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CritPt / Challenge_62_main / Quantum superposition of a single particle can be demonstrated by the…

Problem

Answer published by the source. Consult the official source to check your work against its answer.

code template

Code

import sympy as sp

phi = sp.symbols('phi')
k = sp.symbols('k', integer=True)

def answer(phi, k, k_value):
    r"""
    Return the violation as a function of $\phi$ for the case $k=1$, and the expressions of $T$ and $\phi_{\max}$ in Sympy format for any given k_value.

    Inputs
    ----------
    phi: sympy.Symbol, phase angle $\phi$
    k: sympy.Symbol, $k$
    k_value: int
        A specific value of $k$ used to determine which interval expression to return, but is *not* substituted into the final result.

    Outputs
    ----------
    violation: sympy.Expr, the violation $\delta$ as a function of $\phi$ for the case $k=1 (N=2k+1=3)$.
    T: sympy.Interval, the interval $T \subseteq [0,\pi]$ of phases for which a quantum violation occurs.
        Explicitly set the parameters `left_open` and `right_open`.
    phi_max: sympy.Expr, the phase $\phi_{\max}$ at which the maximal quantum violation occurs.

    The returned expressions of T and phi_max should preserve full symbolic dependence on k, while `k_value` is used internally solely to to select which interval expression to return.
    """

    # ------------------ FILL IN YOUR RESULTS BELOW ------------------
    violation = ...
    T = ...
    phi_max = ...
    # ---------------------------------------------------------------

    return violation, T, phi_max

problem description

# Problem setup: Quantum superposition of a single particle can be demonstrated by the interference pattern observed in a double-slit experiment. Although it has been shown that no genuine triple-slit interference exists for a single quantum particle, an NN-slit experiment can still reveal stronger nonclassical patterns, detectable through violations of the following inequality (Note: A classical theory that allows N1N-1-slit inteference cannot violate such an inequality.): p(0|0,,0)+i=1Np(1|0,,1i,,0)N, p(0|0,\cdots,0) + \sum_{i=1}^N p(1|0,\cdots,1_i,\cdots,0) \leq N, where p(b|a1,,an)p(b|a_1,\cdots,a_n) represents the statistics of getting outcome bb when each path encodes aia_i. To see the violation, a quantum particle is prepared among NN paths, such that |ΨN=1Ni=1N|ei|\Psi_N\rangle = \frac{1}{\sqrt{N}} \sum^N_{i=1} |e_i\rangle with $|e_i\rangle=|0\rangle_{A_1}\cdots|1\rangle_{A_i}\cdots|0⟩_{A_N}$. Each party encodes a local phase on their path ii given input aia_i, by applying the unitary Ui(ϕai)=eiϕai|1i1i|U_i(\phi_{a_i}) = e^{i\phi_{a_i}}|1_i\rangle\langle1_i|. The resulting encoded state is given by $$\sigma_{a_1,\cdots,a_n}= \bigotimes_i U_i(\phi_{a_i}) |\Psi_N\rangle\langle\Psi_N|\bigotimes_i U_i^{\dagger}(\phi_{a_i}),$$ which is measured using a fixed measurement {Πb}b\{\Pi_b\}_b, and the statistics are given by p(b|a1,,an)=Tr[Πbσa1,,an].p(b|a_1,\cdots,a_n)=Tr[\Pi_b \sigma_{a_1,\cdots,a_n}]. We consider a specific encoding strategy for an odd number of paths, N=2k+1N = 2k + 1, as follows: Set ϕ1i=ϕ\phi_{1_i} = \phi for iki \leq k; ϕ1i=π\phi_{1_i} = \pi for i=k+1i = k + 1; ϕ1i=ϕ\phi_{1_i} = -\phi for i>k+1i > k + 1; and ϕi0=0\phi_{i_0} = 0 for all ii. # Main problem: (1) For the case k=1(N=2k+1=3)k=1 (N=2k+1=3), express the violation as a function of ϕ\phi, i.e., the violation is defined as δ=p(0|0,,0)+i=1Np(1|0,,1i,,0)N. \delta= p(0|0,\cdots,0) + \sum_{i=1}^N p(1|0,\cdots,1_i,\cdots,0)- N. (2) Determine the range of ϕ\phi for which a quantum violation occurs, denoted as TT, where T[0,π]T \subseteq [0, \pi] is a function of kk. (3) Determine the value of ϕ\phi at which the maximal quantum violation occurs, denoted as ϕmax\phi_{\max} , as a function of kk.
Plain-text mathematical notation (without MathML)

# Problem setup:
Quantum superposition of a single particle can be demonstrated by the interference pattern observed in a double-slit experiment. Although it has been shown that no genuine triple-slit interference exists for a single quantum particle, an N-slit experiment can still reveal stronger nonclassical patterns, detectable through violations of the following inequality (Note: A classical theory that allows N−1-slit inteference cannot violate such an inequality.):

p(0|0,⋯,0)+∑_(i=1)^(N)p(1|0,⋯,1_(i),⋯,0)≤N,

where p(b|a₁,⋯,a_(n)) represents the statistics of getting outcome b when each path encodes a_(i).

