Lecture Notes on Composite Quantum Systems and Entanglement

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February 5, 2025

Introduction

This lecture introduces the concept of composite quantum systems, focusing on how to describe and analyze systems composed of multiple subsystems. We will explore the reduced density matrix as a tool for describing subsystems and its application in quantum teleportation. Furthermore, we will delve into the crucial concept of entanglement, its quantification using Schmidt decomposition, and methods for detecting it. We will also discuss the purification of mixed states and finally, examine the foundational Einstein-Podolsky-Rosen (EPR) paradox and Bell’s Theorem, which highlight the profound implications of quantum entanglement and challenge classical notions of reality.

Composite Systems and Reduced Density Matrix

In quantum mechanics, a composite system is described by combining two or more subsystems. Consider a system composed of two subsystems, labeled A and B. The Hilbert space of the composite system, denoted as $\mathcal{H}_{AB}$, is the tensor product of the Hilbert spaces of the individual subsystems, $\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B$.

Measurements on Subsystem A

When a measurement is performed on subsystem A while leaving subsystem B undisturbed, the measurement operator acts solely on the Hilbert space $\mathcal{H}_A$. Let $P_A$ be a projective measurement operator acting on subsystem A, and $I_B$ be the identity operator on subsystem B. The measurement operator on the composite system AB is then given by $P_A \otimes I_B$.

The expectation value of this measurement for a composite system in a state described by the density matrix $\rho_{AB}$ is given by: \[\langle P_A \otimes I_B \rangle = \text{Tr}\left[ (P_A \otimes I_B) \rho_{AB}\right]\] This expectation value represents the average outcome of the measurement $P_A$ when performed on subsystem A. To effectively calculate and interpret such measurements, the concept of the partial trace and the reduced density matrix are essential.

Partial Trace: Tracing out Subsystem B

The partial trace is a linear map that effectively "traces out" or removes a subsystem from a composite system’s description. For a composite system AB with a density matrix $\rho_{AB}$ acting on $\mathcal{H}_A \otimes \mathcal{H}_B$, the partial trace over subsystem B, denoted as $\text{Tr}_B$, results in an operator acting only on $\mathcal{H}_A$.

Mathematically, for an operator $O = O_A \otimes O_B$, where $O_A$ acts on $\mathcal{H}_A$ and $O_B$ acts on $\mathcal{H}_B$, the partial trace over subsystem B is defined as: \[\text{Tr}_B [O_A \otimes O_B] = O_A \text{Tr}[O_B]\] For a general operator $M$ on $\mathcal{H}_{AB}$, we can define the partial trace in terms of matrix elements. Let $\{\lvert i \rangle_A\}$ and $\{\lvert k \rangle_B\}$ be orthonormal bases for $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. If the matrix elements of $\rho_{AB}$ in the product basis $\{\lvert i \rangle_A \otimes \lvert k \rangle_B\}$ are given by $\langle i \rvert_A \langle k \rvert_B \rho_{AB}\lvert j \rangle_A \lvert l \rangle_B = (\rho_{AB})_{ik,jl}$, then the matrix elements of the partial trace over B, denoted as $\rho_{A}= \text{Tr}_B[\rho_{AB}]$, in the basis $\{\lvert i \rangle_A\}$ are: \[(\rho_{A})_{ij} = (\text{Tr}_B[\rho_{AB}])_{ij} = \sum_{k} \langle i \rvert_A \langle k \rvert_B \rho_{AB}\lvert j \rangle_A \lvert k \rangle_B\] This operation sums over the basis states of subsystem B, effectively averaging over or ignoring the degrees of freedom of subsystem B.

Reduced Density Matrix: Describing Subsystem A

Definition 1 (Reduced Density Matrix). The reduced density matrix for subsystem A, denoted as $\rho_{A}$, is obtained by taking the partial trace of the density matrix $\rho_{AB}$ of the composite system over subsystem B: \[\rho_{A}= \text{Tr}_B [\rho_{AB}]\] Similarly, the reduced density matrix for subsystem B is obtained by tracing over subsystem A: \[\rho_{B}= \text{Tr}_A [\rho_{AB}]\] The reduced density matrix $\rho_{A}$ provides a complete description of subsystem A when subsystem B is not considered or when we are only interested in measurements performed on subsystem A. All expectation values of observables acting solely on subsystem A can be calculated using $\rho_{A}$.

Returning to the expectation value of the measurement $P_A \otimes I_B$, we can rewrite it using the partial trace: \[\begin{aligned}\langle P_A \otimes I_B \rangle &= \text{Tr}\left[ (P_A \otimes I_B) \rho_{AB}\right] \\&= \text{Tr}_A \text{Tr}_B \left[ (P_A \otimes I_B) \rho_{AB}\right] \\&= \text{Tr}_A \left[ P_A \text{Tr}_B [I_B \rho_{AB}] \right] \\&= \text{Tr}_A \left[ P_A \text{Tr}_B [\rho_{AB}] \right] \\&= \text{Tr}_A \left[ P_A \rho_{A}\right]\end{aligned}\] This derivation explicitly shows that the expectation value of a measurement on subsystem A can be computed solely from the reduced density matrix $\rho_{A}$ and the measurement operator $P_A$ acting on subsystem A’s Hilbert space.

Reduced Density Matrix and Quantum Teleportation

Quantum teleportation is a protocol for transferring an unknown quantum state from one location to another using pre-shared entanglement and classical communication. The reduced density matrix is crucial for understanding the state of the qubit being teleported before the classical communication step is completed.

Quantum Teleportation Protocol and State Evolution

Consider Alice, who wants to teleport an unknown quantum state $\lvert \psi_{\text{in}} \rangle= \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle$ to Bob. They initially share a maximally entangled Bell state, specifically $\lvert \Phi^+ \rangle= \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$. Alice possesses the particle in state $\lvert \psi_{\text{in}} \rangle$ and half of the entangled pair (particle A), while Bob holds the other half of the entangled pair (particle B). The initial state of the composite system, including Alice’s unknown qubit (particle ‘in’) and the entangled pair (particles A and B), is: \[\lvert \Psi_1 \rangle = \lvert \psi_{\text{in}} \rangle\otimes \lvert \Phi^+ \rangle= (\alpha \lvert 0 \rangle_{\text{in}} + \beta \lvert 1 \rangle_{\text{in}}) \otimes \frac{1}{\sqrt{2}} (\lvert 0 \rangle_A \lvert 0 \rangle_B + \lvert 1 \rangle_A \lvert 1 \rangle_B)\] After Alice performs a Bell basis measurement on her particles (‘in’ and A), the combined state of the system collapses. Before Alice communicates her measurement outcome to Bob, the state of Bob’s particle (B), represented by the reduced density matrix, is a mixed state. The state $\lvert \Psi_2 \rangle$ immediately after Alice’s Bell measurement, but before she announces the result, can be written as a superposition of states, each corresponding to a different measurement outcome for Alice and a resulting state for Bob: \[\begin{aligned}\lvert \Psi_2 \rangle = \frac{1}{2} & [\lvert \Phi^+ \rangle_{\text{in}A} (\alpha \lvert 0 \rangle_B + \beta \lvert 1 \rangle_B) + \lvert \Phi^- \rangle_{\text{in}A} (\alpha \lvert 0 \rangle_B - \beta \lvert 1 \rangle_B) \\&+ \lvert \Psi^+ \rangle_{\text{in}A} (\alpha \lvert 1 \rangle_B + \beta \lvert 0 \rangle_B) + \lvert \Psi^- \rangle_{\text{in}A} (\alpha \lvert 1 \rangle_B - \beta \lvert 0 \rangle_B) ]\end{aligned}\] Here, $\lvert \Phi^\pm \rangle_{\text{in}A} = \frac{1}{\sqrt{2}} (\lvert 00 \rangle \pm \lvert 11 \rangle)_{\text{in}A}$ and $\lvert \Psi^\pm \rangle_{\text{in}A} = \frac{1}{\sqrt{2}} (\lvert 01 \rangle \pm \lvert 10 \rangle)_{\text{in}A}$ are the Bell states for particles ‘in’ and A.

