold excitation energy for the transition (ev)
c
c     pcf(6) = ionization energy of the state
c
c     pe = electron energy (ev)
c
c     kermsg = blank if no errors
c
c     pxs = cross section in cm[2]
c
c     written by j. j. smith , iaea atomic and molecular data unit
c     modified by d. humbert, 11 March 2008:  n <==> l
c
c------------------------------------------------------------------------
c
      double precision pe, pcf, pxs
      double precision u3, u5, u9, u, nstar, bc
      integer n, l
      dimension pcf(6)
      character*(*) kermsg
      data ry/13.58/

      a= pcf(1)
      b = pcf(2)
      n = pcf(4)
      l = pcf(3)
c
      kermsg = ' '
      if (kncf .lt. 5) then
c
c        first call to jeexc3 determine energy independent
c        parameters and place in pcf for further use
c
c        determine the excitation energy of the final state
c
         call heexcen (n, l,  3, 0,  eexc, kermsg)
         if (kermsg .ne.  ' ') return
c
c        determine the ionization energy of the final state
c
         call heionen (n, l,  3, 0,  eion, kermsg)
         if ( kermsg .ne.  ' ') return
c
c        place energy independent parameters in coefficient array and
c        update kncf
c
         pcf(5) = eexc
         pcf(6) = eion
         kncf = 6
c
      else if (kncf .eq. 6) then
          eexc = pcf(5)
          eion = pcf(6)
c
      else
          kermsg = ' incorrect number of coefficients passed to jeexc3'
          return
      endif
c
      if (n .ge. 4) then
        nstar = sqrt(ry/eion)
        ac = a /(n**3)
      else
         ac = a
      endif
c
      if (pe .lt. eexc) then
        pxs = 0.0
        return
      endif
c
      u = pe / eexc
      u3 = u**3
      u5 = u**5
      u9 = u**9
c
      if (b .gt. 0.0) then
        bc = b/u9
      else
        bc = 0.0
      endif
c
      pxs = 3.52e-16 * ( ac * (1.0/u3 - 1.0/u5) + bc )
c
      return
c
      end

def heexcen(n, l, mult, sumen):
   """
    This function passes the excitation energy for excited states of  helium 
    taken from tables a.4 and a.5 given by janev et al. (Elementary processes in H-He plasmsas.
 
    n: principal quantum number of the excited electronic state
    l: orbital angular momentum quantum number of the exited state
    mul: the spin multiplicity (2s+1) of the state
    sumen: indicates choice of excitation enegy to be returned.
    """
    enl = np.zeros((4, 4, 2))
    en = np.zeros(7)
    
    # Define the excitation energy values
    enl[0, 0, 0] = 24.588
    enl[1, 0, 0] = 20.614
    enl[1, 0, 1] = 19.818
    enl[1, 1, 0] = 21.217
    enl[1, 1, 1] = 20.963
    enl[2, 0, 0] = 22.919
    enl[2, 0, 1] = 22.717
    enl[2, 1, 0] = 23.086
    enl[2, 1, 1] = 23.006
    enl[2, 2, 0] = 23.073
    enl[2, 2, 1] = 23.072
    enl[3, 0, 0] = 23.672
    enl[3, 0, 1] = 23.529
    enl[3, 1, 0] = 23.741
    enl[3, 1, 1] = 23.706
    enl[3, 2, 0] = 23.735
    enl[3, 2, 1] = 23.735
    enl[3, 3, 0] = 23.736
    enl[3, 3, 1] = 23.736
    en[2:5] = [22.9799, 23.699, 24.07, 24.30, 24.71]
    
    if sumen == 1 or n > 4:
        if n >= 8:
            eexc = enl[0, 0, 0] - 13.58 / (n * n)
        else:
            try:
                eexc = en[n - 1]
            except IndexError:
                raise ValueError('Excitation energy for n value not in table in heexcen')
    else:
        lp = l + 1
        if mult == 1:
            multp = 1
        elif mult == 3:
            multp = 2
        else:
            raise ValueError('Invalid spin multiplicity as input in heexcen')
        
        try:
            eexc = enl[n - 1, lp - 1, multp - 1]
        except IndexError:
            raise ValueError('Excitation energy not in table in heexcen')
    
    return eexc

###############################################################
def heionen(n, l, mult, sumen):
    """
    This function passes the ionization energy for excited states of helium 
    taken from tables a.4 and a.5 given by janev et al.
 
