Open problems in quantum state optimization

This page consists of certain problems in the domain of quantum state optimization that may be framed as computational tasks using the qustop module. Most of these examples are provided as brute-force computational approaches. It may be possible to refine many of these approaches in a more sophisticated manner, and I welcome any such input and feedback from the community.

Two-copy problem

Problem statement:

Let \(n \geq 1\) be an integer, let \(\mathcal{X}\) be a compelx Euclidean space, let \(\rho_i \in \text{D}(\mathcal{X})\) be a pure quantum state represented as a density operator and let

\[\eta = \left\{\left(\frac{1}{n}, \rho_1\right), \ldots, \left(\frac{1}{n}, \rho_n\right) \right\} \subset \mathcal{X}\]

be an ensemble of pure and mutually orthogonal quantum states. Define \(\eta^{\otimes 2}\) as the two-copy ensemble where

\[\eta^{\otimes 2} = \left\{\left(\frac{1}{n}, \rho_1 \otimes \rho_1\right), \ldots, \left(\frac{1}{n}, \rho_n \otimes \rho_n\right) \right\} \subset \mathcal{X} \otimes\mathcal{X}.\]

Question: Does there exist a certain ensemble \(\eta^{\otimes 2}\) such that

\[\text{opt}_{\text{PPT}}\left(\eta^{\otimes 2}\right) < 1 \quad \text{or} \quad \text{opt}_{\text{SEP}}\left(\eta^{\otimes 2}\right) < 1?\]

Computational approach:

The computational approach here is to generate a random ensemble of quantum states and then construct two copies of that ensemble. We take the two-copy ensemble and check whether or not we get a value less than \(1\) for distinguishing via PPT measurements.

 1import pickle
 2
 3import numpy as np
 4from scipy.stats import unitary_group
 5
 6from qustop import Ensemble, OptDist, State
 7
 8
 9def generate_random_two_copy_ensemble(num_states: int) -> Ensemble:
10    # Generate an ensemble of random mutually orthogonal and pure states.
11    # In this case, the ensemble consists of four states, but we could
12    # potentially have any n >= 1 number of states in the ensemble.
13    group = unitary_group.rvs(num_states)
14    states = [np.atleast_2d(group[:, i]).T for i in range(num_states)]
15
16    # Create the two-copy ensemble.
17    dims = [2] * num_states
18    ensemble = [
19        State(np.kron(state, state), dims) for _, state in enumerate(states)
20    ]
21    return Ensemble(ensemble)
22
23
24def run(num_states: int, num_trials: int) -> None:
25    for test in range(num_trials):
26        ensemble_2_copies = generate_random_two_copy_ensemble(num_states)
27
28        # Solve the two-copy PPT distinguishability SDPs.
29        ppt_2_copy = OptDist(ensemble_2_copies, "ppt", "min-error")
30        ppt_2_copy.solve()
31
32        # If the PPT value of the two-copy ensemble is below some threshold
33        # of perfect distinguishability, such an example has been found, in
34        # which case, we want to ensure we capture the values and states!
35        if not np.isclose(ppt_2_copy.value, 1, atol=0.001):
36            print(f"PPT 2-copy: {ppt_2_copy.value}")
37            with open("FOUND.pickle", "wb") as f:
38                pickle.dump([ensemble_2_copies], f)
39            break
40
41        # In any case, print out the two-copy values of the ensembles
42        # as we progress through the trials.
43        print(f"PPT 2-copy: {ppt_2_copy.value}")

One can specify the num_states parameter to determine how many states are in the ensemble and can also modify the num_trials parameter to specify for how many random ensembles should be generated and checked.

