POVM simulation
This notebook shows how we can check how much depolarizing noise a qubit positive operator-valued measure (POVM) can take before it becomes simulable by projective measurements. The general method is described in arXiv:1609.06139. The question of simulability by projective measurements boils down to an SDP problem. Eq. (8) from the paper defines the noisy POVM that we obtain subjecting a POVM $\mathbf{M}$ to a depolarizing channel $\Phi_t$:
\[\left[\Phi_t\left(\mathbf{M}\right)\right]_i := t M_i + (1-t)\frac{\mathrm{tr}(M_i)}{d} \mathbb{1}.\]
If this visibility $t\in[0,1]$ is one, the POVM $\mathbf{M}$ is simulable.
We will use Convex.jl to solve the SDP problem.
using Convex, SCS, LinearAlgebraFor the qubit case, a four outcome qubit POVM $\mathbf{M} \in\mathcal{P}(2,4)$ is simulable if and only if
\[M_{1}=N_{12}^{+}+N_{13}^{+}+N_{14}^{+},\]
\[M_{2}=N_{12}^{-}+N_{23}^{+}+N_{24}^{+},\]
\[M_{3}=N_{13}^{-}+N_{23}^{-}+N_{34}^{+},\]
\[M_{4}=N_{14}^{-}+N_{24}^{-}+N_{34}^{-},\]
where Hermitian operators $N_{ij}^{\pm}$ satisfy $N_{ij}^{\pm}\geq0$ and $N_{ij}^{+}+N_{ij}^{-}=p_{ij}\mathbb{1}$, where $i<j$ , $i,j=1,2,3,4$ and $p_{ij}\geq0$ as well as $\sum_{i<j}p_{ij}=1$, that is, the $p_{ij}$ values form a probability vector. This forms an SDP feasibility problem, which we can rephrase as an optimization problem by adding depolarizing noise to the left-hand side of the above equations and maximizing the visibility $t$:
\[\max_{t\in[0,1]} t\]
such that
\[t\,M_{1}+(1-t)\,\mathrm{tr}(M_{1})\frac{\mathbb{1}}{2}=N_{12}^{+}+N_{13}^{+}+N_{14}^{+},\]
\[t\,M_{2}+(1-t)\,\mathrm{tr}(M_{2})\frac{\mathbb{1}}{2}=N_{12}^{-}+N_{23}^{+}+N_{24}^{+},\]
\[t\,M_{3}+(1-t)\,\mathrm{tr}(M_{3})\frac{\mathbb{1}}{2}=N_{13}^{-}+N_{23}^{-}+N_{34}^{+},\]
\[t\,M_{4}+(1-t)\,\mathrm{tr}(M_{4})\frac{\mathbb{1}}{2}=N_{14}^{-}+N_{24}^{-}+N_{34}^{-}\]
.
We organize these constraints in a function that takes a four-output qubit POVM as its argument:
function get_visibility(K)
noise = real([tr(K[i]) * I(2) / 2 for i in 1:size(K, 1)])
P = [[ComplexVariable(2, 2) for i in 1:2] for j in 1:6]
q = Variable(6, Positive())
t = Variable(1, Positive())
constraints = [isposdef(P[i][j]) for i in 1:6 for j in 1:2]
constraints += sum(q) == 1
constraints += t <= 1
constraints += [P[i][1] + P[i][2] == q[i] * I(2) for i in 1:6]
constraints += t * K[1] + (1 - t) * noise[1] == P[1][1] + P[2][1] + P[3][1]
constraints += t * K[2] + (1 - t) * noise[2] == P[1][2] + P[4][1] + P[5][1]
constraints += t * K[3] + (1 - t) * noise[3] == P[2][2] + P[4][2] + P[6][1]
constraints += t * K[4] + (1 - t) * noise[4] == P[3][2] + P[5][2] + P[6][2]
p = maximize(t, constraints)
solve!(p, SCS.Optimizer; silent_solver = true)
return p.optval
endget_visibility (generic function with 1 method)We check this function using the tetrahedron measurement (see Appendix B in arXiv:quant-ph/0702021). This measurement is non-simulable, so we expect a value below one.
function dp(v)
return I(2) + v[1] * [0 1; 1 0] + v[2] * [0 -im; im 0] + v[3] * [1 0; 0 -1]
end
b = [
1 1 1
-1 -1 1
-1 1 -1
1 -1 -1
] / sqrt(3)
M = [dp(b[i, :]) for i in 1:size(b, 1)] / 4;
get_visibility(M)0.8165487342796811This value matches the one we obtained using PICOS.
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