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import mmf_setup;mmf_setup.nbinit()
import logging;logging.getLogger('matplotlib').setLevel(logging.CRITICAL)
%matplotlib inline
import numpy as np, matplotlib.pyplot as plt

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  • Choose "Trust Notebook" from the "File" menu.
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Entanglement#

One of the key features of quantum mechanics that distinguishes it from classical mechanics is the notion of entanglement. Intuitively, entanglement is a form of correlation between separated measurements of a quantum system.

Entanglement is correlation without causation.

Consider two spin-½ particles in a singlet state:

\[\begin{gather*} \ket{\psi_0} = \frac{\ket{↑↓} - \ket{↓↑}}{\sqrt{2}} = \frac{\ket{01} - \ket{10}}{\sqrt{2}}. \end{gather*}\]

The idea is that Alice has access to the first qubit, and Bob has access to the second qubit. Formally, Alice and evolve/measure operators of the form \(\op{A} = \op{M}\otimes\mat{1}\) and Bob can evolve/measure operators of the form \(\op{B} = \op{1}\otimes\op{M}\): i.e. single-qubit operations. To produce the state \(\ket{\psi_0}\) from, say, \(\ket{00}\), one needs genuine two-qubit operators such as ᴄɴᴏᴛ.

Do It! Find a circuit to produce a singlet state from \(\ket{00}\).

These states are said to be maximally entangled, from the perspective of Alice and Bob.

Hide code cell content 
from qiskit.quantum_info import Statevector, Operator
from qiskit import QuantumCircuit
qc = QuantumCircuit(2)
qc.reset([0, 1])
qc.h(0)
qc.cnot(0,1)
qc.y(0)
display(qc.draw('mpl'))
psi = Statevector.from_label('00')
display(psi.evolve(qc).draw('latex'))
#display(array_to_latex(Operator(qc).data))
../_images/39e09f86fce4646443af584defcc622083f4ba07851fae01edfefc8c528d4324.png
\[\frac{\sqrt{2} i}{2} |01\rangle- \frac{\sqrt{2} i}{2} |10\rangle\]

Bell Inequalities#

Single Particle Entanglement#

This section discusses wavefunctions, which are not a part of the course, but very familiar to physicists. Please ignore for now unless this interests you. Consider it a work it progress that may be migrated to finite-dimensional Hilbert spaces later.

Here we develop the following idea: suppose we have a single particle on the real line, described by a wavefunction \(\psi(x)\). Define the projectors onto the left and right subspaces:

\[\begin{gather*} \op{P}_{L} = \int_{-\infty}^{0} \d{x} \ket{x}\bra{x}, \qquad \op{P}_{R} = \int_{0}^{\infty} \d{x} \ket{x}\bra{x}. \end{gather*}\]

The key property of these projectors is that they are idempotent, and in our case, they are complete (they partition the space into two orthogonal subspaces):

\[\begin{gather*} \op{P}_{R,L}^2 = \op{P}_{R,L}, \qquad \op{P}_{L} + \op{P}_{R} = \op{1}. \end{gather*}\]

This means that the eigenvalues are either 0 or 1. The span of the eigenvectors with eigenvalue 1 form a subspace onto which the projector projects vectors.

These projectors define a Positive Operator Valued Measurement (POVM) with two outcomes: L that the particle is on the left (\(x<0\)) and R that the particle is on the right.

Now we can ask: can we define a notion of entanglement between the left (L) and right (R) subspaces? If we follow the arguments in [Klich, 2006], we can proceed as follow:

  1. Suppose the particle is in state \(\ket{\psi_0}\) with wavefunction \(\psi_0(x) = \braket{x|\psi_0}\). Define the following orthonormal states:

    \[\begin{gather*} \ket{L_0} = \frac{\op{P}_{L}\ket{\psi_0}}{\sqrt{p_L}}, \qquad \ket{R_0} = \frac{\op{P}_{R}\ket{\psi_0}}{\sqrt{p_R}},\\ p_L = \braket{\psi_0|\op{P}_{L}|\psi_0}, \qquad p_R = \braket{\psi_0|\op{P}_{R}|\psi_0}. \end{gather*}\]

    These are the states that a measurement of \(\ket{\psi_0}\) would yield, with probabilities \(p_L\) and \(p_R\) respectively.