Calculating the Reduced Density Matrix for Bob’s Qubit

Theorem 1 (Reduced Density Matrix of Bob’s Qubit in Teleportation). To determine the state of Bob’s qubit before Alice’s classical communication, we calculate the reduced density matrix $\rho_{B}$ by tracing out Alice’s particles (‘in’ and A) from the density matrix of the joint state $\rho_{\text{in}AB} = \lvert \Psi_2 \rangle\langle \Psi_2 \rvert$. \[\rho_{B}= \text{Tr}_{\text{in}A} [\lvert \Psi_2 \rangle\langle \Psi_2 \rvert]\] We perform the partial trace over the Hilbert space of Alice’s qubits, spanned by the Bell basis $\{\lvert \Phi^+ \rangle_{\text{in}A}, \lvert \Phi^- \rangle_{\text{in}A}, \lvert \Psi^+ \rangle_{\text{in}A}, \lvert \Psi^- \rangle_{\text{in}A}\}$. Due to the orthogonality of the Bell basis, the reduced density matrix $\rho_{B}$ is given by: \[\begin{aligned}\rho_{B}&= \text{Tr}_{\text{in}A} [\lvert \Psi_2 \rangle\langle \Psi_2 \rvert] \\&= \sum_{\lvert \chi \rangle_{\text{in}A} \in \text{Bell basis}} \langle \chi \rvert_{\text{in}A} \lvert \Psi_2 \rangle\langle \Psi_2 \rvert \lvert \chi \rangle_{\text{in}A} \\&= \frac{1}{4} \sum_{\lvert \chi \rangle_{\text{in}A} \in \text{Bell basis}} \langle \chi \rvert_{\text{in}A} \left[ \sum_{\lvert \phi \rangle_{\text{in}A}, \lvert \psi \rangle_{\text{in}A} \in \text{Bell basis}} \lvert \phi \rangle_{\text{in}A} \lvert \tilde{\psi}_\phi \rangle_B \langle \tilde{\psi}_\psi \rvert_B \langle \psi \rvert_{\text{in}A} \right] \lvert \chi \rangle_{\text{in}A}\end{aligned}\] where $\lvert \tilde{\psi}_{\Phi^+} \rangle = (\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle)_B$, $\lvert \tilde{\psi}_{\Phi^-} \rangle = (\alpha \lvert 0 \rangle - \beta \lvert 1 \rangle)_B$, $\lvert \tilde{\psi}_{\Psi^+} \rangle = (\alpha \lvert 1 \rangle + \beta \lvert 0 \rangle)_B$, and $\lvert \tilde{\psi}_{\Psi^-} \rangle = (\alpha \lvert 1 \rangle - \beta \lvert 0 \rangle)_B$. Using the orthogonality $\langle \psi \rvert \chi \rangle_{\text{in}A} = \delta_{\psi\chi}$, this simplifies to: \[\begin{aligned}\rho_{B}&= \frac{1}{4} \sum_{\lvert \chi \rangle_{\text{in}A} \in \text{Bell basis}} \lvert \tilde{\psi}_\chi \rangle\langle \tilde{\psi}_\chi \rvert_B \\&= \frac{1}{4} \left[ \lvert \tilde{\psi}_{\Phi^+} \rangle\langle \tilde{\psi}_{\Phi^+} \rvert + \lvert \tilde{\psi}_{\Phi^-} \rangle\langle \tilde{\psi}_{\Phi^-} \rvert + \lvert \tilde{\psi}_{\Psi^+} \rangle\langle \tilde{\psi}_{\Psi^+} \rvert + \lvert \tilde{\psi}_{\Psi^-} \rangle\langle \tilde{\psi}_{\Psi^-} \rvert \right]_B\end{aligned}\] Substituting the expressions for $\lvert \tilde{\psi}_\chi \rangle$ and expanding: \[\begin{aligned}\rho_{B}&= \frac{1}{4} [ (\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle)(\alpha^* \langle 0 \rvert + \beta^* \langle 1 \rvert) + (\alpha \lvert 0 \rangle - \beta \lvert 1 \rangle)(\alpha^* \langle 0 \rvert - \beta^* \langle 1 \rvert) \\&+ (\alpha \lvert 1 \rangle + \beta \lvert 0 \rangle)(\alpha^* \langle 1 \rvert + \beta^* \langle 0 \rvert) + (\alpha \lvert 1 \rangle - \beta \lvert 0 \rangle)(\alpha^* \langle 1 \rvert - \beta^* \langle 0 \rvert) ] \\&= \frac{1}{4} [ 2(|\alpha|^2 \lvert 0 \rangle\langle 0 \rvert + |\beta|^2 \lvert 1 \rangle\langle 1 \rvert) + 2(|\alpha|^2 \lvert 1 \rangle\langle 1 \rvert + |\beta|^2 \lvert 0 \rangle\langle 0 \rvert) ] \\&= \frac{1}{2} (|\alpha|^2 + |\beta|^2) (\lvert 0 \rangle\langle 0 \rvert + \lvert 1 \rangle\langle 1 \rvert)\end{aligned}\] Since $|\alpha|^2 + |\beta|^2 = 1$ due to normalization of $\lvert \psi_{\text{in}} \rangle$, we obtain: \[\rho_{B}= \frac{1}{2} (\lvert 0 \rangle\langle 0 \rvert + \lvert 1 \rangle\langle 1 \rvert) = \frac{1}{2} \mathbb{I}= \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}\] This is the maximally mixed state, represented by half the identity matrix $\mathbb{I}$.

Mixed State Interpretation and No-Communication Theorem

Remark. Remark 1 (Mixed State in Teleportation). The reduced density matrix for Bob’s qubit, $\rho_{B}= \frac{1}{2} \mathbb{I}$, is a mixed state. This implies that before Alice communicates her measurement outcome, Bob’s qubit is in a completely random, unpolarized state. Bob cannot gain any information about the original state $\lvert \psi_{\text{in}} \rangle$ by performing measurements on his qubit alone until he receives classical information from Alice about her measurement result.

Remark. Remark 2 (Verification of Mixed State). To confirm that $\rho_{B}$ is a mixed state and not a pure state, we can check if $\rho_{B}^2 = \rho_{B}$. \[\rho_{B}^2 = \left(\frac{1}{2} \mathbb{I}\right)^2 = \frac{1}{4} \mathbb{I}^2 = \frac{1}{4} \mathbb{I}\neq \rho_{B}\] Since $\rho_{B}^2 \neq \rho_{B}$, $\rho_{B}$ is indeed a mixed state. This result is consistent with the no-communication theorem of quantum mechanics, which states that entanglement cannot be used to send classical information faster than light. Bob’s state is random until he learns Alice’s measurement outcome, at which point he can perform a correction operation to retrieve the teleported state.

Entanglement and Schmidt Decomposition

Entanglement is a quintessential quantum mechanical phenomenon where two or more subsystems are intrinsically linked, such that their quantum states are interdependent, regardless of the physical distance separating them. This section formalizes the definition of entanglement for bipartite systems and introduces the Schmidt decomposition as a tool for detecting and quantifying entanglement in pure states.

Defining Entanglement for Bipartite Systems

For a bipartite system composed of subsystems A and B, entanglement is defined differently for pure and mixed states.

Pure States:

A pure state $\lvert \psi \rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$ is entangled if it cannot be expressed as a product of pure states of the individual subsystems, i.e., it cannot be written in the form $\lvert \psi \rangle_{AB} = \lvert \phi \rangle_A \otimes \lvert \chi \rangle_B$, where $\lvert \phi \rangle_A \in \mathcal{H}_A$ and $\lvert \chi \rangle_B \in \mathcal{H}_B$. If a pure state can be written as a product state, it is called a separable state or unentangled state.