For instance, here we can randomly generate a 4-state two-copy ensemble 5 different times:

>>> run(num_states=4, num_trials=5)
PPT 2-copy: 0.9999999997183334
PPT 2-copy: 1.0000000011197347
PPT 2-copy: 1.000000005640175
PPT 2-copy: 1.0000000003272558
PPT 2-copy: 0.9999999981951567

Entanglement cost of distinguishing four arbitrary two-qubit ensemble with resource state

Problem statement:

Let \(\mathbb{C}^2 = \mathcal{X}_1 = \mathcal{X}_2 = \mathcal{Y}_1 = \mathcal{Y}_2\) be complex Euclidean spaces and let \(\mathcal{X} = \mathcal{X}_1 \otimes \mathcal{X}_2\) and \(\mathcal{Y} = \mathcal{Y}_1 \otimes \mathcal{Y}_2\). Define the following orthogonal two-qubit basis

\[\begin{split}\begin{equation} \begin{aligned} | \psi_0 \rangle = \alpha | 00 \rangle + \beta | 11 \rangle, \quad | \psi_1 \rangle = \beta | 00 \rangle - \alpha | 11 \rangle, \\ | \psi_2 \rangle = \alpha | 01 \rangle + \beta | 10 \rangle, \quad | \psi_3 \rangle = \beta | 01 \rangle - \alpha | 10 \rangle, \end{aligned} \end{equation}\end{split}\]

where \(| \psi_i \rangle \in \mathcal{X}_1 \otimes \mathcal{Y}_1\) for all \(i \in [0,1,2,3]\) and where

\[\alpha = \sqrt{\frac{1+n}{2}} \quad \text{and} \quad \beta = \sqrt{\frac{1-n}{2}}.\]

Define the resource state

\[| \tau_{\epsilon} \rangle = \sqrt{\frac{1+\epsilon}{2}} | 00 \rangle + \sqrt{\frac{1-\epsilon}{2}} | 11 \rangle \in \mathcal{X}_2 \otimes \mathcal{Y}_2\]

for some choice of \(\epsilon \in [0,1]\).

Consider the ensemble

\[\eta = \left\{ | \psi_0 \rangle \otimes | \tau_{\epsilon} \rangle, | \psi_1 \rangle \otimes | \tau_{\epsilon} \rangle, | \psi_2 \rangle \otimes | \tau_{\epsilon} \rangle, | \psi_3 \rangle \otimes | \tau_{\epsilon} \rangle \right\} \subset \mathcal{X} \otimes \mathcal{Y}.\]

Question: Assuming a uniform distribution \(p_1 = p_2 = p_3 = p_4 = 1/4\), with any state from \(\eta\) being selected, what is the closed-form entanglement cost of distinguishing \(\eta\) via PPT measurements for any choice of \(\alpha\) and \(\beta\)?

Note

When \(\alpha = \beta = \frac{1}{\sqrt{2}}\), the ensemble \(\eta\) consists of the Bell states. In this case, it is known that the closed-form entanglement cost is

\[\frac{1}{2} \left(1 + \sqrt{1-\epsilon^2}\right).\]

Follow-up Question: Assuming a uniform distribution \(p_1 = p_2 = p_3 = p_4 = 1/4\), with any state from \(\eta\) being selected, what is the closed-form entanglement cost of distinguishing \(\eta\) via separable measurements for any choice of \(\alpha\) and \(\beta\)?

This question will most likely be more difficult than the PPT case, but it’s still worth asking.

Computational approach:

Consider the following snippet of code that considers this problem:

 1import numpy as np
 2from toqito.states import basis
 3
 4from qustop import Ensemble, OptDist, State
 5
 6# Define the |0> and |1> basis states:
 7e_0, e_1 = basis(2, 0), basis(2, 1)
 8
 9# Define constants "n" and "epsilon":
10n = 0.5
11eps = 0.0
12
13# Parameters alpha and beta are defined in terms of "n"
14alpha, beta = np.sqrt((1 + n) / 2), np.sqrt((1 - n) / 2)
15
16# Define the two-qubit ensemble states:
17psi_0 = alpha * np.kron(e_0, e_0) + beta * np.kron(e_1, e_1)
18psi_1 = beta * np.kron(e_0, e_0) - alpha * np.kron(e_1, e_1)
19psi_2 = alpha * np.kron(e_0, e_1) + beta * np.kron(e_1, e_0)
20psi_3 = beta * np.kron(e_0, e_1) - alpha * np.kron(e_1, e_0)
21
22# Define the resource state:
23tau_state = np.sqrt((1 + eps) / 2) * np.kron(e_0, e_0) + np.sqrt(
24    (1 - eps) / 2
25) * np.kron(e_1, e_1)
26tau = tau_state * tau_state.conj().T
27
28# Create the ensemble to distinguish:
29dims = [2, 2, 2, 2]
30rho_0 = State(np.kron(psi_0 * psi_0.conj().T, tau), dims)
31rho_1 = State(np.kron(psi_1 * psi_1.conj().T, tau), dims)
32rho_2 = State(np.kron(psi_2 * psi_2.conj().T, tau), dims)
33rho_3 = State(np.kron(psi_3 * psi_3.conj().T, tau), dims)
34ensemble = Ensemble([rho_0, rho_1, rho_2, rho_3])
35
36# Determine the optimal value of distinguishing the ensemble via PPT
37# measurements:
38ppt_res = OptDist(ensemble, "ppt", "min-error")
39ppt_res.solve()
40
41# Print value of "n", "eps", and the optimal value of distinguishing via PPT
42# measurements:
43print(f"For n = {n} and eps={eps}, the PPT value is {ppt_res.value}")

