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Slides for my University of Idaho Colloquium on 6 March 2023 about Entanglement and the 2022 Nobel Prize in physics.

Path: UI_Colloquium/UI_Colloquium.slides

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Is Quantum Physics Fundamentally Different than Classical Physics?Tutorial on the 2022 Nobel Prize

Michael McNeil ForbesWashington State University and the University of Washington

Experimental ValidationClosing loopholes in tests of Bell's Inequalities

2023 Nobel Prize 

Abstract

The 2022 Nobel Prize in physics was awarded to Alain Aspect, John F. Clauser, and Anton Zeilinger
“for experiments with entangled photons, establishing the violation of Bell inequalities and pioneering quantum information science”.
The inequalities set forth by John Bell provide a formal proof supporting the view of Bohr and Schrödinger, that quantum mechanics is fundamentally incompatible with the local realist model of classical physics favored by Einstein, Podolsky, and Rosen (EPR).  The 2022 Nobel Prize recognizes the experimental efforts made to test the violations of these inequalities, supporting the predictions of the quantum theory.
In this colloquium I will provide a pedagogical but rigorous explanation of a modern formulation of quantum theory, demonstrating a form of the Bell inequalities due to John Clauser, Michael Horne, Abner Shimony, and Richard Holt (CHSH), whose violation supports the reality of quantum entanglement – what Schrödinger called “the characteristic trait of quantum mechanics, the one that enforces its entire departure from classical lines of thought”.
I will discuss aspects of the experiments, what the results mean for our interpretation of reality, and try to shed some light on the question posed in my title.

This cell contains math definitions. \renewcommand{\d}{\mathrm{d}}
\renewcommand{\abs}[1]{\lvert#1\rvert}
\renewcommand{\I}{\mathrm{i}}
\renewcommand{\op}[1]{\bm{\hat{#1}}}
\renewcommand{\mat}[1]{\bm{#1}}
\renewcommand{\smat}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}
\renewcommand{\ua}{\uparrow}
\renewcommand{\da}{\downarrow}
\renewcommand{\la}{\leftarrow}
\renewcommand{\ra}{\roghtarrow}

%%writefile my_code.py
import numpy as np
import matplotlib.pyplot as plt
from qiskit.visualization import plot_bloch_vector


def detector(n=np.inf, label=None):
    """Draw a detector at angle np.pi/n"""
    theta = np.pi/n
    c = "C0"
    if label is None:
        if np.isinf(n):
            c = "g"
            label = r"$\mathbf{\hat{A}}_1=\mathbf{\hat{R}}(0)=\mathbf{\hat{Z}}}$"
        elif n == 2:
            c = "g"
            label = r"$\mathbf{\hat{A}}_2=\mathbf{\hat{R}}(\pi/2)=\mathbf{\hat{X}}$"
        else:
            if n == 4:
                c = "b"
                pre = r"\mathbf{\hat{B}}_1="
            elif n == -4:
                c = "b"
                pre = r"\mathbf{\hat{B}}_2="
            else:
                pre = ""
            if n < 0:
                label = fr"${pre}\mathbf{{\hat{{R}}}}(-\pi/{-n})$"
            else:
                label = fr"${pre}\mathbf{{\hat{{R}}}}(\pi/{n})$"
    fig, ax = plt.subplots(figsize=(2,2))
    z = np.linspace(-1, 1, 500)
    th = np.pi * z
    ph = np.exp(1j*(np.pi/2-theta))
    ls = (z + 0.03j)*ph, (z - 0.03j)*ph
    ax.plot(np.cos(th), np.sin(th), 'k-')
    [ax.plot(_l.real, _l.imag, c+'-') for _l in ls]
    ax.text(0.7,0.8, label)
    ax.axis('off')


def get_vec(psi):
    """Return the Bloch vector from psi."""
    gamma = np.angle(psi[0])
    psi = np.exp(-1j*gamma) * np.asarray(psi)
    phi = np.angle(psi[1]) - np.angle(psi[0])
    theta = 2*np.arctan2(psi[0].real,
                         (np.exp(-1j*phi)*psi[1]).real)
    x = np.sin(theta)*np.cos(phi)
    y = np.sin(theta)*np.sin(phi)
    z = np.cos(theta)
    return (x, y, z)

Overwriting my_code.py

Outline

What is Quantum Mechanics?
Qubits: Spin 1/2, ∣ψ⟩=a∣↑⟩+b∣↓⟩\ket{\psi} = a\ket{\uparrow} + b\ket{\downarrow}∣ψ⟩=a∣↑⟩+b∣↓⟩ 
Two particles: ∣ψ⟩=a∣↑↑⟩+b∣↑↓⟩+c∣↓↑⟩+d∣↓↓⟩\ket{\psi} = a\ket{\uparrow\uparrow} + b\ket{\uparrow\downarrow} + c\ket{\downarrow\uparrow} + d\ket{\downarrow\downarrow}∣ψ⟩=a∣↑↑⟩+b∣↑↓⟩+c∣↓↑⟩+d∣↓↓⟩
The Einstein-Podolski-Rosen (EPR) "paradox". Spooky action at a distance.
The Bell inequalities (CHSH formulation)
Experimental tests: the 2022 Nobel Prize
Discussion: Teleportation, Delayed-Choice Quantum Eraser (no retrocausality) Superdeterminism
Questions?

EPR

What is Classical Mechanics?

Kinematics: How do we describe the behaviour of a particle?
x(t):v(t)=x˙(t),a(t)=x¨(t).x(t): \qquad v(t)=\dot{x}(t), \qquad a(t) = \ddot{x}(t).x(t):v(t)=x˙(t),a(t)=x¨(t).
Dynamics: How do we see if the behaviour is physical?
F(x,v,t)=mx¨.(Newton’s Law)F(x, v, t) = m\ddot{x}.  \tag{Newton's Law}F(x,v,t)=mx¨.(Newton’s Law)Alternative formulations: Lagrangian and Hamiltonian.  I.e. δS[x]δx=0,S[x]=∫dx  L(x,x˙,t)=∫dx  (mx˙22−V(x,t)).
   \frac{\delta S[x]}{\delta x} = 0, \qquad
   S[x] = \int\d{x}\; L(x, \dot{x}, t) 
          = \int\d{x}\; \Bigl(\frac{m\dot{x}^2}{2} - V(x, t)\Bigr).
 δxδS[x]​=0,S[x]=∫dxL(x,x˙,t)=∫dx(2mx˙2​−V(x,t)).