Mixed States:

A mixed state $\rho_{AB}$ is separable if it can be represented as a convex combination of product states: \[\rho_{AB}= \sum_i p_i \rho_{A,i} \otimes \rho_{B,i}\] where $p_i \geq 0$ with $\sum_i p_i = 1$, and $\rho_{A,i}$ and $\rho_{B,i}$ are density matrices for subsystems A and B, respectively. If a mixed state cannot be expressed in this form, it is considered entangled.

Schmidt Decomposition for Pure Bipartite States

Theorem 2 (Schmidt Decomposition Theorem). For any pure state $\lvert \psi \rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$, there exist orthonormal bases $\{\lvert i \rangle_A\}$ for $\mathcal{H}_A$ and $\{\lvert i \rangle_B\}$ for $\mathcal{H}_B$ such that $\lvert \psi \rangle_{AB}$ can be written as: \[\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B\] where $\lambda_i$ are positive real numbers called Schmidt coefficients, which satisfy $\sum_{i=1}^{r} \lambda_i = 1$. The number of non-zero Schmidt coefficients, $r$, is called the Schmidt rank of the state. The square roots $\sqrt{\lambda_i}$ are the singular values of the coefficient matrix when the state $\lvert \psi \rangle_{AB}$ is expanded in any product basis.

To apply Schmidt decomposition in practice, we first express the bipartite state $\lvert \psi \rangle_{AB}$ in a product basis $\{\lvert j \rangle_A \otimes \lvert k \rangle_B\}$: \[\lvert \psi \rangle_{AB} = \sum_{j,k} C_{jk} \lvert j \rangle_A \otimes \lvert k \rangle_B\] We then form a coefficient matrix $C$ with entries $C_{jk}$. The Schmidt decomposition is closely related to the singular value decomposition of the matrix $C$. The Schmidt rank $r$ is equal to the rank of the matrix $C$.

Schmidt Number and Entanglement Quantification

The Schmidt number, which is the Schmidt rank $r$, serves as a measure of entanglement for pure bipartite states.

Schmidt rank $r = 1$: The state has only one non-zero Schmidt coefficient and is a product state. Such a state is separable and therefore, not entangled.
Schmidt rank $r > 1$: The state has more than one non-zero Schmidt coefficient and is an entangled state. A higher Schmidt number generally indicates a higher degree of entanglement.

Examples of Entanglement Analysis using Schmidt Decomposition

We now illustrate the Schmidt decomposition method with examples to determine whether given bipartite states are entangled.

Example 1 (Bell State $\lvert \Phi^+ \rangle = \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$). Consider the Bell state $\lvert \Phi^+ \rangle = \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$. In the computational basis $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ for both subsystems, the coefficient matrix $C$ is: \[C = \begin{pmatrix} 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} \end{pmatrix}\] The rank of the matrix $C$ is 2, as there are two linearly independent rows (or columns). Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the Bell state $\lvert \Phi^+ \rangle$ is entangled.

Example 2 (Separable State $\lvert \psi \rangle = \frac{1}{2} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle)$). Consider the state $\lvert \psi \rangle = \frac{1}{2} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle)$. The coefficient matrix $C$ in the computational basis is: \[C = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}\] The rank of the matrix $C$ is 1, because the rows (and columns) are linearly dependent (the second row is identical to the first). Thus, the Schmidt rank is $r=1$. Since $r = 1$, this state is separable, meaning it is not entangled. Indeed, this state can be written as a product state: \[\lvert \psi \rangle = \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right) \otimes \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right)\]

Example 3 (Entangled State $\lvert \phi \rangle = \frac{1}{\sqrt{3}} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle)$). Consider the state $\lvert \phi \rangle = \frac{1}{\sqrt{3}} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle)$. The coefficient matrix $C$ is: \[C = \begin{pmatrix} 1/\sqrt{3} & 1/\sqrt{3} \\ 1/\sqrt{3} & 0 \end{pmatrix}\] The rank of the matrix $C$ is 2, as the rows are linearly independent. Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the state $\lvert \phi \rangle$ is entangled.

Entanglement Detection via Rank of Coefficient Matrix

Remark. Remark 3 (Entanglement Criterion for Pure Bipartite Qubit States). As demonstrated by the examples, the rank of the coefficient matrix $C$ provides a direct method to determine if a pure bipartite state is entangled. For a pure bipartite state of qubits, the criterion simplifies to:

$\text{Rank}(C) = 1$: The state is separable (not entangled).
$\text{Rank}(C) = 2$: The state is entangled.

For qubit systems, a Schmidt rank greater than 1 is always equal to 2 because the dimension of the subsystems is 2, leading to at most a $2 \times 2$ coefficient matrix for a bipartite state.

Purification of Mixed States

A mixed state density matrix $\rho_{A}$ represents a statistical ensemble of pure quantum states. Purification is a conceptual and mathematical procedure that allows us to represent a mixed state $\rho_{A}$ of a system A as the reduced density matrix of a pure state defined on a larger composite system AR, where R is an auxiliary reference system. This process highlights that any mixed state can be viewed as arising from a pure state by disregarding or "tracing out" the degrees of freedom of an environment (in this case, the reference system R).

Spectral Decomposition of the Density Operator

The foundation of purification lies in the spectral decomposition of density operators. Any density operator $\rho_{A}$ is, by definition, a Hermitian, positive semi-definite operator with trace equal to one. Consequently, $\rho_{A}$ can be diagonalized in an orthonormal basis $\{\lvert i \rangle_A\}$ of the Hilbert space $\mathcal{H}_A$. The spectral decomposition of $\rho_{A}$ is given by: \[\rho_{A}= \sum_i p_i \lvert i \rangle\langle i \rvert_A\] where $p_i$ are the non-negative eigenvalues of $\rho_{A}$, representing the probabilities of the system being in the corresponding eigenstate $\lvert i \rangle_A$, and they satisfy $\sum_i p_i = 1$.

Purification Procedure: Constructing a Pure State

Definition 2 (Purification of a Mixed State). To purify the mixed state $\rho_{A}$, we introduce a reference system R, which is chosen to have a Hilbert space $\mathcal{H}_R$ of at least the same dimension as the support of $\rho_{A}$. We then construct a pure state $\lvert \Psi \rangle_{AR}$ for the composite system AR in the Hilbert space $\mathcal{H}_{AR} = \mathcal{H}_A \otimes \mathcal{H}_R$ as follows: \[\lvert \Psi \rangle_{AR} = \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R\] Here, $\{\lvert i \rangle_A\}$ are the eigenstates from the spectral decomposition of $\rho_{A}$, and $\{\lvert i \rangle_R\}$ is an orthonormal basis for the reference system R, chosen to be in one-to-one correspondence with the eigenstates $\{\lvert i \rangle_A\}$. The coefficients $\sqrt{p_i}$ are the square roots of the eigenvalues of $\rho_{A}$.

