#!/usr/local/bin/python

#Solve the finite 1-D well
#The constants here are for the (Ga,Al)As well in the problem sheet 2,
#but can be redefined for any similar problem.
#By Graham Lee

#How this works:
#1. define a value of kappa
#2. use the energy relation to define k
#3. see if this is consistent with the tan() relation
#4. rinse
#5. repeat
#6. ???
#7. PROFIT!!!!

# The program works in SI units, converting to eV to report the energies.
# Good job Python supports infinite precision :-)

import math

effMassA=0.0632
effMassB=0.091

elecMass=9.109e-31
elecCharge=1.602e-19
hBar=1.055e-34

a=1.05e-8

kappa0=1.222e4
#Based on a crude estimation of the wavevector at the top of the well :)

deltaE=0.149*elecCharge

print "Here we go...."
for i in range(1,50000):
    kappa=i*kappa0
    k=math.sqrt((-1.0*effMassA*kappa*kappa/effMassB)+2.0*effMassA*elecMass*deltaE/(hBar*hBar))
    if math.floor(1000.0*math.tan(k*a/2.0))==math.floor(1000.0*effMassA*kappa/(effMassB*k)):
        energy=hBar*hBar*k*k/(2.0*effMassA*elecMass*elecCharge)
        print "Even solution found at k="+str(k)+"m^-1"
        print "                   kappa="+str(kappa)+"m^-1"
        print "Energy of this solution ="+str(energy)+"eV"
    if math.floor(1000.0*math.tan(k*a/2.0))==math.floor(-1000.0*effMassB*k/(effMassA*kappa)):
        energy=hBar*hBar*k*k/(2.0*effMassA*elecMass*elecCharge)
        print "Odd solution found at  k="+str(k)+"m^-1"
        print "                   kappa="+str(kappa)+"m^-1"
        print "Energy of this solution ="+str(energy)+"eV"
print "Done."
raw_input()