What is Quantum Mechanics?

Kinematics: How do we describe the behaviour of a particle?
ψ(x,t):p(x,t)=∣ψ(x,t)∣2.(Wavefunction)\psi(x, t): \qquad p(x, t) = \abs{\psi(x, t)}^2. \tag{Wavefunction}ψ(x,t):p(x,t)=∣ψ(x,t)∣2.(Wavefunction)
Dynamics: How do we see if the behaviour is physical?
iℏψ˙(x,t)=(−ℏ2∇22m+V(x,t))⏟H^(Hamiltonian)ψ(x,t).(Schro¨dinger Eq.)\I\hbar \dot{\psi}(x, t) = \underbrace{\biggl(\frac{-\hbar^2\nabla^2}{2m} + V(x, t)\biggr)}_{\op{H} \quad \text{(Hamiltonian)}}\psi(x, t).
            \qquad \tag{Schrödinger Eq.}iℏψ˙​(x,t)=H^(Hamiltonian)(2m−ℏ2∇2​+V(x,t))​​ψ(x,t).(Schro¨dinger Eq.)

What is Quantum Mechanics?

Linear Algebra (Heisenberg's Matrix Mechanics)

[unavailable png image]

∣ψ⟩=(0.03300.24390.66290.66290.24390.0330)\ket{\psi} = \begin{pmatrix}0.0330\\ 0.2439\\ 0.6629\\ 0.6629\\ 0.2439\\ 0.0330\end{pmatrix}∣ψ⟩=​0.03300.24390.66290.66290.24390.0330​​

x=(−10−6−22610)\mat{x} = \begin{pmatrix}-10&&&&&\\&-6&&&&\\&&-2&&&&\\&&&2&&\\&&&&6&\\&&&&&10\end{pmatrix}x=​−10​−6​−2​2​6​10​​

p=−iℏ4(−11−11−11−11−111−1)\mat{p} =
\frac{-\I\hbar}{4}
\begin{pmatrix}
-1&1&&&&\\
&-1&1&&&\\
&&-1&1&&\\
&&&-1&1&\\
&&&&-1&1\\
1&&&&&-1
\end{pmatrix}
p=4−iℏ​​−11​1−1​1−1​1−1​1−1​1−1​​

Wavefunction ψ(x)\psi(x)ψ(x) is a vector ∣ψ⟩\ket{\psi}∣ψ⟩ (in Hilbert space).
"Observables" become operators: x→x^=δ(x−y)x,p→p^=δ(x−y)(−iℏ∂x).\begin{align*}
    x \rightarrow \op{x} &= \delta(x-y)x,\\
    p \rightarrow \op{p} &= \delta(x-y)(-\I\hbar \partial_x).
  \end{align*}x→x^p→p^​​=δ(x−y)x,=δ(x−y)(−iℏ∂x​).​

What is Quantum Mechanics? 

Eigenvectors and Eigenvalues

x=(−10−6−22610)\mat{x} = \begin{pmatrix}-10&&&&&\\&-6&&&&\\&&-2&&&&\\&&&2&&\\&&&&6&\\&&&&&10\end{pmatrix}x=​−10​−6​−2​2​6​10​​

p=−iℏ4(−11−11−11−11−111−1)\mat{p} =
\frac{-\I\hbar}{4}
\begin{pmatrix}
-1&1&&&&\\
&-1&1&&&\\
&&-1&1&&\\
&&&-1&1&\\
&&&&-1&1\\
1&&&&&-1
\end{pmatrix}
p=4−iℏ​​−11​1−1​1−1​1−1​1−1​1−1​​

∣−10⟩x=(100000),∣−6⟩x=(010000),⋯x^∣xn⟩=∣xn⟩xxn,x^∣−10⟩x=−10∣−10⟩x,⋯\ket{-10}_x = \begin{pmatrix}1\\0\\0\\0\\0\\0\end{pmatrix}, \qquad
  \ket{-6}_x = \begin{pmatrix}0\\1\\0\\0\\0\\0\end{pmatrix}, \qquad\cdots\\
  \op{x}\ket{x_n} = \ket{x_n}_x x_n, \qquad 
  \op{x}\ket{-10} _x= -10\ket{-10}_x, \qquad
  \cdots∣−10⟩x​=​100000​​,∣−6⟩x​=​010000​​,⋯x^∣xn​⟩=∣xn​⟩x​xn​,x^∣−10⟩x​=−10∣−10⟩x​,⋯

⟨x∣kn⟩p∝eiknx,kn=2πn24∣n∈{−3,−2,−1,0,1,2}\braket{x|k_n}_{p}  \propto e^{\I k_n x}, \qquad
  k_n = \left.\frac{2\pi n}{24}\right|_{n \in \{-3, -2, -1, 0, 1, 2\}}⟨x∣kn​⟩p​∝eikn​x,kn​=242πn​​n∈{−3,−2,−1,0,1,2}​

Measurementaka "Nobody understands quantum mechanics"

∣ψ⟩=(0.03300.24390.66290.66290.24390.0330)\ket{\psi} = \begin{pmatrix}0.0330\\ 0.2439\\ 0.6629\\ 0.6629\\ 0.2439\\ 0.0330\end{pmatrix}∣ψ⟩=​0.03300.24390.66290.66290.24390.0330​​

x=(−10−6−22610)\mat{x} = \begin{pmatrix}-10&&&&&\\&-6&&&&\\&&-2&&&&\\&&&2&&\\&&&&6&\\&&&&&10\end{pmatrix}x=​−10​−6​−2​2​6​10​​

[unavailable png image]

