def dnorm(A, B=None, solver="CVXOPT", verbose=False, force_solve=False,
sparse=True):
"""
Calculates the diamond norm of the quantum map q_oper, using
the simplified semidefinite program of [Wat12]_.
The diamond norm SDP is solved by using CVXPY_.
Parameters
----------
A : Qobj
Quantum map to take the diamond norm of.
B : Qobj or None
If provided, the diamond norm of :math:`A - B` is taken instead.
solver : str
Solver to use with CVXPY. One of "CVXOPT" (default) or "SCS". The
latter tends to be significantly faster, but somewhat less accurate.
verbose : bool
If True, prints additional information about the solution.
force_solve : bool
If True, forces dnorm to solve the associated SDP, even if a special
case is known for the argument.
sparse : bool
Whether to use sparse matrices in the convex optimisation problem.
Default True.
Returns
-------
dn : float
Diamond norm of q_oper.
Raises
------
ImportError
If CVXPY cannot be imported.
.. _cvxpy: http://www.cvxpy.org/en/latest/
"""
if cvxpy is None: # pragma: no cover
raise ImportError("dnorm() requires CVXPY to be installed.")
# We follow the strategy of using Watrous' simpler semidefinite
# program in its primal form. This is the same strategy used,
# for instance, by both pyGSTi and SchattenNorms.jl. (By contrast,
# QETLAB uses the dual problem.)
# Check if A and B are both unitaries. If so, then we can without
# loss of generality choose B to be the identity by using the
# unitary invariance of the diamond norm,
# || A - B ||_♢ = || A B⁺ - I ||_♢.
# Then, using the technique mentioned by each of Johnston and
# da Silva,
# || A B⁺ - I ||_♢ = max_{i, j} | \lambda_i(A B⁺) - \lambda_j(A B⁺) |,
# where \lambda_i(U) is the ith eigenvalue of U.
if (
# There's a lot of conditions to check for this path.
not force_solve and B is not None and
# Only check if they aren't superoperators.
A.type == "oper" and B.type == "oper" and
# The difference of unitaries optimization is currently
# only implemented for d == 2. Much of the code below is more general,
# though, in anticipation of generalizing the optimization.
A.shape[0] == 2
):
# Make an identity the same size as A and B to
# compare against.
I = qeye(A.dims[0])
# Compare to B first, so that an error is raised
# as soon as possible.
Bd = B.dag()
if (
(B * Bd - I).norm() < 1e-6 and
(A * A.dag() - I).norm() < 1e-6
):
# Now we are on the fast path, so let's compute the
# eigenvalues, then find the diameter of the smallest circle
# containing all of them.
#
# For now, this is only implemented for dim = 2, such that
# generalizing here will allow for generalizing the optimization.
# A reasonable approach would probably be to use Welzl's algorithm
# (https://en.wikipedia.org/wiki/Smallest-circle_problem).
U = A * B.dag()
eigs = U.eigenenergies()
eig_distances = np.abs(eigs[:, None] - eigs[None, :])
return np.max(eig_distances)
# Force the input superoperator to be a Choi matrix.
J = to_choi(A)
if B is not None:
J -= to_choi(B)
# Watrous 2012 also points out that the diamond norm of Lambda
# is the same as the completely-bounded operator-norm (∞-norm)
# of the dual map of Lambda. We can evaluate that norm much more
# easily if Lambda is completely positive, since then the largest
# eigenvalue is the same as the largest singular value.
if not force_solve and J.iscp:
S_dual = to_super(J.dual_chan())
vec_eye = operator_to_vector(qeye(S_dual.dims[1][1]))
op = vector_to_operator(S_dual * vec_eye)
# The 2-norm was not implemented for sparse matrices as of the time
# of this writing. Thus, we must yet again go dense.
return la.norm(op.data.todense(), 2)
# If we're still here, we need to actually solve the problem.
# Assume square...
dim = np.prod(J.dims[0][0])
J_dat = J.data
if not sparse:
# The parameters and constraints only depend on the dimension, so
# we can cache them efficiently.
problem, Jr, Ji = dnorm_problem(dim)
# Load the parameters with the Choi matrix passed in.
Jr.value = sp.csr_matrix((J_dat.data.real, J_dat.indices,
J_dat.indptr),
shape=J_dat.shape).toarray()
Ji.value = sp.csr_matrix((J_dat.data.imag, J_dat.indices,
J_dat.indptr),
shape=J_dat.shape).toarray()
else:
# The parameters do not depend solely on the dimension,
# so we can not cache them efficiently.
problem = dnorm_sparse_problem(dim, J_dat)
problem.solve(solver=solver, verbose=verbose)
return problem.value
def dnorm(A, B=None, solver="CVXOPT", verbose=False, force_solve=False):
"""
Calculates the diamond norm of the quantum map q_oper, using
the simplified semidefinite program of [Wat12]_.
