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alvinquantum opened this issue Jan 10, 2022 · 9 comments · May be fixed by #7292
Open

Fidelity calculation for density matrices improvement #4819

alvinquantum opened this issue Jan 10, 2022 · 9 comments · May be fixed by #7292
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@anonymousr007
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area/accuracy area/qis kind/bug-report Something doesn't seem to work. triage/accepted A consensus emerged that this bug report, feature request, or other action should be worked on
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2025-T1: Performa...

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@alvinquantum
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alvinquantum commented Jan 10, 2022 •
edited by vtomole
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Cirq/cirq-core/cirq/qis/measures.py

Lines 239 to 242 in 782c14e

state1_sqrt = _sqrt_positive_semidefinite_matrix(state1)
eigs = scipy.linalg.eigvalsh(state1_sqrt @ state2 @ state1_sqrt)
trace = np.sum(np.sqrt(np.abs(eigs)))
return trace ** 2

The current implementation uses the direct definition of fidelity, which works well for small dimensions. From my experience, I was getting values significantly greater than 1. For the fidelity between rho1 and rho2, I found that squaring the sum of the singular values of the product of square_root(rho1) and square_root(rho2) gives a more accurate calculation. This equation is equal to fidelity. The idea is not mine and came from qutip.

This is my version of the implementation that I use.

import scipy
def get_fidelity(rho1, rho2):
    rho1_sqrt=scipy.linalg.sqrtm(rho1)
    rho2_sqrt=scipy.linalg.sqrtm(rho2)
    return scipy.linalg.svdvals(rho1_sqrt @ rho2_sqrt).sum()**2
@tanujkhattar tanujkhattar added area/qis triage/discuss Needs decision / discussion, bring these up during Cirq Cynque labels Jan 10, 2022
@tanujkhattar
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This looks like a reasonable request to me.

@viathor What do you think?

For completeness, the modified formula can be derived as follows:

@viathor
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viathor commented Jan 10, 2022

I agree the request is reasonable.

Ideally, the PR implementing this would include a unit test demonstrating improved numerical stability. It should fail for the current implementation and succeed for the new one. @alvinquantum, any of the cases you encountered where fidelity exceeded 1 in cirq and stayed <=1 in your code can easily become such a unit test. Also, it'd be nice to have some evaluation of the effect on performance, e.g. ipython's %timeit output posted here would work.

@tanujkhattar Your derivation LGTM with a nit that |A| = sqrt(A.T.conj() @ A).

@tanujkhattar tanujkhattar added area/performance triage/accepted A consensus emerged that this bug report, feature request, or other action should be worked on and removed triage/discuss Needs decision / discussion, bring these up during Cirq Cynque labels Jan 10, 2022
@alvinquantum
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alvinquantum commented Jan 11, 2022 •
edited
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I think this is a working example. Here cirq.fidelity gives around 1.09 while get_fidelity gives 1 with a 10^-5 tolerance. The code starts with a 10 qubit state, traces out the 10th system, and then calculates the fidelity of the reduced state with itself. The get_fidelity can also be improved since it tends to give values greater than 1, but from my experience those values have been greater by around 10^-5.

from cirq.contrib.qasm_import import circuit_from_qasm
import scipy
import cirq
import numpy as np
from cirq import partial_trace, DensityMatrixSimulator

def get_fidelity(rho1, rho2):
    '''Fidelity through singular values of sqrt(rho1)*sqrt(rho2)'''
    rho1_sqrt=scipy.linalg.sqrtm(rho1)
    rho2_sqrt=scipy.linalg.sqrtm(rho2)
    return scipy.linalg.svdvals(rho1_sqrt @ rho2_sqrt).sum()**2