Px=−6=0.24392P_{x=-6}=0.2439^2Px=−6​=0.24392

Px=2=0.66292P_{x=2}=0.6629^2Px=2​=0.66292

import numpy as np, matplotlib.pyplot as plt
from IPython.display import Math

N = 6
x = np.linspace(-10, 10, N)
X = np.linspace(-10, 10, 200)
sigma = 4
def psi(x, x_=0, sigma=sigma):
    return (np.exp(-(x-x_)**2/sigma**2/2) / np.sqrt(np.pi) * 2 * sigma)
style_dict = {'text.usetex': True}
style_dict = {}
with plt.style.context(style_dict, after_reset=True):
    fig, ax = plt.subplots(figsize=(5,2))
    plt.plot(X, psi(X), '-')
    plt.plot(x, psi(x), 'o')
    ax.set(xlabel='$x$', ylabel=r'$\psi(x)$');
    ket_ = Math(r"$\ket{\psi} = \begin{pmatrix}" 
                + r"\\ ".join(
            map("{:.4f}".format, psi(x) / np.linalg.norm(psi(x))))
                + r"\end{pmatrix}$")
    #print(ket_.data)
    X_ = Math(
        r"$\mat{{X}} = \begin{{pmatrix}}{}\end{{pmatrix}}$".format(
            r"\\".join(["&" * i + f"{x[i]:g}" + "&"* (N-i-1)
                        for i in range(N)])))
    #print(X_.data)
    #print(np.diff(x).mean())

Px=−2=0.66292P_{x=-2}=0.6629^2Px=−2​=0.66292

Many Particles?

Classical: NNN variables x1(t)x_1(t)x1​(t), x2(t)x_2(t)x2​(t), x3(t)x_3(t)x3​(t), ...
NN1NN_1NN1​ times as much information.
Quantum: Wavefunction of NNN variables ψ(x1,x2,x3,⋯ )\psi(x_1, x_2, x_3, \cdots)ψ(x1​,x2​,x3​,⋯)
Tensor product.
Exponential amount of information: N1NN_1^NN1N​.
Hilbert space is big!
Note: 1 particle in 2D is mathematically equivalent to 2 particles in 1D.

Postulates of Quantum Mechanics

The state ∣ψ⟩\ket{\psi}∣ψ⟩ of an isolated systems is a complex unit vector ⟨ψ∣ψ⟩=1\braket{\psi|\psi} = 1⟨ψ∣ψ⟩=1. (Hilbert space.) ∣ψ⟩≡α∣ψ⟩\ket{\psi} \equiv \alpha \ket{\psi}∣ψ⟩≡α∣ψ⟩. (phase equivalence)
States evolve under unitary operations: ∣ψ⟩→U∣ψ⟩\ket{\psi} \rightarrow \mat{U}\ket{\psi}∣ψ⟩→U∣ψ⟩ where U†U=1\mat{U}^\dagger\mat{U} = \mat{1}U†U=1.
Measurements are a set of operators {M^m}\{\op{M}_{m}\}{M^m​} such that ∑mM^m†M^m=1^\sum_{m}\op{M}_m^\dagger\op{M}_m = \op{1}∑m​M^m†​M^m​=1^.  A measurement made on the state ∣ψ⟩\ket{\psi}∣ψ⟩ will produce outcome mmm with probability pmp_mpm​, transforming the state as follows:
pm=⟨ψ∣M^m†M^m∣ψ⟩,∣ψ⟩→M^m∣ψ⟩pm.p_m = \braket{\psi|\op{M}_m^\dagger\op{M}_m|\psi}, \qquad
     \ket{\psi} \rightarrow \frac{\op{M}_m\ket{\psi}}{\sqrt{p_m}}.pm​=⟨ψ∣M^m†​M^m​∣ψ⟩,∣ψ⟩→pm​​M^m​∣ψ⟩​.
Composite systems have states in the tensor product of the component Hilbert spaces.
∣ψ⟩=∑ijaij∣i⟩A⊗∣j⟩B⏟∣ij⟩.\ket{\psi} = \sum_{ij}a_{ij}\underbrace{\ket{i}_A\otimes\ket{j}_B}_{\ket{ij}}.∣ψ⟩=ij∑​aij​∣ij⟩∣i⟩A​⊗∣j⟩B​​​.

Nielson and Chuang

Qubits (Spin 1/2)

∣↑⟩=∣0⟩=(10),∣↓⟩=∣1⟩=(01)\renewcommand{\smat}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}
  \ket{\uparrow} = \ket{0} = \smat{1\\0}, \qquad
  \ket{\downarrow} = \ket{1} = (\begin{smallmatrix}0\\1\end{smallmatrix})∣↑⟩=∣0⟩=(10​),∣↓⟩=∣1⟩=(01​)Z^=(100−1),Z^∣↑⟩=∣↑⟩,Z^∣↓⟩=−∣↓⟩X^=(0110),X^∣→⟩⏟12(11)=∣→⟩,X^∣←⟩⏟12(1−1)=−∣←⟩.\begin{align*}
  \op{Z} &= \begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}, &
  \op{Z}\ket{\uparrow} &= \ket{\uparrow}, & 
  \op{Z}\ket{\downarrow} &= -\ket{\downarrow}\\
  \op{X} &= \begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}, &
  \op{X}\underbrace{\ket{\rightarrow}}_{\tfrac{1}{\sqrt{2}}\smat{1\\1}} &= \ket{\rightarrow}, &
  \op{X}\underbrace{\ket{\leftarrow} }_{\tfrac{1}{\sqrt{2}}\smat{1\\-1}}&= -\ket{\leftarrow}.
\end{align*}Z^X^​=(10​0−1​),=(01​10​),​Z^∣↑⟩X^2​1​(11​)∣→⟩​​​=∣↑⟩,=∣→⟩,​Z^∣↓⟩X^2​1​(1−1​)∣←⟩​​​=−∣↓⟩=−∣←⟩.​

[unavailable png image][unavailable png image][unavailable png image][unavailable png image]