The diamond norm SDP is solved by using CVXPY_.
Parameters
----------
A : Qobj
Quantum map to take the diamond norm of.
B : Qobj or None
If provided, the diamond norm of :math:`A - B` is
taken instead.
solver : str
Solver to use with CVXPY. One of "CVXOPT" (default)
or "SCS". The latter tends to be significantly faster,
but somewhat less accurate.
verbose : bool
If True, prints additional information about the
solution.
force_solve : bool
If True, forces dnorm to solve the associated SDP, even if a special
case is known for the argument.
Returns
-------
dn : float
Diamond norm of q_oper.
Raises
------
ImportError
If CVXPY cannot be imported.
.. _cvxpy: http://www.cvxpy.org/en/latest/
"""
if cvxpy is None: # pragma: no cover
raise ImportError("dnorm() requires CVXPY to be installed.")
# We follow the strategy of using Watrous' simpler semidefinite
# program in its primal form. This is the same strategy used,
# for instance, by both pyGSTi and SchattenNorms.jl. (By contrast,
# QETLAB uses the dual problem.)
# Check if A and B are both unitaries. If so, then we can without
# loss of generality choose B to be the identity by using the
# unitary invariance of the diamond norm,
# || A - B ||_♢ = || A B⁺ - I ||_♢.
# Then, using the technique mentioned by each of Johnston and
# da Silva,
# || A B⁺ - I ||_♢ = max_{i, j} | \lambda_i(A B⁺) - \lambda_j(A B⁺) |,
# where \lambda_i(U) is the ith eigenvalue of U.
if (
# There's a lot of conditions to check for this path.
not force_solve and B is not None and
# Only check if they aren't superoperators.
A.type == "oper" and B.type == "oper" and
# The difference of unitaries optimization is currently
# only implemented for d == 2. Much of the code below is more general,
# though, in anticipation of generalizing the optimization.
A.shape[0] == 2
):
# Make an identity the same size as A and B to
# compare against.
I = qeye(A.dims[0])
# Compare to B first, so that an error is raised
# as soon as possible.
Bd = B.dag()
if (
(B * Bd - I).norm() < 1e-6 and
(A * A.dag() - I).norm() < 1e-6
):
# Now we are on the fast path, so let's compute the
# eigenvalues, then find the diameter of the smallest circle
# containing all of them.
#
# For now, this is only implemented for dim = 2, such that
# generalizing here will allow for generalizing the optimization.
# A reasonable approach would probably be to use Welzl's algorithm
# (https://en.wikipedia.org/wiki/Smallest-circle_problem).
U = A * B.dag()
eigs = U.eigenenergies()
eig_distances = np.abs(eigs[:, None] - eigs[None, :])
return np.max(eig_distances)
# Force the input superoperator to be a Choi matrix.
J = to_choi(A)
if B is not None:
J -= to_choi(B)
# Watrous 2012 also points out that the diamond norm of Lambda
# is the same as the completely-bounded operator-norm (∞-norm)
# of the dual map of Lambda. We can evaluate that norm much more
# easily if Lambda is completely positive, since then the largest
# eigenvalue is the same as the largest singular value.
if not force_solve and J.iscp:
S_dual = to_super(J.dual_chan())
vec_eye = operator_to_vector(qeye(S_dual.dims[1][1]))
op = vector_to_operator(S_dual * vec_eye)
# The 2-norm was not implemented for sparse matrices as of the time
# of this writing. Thus, we must yet again go dense.
return la.norm(op.data.todense(), 2)
# If we're still here, we need to actually solve the problem.
# Assume square...
dim = np.prod(J.dims[0][0])
# The constraints only depend on the dimension, so
# we can cache them efficiently.
problem, Jr, Ji, X, rho0, rho1 = dnorm_problem(dim)
# Load the parameters with the Choi matrix passed in.
J_dat = J.data
Jr.value = sp.csr_matrix((J_dat.data.real, J_dat.indices, J_dat.indptr),
shape=J_dat.shape)
Ji.value = sp.csr_matrix((J_dat.data.imag, J_dat.indices, J_dat.indptr),
shape=J_dat.shape)
# Finally, set up and run the problem.
problem.solve(solver=solver, verbose=verbose)
return problem.value