Theorem 3 (Verification of Purification). The density matrix for this composite system AR in the pure state $\lvert \Psi \rangle_{AR}$ is given by $\rho_{AR} = \lvert \Psi \rangle\langle \Psi \rvert_{AR}$. To verify that $\lvert \Psi \rangle_{AR}$ is indeed a purification of $\rho_{A}$, we compute the reduced density matrix for system A by taking the partial trace of $\rho_{AR}$ over the reference system R: \[\begin{aligned}\rho'_A &= \text{Tr}_R [\rho_{AR}] \\&= \text{Tr}_R [\lvert \Psi \rangle\langle \Psi \rvert_{AR}] \\&= \text{Tr}_R \left[ \left( \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R \right) \left( \sum_j \sqrt{p_j} \langle j \rvert_A \otimes \langle j \rvert_R \right) \right] \\&= \text{Tr}_R \left[ \sum_{i,j} \sqrt{p_i p_j} (\lvert i \rangle\langle j \rvert_A \otimes \lvert i \rangle\langle j \rvert_R) \right] \\&= \sum_{i,j} \sqrt{p_i p_j} \lvert i \rangle\langle j \rvert_A \text{Tr}_R [\lvert i \rangle\langle j \rvert_R] \\&= \sum_{i,j} \sqrt{p_i p_j} \lvert i \rangle\langle j \rvert_A \langle j \rvert i \rangle_R\end{aligned}\] Since $\{\lvert i \rangle_R\}$ is an orthonormal basis, $\langle j \rvert i \rangle_R = \delta_{ij}$. Thus, the expression simplifies to: \[\begin{aligned}\rho'_A &= \sum_{i,j} \sqrt{p_i p_j} \lvert i \rangle\langle j \rvert_A \delta_{ij} \\&= \sum_i \sqrt{p_i p_i} \lvert i \rangle\langle i \rvert_A \\&= \sum_i p_i \lvert i \rangle\langle i \rvert_A \\&= \rho_{A}\end{aligned}\] Therefore, by tracing out the reference system R from the pure state $\lvert \Psi \rangle_{AR}$, we recover the original mixed state $\rho_{A}$. This confirms that $\lvert \Psi \rangle_{AR}$ is a purification of $\rho_{A}$. The purification procedure demonstrates that any mixed state can be regarded as part of a larger pure state, providing a deeper understanding of mixed states in quantum mechanics.

The Einstein-Podolsky-Rosen Paradox and Bell’s Theorem

The Einstein-Podolsky-Rosen (EPR) paradox, introduced in their seminal 1935 paper, and Bell’s Theorem, developed by John Stewart Bell in 1964, are pivotal in understanding the conceptual challenges posed by quantum mechanics, particularly concerning entanglement and the nature of reality. These concepts highlight the profound departure of quantum mechanics from classical physics.

The EPR Paradox: Challenging Completeness of Quantum Mechanics

The EPR paradox emerges from considering the implications of quantum entanglement for spatially separated particles. EPR’s thought experiment involved entangled particles, though their original paper focused on position and momentum. A more illustrative example, and closer to the transcript’s discussion, involves spin-entangled particles. Consider a source thatemits pairs of particles in the spin singlet state: \[\lvert \Psi^- \rangle = \frac{1}{\sqrt{2}} (\lvert 0 \rangle_A \lvert 1 \rangle_B - \lvert 1 \rangle_A \lvert 0 \rangle_B)\] where particle A is sent to Alice and particle B is sent to Bob, who are assumed to be far apart, such that no physical signal can travel between them faster than light.

EPR’s Criterion of Physical Reality

Central to the EPR paradox is their definition of an "element of reality." They posited: "If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity."

Imagine Alice measures the spin of particle A along the z-axis ($\sigma_z^A$). If she obtains spin-up (outcome $\lvert 0 \rangle_A$), due to the entangled nature of the singlet state, she can predict with certainty that a measurement of the spin of particle B along the z-axis ($\sigma_z^B$) will yield spin-down (outcome $\lvert 1 \rangle_B$). Crucially, this prediction is made without performing any measurement on particle B, thus, according to EPR, without disturbing it. Applying their criterion, EPR concluded that the spin of particle B along the z-axis must be an "element of reality" even before Bob performs any measurement.

The Argument for Incompleteness and Hidden Variables

EPR extended this argument by considering that Alice could have chosen to measure the spin of particle A along the x-axis ($\sigma_x^A$) instead. Depending on her outcome, she could similarly predict with certainty the outcome of a $\sigma_x^B$ measurement by Bob. This led them to argue that both the spin along the z-axis and the spin along the x-axis (and indeed, along any axis) for particle B are simultaneously "elements of reality."

However, quantum mechanics dictates that spin components along different axes are incompatible observables; they cannot have simultaneously definite values. This apparent contradiction led EPR to conclude that quantum mechanics provides an incomplete description of physical reality. They hypothesized that there must be some "hidden variables" not accounted for in quantum mechanics that predetermine the outcomes of all possible measurements, thus restoring determinism and locality. These hidden variables would ensure that particles possess definite properties independently of measurement, consistent with classical intuition.

Bell’s Theorem: Testing Local Realism

Theorem 4 (Bell’s Theorem). Bell’s Theorem, formulated by John Stewart Bell, provides a way to experimentally test the validity of local hidden variable theories against the predictions of quantum mechanics. Bell formalized the concept of "local realism," which underpins the EPR argument. Local realism combines two key assumptions:

Realism: Physical properties of systems have definite values at all times, independent of observation.
Locality: The outcome of a measurement on one particle is independent of measurement choices made on a spatially separated particle. There is no instantaneous influence between distant events.

Bell derived inequalities, known as Bell’s inequalities, that must be satisfied by any theory adhering to local realism. He considered scenarios where Alice and Bob independently choose to measure spin along different directions on entangled particles. Quantum mechanics, however, predicts correlations between measurement outcomes for certain entangled states that violate Bell’s inequalities.

In essence, Bell’s theorem demonstrates that if nature is governed by local realism, the correlations between measurements on entangled particles must be weaker than what quantum mechanics predicts. Numerous experimental tests of Bell’s inequalities, starting with Alain Aspect’s experiments in the 1980s and continuing with increasingly sophisticated experiments, have consistently shown violations of Bell’s inequalities.

These experimental violations strongly support quantum mechanics and rule out local realistic hidden variable theories as a complete description of nature. The implication is profound: at least one of the core assumptions of local realism—realism or locality—must be abandoned in the quantum realm. While interpretations vary, the most widely accepted conclusion is that locality, in the classical sense, is violated by quantum mechanics, suggesting the existence of non-local correlations between entangled particles. The transcript’s mention of measurements along directions A and B and the expectation value of $\sigma_1 \cdot A \otimes \sigma_2 \cdot B$ alludes to the mathematical framework used to derive and test Bell’s inequalities, which quantify these correlations and demonstrate the conflict with local realism.

Conclusion

This lecture explored fundamental concepts related to composite quantum systems and the intriguing phenomenon of entanglement. We introduced the reduced density matrix as a crucial tool for describing the state of subsystems within a larger composite system, and illustrated its application in the context of quantum teleportation, demonstrating how it describes the state of Bob’s qubit before classical communication from Alice. We then delved into the nature of entanglement, using Schmidt decomposition as a method to quantify and detect entanglement in pure bipartite states, providing illustrative examples. The concept of purification revealed how mixed states can be understood as arising from pure states of larger systems by tracing out environmental degrees of freedom. Finally, we discussed the profound implications of entanglement through the Einstein-Podolsky-Rosen paradox and Bell’s Theorem, underscoring the fundamental conflict between quantum mechanics and classical notions of local realism and highlighting the non-local nature of quantum correlations.

Key takeaways from this lecture are:

The reduced density matrix is an indispensable tool for describing the quantum state of subsystems and for making predictions about local measurements on composite systems.
Entanglement is a uniquely quantum correlation that defies classical explanation and represents a fundamental resource in quantum information processing.
Schmidt decomposition provides a practical method to determine and quantify entanglement in pure bipartite states by analyzing the rank of the coefficient matrix.
Purification offers a conceptual framework to understand mixed states as partial descriptions of larger pure states, linking mixed states to entanglement with an environment.
Bell’s Theorem and its experimental verification present a compelling challenge to local realism, demonstrating the non-local nature of quantum mechanics and the profound implications of entanglement for our understanding of physical reality.

Further study should delve into the experimental verifications of Bell’s inequalities, explore the diverse interpretations of quantum mechanics in light of non-locality, and investigate the burgeoning field of quantum technologies that harness entanglement for applications such as quantum computation, quantum communication, and quantum cryptography.