Stern–Gerlach

Measuring Z^\op{Z}Z^

∣↑⟩=(10)\ket{↑} = \smat{1\\0}∣↑⟩=(10​): (+1,100%)(+1, 100\%)(+1,100%)
∣↓⟩=(01)\ket{↓} = \smat{0\\1}∣↓⟩=(01​): (−1,100%)(-1, 100\%)(−1,100%)
∣→⟩=12(11)≡∣↑⟩+∣↓⟩\ket{→} = \tfrac{1}{\sqrt{2}}\smat{1\\1} \equiv \ket{↑} + \ket{↓}∣→⟩=2​1​(11​)≡∣↑⟩+∣↓⟩: (+1,50%)(+1, 50\%)(+1,50%), (−1,50%)(-1, 50\%)(−1,50%)
∣←⟩=12(1−1)≡∣↑⟩−∣↓⟩\ket{←} = \tfrac{1}{\sqrt{2}}\smat{1\\-1} \equiv \ket{↑} - \ket{↓}∣←⟩=2​1​(1−1​)≡∣↑⟩−∣↓⟩: (+1,50%)(+1, 50\%)(+1,50%), (−1,50%)(-1, 50\%)(−1,50%)

Intererence

Particle Intererence

Measure state where angle between detector and state is θ\thetaθ:
P1=cos⁡2θ2,P−1=sin⁡2θ2.P_1 = \cos^2\tfrac{\theta}{2},\qquad
  P_{-1} = \sin^2\tfrac{\theta}{2}.P1​=cos22θ​,P−1​=sin22θ​.

∣θ=0⟩=∣↑⟩,∣θ=π/4⟩=∣↗⟩,∣θ=π/2⟩=∣→⟩,∣θ=3π/4⟩=∣↘⟩,∣θ=π⟩=∣↓⟩,∣θ=−3π/4⟩=∣↙⟩,∣θ=−π/2⟩=∣←⟩,∣θ=−π/4⟩=∣↖⟩.\begin{align*}
  \ket{\theta=0} &= \ket{↑},\\
  \ket{\theta=\pi/4} &= \ket{↗},\\
  \ket{\theta=\pi/2} &= \ket{→},\\
  \ket{\theta=3\pi/4} &=\ket{↘},\\
  \ket{\theta=\pi} &= \ket{↓},\\
  \ket{\theta=-3\pi/4} &=\ket{↙},\\
  \ket{\theta=-\pi/2} &= \ket{←},\\
  \ket{\theta=-\pi/4} &= \ket{↖}.
\end{align*}∣θ=0⟩∣θ=π/4⟩∣θ=π/2⟩∣θ=3π/4⟩∣θ=π⟩∣θ=−3π/4⟩∣θ=−π/2⟩∣θ=−π/4⟩​=∣↑⟩,=∣↗⟩,=∣→⟩,=∣↘⟩,=∣↓⟩,=∣↙⟩,=∣←⟩,=∣↖⟩.​

∣θ⟩=(cos⁡(θ/2)sin⁡(θ/2))\ket{\theta} = \begin{pmatrix}\cos(\theta/2)\\ \sin(\theta/2)\end{pmatrix}∣θ⟩=(cos(θ/2)sin(θ/2)​)

[unavailable png image]

Qubits (Bloch Sphere)

Expectation Value

⟨θ∣Z^∣θ⟩=+1cos⁡2θ2−1sin⁡2θ2=−cos⁡θ=−a⃗⋅b⃗.\begin{align*}
  \braket{\theta|\op{Z}|\theta} &= +1\cos^2\tfrac{\theta}{2} -1\sin^2\tfrac{\theta}{2} \\ 
  &= -\cos\theta =-\vec{a}\cdot\vec{b}.
\end{align*}⟨θ∣Z^∣θ⟩​=+1cos22θ​−1sin22θ​=−cosθ=−a⋅b.​

∣θ⟩=(cos⁡(θ/2)sin⁡(θ/2))\ket{\theta} = \begin{pmatrix}\cos(\theta/2)\\ \sin(\theta/2)\end{pmatrix}∣θ⟩=(cos(θ/2)sin(θ/2)​)

Stern–Gerlach

Measuring Z^\op{Z}Z^

∣↑⟩\ket{↑}∣↑⟩: (+1,100%)(+1, 100\%)(+1,100%)
∣↓⟩\ket{↓}∣↓⟩: (−1,100%)(-1, 100\%)(−1,100%)
∣→⟩\ket{→}∣→⟩: (+1,50%)(+1, 50\%)(+1,50%), (−1,50%)(-1, 50\%)(−1,50%)
∣←⟩\ket{←}∣←⟩: (+1,50%)(+1, 50\%)(+1,50%), (−1,50%)(-1, 50\%)(−1,50%)
∣↗︎⟩\ket{↗︎}∣↗︎⟩: (+1,85%)(+1, 85\%)(+1,85%), (−1,15%)(-1, 15\%)(−1,15%)
∣↙︎⟩\ket{↙︎}∣↙︎⟩: (+1,15%)(+1, 15\%)(+1,15%), (−1,85%)(-1, 85\%)(−1,85%)
∣θ⟩\ket{θ}∣θ⟩: (+1,cos⁡2θ2)(+1, \cos^2\tfrac{\theta}{2})(+1,cos22θ​), (−1,sin⁡2θ2)(-1, \sin^2\tfrac{\theta}{2})(−1,sin22θ​)

Qubits (Spin 1/2)

