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blochsolve
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@ -1,5 +1,11 @@
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# fdfd_tools
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** DEPRECATED **
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The functionality in this module is now provided by [meanas](https://mpxd.net/code/jan/meanas).
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-----------------------
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**fdfd_tools** is a python package containing utilities for
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creating and analyzing 2D and 3D finite-difference frequency-domain (FDFD)
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electromagnetic simulations.
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68
examples/bloch.py
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68
examples/bloch.py
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@ -0,0 +1,68 @@
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import numpy, scipy, gridlock, fdfd_tools
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from fdfd_tools import bloch
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from numpy.linalg import norm
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import logging
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logging.basicConfig(level=logging.DEBUG)
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logger = logging.getLogger(__name__)
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dx = 40
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x_period = 400
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y_period = z_period = 2000
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g = gridlock.Grid([numpy.arange(-x_period/2, x_period/2, dx),
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numpy.arange(-1000, 1000, dx),
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numpy.arange(-1000, 1000, dx)],
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shifts=numpy.array([[0,0,0]]),
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initial=1.445**2,
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periodic=True)
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g.draw_cuboid([0,0,0], [200e8, 220, 220], eps=3.47**2)
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#x_period = y_period = z_period = 13000
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#g = gridlock.Grid([numpy.arange(3), ]*3,
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# shifts=numpy.array([[0, 0, 0]]),
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# initial=2.0**2,
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# periodic=True)
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g2 = g.copy()
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g2.shifts = numpy.zeros((6,3))
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g2.grids = [numpy.zeros(g.shape) for _ in range(6)]
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epsilon = [g.grids[0],] * 3
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reciprocal_lattice = numpy.diag(1e6/numpy.array([x_period, y_period, z_period])) #cols are vectors
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#print('Finding k at 1550nm')
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#k, f = bloch.find_k(frequency=1/1550,
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# tolerance=(1/1550 - 1/1551),
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# direction=[1, 0, 0],
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# G_matrix=reciprocal_lattice,
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# epsilon=epsilon,
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# band=0)
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#
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#print("k={}, f={}, 1/f={}, k/f={}".format(k, f, 1/f, norm(reciprocal_lattice @ k) / f ))
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print('Finding f at [0.25, 0, 0]')
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for k0x in [.25]:
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k0 = numpy.array([k0x, 0, 0])
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kmag = norm(reciprocal_lattice @ k0)
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tolerance = (1e6/1550) * 1e-4/1.5 # df = f * dn_eff / n
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logger.info('tolerance {}'.format(tolerance))
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n, v = bloch.eigsolve(4, k0, G_matrix=reciprocal_lattice, epsilon=epsilon, tolerance=tolerance)
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v2e = bloch.hmn_2_exyz(k0, G_matrix=reciprocal_lattice, epsilon=epsilon)
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v2h = bloch.hmn_2_hxyz(k0, G_matrix=reciprocal_lattice, epsilon=epsilon)
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ki = bloch.generate_kmn(k0, reciprocal_lattice, g.shape)
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z = 0
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e = v2e(v[0])
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for i in range(3):
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g2.grids[i] += numpy.real(e[i])
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g2.grids[i+3] += numpy.imag(e[i])
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f = numpy.sqrt(numpy.real(numpy.abs(n))) # TODO
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print('k0x = {:3g}\n eigval = {}\n f = {}\n'.format(k0x, n, f))
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n_eff = norm(reciprocal_lattice @ k0) / f
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print('kmag/f = n_eff = {} \n wl = {}\n'.format(n_eff, 1/f ))
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@ -359,6 +359,8 @@ def eigsolve(num_modes: int,
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"""
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h_size = 2 * epsilon[0].size
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kmag = norm(G_matrix @ k0)
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'''
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Generate the operators
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'''
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@ -390,7 +392,7 @@ def eigsolve(num_modes: int,
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onto the space orthonormal to Z. If approx_grad is True, the approximate
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inverse of the maxwell operator is used to precondition the gradient.
