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Robust way to identify eigenvalue using finer c steps

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Nuwan Yapa 2025-04-27 20:38:11 -04:00
parent 65d0f44b69
commit 3d7b964cde
1 changed files with 14 additions and 5 deletions
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19
calculations/PMM.py
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@ -1,14 +1,15 @@
#%% #%%
import pandas as pd import pandas as pd
import torch import torch
import numpy as np
#%% #%%
df = pd.read_csv('../temp/2body_data.csv') df = pd.read_csv('../temp/2body_data.csv').sort_values(by='c')
df['E'] = df['re_E'] + 1j * df['im_E'] df['E'] = df['re_E'] + 1j * df['im_E']
train_data = df[df['re_E'] < 0] train_data = df[df['re_E'] < 0]
target_data = df[df['re_E'] > 0] target_data = df[df['re_E'] > 0]
train_cs = torch.tensor(train_data['c'].to_numpy(), dtype=torch.float64) train_cs = train_data['c'].to_numpy()
train_Es = torch.tensor(train_data['E'].to_numpy(), dtype=torch.complex128) train_Es = torch.tensor(train_data['E'].to_numpy(), dtype=torch.complex128)
#%% #%%
@ -24,15 +25,23 @@ H1 = (H1 + torch.transpose(H1, 0, 1)).requires_grad_() # symmetric
#%% #%%
# training # training
# generate a set of c values to follow by subdividing the training cs
subdivisions = 3
c_steps = np.concatenate([np.linspace(start, stop, subdivisions, endpoint=False) for (start, stop) in zip(train_cs, train_cs[1:])])
c_steps = np.append(c_steps, train_cs[-1])
lr = 0.05 lr = 0.05
epochs = 100000 epochs = 100000
for epoch in range(epochs): for epoch in range(epochs):
Es = torch.empty(len(train_data), dtype=torch.complex128) Es = torch.empty(len(train_data), dtype=torch.complex128)
for (index, (c, E)) in enumerate(zip(train_cs, train_Es)): current_E = 0.0 # start at the threshold
for c in c_steps:
H = H0 + c * H1 H = H0 + c * H1
evals = torch.linalg.eigvals(H) evals = torch.linalg.eigvals(H)
i = torch.argmin(torch.abs(evals - E)) # TODO: more robust way to identify the eigenvector current_E = evals[torch.argmin(torch.abs(evals - current_E))]
Es[index]= evals[i] if np.any(c == train_cs):
index = np.where(c == train_cs)[0][0]
Es[index] = current_E
loss = ((Es - train_Es).abs() ** 2).sum() loss = ((Es - train_Es).abs() ** 2).sum()