∣θ=0⟩=∣↑⟩=(10),(cos⁡(π/8)sin⁡(π/8))=∣θ=π/4⟩=∣↗⟩≈(0.90.4),(cos⁡(π/4)sin⁡(π/4))=∣θ=π/2⟩=∣→⟩≈(0.70.7),(cos⁡(3π/8)sin⁡(3π/8))=∣θ=3π/4⟩=∣↘⟩≈(0.40.9),(cos⁡(π/2)sin⁡(π/2))=∣θ=π⟩=∣↓⟩=(01),(cos⁡(5π/8)sin⁡(5π/8))=∣θ=5π/4⟩=∣↙⟩≈(−0.40.9),(cos⁡(3π/4)sin⁡(3π/4))=∣θ=3π/2⟩=∣←⟩≈(−0.70.7),(cos⁡(7π/8)sin⁡(7π/8))=∣θ=7π/4⟩=∣↖⟩≈(−0.90.4),(cos⁡(π)sin⁡(π))=∣θ=2π⟩=−∣↑⟩≈(−10).\begin{align*}
  \ket{\theta=0} &= \ket{↑} = \smat{1\\0},\\
  \smat{\cos(\pi/8)\\\sin(\pi/8)} = \ket{\theta=\pi/4} &= \ket{↗} \approx \smat{0.9\\0.4},\\
  \smat{\cos(\pi/4)\\ \sin(\pi/4)} = \ket{\theta=\pi/2} &= \ket{→} \approx \smat{0.7\\0.7},\\
  \smat{\cos(3\pi/8)\\ \sin(3\pi/8)} = \ket{\theta=3\pi/4} &=\ket{↘} \approx \smat{0.4\\0.9},\\
  \smat{\cos(\pi/2)\\ \sin(\pi/2)} = \ket{\theta=\pi} &= \ket{↓} = \smat{0\\1},\\
  \smat{\cos(5\pi/8)\\ \sin(5\pi/8)} = \ket{\theta=5\pi/4} &=\ket{↙} \approx \smat{-0.4\\0.9},\\
  \smat{\cos(3\pi/4)\\ \sin(3\pi/4)} = \ket{\theta=3\pi/2} &= \ket{←} \approx \smat{-0.7\\0.7},\\
  \smat{\cos(7\pi/8)\\ \sin(7\pi/8)} = \ket{\theta=7\pi/4} &= \ket{↖} \approx \smat{-0.9\\0.4},\\
  \smat{\cos(\pi)\\ \sin(\pi)} = \ket{\theta=2\pi} &= -\ket{↑} \approx \smat{-1\\0}.
\end{align*}∣θ=0⟩(cos(π/8)sin(π/8)​)=∣θ=π/4⟩(cos(π/4)sin(π/4)​)=∣θ=π/2⟩(cos(3π/8)sin(3π/8)​)=∣θ=3π/4⟩(cos(π/2)sin(π/2)​)=∣θ=π⟩(cos(5π/8)sin(5π/8)​)=∣θ=5π/4⟩(cos(3π/4)sin(3π/4)​)=∣θ=3π/2⟩(cos(7π/8)sin(7π/8)​)=∣θ=7π/4⟩(cos(π)sin(π)​)=∣θ=2π⟩​=∣↑⟩=(10​),=∣↗⟩≈(0.90.4​),=∣→⟩≈(0.70.7​),=∣↘⟩≈(0.40.9​),=∣↓⟩=(01​),=∣↙⟩≈(−0.40.9​),=∣←⟩≈(−0.70.7​),=∣↖⟩≈(−0.90.4​),=−∣↑⟩≈(−10​).​

∣θ⟩=(cos⁡(θ/2)sin⁡(θ/2))\ket{\theta} = \begin{pmatrix}\cos(\theta/2)\\ \sin(\theta/2)\end{pmatrix}∣θ⟩=(cos(θ/2)sin(θ/2)​)

Two Qubits (Tensor Product)

∣↑↑⟩=(1000),∣↑↓⟩=(0100),∣↓↑⟩=(0010),∣↓↓⟩=(0001).\ket{\uparrow\uparrow} = \smat{1\\0\\0\\0}, \qquad
  \ket{\uparrow\downarrow} = \smat{0\\1\\0\\0}, \qquad
  \ket{\downarrow\uparrow} = \smat{0\\0\\1\\0}, \qquad
  \ket{\downarrow\downarrow} = \smat{0\\0\\0\\1}.∣↑↑⟩=(1000​),∣↑↓⟩=(0100​),∣↓↑⟩=(0010​),∣↓↓⟩=(0001​).

Z^A=Z^⊗1^=(1−1)=(11−1−1),Z^B=1^⊗Z^=(ZZ)=(1−11−1),X^A=X^⊗1^=(11)=(1111),X^B=1^⊗X^=(XX)=(1111).\begin{align*}
  \op{Z}_A = \op{Z}\otimes\op{1} &= 
 \begin{pmatrix} 
  \mat{1}\\
  & -\mat{1}\end{pmatrix}
  =
  \begin{pmatrix} 
  1 \\
  & 1 \\
  & & -1\\
  & & & -1\end{pmatrix}, &
  \op{Z}_B = \op{1}\otimes\op{Z} &= 
  \begin{pmatrix} 
  \mat{Z}\\
  & \mat{Z}
  \end{pmatrix} 
  = 
  \begin{pmatrix} 
    1 \\
  & -1 \\
  & & 1\\
  & & & -1\end{pmatrix}, 
  \\
  \op{X}_A = \op{X}\otimes\op{1} &= 
  \begin{pmatrix} 
  & \mat{1}\\
  \mat{1}
  \end{pmatrix} 
  =
  \begin{pmatrix} 
  & & 1 \\
  & & & 1 \\
  1\\
  & 1
  \end{pmatrix},&
  \op{X}_B = \op{1}\otimes\op{X} &= 
  \begin{pmatrix} 
  \mat{X}\\
  & \mat{X}
  \end{pmatrix} 
  =
  \begin{pmatrix} 
  & 1 \\
  1& \\
  & & & 1\\
  & &1
  \end{pmatrix}.
\end{align*}Z^A​=Z^⊗1^X^A​=X^⊗1^​=(1​−1​)=​1​1​−1​−1​​,=(1​1)=​1​1​11​,​Z^B​=1^⊗Z^X^B​=1^⊗X^​=(Z​Z​)=​1​−1​1​−1​​,=(X​X​)=​1​1​1​1​​.​