    n: principal quantum number of the excited electronic state
    l: orbital angular momentum quantum number of the exited state
    mult: the spin multiplicity (2s+1) of the state
    sumen: indicates choice of ionization energy to be returned.
    """
    enl = np.zeros((4, 4, 2))
    en = np.zeros(7)
    
    # Define the ionization energy values
    enl[0, 0, 0] = 24.588
    enl[1, 0, 0] = 3.973
    enl[1, 0, 1] = 4.769
    enl[1, 1, 0] = 3.371
    enl[1, 1, 1] = 3.625
    enl[2, 0, 0] = 1.699
    enl[2, 0, 1] = 1.871
    enl[2, 1, 0] = 1.502
    enl[2, 1, 1] = 1.582
    enl[2, 2, 0] = 1.515
    enl[2, 2, 1] = 1.515
    enl[3, 0, 0] = 0.9155
    enl[3, 0, 1] = 0.9951
    enl[3, 1, 0] = 0.8470
    enl[3, 1, 1] = 0.8812
    enl[3, 2, 0] = 0.8529
    enl[3, 2, 1] = 0.8530
    enl[3, 3, 0] = 0.8520
    enl[3, 3, 1] = 0.8520
    en[2:5] = [1.609, 0.8811, 0.52, 0.37, 0.29]
    
    eion = 0.0
    
    if sumen == 1:
        if n >= 8:
            eion = 13.58 / (n * n)
        else:
            try:
                eion = en[n - 1]
            except IndexError:
                raise ValueError('Ionization energy for n value not in table in heionen')
    else:
        lp = l + 1
        if mult == 1:
            multp = 1
        elif mult == 3:
            multp = 2
        else:
            raise ValueError('Invalid spin multiplicity as input in heionen')
        
        try:
            eion = enl[n - 1, lp - 1, multp - 1]
        except IndexError:
            raise ValueError('Ionization energy not in table in heionen')
    
    return eion

##############################################################
def jeexc3(pe, pcf, kncf):
    """
    This is a function to calculate a cross section in cm^2 as a function of
    electron energy in eV for dipole forbidden transitions with spin change.

    pe: electron energy (eV)
    pcf: coefficient data array
    kncf: number of coefficients
    """

    a = pcf[0]
    b = pcf[1]
    n = pcf[3]
    l = pcf[2]

    ry = 13.58

    # Initialize energy-independent parameters
    eexc = 0.0
    eion = 0.0

    if kncf < 5:
        # First call to jeexc3 determines energy-independent parameters
        # and places them in pcf for further use

        try:
            eexc = heexcen(n, l, 3, 0)
            eion = heionen(n, l, 3, 0)
        except ValueError:
            raise ValueError("Error calculating excitation or ionization energy")

        # Place energy-independent parameters in the coefficient array and update kncf
        pcf[4] = eexc
        pcf[5] = eion
        kncf = 6

    elif kncf == 6:
        eexc = pcf[4]
        eion = pcf[5]
    else:
        raise ValueError("Incorrect number of coefficients passed to jeexc3")

    if n >= 4:
        nstar = np.sqrt(ry / eion)
        ac = a / (n ** 3)
    else:
        ac = a

    if pe < eexc:
        pxs = 0.0
    else:
        u = pe / eexc
        u3 = u ** 3
        u5 = u ** 5
        u9 = u ** 9

        if b > 0.0:
            bc = b / u9
        else:
            bc = 0.0

        pxs = 3.52e-16 * (ac * (1.0 / u3 - 1.0 / u5) + bc)

    return pxs