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
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
Dear Santu,
At least, loop with
'local_S' : 0.5, 'L' : 40, 'J' : -1 , 'Jxy' : 0,
generates consistent results with spinmc.
As for qwl, it produces results larger than the others, but I don't know if the difference is significant as it does not give the statistical error.
Best, Synge
2018/08/27 17:24、S Baidya santubaidya2009@gmail.comのメール:
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
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Dear ALPS community,
I was wondering if it is possible to define an interaction involving three different lattice sites (for a 1D lattice model to perform dmrg calculations). To clarify, an example would be to construct the term \sum_i n_i n_{i+1} n_{i-1}. I checked in the doc if anyone had already the same kind of problem but it seems not. I guess that the solution is to build a bond term in the definition of the Hamiltonian by specifying the source and targets and to modify accordingly the lattice. Or is there a trick to specify one source (i) and two targets (i+1, i-1) ?
Thanks in advance, Julien
Thak you for your reply. I did use Loop and I found the two results comparable. But qwl showed me different result.
Thanks,
Santu
On Wed, 29 Aug 2018 at 12:04, Synge Todo wistaria@phys.s.u-tokyo.ac.jp wrote:
Dear Santu,
At least, loop with
'local_S' : 0.5, 'L' : 40, 'J' : -1 , 'Jxy' : 0,
generates consistent results with spinmc.
As for qwl, it produces results larger than the others, but I don't know if the difference is significant as it does not give the statistical error.
Best, Synge
2018/08/27 17:24、S Baidya santubaidya2009@gmail.comのメール:
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
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