Entanglement 

Start with ∣↑↑⟩\ket{\uparrow\uparrow}∣↑↑⟩. Alice does stuff U^A⊗1^\op{U}_A\otimes \op{1}U^A​⊗1^ and Bob does stuff 1^⊗U^B\op{1}\otimes\op{U}_B1^⊗U^B​. They generate no entanglement: State remains a product:
∣ψ⟩=(a∣↑⟩+b∣↓⟩)⏞U^A∣↑⟩⊗(c∣↑⟩+d∣↓⟩)⏞U^B∣↑⟩\ket{\psi} = \overbrace{(a\ket{\ua} + b \ket{\da})}^{\op{U}_A\ket{\ua}}
  \otimes \overbrace{(c\ket{\ua} + d\ket{\da})}^{\op{U}_B\ket{\ua}}∣ψ⟩=(a∣↑⟩+b∣↓⟩)​U^A​∣↑⟩​⊗(c∣↑⟩+d∣↓⟩)​U^B​∣↑⟩​Bell States cannot be written as products:
∣↑↑⟩+∣↓↓⟩2,∣↑↑⟩−∣↓↓⟩2,∣↑↓⟩+∣↓↑⟩2,∣↑↓⟩−∣↓↑⟩2.\frac{\ket{\ua\ua} + \ket{\da\da}}{\sqrt{2}}, \qquad
  \frac{\ket{\ua\ua} - \ket{\da\da}}{\sqrt{2}}, \qquad
  \frac{\ket{\ua\da} + \ket{\da\ua}}{\sqrt{2}}, \qquad
  \frac{\ket{\ua\da} - \ket{\da\ua}}{\sqrt{2}}.2​∣↑↑⟩+∣↓↓⟩​,2​∣↑↑⟩−∣↓↓⟩​,2​∣↑↓⟩+∣↓↑⟩​,2​∣↑↓⟩−∣↓↑⟩​.

Entanglement: Singlet

∣↑↓⟩−∣↓↑⟩2≡∣→←⟩−∣←→⟩2≡∣↖︎↘︎⟩−∣↘︎↖︎⟩2≡∣↗︎↙︎⟩−∣↙︎↗︎⟩2.\renewcommand{\singlet}[2]{\frac{\ket{#1#2}-\ket{#2#1}}{\sqrt{2}}}
  \singlet{↑}{↓} \equiv
  \singlet{→}{←} \equiv
  \singlet{↖︎}{↘︎}\equiv
  \singlet{↗︎}{↙︎}.2​∣↑↓⟩−∣↓↑⟩​≡2​∣→←⟩−∣←→⟩​≡2​∣↖︎↘︎⟩−∣↘︎↖︎⟩​≡2​∣↗︎↙︎⟩−∣↙︎↗︎⟩​.

import numpy as np
from numpy import kron

def normalize(psi):
    return np.divide(psi, np.linalg.norm(psi))

A1 = [normalize([1, 0]), normalize([0, 1])]
A2 = [normalize([1, 1]), normalize([1, -1])]
th = np.pi/4; B1 = [normalize([np.cos(th), np.sin(th)]), normalize([np.sin(th), -np.cos(th)])]
th = 3*np.pi/4; B2 = [normalize([np.cos(th), np.sin(th)]), normalize([np.sin(th), -np.cos(th)])]
print("|↑↓⟩-|↓↑⟩ = {}".format(kron(A1[0], A1[1]) - kron(A1[1], A1[0])))
print("|→←⟩-|←→⟩ = {}".format(kron(A2[0], A2[1]) - kron(A2[1], A2[0])))
print("|↗︎↙︎⟩-|↙︎↗︎⟩ = {}".format(kron(B1[0], B1[1]) - kron(B1[1], B1[0])))
print("|↖︎↘︎⟩-|↘︎↖︎⟩ = {}".format(kron(B2[0], B2[1]) - kron(B2[1], B2[0])))

|↑↓⟩-|↓↑⟩ = [ 0.  1. -1.  0.]
|→←⟩-|←→⟩ = [ 0. -1.  1.  0.]
|↗︎↙︎⟩-|↙︎↗︎⟩ = [ 0. -1.  1.  0.]
|↖︎↘︎⟩-|↘︎↖︎⟩ = [ 0. -1.  1.  0.]

Einstein-Podolsky-Rosen (EPR)

∣↑↓⟩−∣↓↑⟩2\singlet{↑}{↓}2​∣↑↓⟩−∣↓↑⟩​

Hidden Variable λ

R^(a⃗)\op{R}(\vec{a})R^(a)

R^(b⃗)\op{R}(\vec{b})R^(b)

Local Realism: E(a⃗,b⃗)=⟨R(a⃗)R(b⃗)⟩=∫p(λ)A(a⃗,λ)B(b⃗,λ)E(\vec{a},\vec{b})=\braket{R(\vec{a})R(\vec{b})} = \int p(λ)A(\vec{a},λ)B(\vec{b},λ)E(a,b)=⟨R(a)R(b)⟩=∫p(λ)A(a,λ)B(b,λ).
Quantum: E(a⃗,b⃗)=⟨R(a⃗)R(b⃗)⟩=−a⃗⋅b⃗\qquad E(\vec{a},\vec{b})=\braket{R(\vec{a})R(\vec{b})}
= -\vec{a}\cdot\vec{b}E(a,b)=⟨R(a)R(b)⟩=−a⋅b

Bell Inequalities 

E(a⃗,b⃗)=⟨R(a⃗)R(b⃗)⟩=∫p(λ)A(a⃗,λ)B(b⃗,λ),A(a⃗,λ)=±1,B(b⃗,λ)=±1.E(\vec{a}, \vec{b}) = \braket{R(\vec{a})R(\vec{b})} = \int p(λ)A(\vec{a},λ)B(\vec{b},λ),
  \qquad A(\vec{a},λ)=\pm1, \qquad B(\vec{b},λ)=\pm 1.E(a,b)=⟨R(a)R(b)⟩=∫p(λ)A(a,λ)B(b,λ),A(a,λ)=±1,B(b,λ)=±1.Bell proved:
∣E(a⃗,b⃗)−E(a⃗,c⃗)∣≤1+E(b⃗,c⃗).\abs{E(\vec{a},\vec{b}) - E(\vec{a},\vec{c})} \leq 1 + E(\vec{b},\vec{c}).∣E(a,b)−E(a,c)∣≤1+E(b,c).Take a⃗=↑\vec{a} = ↑a=↑, b⃗=→\vec{b}=→b=→, and c⃗=↗︎\vec{c}=↗︎c=↗︎:
∣E(↑,→)⏞0−E(↑,↗︎)⏞−1/2∣≰1+E(→,↗︎)⏞−1/20.707≈12≰1−12≈0.293\begin{align*}
  \abs{\overbrace{E(↑,→)}^{0} - \overbrace{E(↑,↗︎)}^{-1/\sqrt{2}}} &≰ 1 + \overbrace{E(→,↗︎)}^{-1/\sqrt{2}}\\
  0.707\approx \frac{1}{\sqrt{2}} &≰ 1-\frac{1}{\sqrt{2}}\approx 0.293
\end{align*}∣E(↑,→)​0​−E(↑,↗︎)​−1/2​​∣0.707≈2​1​​≰1+E(→,↗︎)​−1/2​​≰1−2​1​≈0.293​