Algorithm for Schmidt Decomposition

Algorithm 1 (Schmidt Decomposition Algorithm). Input: A pure bipartite state $\lvert \psi \rangle_{AB} = \sum_{j,k} C_{jk} \lvert j \rangle_A \otimes \lvert k \rangle_B$ in a product basis $\{\lvert j \rangle_A \otimes \lvert k \rangle_B\}$.

Output: Schmidt decomposition $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$, Schmidt coefficients $\lambda_i$, and Schmidt rank $r$.

Steps:

Construct the coefficient matrix $C$ with entries $C_{jk}$.
Compute the Singular Value Decomposition (SVD) of $C = U S V^\dagger$, where $U$ and $V$ are unitary matrices, and $S$ is a diagonal matrix with singular values $s_1, s_2, \dots, s_r, 0, \dots, 0$ on the diagonal.
The Schmidt coefficients are $\lambda_i = s_i^2$.
The Schmidt rank $r$ is the number of non-zero singular values (or Schmidt coefficients).
The orthonormal basis $\{\lvert i \rangle_A\}$ for subsystem A is given by the columns of $U$.
The orthonormal basis $\{\lvert i \rangle_B\}$ for subsystem B is given by the columns of $V$.
The Schmidt decomposition is then $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} s_i \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$, where $\lvert u_i \rangle_A$ are columns of $U$ and $\lvert v_i \rangle_B$ are columns of $V$. Using $\sqrt{\lambda_i} = s_i$, we get $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$.

Note: In the main text, for simplicity of notation, we used the same index for bases in both subsystems in the Schmidt decomposition $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$. In general, the bases can be different, as shown in the algorithm output using $U$ and $V$ matrices from SVD. For square coefficient matrices and when using SVD directly to find Schmidt decomposition, using different bases $\{\lvert u_i \rangle_A\}$ and $\{\lvert v_i \rangle_B\}$ from SVD is more accurate. For the purpose of entanglement detection via Schmidt rank, the rank of the coefficient matrix $C$ is sufficient.

Mathematical Statements in Tcolorboxes

Definitions

Definition: The reduced density matrix for subsystem A, denoted as $\rho_{A}$, is obtained by taking the partial trace of the density matrix $\rho_{AB}$ of the composite system over subsystem B: \[\rho_{A}= \text{Tr}_B [\rho_{AB}]\] Similarly, the reduced density matrix for subsystem B is obtained by tracing over subsystem A: \[\rho_{B}= \text{Tr}_A [\rho_{AB}]\] The reduced density matrix $\rho_{A}$ provides a complete description of subsystem A when subsystem B is not considered or when we are only interested in measurements performed on subsystem A. All expectation values of observables acting solely on subsystem A can be calculated using $\rho_{A}$.

Definition: To purify the mixed state $\rho_{A}$, we introduce a reference system R, which is chosen to have a Hilbert space $\mathcal{H}_R$ of at least the same dimension as the support of $\rho_{A}$. We then construct a pure state $\lvert \Psi \rangle_{AR}$ for the composite system AR in the Hilbert space $\mathcal{H}_{AR} = \mathcal{H}_A \otimes \mathcal{H}_R$ as follows: \[\lvert \Psi \rangle_{AR} = \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R\] Here, $\{\lvert i \rangle_A\}$ are the eigenstates from the spectral decomposition of $\rho_{A}$, and $\{\lvert i \rangle_R\}$ is an orthonormal basis for the reference system R, chosen to be in one-to-one correspondence with the eigenstates $\{\lvert i \rangle_A\}$. The coefficients $\sqrt{p_i}$ are the square roots of the eigenvalues of $\rho_{A}$.

Theorems

Theorem: In the quantum teleportation protocol described, the reduced density matrix $\rho_{B}$ of Bob’s qubit, before Alice sends classical information, is the maximally mixed state: \[\rho_{B}= \text{Tr}_{\text{in}A} [\lvert \Psi_2 \rangle\langle \Psi_2 \rvert] = \frac{1}{2} (\lvert 0 \rangle\langle 0 \rvert + \lvert 1 \rangle\langle 1 \rvert) = \frac{1}{2} \mathbb{I}= \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}\] This indicates that Bob’s qubit is in a completely random state until he receives classical information from Alice.

Theorem: (Schmidt Decomposition) For any pure state $\lvert \psi \rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$, there exist orthonormal bases $\{\lvert i \rangle_A\}$ for $\mathcal{H}_A$ and $\{\lvert i \rangle_B\}$ for $\mathcal{H}_B$ such that $\lvert \psi \rangle_{AB}$ can be written as: \[\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B\] where $\lambda_i$ are positive real numbers called Schmidt coefficients, which satisfy $\sum_{i=1}^{r} \lambda_i = 1$. The number of non-zero Schmidt coefficients, $r$, is called the Schmidt rank of the state. The square roots $\sqrt{\lambda_i}$ are the singular values of the coefficient matrix when the state $\lvert \psi \rangle_{AB}$ is expanded in any product basis.

Theorem: (Bell’s Theorem) Bell’s Theorem demonstrates the incompatibility between quantum mechanics and local realism. It states that any theory adhering to local realism must satisfy certain inequalities (Bell’s inequalities). Quantum mechanics predicts violations of these inequalities for entangled states, which has been experimentally confirmed, thus challenging the assumptions of local realism. Local realism combines two key assumptions:

Realism: Physical properties of systems have definite values at all times, independent of observation.
Locality: The outcome of a measurement on one particle is independent of measurement choices made on a spatially separated particle. There is no instantaneous influence between distant events.

Bell derived inequalities that must be satisfied by local realistic theories, and showed that quantum mechanics violates them.

Theorem: (Verification of Purification) Given a mixed state $\rho_{A}$ and its purification $\lvert \Psi \rangle_{AR} = \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R$, tracing out the reference system R from the pure state density matrix $\rho_{AR} = \lvert \Psi \rangle\langle \Psi \rvert_{AR}$ recovers the original mixed state $\rho_{A}$: \[\text{Tr}_R [\rho_{AR}] = \rho_{A}\] This confirms that $\lvert \Psi \rangle_{AR}$ is indeed a valid purification of $\rho_{A}$.

Examples

Example: Consider the Bell state $\lvert \Phi^+ \rangle = \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$. In the computational basis $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ for both subsystems, the coefficient matrix $C$ is: \[C = \begin{pmatrix} 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} \end{pmatrix}\] The rank of the matrix $C$ is 2, as there are two linearly independent rows (or columns). Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the Bell state $\lvert \Phi^+ \rangle$ is entangled.

Example: Consider the state $\lvert \psi \rangle = \frac{1}{2} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle)$. The coefficient matrix $C$ in the computational basis is: \[C = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}\] The rank of the matrix $C$ is 1, because the rows (and columns) are linearly dependent (the second row is identical to the first). Thus, the Schmidt rank is $r=1$. Since $r = 1$, this state is separable, meaning it is not entangled. Indeed, this state can be written as a product state: \[\lvert \psi \rangle = \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right) \otimes \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right)\]

Example: Consider the state $\lvert \phi \rangle = \frac{1}{\sqrt{3}} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle)$. The coefficient matrix $C$ is: \[C = \begin{pmatrix} 1/\sqrt{3} & 1/\sqrt{3} \\ 1/\sqrt{3} & 0 \end{pmatrix}\] The rank of the matrix $C$ is 2, as the rows are linearly independent. Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the state $\lvert \phi \rangle$ is entangled.

Remarks

Remark: The reduced density matrix for Bob’s qubit, $\rho_{B}= \frac{1}{2} \mathbb{I}$, is a mixed state. This implies that before Alice communicates her measurement outcome, Bob’s qubit is in a completely random, unpolarized state. Bob cannot gain any information about the original state $\lvert \psi_{\text{in}} \rangle$ by performing measurements on his qubit alone until he receives classical information from Alice about her measurement result.