if __name__ == "__main__":
    circ = circuit_from_qasm(
                    '''OPENQASM 2.0;
                    include "qelib1.inc";
                    qreg q[10];
                    rx(1.0349347) q[0];
                    ry(1.1344214) q[0];
                    rz(0.49872878) q[0];
                    rx(1.2483498) q[1];
                    ry(1.5452867) q[1];
                    rz(1.1293869) q[1];
                    rx(1.3588455) q[2];
                    ry(0.49602024) q[2];
                    rz(0.21943984) q[2];
                    rx(1.9583295) q[3];
                    ry(1.8588403) q[3];
                    rz(1.2323985) q[3];
                    rx(0.027308104) q[4];
                    ry(1.0267784) q[4];
                    rz(1.0619419) q[4];
                    rx(0.89107072) q[5];
                    ry(0.86430418) q[5];
                    rz(0.98182118) q[5];
                    rx(0.97120279) q[6];
                    ry(1.5382162) q[6];
                    rz(1.9003976) q[6];
                    rx(0.34264218) q[7];
                    ry(0.21962922) q[7];
                    rz(1.6263407) q[7];
                    rx(0.62256334) q[8];
                    ry(0.47124841) q[8];
                    rz(0.61290254) q[8];
                    rx(0.0017523425) q[9];
                    ry(1.7104261) q[9];
                    rz(0.13981947) q[9];
                    ''')

    init_qubits=10
    final_qubits=init_qubits-1
    keep_idxs=list(range(final_qubits))
    simulator=DensityMatrixSimulator()
    rho=simulator.simulate(circ).final_density_matrix
    rho_reshaped=np.reshape(rho, (2,2)*init_qubits)
    rho_reduced=np.reshape(partial_trace(rho_reshaped, keep_idxs),(2**final_qubits, 2**final_qubits))
    cirq_fidelity=cirq.fidelity(rho_reduced, rho_reduced, validate=False, qid_shape=(2,)*final_qubits)
    print(f"cirq.fidelity: {cirq_fidelity}") #approx. 1.0892252935292532
    print(f"get_fidelity: {get_fidelity(rho_reduced, rho_reduced)}") #approx 1.0000061793639072

@viathor
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viathor commented Jan 12, 2022

Thanks, @alvinquantum! This is helpful.

Note: this is good basis for a unit test, but should use cirq gates rather than converting from qasm (because ideally the unit test shouldn't be sensitive to bugs in our qasm converter!).

@viathor viathor added the kind/bug-report Something doesn't seem to work. label Jan 12, 2022
@anonymousr007
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Hi Google Quantum Team, Is this issue still open? If, yes please tell me more about it.

@tanujkhattar
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@anonymousr007 Yes, the issue is still open.

The goal is to replace the current implementation of calculating fidelity of two density matrices with the improved implementation that has lower numerical error.

Current implementation:

Cirq/cirq-core/cirq/qis/measures.py

Lines 239 to 242 in 782c14e

state1_sqrt = _sqrt_positive_semidefinite_matrix(state1)
eigs = scipy.linalg.eigvalsh(state1_sqrt @ state2 @ state1_sqrt)
trace = np.sum(np.sqrt(np.abs(eigs)))
return trace ** 2

New implementation:

def get_fidelity(rho1, rho2):
    rho1_sqrt=scipy.linalg.sqrtm(rho1)
    rho2_sqrt=scipy.linalg.sqrtm(rho2)
    return scipy.linalg.svdvals(rho1_sqrt @ rho2_sqrt).sum()**2

The PR implementing the change should also add a unit test demonstrating improved numerical stability of the new implementation, s.t. the current implementation fails but the new one passes.

@anonymousr007
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Please, assign this issue to me. I want to work on this.

@verult
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verult commented Mar 28, 2022

Marking as after 1.0 as this is a feature addition.

@RevanthGundala RevanthGundala linked a pull request Apr 21, 2025 that will close this issue
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Labels
area/accuracy area/qis kind/bug-report Something doesn't seem to work. triage/accepted A consensus emerged that this bug report, feature request, or other action should be worked on
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Cirq 2025 roadmap
Status: No status
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2025-T1: Performance improvements for...
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Use SVD-based fidelity for density matrices and add numerical stability test RevanthGundala/Cirq
7 participants
@mhucka @verult @MichaelBroughton @tanujkhattar @viathor @anonymousr007 @alvinquantum