Is it possible to determine a closed form of the optimal value of distinguishing via PPT measurements for any choice of n and eps? It may be illuminating to alter these values and see if any obvious closed-form expressions emerge.

It is also possible to alter the ppt argument to sep to get a sense on the follow-up question. However, keep in mind that the computational technique used to compute separable values is an upper bound–a hierarchy of SDPs that will eventually converge, but may take substantial computational resources and time to do so.

Entanglement cost of distinguishing three arbitrary two-qubit ensemble with resource state

Problem statement:

Let \(\mathbb{C}^2 = \mathcal{X}_1 = \mathcal{X}_2 = \mathcal{Y}_1 = \mathcal{Y}_2\) be complex Euclidean spaces and let \(\mathcal{X} = \mathcal{X}_1 \otimes \mathcal{X}_2\) and \(\mathcal{Y} = \mathcal{Y}_1 \otimes \mathcal{Y}_2\). Define the following orthogonal two-qubit basis

\[\begin{split}\begin{equation} \begin{aligned} | \psi_0 \rangle = \alpha | 00 \rangle + \beta | 11 \rangle, \quad | \psi_1 \rangle = \beta | 00 \rangle - \alpha | 11 \rangle, \\ | \psi_2 \rangle = \alpha | 01 \rangle + \beta | 10 \rangle, \quad \end{aligned} \end{equation}\end{split}\]

where \(| \psi_i \rangle \in \mathcal{X}_1 \otimes \mathcal{Y}_1\) for all \(i \in [0,1,2]\) and where

\[\alpha = \sqrt{\frac{1+n}{2}} \quad \text{and} \quad \beta = \sqrt{\frac{1-n}{2}}.\]

Define the resource state

\[| \tau_{\epsilon} \rangle = \sqrt{\frac{1+\epsilon}{2}} | 00 \rangle + \sqrt{\frac{1-\epsilon}{2}} | 11 \rangle \in \mathcal{X}_2 \otimes \mathcal{Y}_2\]

for some choice of \(\epsilon \in [0,1]\).

Consider the ensemble

\[\begin{split}\eta = \left\{ | \psi_0 \rangle \otimes | \tau_{\epsilon} \rangle, \quad | \psi_1 \rangle \otimes | \tau_{\epsilon} \rangle, \\ | \psi_2 \rangle \otimes | \tau_{\epsilon} \rangle, \right\} \subset \mathcal{X} \otimes \mathcal{Y}.\end{split}\]

Question: Assuming a uniform distribution \(p_1 = p_2 = p_3 = 1/3\), with any state from \(\eta\) being selected, what is the closed-form entanglement cost of distinguishing \(\eta\) via separable measurements for any choice of \(\alpha\) and \(\beta\)?

Note

When \(\alpha = \beta = \frac{1}{\sqrt{2}}\), the ensemble \(\eta\) consists of three out of the four the Bell states. In this case, it is known that the closed-form entanglement cost for separable measurements is

\[\frac{1}{3} \left(2 + \sqrt{1-\epsilon^2}\right).\]

A simpler version of this question is to determine \(\text{opt}_{\text{PPT}}(\eta)\) for \(n=0\) for any \(\epsilon\).