Quantum: E(a⃗)(b⃗)=−a⃗⋅b⃗E(\vec{a})(\vec{b}) = -\vec{a}\cdot\vec{b}E(a)(b)=−a⋅b

Quantum Calculation

WLoG: Alice measures Z^\op{Z}Z^, Bob measures R^(θ)\op{R}(\theta)R^(θ).
P1=(↑)(A)=0.5,∣ψ⟩B=∣↓⟩,{P1(B)=sin⁡2θ2,P−1(B)=cos⁡2θ2,P−1=(↓)(A)=0.5,∣ψ⟩B=∣↑⟩,{P1(B)=cos⁡2θ2,P−1(B)=sin⁡2θ2.P^{(A)}_{1=(↑)} = 0.5, \qquad \ket{\psi}_B = \ket{↓}, \qquad
  \begin{cases}
    P^{(B)}_{1} = \sin^2\frac{\theta}{2},\\
    P^{(B)}_{-1} = \cos^2\frac{\theta}{2},
  \end{cases}\\
  P^{(A)}_{-1=(↓)} = 0.5, \qquad \ket{\psi}_B = \ket{↑}, \qquad
  \begin{cases}
    P^{(B)}_{1} = \cos^2\frac{\theta}{2},\\
    P^{(B)}_{-1} = \sin^2\frac{\theta}{2}.
  \end{cases}P1=(↑)(A)​=0.5,∣ψ⟩B​=∣↓⟩,{P1(B)​=sin22θ​,P−1(B)​=cos22θ​,​P−1=(↓)(A)​=0.5,∣ψ⟩B​=∣↑⟩,{P1(B)​=cos22θ​,P−1(B)​=sin22θ​.​Thus
E(0,θ)=sin⁡2θ2−cos⁡2θ2−cos⁡2θ2+sin⁡2θ22=−(cos⁡2θ2−sin⁡2θ2)=−cos⁡θ=−a⃗⋅b⃗.E(0,θ) = \frac{\sin^2\frac{\theta}{2} - \cos^2\frac{\theta}{2} - \cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}}{2}
    = - \Bigl(\cos^2\tfrac{\theta}{2} - \sin^2\tfrac{\theta}{2}\Bigr)
    = -\cos\theta = -\vec{a}\cdot\vec{b}.E(0,θ)=2sin22θ​−cos22θ​−cos22θ​+sin22θ​​=−(cos22θ​−sin22θ​)=−cosθ=−a⋅b.

∣↑↓⟩−∣↓↑⟩2≡∣→←⟩−∣←→⟩2≡∣↖︎↘︎⟩−∣↘︎↖︎⟩2≡∣↗︎↙︎⟩−∣↙︎↗︎⟩2.\renewcommand{\singlet}[2]{\frac{\ket{#1#2}-\ket{#2#1}}{\sqrt{2}}}
  \singlet{↑}{↓} \equiv
  \singlet{→}{←} \equiv
  \singlet{↖︎}{↘︎}\equiv
  \singlet{↗︎}{↙︎}.2​∣↑↓⟩−∣↓↑⟩​≡2​∣→←⟩−∣←→⟩​≡2​∣↖︎↘︎⟩−∣↘︎↖︎⟩​≡2​∣↗︎↙︎⟩−∣↙︎↗︎⟩​.

E(a⃗,b⃗)=⟨R(a⃗)R(b⃗)⟩=∫p(λ)A(a⃗,λ)B(b⃗,λ),A(a⃗,λ)=±1,B(b⃗,λ)=±1.E(\vec{a}, \vec{b}) = \braket{R(\vec{a})R(\vec{b})} = \int p(λ)A(\vec{a},λ)B(\vec{b},λ),
  \qquad A(\vec{a},λ)=\pm1, \qquad B(\vec{b},λ)=\pm 1.E(a,b)=⟨R(a)R(b)⟩=∫p(λ)A(a,λ)B(b,λ),A(a,λ)=±1,B(b,λ)=±1.Bell proved:
∣E(a⃗,b⃗)−E(a⃗,c⃗)∣≤1+E(b⃗,c⃗).\abs{E(\vec{a},\vec{b}) - E(\vec{a},\vec{c})} \leq 1 + E(\vec{b},\vec{c}).∣E(a,b)−E(a,c)∣≤1+E(b,c).Take a⃗=↑\vec{a} = ↑a=↑, b⃗=→\vec{b}=→b=→, and c⃗=↗︎\vec{c}=↗︎c=↗︎, and use quantum E(a⃗)(b⃗)=−a⃗⋅b⃗=−cos⁡θE(\vec{a})(\vec{b}) = -\vec{a}\cdot\vec{b} = -\cos\thetaE(a)(b)=−a⋅b=−cosθ:
E(↑,→)=−cos⁡(90°)=0,E(↑,↗︎)=E(→,↗︎)=−cos⁡(45°)=−12.E(↑,→) = -\cos(90°) = 0, \qquad
  E(↑,↗︎) = E(→,↗︎) = -\cos(45°) = - \frac{1}{\sqrt{2}}.E(↑,→)=−cos(90°)=0,E(↑,↗︎)=E(→,↗︎)=−cos(45°)=−2​1​.∣E(↑,→)⏞0−E(↑,↗︎)⏞−1/2∣≰1+E(→,↗︎)⏞−1/20.707≈12≰1−12≈0.293\begin{align*}
  \abs{\overbrace{E(↑,→)}^{0} - \overbrace{E(↑,↗︎)}^{-1/\sqrt{2}}} &≰ 1 + \overbrace{E(→,↗︎)}^{-1/\sqrt{2}}\\
  0.707\approx \frac{1}{\sqrt{2}} &≰ 1-\frac{1}{\sqrt{2}}\approx 0.293
\end{align*}∣E(↑,→)​0​−E(↑,↗︎)​−1/2​​∣0.707≈2​1​​≰1+E(→,↗︎)​−1/2​​≰1−2​1​≈0.293​

Bell Inequalities 

Clauser-Horne-Shimony-Holt (CHSH)

Consider four measurements with values ±1\pm 1±1 – two for Alice A1A_1A1​, A2A_2A2​ and two for Bob B1B_1B1​, B2B_2B2​:
Ai,Bi=±1Q=A1(B1+B2)+A2(B1−B2)=±2.A_i, B_i = \pm 1\\
  Q = A_1(B_1+B_2) + A_2(B_1-B_2) = \pm 2.Ai​,Bi​=±1Q=A1​(B1​+B2​)+A2​(B1​−B2​)=±2.