Remark: To confirm that $\rho_{B}$ is a mixed state and not a pure state, we can check if $\rho_{B}^2 = \rho_{B}$. \[\rho_{B}^2 = \left(\frac{1}{2} \mathbb{I}\right)^2 = \frac{1}{4} \mathbb{I}^2 = \frac{1}{4} \mathbb{I}\neq \rho_{B}\] Since $\rho_{B}^2 \neq \rho_{B}$, $\rho_{B}$ is indeed a mixed state. This result is consistent with the no-communication theorem of quantum mechanics, which states that entanglement cannot be used to send classical information faster than light. Bob’s state is random until he learns Alice’s measurement outcome, at which point he can perform a correction operation to retrieve the teleported state.

Remark: As demonstrated by the examples, the rank of the coefficient matrix $C$ provides a direct method to determine if a pure bipartite state is entangled. For a pure bipartite state of qubits, the criterion simplifies to:

$\text{Rank}(C) = 1$: The state is separable (not entangled).
$\text{Rank}(C) = 2$: The state is entangled.

For qubit systems, a Schmidt rank greater than 1 is always equal to 2 because the dimension of the subsystems is 2, leading to at most a $2 \times 2$ coefficient matrix for a bipartite state.

Algorithms

Algorithm: Schmidt Decomposition Algorithm

Input: A pure bipartite state $\lvert \psi \rangle_{AB} = \sum_{j,k} C_{jk} \lvert j \rangle_A \otimes \lvert k \rangle_B$ in a product basis $\{\lvert j \rangle_A \otimes \lvert k \rangle_B\}$.

Steps:

Construct the coefficient matrix $C$ with entries $C_{jk}$.
Compute the Singular Value Decomposition (SVD) of $C = U S V^\dagger$, where $U$ and $V$ are unitary matrices, and $S$ is a diagonal matrix with singular values $s_1, s_2, \dots, s_r, 0, \dots, 0$ on the diagonal.
The Schmidt coefficients are $\lambda_i = s_i^2$.
The Schmidt rank $r$ is the number of non-zero singular values (or Schmidt coefficients).
The orthonormal basis $\{\lvert i \rangle_A\}$ for subsystem A is given by the columns of $U$.
The orthonormal basis $\{\lvert i \rangle_B\}$ for subsystem B is given by the columns of $V$.
The Schmidt decomposition is then $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} s_i \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$, where $\lvert u_i \rangle_A$ are columns of $U$ and $\lvert v_i \rangle_B$ are columns of $V$. Using $\sqrt{\lambda_i} = s_i$, we get $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$.