To see the violation, a quantum particle is prepared among N paths, such that |Ψ_(N)⟩=(1)/(√(N))∑_(i=1)^(N)|e_(i)⟩  with $|e_i\rangle=|0\rangle_{A_1}\cdots|1\rangle_{A_i}\cdots|0⟩_{A_N}$. Each party encodes a local phase on their path i given input a_(i), by applying the unitary U_(i)(ϕ_(a_(i)))=e^(iϕ_(a_(i)))|1_(i)⟩⟨1_(i)|. The resulting encoded state is given by

$$\sigma_{a_1,\cdots,a_n}= \bigotimes_i U_i(\phi_{a_i}) |\Psi_N\rangle\langle\Psi_N|\bigotimes_i U_i^{\dagger}(\phi_{a_i}),$$

which is measured using a fixed measurement {Π_(b)}_(b), and the statistics are given by

p(b|a₁,⋯,a_(n))=Tr[Π_(b)σ_(a₁,⋯,a_(n))].

We consider a specific encoding strategy for an odd number of paths, N=2k+1, as follows:
Set ϕ_(1_(i))=ϕ for i≤k; ϕ_(1_(i))=π for i=k+1; ϕ_(1_(i))=−ϕ for i>k+1; and ϕ_(i₀)=0 for all i.




# Main problem:

(1) For the case k=1(N=2k+1=3), express the violation as a function of ϕ, i.e., the violation is defined as

δ=p(0|0,⋯,0)+∑_(i=1)^(N)p(1|0,⋯,1_(i),⋯,0)−N.

(2) Determine the range of ϕ for which a quantum violation occurs, denoted as T, where T⊆[0,π] is a function of k.

(3) Determine the value of ϕ at which the maximal quantum violation occurs, denoted as ϕ_(max), as a function of k.
Original LaTeX notation

# Problem setup:
Quantum superposition of a single particle can be demonstrated by the interference pattern observed in a double-slit experiment. Although it has been shown that no genuine triple-slit interference exists for a single quantum particle, an $N$-slit experiment can still reveal stronger nonclassical patterns, detectable through violations of the following inequality (Note: A classical theory that allows $N-1$-slit inteference cannot violate such an inequality.):

$$
p(0|0,\cdots,0) + \sum_{i=1}^N p(1|0,\cdots,1_i,\cdots,0) \leq N,
$$

where $p(b|a_1,\cdots,a_n)$ represents the statistics of getting outcome $b$ when each path encodes $a_i$.

To see the violation, a quantum particle is prepared among $N$ paths, such that $|\Psi_N\rangle = \frac{1}{\sqrt{N}} \sum^N_{i=1} |e_i\rangle$  with $|e_i\rangle=|0\rangle_{A_1}\cdots|1\rangle_{A_i}\cdots|0⟩_{A_N}$. Each party encodes a local phase on their path $i$ given input $a_i$, by applying the unitary $U_i(\phi_{a_i}) = e^{i\phi_{a_i}}|1_i\rangle\langle1_i|$. The resulting encoded state is given by

$$\sigma_{a_1,\cdots,a_n}= \bigotimes_i U_i(\phi_{a_i}) |\Psi_N\rangle\langle\Psi_N|\bigotimes_i U_i^{\dagger}(\phi_{a_i}),$$

which is measured using a fixed measurement $\{\Pi_b\}_b$, and the statistics are given by

$$p(b|a_1,\cdots,a_n)=Tr[\Pi_b \sigma_{a_1,\cdots,a_n}].$$

We consider a specific encoding strategy for an odd number of paths, $N = 2k + 1$, as follows:
Set $\phi_{1_i} = \phi$ for $i \leq k$; $\phi_{1_i} = \pi$ for $i = k + 1$; $\phi_{1_i} = -\phi$ for $i > k + 1$; and $\phi_{i_0} = 0$ for all $i$.




# Main problem:

(1) For the case $k=1 (N=2k+1=3)$, express the violation as a function of $\phi$, i.e., the violation is defined as

$$
\delta= p(0|0,\cdots,0) + \sum_{i=1}^N p(1|0,\cdots,1_i,\cdots,0)- N.
$$

(2) Determine the range of $\phi$ for which a quantum violation occurs, denoted as $T$, where $T \subseteq [0, \pi]$ is a function of $k$.

(3) Determine the value of $\phi$ at which the maximal quantum violation occurs, denoted as $\phi_{\max} $, as a function of $k$.

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