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"""
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z = Z.reshape(y_shape)
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z = Z.view(dtype=complex).reshape(y_shape)
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U = numpy.linalg.inv(z.conj().T @ z)
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zU = z @ U
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AzU = scipy_op @ zU
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@ -400,27 +402,49 @@ def eigsolve(num_modes: int,
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df_dy = scipy_iop @ (AzU - zU @ zTAzU)
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else:
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df_dy = (AzU - zU @ zTAzU)
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return numpy.abs(f), numpy.sign(f) * numpy.real(df_dy).ravel()
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df_dy_flat = df_dy.view(dtype=float).ravel()
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return numpy.abs(f), numpy.sign(f) * df_dy_flat
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'''
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Use the conjugate gradient method and the approximate gradient calculation to
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quickly find approximate eigenvectors.
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'''
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result = scipy.optimize.minimize(rayleigh_quotient,
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numpy.random.rand(*y_shape),
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numpy.random.rand(*y_shape, 2),
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jac=True,
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method='CG',
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tol=1e-5,
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options={'maxiter': 30, 'disp':True})
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method='L-BFGS-B',
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tol=1e-20,
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options={'maxiter': 2000, 'gtol':0, 'ftol':1e-20 , 'disp':True})#, 'maxls':80, 'm':30})
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result = scipy.optimize.minimize(lambda y: rayleigh_quotient(y, True),
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result.x,
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jac=True,
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method='L-BFGS-B',
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tol=1e-20,
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options={'maxiter': 2000, 'gtol':0, 'disp':True})
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result = scipy.optimize.minimize(lambda y: rayleigh_quotient(y, False),
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result.x,
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jac=True,
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method='CG',
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tol=1e-13,
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options={'maxiter': 100, 'disp':True})
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method='L-BFGS-B',
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tol=1e-20,
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options={'maxiter': 2000, 'gtol':0, 'disp':True})
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z = result.x.reshape(y_shape)
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for i in range(20):
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result = scipy.optimize.minimize(lambda y: rayleigh_quotient(y, False),
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result.x,
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jac=True,
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method='L-BFGS-B',
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tol=1e-20,
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options={'maxiter': 70, 'gtol':0, 'disp':True})
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if result.nit == 0:
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# We took 0 steps, so re-running won't help
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break
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z = result.x.view(dtype=complex).reshape(y_shape)
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'''
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Recover eigenvectors from Z
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@ -436,25 +460,13 @@ def eigsolve(num_modes: int,
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v = eigvecs[:, i]
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n = eigvals[i]
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v /= norm(v)
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logger.info('eigness {}: {}'.format(i, norm(scipy_op @ v - (v.conj() @ (scipy_op @ v)) * v )))
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eigness = norm(scipy_op @ v - (v.conj() @ (scipy_op @ v)) * v )
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f = numpy.sqrt(-numpy.real(n))
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df = numpy.sqrt(-numpy.real(n + eigness))
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neff_err = kmag * (1/df - 1/f)
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logger.info('eigness {}: {}\n neff_err: {}'.format(i, eigness, neff_err))
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ev2 = eigvecs.copy()
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for i in range(len(eigvals)):
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logger.info('Refining eigenvector {}'.format(i))
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eigvals[i], ev2[:, i] = rayleigh_quotient_iteration(scipy_op,
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guess_vector=eigvecs[:, i],
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iterations=40,
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tolerance=tolerance * numpy.real(numpy.sqrt(eigvals[i])) * 2,
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solver = lambda A, b: spalg.bicgstab(A, b, maxiter=200)[0])
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eigvecs = ev2
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order = numpy.argsort(numpy.abs(eigvals))
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for i in range(len(eigvals)):
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v = eigvecs[:, i]
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n = eigvals[i]
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v /= norm(v)
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logger.info('eigness {}: {}'.format(i, norm(scipy_op @ v - (v.conj() @ (scipy_op @ v)) * v )))
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return eigvals[order], eigvecs.T[order]
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@ -53,7 +53,7 @@ def rayleigh_quotient_iteration(operator: sparse.spmatrix or spalg.LinearOperato
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:return: (eigenvalue, eigenvector)
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"""
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try:
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_test = operator - sparse.eye(operator.shape)
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_test = operator - sparse.eye(operator.shape[0])
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shift = lambda eigval: eigval * sparse.eye(operator.shape[0])
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if solver is None:
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solver = spalg.spsolve
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