Artificial Single-Particle Entanglement#

Consider a single particle with wavefunction \(\psi(x)\) factored as:

\[\begin{gather*} \psi(x) = \begin{cases} \psi_L(-x) & x \leq 0\\ \psi_R(x) & x > 0. \end{cases}. \end{gather*}\]

We can think of this as a tensor product of a qubit with a particle moving on the positive half-line \(x\geq 0\). Thus, we can write, for \(x\geq0\):

\[\begin{gather*} \psi_{\sigma}(x) = \begin{pmatrix} \psi_{L}(x)\\ \psi_{R}(x) \end{pmatrix} \in \overbrace{\mathbb{SP}_2}^{A} \otimes \overbrace{\mathbb{H}^{+}_{\infty/2}}^{B}. \end{gather*}\]

Note that this wavefunction has two indices: an index \(\sigma \in \{L, R\}\), and a “spatial index” \(x\in[0, \infty)\). This demonstrates the tensor-product structure of the space. The full density matrix \(\mat{\rho}\) thus has four indices grouped into pairs:

\[\begin{gather*} [\mat{\rho}]_{\sigma,x;\sigma',x'}=\psi_{\sigma}(x)\psi^*_{\sigma'}(x'). \end{gather*}\]

Thus, we can apply standard techniques and take the partial trace to obtain a \(2\times 2\) reduced density matrix

\[\begin{gather*} [\mat{\rho_{A}}]_{\sigma \sigma'} = \Tr_{B}(\mat{\rho}) = \int \d{x}\psi_{\sigma}(x)\psi_{\sigma'}(x)\\ = \int_0^{\infty} \d{x} \begin{pmatrix} \psi_L(x)\psi_L^*(x) & \psi_L(x)\psi_R^*(x)\\ \psi_R(x)\psi_L^*(x) & \psi_R(x)\psi_R^*(x) \end{pmatrix}. \end{gather*}\]

Let’s consider a few examples. What about a wavefunction that has support only on the right \(\psi_L(x) = 0\) or \(\psi(x) = \Theta(x)\psi(x)\) or only on the left \(\psi_R(x) = 0\) or \(\psi(x) = \Theta(-x)\psi(x)\)?

\[\begin{gather*} \left.\mat{\rho_{A}}\right|_{\psi_L(x) = 0} = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \qquad \left.\mat{\rho_{A}}\right|_{\psi_R(x) = 0} = \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}. \end{gather*}\]

In these cases, it is obvious that the qubit exhibits no entanglement. These are pure density matrices with eigenvalues \(\lambda \in \{0, 1\}\), hence the von Neumann entropy is zero.

If we the wavefunction is even \(\psi(x) = \psi(-x)\), then

\[\begin{gather*} \mat{\rho_{A}} = \frac{1}{2} \begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \end{gather*}\]

This looks more promising, but the eigenvalues are still \(0\) and \(1\), so this not entangled either.

Instructor Notes

This discussion needs much more attention – student’s don’t obviously make the connection between eigenvectors, eigenvalues, von Neumann entropy, SVD, etc.

Mathematics students don’t like viewing a vector as a matrix.

\[\begin{gather*} \psi_{ij} = U_{ik}D_{kk}V^*_{jk} = \sum_{k} \sigma_k \overbrace{U_{ik}}^{\ket{a_k}}\overbrace{V^*_{jk}}^{\ket{b_k}}\\ \psi = \begin{pmatrix} \psi_{00}\\ \psi_{01}\\ \psi_{10}\\ \psi_{11} \end{pmatrix} \equiv \begin{pmatrix} \psi_{00} & \psi_{01}\\ \psi_{10} & \psi_{11} \end{pmatrix}\\ \psi_{ij} = \ket{a}\otimes\ket{b}\\ \end{gather*}\]

Bell#

\[\begin{gather*} \ket{\psi} = \frac{\ket{00} + \ket{11}}{\sqrt{2}} \neq \ket{a}\otimes\ket{b}\\ \mat{\rho} = \ket{\psi}\bra{\psi} = \frac{1}{2}\bigl(\ket{00} + \ket{11}\bigr)\bigl(\bra{00} + \bra{11}\bigr) =\\ = \frac{1}{2} \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix},\\ \mat{\rho_A} = \frac{1}{2} \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}. \end{gather*}\]

Positive Operator Valued Measurement (POVM)#