Can anyone please at least give a hint of the answer of my question.
Regards,
Santu Baidya
On Mon, 27 Aug 2018 at 17:24, S Baidya santubaidya2009@gmail.com wrote:
Dear ALPS users,
I calculated specific heat capacity for square lattice with similar parameters, for spinmc (classical monte carlo) and qwl (qmc) for a square lattice.
For spinmc,
#prepare the input parameters parms = [] for l in [40]: for t in [5.0,4.5,4.0,3.5,3.0,2.9,2.8,2.7]: parms.append( { 'LATTICE' : "square lattice", 'T' : t, 'S' : 0.5, 'J' : 1 , 'THERMALIZATION' : 1000, 'SWEEPS' : 400000, 'UPDATE' : "cluster", 'MODEL' : "Ising", 'L' : l } ) for t in [2.6, 2.5, 2.4, 2.3, 2.2, 2.1, 2.0, 1.9, 1.8, 1.7, 1.6, 1.5, 1.2]: parms.append( { 'LATTICE' : "square lattice", 'T' : t, 'S' : 0.5, 'J' : 1 , 'THERMALIZATION' : 1000, 'SWEEPS' : 40000, 'UPDATE' : "cluster", 'MODEL' : "Ising", 'L' : l } )
#write the input file and run the simulation input_file = pyalps.writeInputFiles('parm7a',parms) pyalps.runApplication('spinmc',input_file,Tmin=5)
For qwl,
#prepare the input parameters parms = [{ 'LATTICE' : "square lattice", 'MODEL' : "spin", 'local_S' : 0.5, 'L' : 40, 'J' : -1 , 'Jxy' : 0, 'CUTOFF' : 1000 }]
#write the input file and run the simulation input_file = pyalps.writeInputFiles('parm6a',parms) res = pyalps.runApplication('qwl',input_file)
#run the evaluation and load all the plots data = pyalps.evaluateQWL(pyalps.getResultFiles(prefix='parm6a'),DELTA_T=0.1, T_MIN=0.1, T_MAX=10.0)
I hoped that the two results should not show completely different specific heat per site curve with temperature. But, they are totally different. Can anyone please tell me if it is expected that classical and quantum result are completely different.
Thanking you,
Santu Baidya