[A1*(B1+B2) + A2*(B1-B2) for A1 in [-1, 1] for A2 in [-1, 1]
                         for B1 in [-1, 1] for B2 in [-1, 1]]

[2, 2, -2, -2, 2, -2, 2, -2, -2, 2, -2, 2, -2, -2, 2, 2]

S=∣⟨Q⟩∣=∣⟨A1↑B1↗︎⟩⏟−1/2+⟨A1↑B2↖︎⟩⏟−1/2+⟨A2→B1↗︎⟩⏟−1/2−⟨A2→B2↖︎⟩⏟+1/2∣≤2\begin{align*}
  S = \abs{\braket{Q}} 
      &= \abs{
      \underbrace{\braket{\overset{↑}{A_1}\overset{↗︎}{B_1}}}_{-1/\sqrt{2}}
      + \underbrace{\braket{\overset{↑}{A_1}\overset{↖︎}{B_2}}}_{-1/\sqrt{2}}
      + \underbrace{\braket{\overset{→}{A_2}\overset{↗︎}{B_1}}}_{-1/\sqrt{2}}
      - \underbrace{\braket{\overset{→}{A_2}\overset{↖︎}{B_2}}}_{+1/\sqrt{2}}} \leq 2
\end{align*}S=∣⟨Q⟩∣​=∣−1/2​⟨A1​↑​B1​↗︎​⟩​​+−1/2​⟨A1​↑​B2​↖︎​⟩​​+−1/2​⟨A2​→​B1​↗︎​⟩​​−+1/2​⟨A2​→​B2​↖︎​⟩​​∣≤2​

Quantum gives: 4/2≈2.82≰24/\sqrt{2} \approx 2.82 ≰ 24/2​≈2.82≰2

2023 Nobel Prize 

Experimental ValidationClosing loopholes

Coincidence Detection

Discussion 

Local Realism: E(a⃗,b⃗)=⟨R(a⃗)R(b⃗)⟩=∫p(λ)A(a⃗,λ)B(b⃗,λ)
   E(\vec{a},\vec{b})=\braket{R(\vec{a})R(\vec{b})} = \int p(λ)A(\vec{a},λ)B(\vec{b},λ)
 E(a,b)=⟨R(a)R(b)⟩=∫p(λ)A(a,λ)B(b,λ)
Counterfactuals: S=∣⟨Q⟩∣=∣⟨A1↑B1↗︎⟩⏟−1/2+⟨A1↑B2↖︎⟩⏟−1/2+⟨A2→B1↗︎⟩⏟−1/2−⟨A2→B2↖︎⟩⏟+1/2∣≤2\begin{align*}
S = \abs{\braket{Q}} 
    &= \abs{
    \underbrace{\braket{\overset{↑}{A_1}\overset{↗︎}{B_1}}}_{-1/\sqrt{2}}
    + \underbrace{\braket{\overset{↑}{A_1}\overset{↖︎}{B_2}}}_{-1/\sqrt{2}}
    + \underbrace{\braket{\overset{→}{A_2}\overset{↗︎}{B_1}}}_{-1/\sqrt{2}}
    - \underbrace{\braket{\overset{→}{A_2}\overset{↖︎}{B_2}}}_{+1/\sqrt{2}}} \leq 2
 \end{align*}S=∣⟨Q⟩∣​=∣−1/2​⟨A1​↑​B1​↗︎​⟩​​+−1/2​⟨A1​↑​B2​↖︎​⟩​​+−1/2​⟨A2​→​B1​↗︎​⟩​​−+1/2​⟨A2​→​B2​↖︎​⟩​​∣≤2​
Superdeterminism  p(λ)→p(λ,a⃗,b⃗)p(λ)→p(λ,\vec{a},\vec{b})p(λ)→p(λ,a,b)

Teleportation

Quantum Eraser

No Retrocausality!

WSU: Physics 455/555

Quantum Technologies and Computing

https://physics-555-quantum-technologies.readthedocs.io/en/latest/

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Is Quantum Physics Fundamentally Different than Classical Physics?

Tutorial on the 2022 Nobel Prize

Michael McNeil Forbes

Experimental Validation

2023 Nobel Prize

Abstract

Outline

EPR

What is Classical Mechanics?

What is Quantum Mechanics?

What is Quantum Mechanics?

Linear Algebra (Heisenberg's Matrix Mechanics)

What is Quantum Mechanics?

Eigenvectors and Eigenvalues

Measurement

Many Particles?

Postulates of Quantum Mechanics

Qubits (Spin 1/2)

Stern–Gerlach

Measuring Z^\op{Z}Z^

Intererence

Particle Intererence

Qubits (Bloch Sphere)

Expectation Value

Stern–Gerlach

Measuring Z^\op{Z}Z^

Qubits (Spin 1/2)

Two Qubits (Tensor Product)

Entanglement

Entanglement: Singlet

Einstein-Podolsky-Rosen (EPR)

Hidden Variable λ

Bell Inequalities

Quantum Calculation

Bell Inequalities

Clauser-Horne-Shimony-Holt (CHSH)

2023 Nobel Prize

Experimental Validation

Coincidence Detection

Discussion

Teleportation

Quantum Eraser

No Retrocausality!

WSU: Physics 455/555

Quantum Technologies and Computing

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all in one place. Commercial Alternative to JupyterHub.

Measuring $\op{Z}$

Measuring $\op{Z}$