Since \(\text{opt}_{\text{SEP}}(\eta) < \text{opt}_{\text{PPT}}(\eta)\), we have a lower bound on \(\text{opt}_{\text{PPT}}(\eta)\) when \(n=0\). It also holds that \(\text{opt}_{\text{POS}}(\eta) = 1\) for all \(n\) and \(\epsilon\), so we have an obvious upper bound.

Computational approach:

Consider the following snippet of code that considers this problem:

 1import numpy as np
 2from toqito.states import basis
 3
 4from qustop import Ensemble, OptDist, State
 5
 6# Define the |0> and |1> basis states:
 7e_0, e_1 = basis(2, 0), basis(2, 1)
 8
 9# Define constants "n" and "epsilon":
10n = 0.0
11eps = 0.75
12
13# Parameters alpha and beta are defined in terms of "n"
14alpha, beta = np.sqrt((1 + n) / 2), np.sqrt((1 - n) / 2)
15
16# Define the two-qubit ensemble states:
17psi_0 = alpha * np.kron(e_0, e_0) + beta * np.kron(e_1, e_1)
18psi_1 = beta * np.kron(e_0, e_0) - alpha * np.kron(e_1, e_1)
19psi_2 = alpha * np.kron(e_0, e_1) + beta * np.kron(e_1, e_0)
20
21# Define the resource state:
22tau_state = np.sqrt((1 + eps) / 2) * np.kron(e_0, e_0) + np.sqrt(
23    (1 - eps) / 2
24) * np.kron(e_1, e_1)
25tau = tau_state * tau_state.conj().T
26
27# Create the ensemble to distinguish:
28dims = [2, 2, 2, 2]
29rho_0 = State(np.kron(psi_0 * psi_0.conj().T, tau), dims)
30rho_1 = State(np.kron(psi_1 * psi_1.conj().T, tau), dims)
31rho_2 = State(np.kron(psi_2 * psi_2.conj().T, tau), dims)
32ensemble = Ensemble([rho_0, rho_1, rho_2])
33
34# Determine the optimal value of distinguishing the ensemble via PPT
35# measurements:
36ppt_res = OptDist(ensemble, "ppt", "min-error")
37ppt_res.solve()
38
39# Print value of "n", "eps", and the optimal value of distinguishing via PPT
40# measurements:
41print(f"For n = {n} and eps={eps}, the PPT value is {ppt_res.value}")
42sep_eq = 1 / 3 * (2 + np.sqrt(1 - eps ** 2))
43eq = 1 / 3 * (2 + np.sqrt(1.219 - eps ** 2))
44print(f"EQ:{eq}")
45print(f"SEP_EQ: {sep_eq}")

Separation between LOCC and separable value

Problem statement:

Does there exist an ensemble, \(\eta\), fully composed of orthogonal maximally entangled states for which the following strict inequality holds:

\[\text{opt}_{\text{LOCC}}(\eta) < \text{opt}_{\text{SEP}}(\eta)?\]

The mathematical structure of LOCC is difficult to generalize, but it is possible to computationally analyze the separable case via a hierarchy of semidefinite programs.

The antidistinguishability conjecture

A set of \(n\) pure quantum states \(| \rho_1 \rangle, \ldots, | \rho_n \rangle\) are antidistinguishable if there exists an \(n\)-outcome measurement that never outputs the result \(i\) on \(| \rho_i \rangle\).

Problem statement:

Let \(| \rho_1 \rangle, \ldots, | \rho_d \rangle\) be \(d\) pure quantum states each of dimension \(d\). If

\[| \langle \rho_i | \rho_j\rangle| \leq (d-2)/(d-1)\]

for all \(i \not= j\), then the states are antidistinguishable.

Question: Does this conjecture hold for all integers \(d\)?

Note

This conjecture is known to hold when \(d=2\) and \(d=3\).

Computational approach:

We can randomly generate an ensemble of \(d\) pure states of dimension \(d\) using the following function.

X

Next, we check whether the states are antidistinguishable. This is equivalent to checking whether the optimal value of the semidefinite program for the problem for the problem of unambiguous quantum state exclusion from arXiv:1306.4683 yields a non-zero result.

If the states are antidistinguishable, check the bound in the conjecture:

  • If the bound is satisfied it implies a satisfaction of the conjecture, so we go back to the beginning and try to generate another random ensemble.

  • If the bound is not satisfied, this implies a violation of the conjecture.