--- title: "Lecture Notes on Composite Quantum Systems and Entanglement" author: "Your Name" date: "2025-02-05" format: html: toc: true # Table of Contents toc-depth: 2 code-tools: true theme: cosmo # Or "journal" for Distill-like minimalism --- # Introduction This lecture introduces the concept of composite quantum systems, focusing on how to describe and analyze systems composed of multiple subsystems. We will explore the reduced density matrix as a tool for describing subsystems and its application in quantum teleportation. Furthermore, we will delve into the crucial concept of entanglement, its quantification using Schmidt decomposition, and methods for detecting it. We will also discuss the purification of mixed states and finally, examine the foundational Einstein-Podolsky-Rosen (EPR) paradox and Bell's Theorem, which highlight the profound implications of quantum entanglement and challenge classical notions of reality. # Composite Systems and Reduced Density Matrix In quantum mechanics, a composite system is described by combining two or more subsystems. Consider a system composed of two subsystems, labeled A and B. The Hilbert space of the composite system, denoted as $\mathcal{H}_{AB}$, is the tensor product of the Hilbert spaces of the individual subsystems, $\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B$. ## Measurements on Subsystem A When a measurement is performed on subsystem A while leaving subsystem B undisturbed, the measurement operator acts solely on the Hilbert space $\mathcal{H}_A$. Let $P_A$ be a projective measurement operator acting on subsystem A, and $I_B$ be the identity operator on subsystem B. The measurement operator on the composite system AB is then given by $P_A \otimes I_B$. The expectation value of this measurement for a composite system in a state described by the density matrix $\rho_{AB}$ is given by: $$\langle P_A \otimes I_B \rangle = \text{Tr}\left[ (P_A \otimes I_B) \rho_{AB}\right]$$ This expectation value represents the average outcome of the measurement $P_A$ when performed on subsystem A. To effectively calculate and interpret such measurements, the concept of the partial trace and the reduced density matrix are essential. ## Partial Trace: Tracing out Subsystem B The partial trace is a linear map that effectively \"traces out\" or removes a subsystem from a composite system's description. For a composite system AB with a density matrix $\rho_{AB}$ acting on $\mathcal{H}_A \otimes \mathcal{H}_B$, the partial trace over subsystem B, denoted as $\text{Tr}_B$, results in an operator acting only on $\mathcal{H}_A$. Mathematically, for an operator $O = O_A \otimes O_B$, where $O_A$ acts on $\mathcal{H}_A$ and $O_B$ acts on $\mathcal{H}_B$, the partial trace over subsystem B is defined as: $$\text{Tr}_B [O_A \otimes O_B] = O_A \text{Tr}[O_B]$$ For a general operator $M$ on $\mathcal{H}_{AB}$, we can define the partial trace in terms of matrix elements. Let $\{\lvert i \rangle_A\}$ and $\{\lvert k \rangle_B\}$ be orthonormal bases for $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. If the matrix elements of $\rho_{AB}$ in the product basis $\{\lvert i \rangle_A \otimes \lvert k \rangle_B\}$ are given by $\langle i \rvert_A \langle k \rvert_B \rho_{AB}\lvert j \rangle_A \lvert l \rangle_B = (\rho_{AB})_{ik,jl}$, then the matrix elements of the partial trace over B, denoted as $\rho_{A}= \text{Tr}_B[\rho_{AB}]$, in the basis $\{\lvert i \rangle_A\}$ are: $$(\rho_{A})_{ij} = (\text{Tr}_B[\rho_{AB}])_{ij} = \sum_{k} \langle i \rvert_A \langle k \rvert_B \rho_{AB}\lvert j \rangle_A \lvert k \rangle_B$$ This operation sums over the basis states of subsystem B, effectively averaging over or ignoring the degrees of freedom of subsystem B. ## Reduced Density Matrix: Describing Subsystem A ::: definition **Definition 1** (Reduced Density Matrix). *The reduced density matrix for subsystem A, denoted as $\rho_{A}$, is obtained by taking the partial trace of the density matrix $\rho_{AB}$ of the composite system over subsystem B: $$\rho_{A}= \text{Tr}_B [\rho_{AB}]$$ Similarly, the reduced density matrix for subsystem B is obtained by tracing over subsystem A: $$\rho_{B}= \text{Tr}_A [\rho_{AB}]$$ The reduced density matrix $\rho_{A}$ provides a complete description of subsystem A when subsystem B is not considered or when we are only interested in measurements performed on subsystem A. All expectation values of observables acting solely on subsystem A can be calculated using $\rho_{A}$.* ::: Returning to the expectation value of the measurement $P_A \otimes I_B$, we can rewrite it using the partial trace: $$\begin{aligned}\langle P_A \otimes I_B \rangle &= \text{Tr}\left[ (P_A \otimes I_B) \rho_{AB}\right] \\&= \text{Tr}_A \text{Tr}_B \left[ (P_A \otimes I_B) \rho_{AB}\right] \\&= \text{Tr}_A \left[ P_A \text{Tr}_B [I_B \rho_{AB}] \right] \\&= \text{Tr}_A \left[ P_A \text{Tr}_B [\rho_{AB}] \right] \\&= \text{Tr}_A \left[ P_A \rho_{A}\right]\end{aligned}$$ This derivation explicitly shows that the expectation value of a measurement on subsystem A can be computed solely from the reduced density matrix $\rho_{A}$ and the measurement operator $P_A$ acting on subsystem A's Hilbert space. # Reduced Density Matrix and Quantum Teleportation Quantum teleportation is a protocol for transferring an unknown quantum state from one location to another using pre-shared entanglement and classical communication. The reduced density matrix is crucial for understanding the state of the qubit being teleported before the classical communication step is completed. ## Quantum Teleportation Protocol and State Evolution Consider Alice, who wants to teleport an unknown quantum state $\lvert \psi_{\text{in}} \rangle= \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle$ to Bob. They initially share a maximally entangled Bell state, specifically $\lvert \Phi^+ \rangle= \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$. Alice possesses the particle in state $\lvert \psi_{\text{in}} \rangle$ and half of the entangled pair (particle A), while Bob holds the other half of the entangled pair (particle B). The initial state of the composite system, including Alice's unknown qubit (particle 'in') and the entangled pair (particles A and B), is: $$\lvert \Psi_1 \rangle = \lvert \psi_{\text{in}} \rangle\otimes \lvert \Phi^+ \rangle= (\alpha \lvert 0 \rangle_{\text{in}} + \beta \lvert 1 \rangle_{\text{in}}) \otimes \frac{1}{\sqrt{2}} (\lvert 0 \rangle_A \lvert 0 \rangle_B + \lvert 1 \rangle_A \lvert 1 \rangle_B)$$ After Alice performs a Bell basis measurement on her particles ('in' and A), the combined state of the system collapses. Before Alice communicates her measurement outcome to Bob, the state of Bob's particle (B), represented by the reduced density matrix, is a mixed state. The state $\lvert \Psi_2 \rangle$ immediately after Alice's Bell measurement, but before she announces the result, can be written as a superposition of states, each corresponding to a different measurement outcome for Alice and a resulting state for Bob: $$\begin{aligned}\lvert \Psi_2 \rangle = \frac{1}{2} & [\lvert \Phi^+ \rangle_{\text{in}A} (\alpha \lvert 0 \rangle_B + \beta \lvert 1 \rangle_B) + \lvert \Phi^- \rangle_{\text{in}A} (\alpha \lvert 0 \rangle_B - \beta \lvert 1 \rangle_B) \\&+ \lvert \Psi^+ \rangle_{\text{in}A} (\alpha \lvert 1 \rangle_B + \beta \lvert 0 \rangle_B) + \lvert \Psi^- \rangle_{\text{in}A} (\alpha \lvert 1 \rangle_B - \beta \lvert 0 \rangle_B) ]\end{aligned}$$ Here, $\lvert \Phi^\pm \rangle_{\text{in}A} = \frac{1}{\sqrt{2}} (\lvert 00 \rangle \pm \lvert 11 \rangle)_{\text{in}A}$ and $\lvert \Psi^\pm \rangle_{\text{in}A} = \frac{1}{\sqrt{2}} (\lvert 01 \rangle \pm \lvert 10 \rangle)_{\text{in}A}$ are the Bell states for particles 'in' and A. ## Calculating the Reduced Density Matrix for Bob's Qubit ::: theorem **Theorem 1** (Reduced Density Matrix of Bob's Qubit in Teleportation). *To determine the state of Bob's qubit before Alice's classical communication, we calculate the reduced density matrix $\rho_{B}$ by tracing out Alice's particles ('in' and A) from the density matrix of the joint state $\rho_{\text{in}AB} = \lvert \Psi_2 \rangle\langle \Psi_2 \rvert$. $$\rho_{B}= \text{Tr}_{\text{in}A} [\lvert \Psi_2 \rangle\langle \Psi_2 \rvert]$$ We perform the partial trace over the Hilbert space of Alice's qubits, spanned by the Bell basis $\{\lvert \Phi^+ \rangle_{\text{in}A}, \lvert \Phi^- \rangle_{\text{in}A}, \lvert \Psi^+ \rangle_{\text{in}A}, \lvert \Psi^- \rangle_{\text{in}A}\}$. Due to the orthogonality of the Bell basis, the reduced density matrix $\rho_{B}$ is given by: $$\begin{aligned}\rho_{B}&= \text{Tr}_{\text{in}A} [\lvert \Psi_2 \rangle\langle \Psi_2 \rvert] \\&= \sum_{\lvert \chi \rangle_{\text{in}A} \in \text{Bell basis}} \langle \chi \rvert_{\text{in}A} \lvert \Psi_2 \rangle\langle \Psi_2 \rvert \lvert \chi \rangle_{\text{in}A} \\&= \frac{1}{4} \sum_{\lvert \chi \rangle_{\text{in}A} \in \text{Bell basis}} \langle \chi \rvert_{\text{in}A} \left[ \sum_{\lvert \phi \rangle_{\text{in}A}, \lvert \psi \rangle_{\text{in}A} \in \text{Bell basis}} \lvert \phi \rangle_{\text{in}A} \lvert \tilde{\psi}_\phi \rangle_B \langle \tilde{\psi}_\psi \rvert_B \langle \psi \rvert_{\text{in}A} \right] \lvert \chi \rangle_{\text{in}A}\end{aligned}$$ where $\lvert \tilde{\psi}_{\Phi^+} \rangle = (\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle)_B$, $\lvert \tilde{\psi}_{\Phi^-} \rangle = (\alpha \lvert 0 \rangle - \beta \lvert 1 \rangle)_B$, $\lvert \tilde{\psi}_{\Psi^+} \rangle = (\alpha \lvert 1 \rangle + \beta \lvert 0 \rangle)_B$, and $\lvert \tilde{\psi}_{\Psi^-} \rangle = (\alpha \lvert 1 \rangle - \beta \lvert 0 \rangle)_B$. Using the orthogonality $\langle \psi \rvert \chi \rangle_{\text{in}A} = \delta_{\psi\chi}$, this simplifies to: $$\begin{aligned}\rho_{B}&= \frac{1}{4} \sum_{\lvert \chi \rangle_{\text{in}A} \in \text{Bell basis}} \lvert \tilde{\psi}_\chi \rangle\langle \tilde{\psi}_\chi \rvert_B \\&= \frac{1}{4} \left[ \lvert \tilde{\psi}_{\Phi^+} \rangle\langle \tilde{\psi}_{\Phi^+} \rvert + \lvert \tilde{\psi}_{\Phi^-} \rangle\langle \tilde{\psi}_{\Phi^-} \rvert + \lvert \tilde{\psi}_{\Psi^+} \rangle\langle \tilde{\psi}_{\Psi^+} \rvert + \lvert \tilde{\psi}_{\Psi^-} \rangle\langle \tilde{\psi}_{\Psi^-} \rvert \right]_B\end{aligned}$$ Substituting the expressions for $\lvert \tilde{\psi}_\chi \rangle$ and expanding: $$\begin{aligned}\rho_{B}&= \frac{1}{4} [ (\alpha \lvert 0 \rangle + \beta \lvert 1 \rangle)(\alpha^* \langle 0 \rvert + \beta^* \langle 1 \rvert) + (\alpha \lvert 0 \rangle - \beta \lvert 1 \rangle)(\alpha^* \langle 0 \rvert - \beta^* \langle 1 \rvert) \\&+ (\alpha \lvert 1 \rangle + \beta \lvert 0 \rangle)(\alpha^* \langle 1 \rvert + \beta^* \langle 0 \rvert) + (\alpha \lvert 1 \rangle - \beta \lvert 0 \rangle)(\alpha^* \langle 1 \rvert - \beta^* \langle 0 \rvert) ] \\&= \frac{1}{4} [ 2(|\alpha|^2 \lvert 0 \rangle\langle 0 \rvert + |\beta|^2 \lvert 1 \rangle\langle 1 \rvert) + 2(|\alpha|^2 \lvert 1 \rangle\langle 1 \rvert + |\beta|^2 \lvert 0 \rangle\langle 0 \rvert) ] \\&= \frac{1}{2} (|\alpha|^2 + |\beta|^2) (\lvert 0 \rangle\langle 0 \rvert + \lvert 1 \rangle\langle 1 \rvert)\end{aligned}$$ Since $|\alpha|^2 + |\beta|^2 = 1$ due to normalization of $\lvert \psi_{\text{in}} \rangle$, we obtain: $$\rho_{B}= \frac{1}{2} (\lvert 0 \rangle\langle 0 \rvert + \lvert 1 \rangle\langle 1 \rvert) = \frac{1}{2} \mathbb{I}= \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}$$ This is the maximally mixed state, represented by half the identity matrix $\mathbb{I}$.* ::: ## Mixed State Interpretation and No-Communication Theorem ::: remark **Remark 1** (Mixed State in Teleportation). *The reduced density matrix for Bob's qubit, $\rho_{B}= \frac{1}{2} \mathbb{I}$, is a mixed state. This implies that before Alice communicates her measurement outcome, Bob's qubit is in a completely random, unpolarized state. Bob cannot gain any information about the original state $\lvert \psi_{\text{in}} \rangle$ by performing measurements on his qubit alone until he receives classical information from Alice about her measurement result.* ::: ::: remark **Remark 2** (Verification of Mixed State). *To confirm that $\rho_{B}$ is a mixed state and not a pure state, we can check if $\rho_{B}^2 = \rho_{B}$. $$\rho_{B}^2 = \left(\frac{1}{2} \mathbb{I}\right)^2 = \frac{1}{4} \mathbb{I}^2 = \frac{1}{4} \mathbb{I}\neq \rho_{B}$$ Since $\rho_{B}^2 \neq \rho_{B}$, $\rho_{B}$ is indeed a mixed state. This result is consistent with the no-communication theorem of quantum mechanics, which states that entanglement cannot be used to send classical information faster than light. Bob's state is random until he learns Alice's measurement outcome, at which point he can perform a correction operation to retrieve the teleported state.* ::: # Entanglement and Schmidt Decomposition Entanglement is a quintessential quantum mechanical phenomenon where two or more subsystems are intrinsically linked, such that their quantum states are interdependent, regardless of the physical distance separating them. This section formalizes the definition of entanglement for bipartite systems and introduces the Schmidt decomposition as a tool for detecting and quantifying entanglement in pure states. ## Defining Entanglement for Bipartite Systems For a bipartite system composed of subsystems A and B, entanglement is defined differently for pure and mixed states. #### Pure States: A pure state $\lvert \psi \rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$ is entangled if it cannot be expressed as a product of pure states of the individual subsystems, i.e., it cannot be written in the form $\lvert \psi \rangle_{AB} = \lvert \phi \rangle_A \otimes \lvert \chi \rangle_B$, where $\lvert \phi \rangle_A \in \mathcal{H}_A$ and $\lvert \chi \rangle_B \in \mathcal{H}_B$. If a pure state can be written as a product state, it is called a separable state or unentangled state. #### Mixed States: A mixed state $\rho_{AB}$ is separable if it can be represented as a convex combination of product states: $$\rho_{AB}= \sum_i p_i \rho_{A,i} \otimes \rho_{B,i}$$ where $p_i \geq 0$ with $\sum_i p_i = 1$, and $\rho_{A,i}$ and $\rho_{B,i}$ are density matrices for subsystems A and B, respectively. If a mixed state cannot be expressed in this form, it is considered entangled. ## Schmidt Decomposition for Pure Bipartite States ::: theorem **Theorem 2** (Schmidt Decomposition Theorem). *For any pure state $\lvert \psi \rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$, there exist orthonormal bases $\{\lvert i \rangle_A\}$ for $\mathcal{H}_A$ and $\{\lvert i \rangle_B\}$ for $\mathcal{H}_B$ such that $\lvert \psi \rangle_{AB}$ can be written as: $$\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$$ where $\lambda_i$ are positive real numbers called Schmidt coefficients, which satisfy $\sum_{i=1}^{r} \lambda_i = 1$. The number of non-zero Schmidt coefficients, $r$, is called the Schmidt rank of the state. The square roots $\sqrt{\lambda_i}$ are the singular values of the coefficient matrix when the state $\lvert \psi \rangle_{AB}$ is expanded in any product basis.* ::: To apply Schmidt decomposition in practice, we first express the bipartite state $\lvert \psi \rangle_{AB}$ in a product basis $\{\lvert j \rangle_A \otimes \lvert k \rangle_B\}$: $$\lvert \psi \rangle_{AB} = \sum_{j,k} C_{jk} \lvert j \rangle_A \otimes \lvert k \rangle_B$$ We then form a coefficient matrix $C$ with entries $C_{jk}$. The Schmidt decomposition is closely related to the singular value decomposition of the matrix $C$. The Schmidt rank $r$ is equal to the rank of the matrix $C$. ## Schmidt Number and Entanglement Quantification The Schmidt number, which is the Schmidt rank $r$, serves as a measure of entanglement for pure bipartite states. - **Schmidt rank** $r = 1$: The state has only one non-zero Schmidt coefficient and is a product state. Such a state is separable and therefore, not entangled. - **Schmidt rank** $r > 1$: The state has more than one non-zero Schmidt coefficient and is an entangled state. A higher Schmidt number generally indicates a higher degree of entanglement. ## Examples of Entanglement Analysis using Schmidt Decomposition We now illustrate the Schmidt decomposition method with examples to determine whether given bipartite states are entangled. ::: example **Example 1** (Bell State $\lvert \Phi^+ \rangle = \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$). *Consider the Bell state $\lvert \Phi^+ \rangle = \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$. In the computational basis $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ for both subsystems, the coefficient matrix $C$ is: $$C = \begin{pmatrix} 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} \end{pmatrix}$$ The rank of the matrix $C$ is 2, as there are two linearly independent rows (or columns). Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the Bell state $\lvert \Phi^+ \rangle$ is entangled.* ::: ::: example **Example 2** (Separable State $\lvert \psi \rangle = \frac{1}{2} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle)$). *Consider the state $\lvert \psi \rangle = \frac{1}{2} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle)$. The coefficient matrix $C$ in the computational basis is: $$C = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$$ The rank of the matrix $C$ is 1, because the rows (and columns) are linearly dependent (the second row is identical to the first). Thus, the Schmidt rank is $r=1$. Since $r = 1$, this state is separable, meaning it is not entangled. Indeed, this state can be written as a product state: $$\lvert \psi \rangle = \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right) \otimes \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right)$$* ::: ::: example **Example 3** (Entangled State $\lvert \phi \rangle = \frac{1}{\sqrt{3}} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle)$). *Consider the state $\lvert \phi \rangle = \frac{1}{\sqrt{3}} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle)$. The coefficient matrix $C$ is: $$C = \begin{pmatrix} 1/\sqrt{3} & 1/\sqrt{3} \\ 1/\sqrt{3} & 0 \end{pmatrix}$$ The rank of the matrix $C$ is 2, as the rows are linearly independent. Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the state $\lvert \phi \rangle$ is entangled.* ::: ## Entanglement Detection via Rank of Coefficient Matrix ::: remark **Remark 3** (Entanglement Criterion for Pure Bipartite Qubit States). *As demonstrated by the examples, the rank of the coefficient matrix $C$ provides a direct method to determine if a pure bipartite state is entangled. For a pure bipartite state of qubits, the criterion simplifies to:* - *$\text{Rank}(C) = 1$: The state is separable (not entangled).* - *$\text{Rank}(C) = 2$: The state is entangled.* *For qubit systems, a Schmidt rank greater than 1 is always equal to 2 because the dimension of the subsystems is 2, leading to at most a $2 \times 2$ coefficient matrix for a bipartite state.* ::: # Purification of Mixed States A mixed state density matrix $\rho_{A}$ represents a statistical ensemble of pure quantum states. Purification is a conceptual and mathematical procedure that allows us to represent a mixed state $\rho_{A}$ of a system A as the reduced density matrix of a pure state defined on a larger composite system AR, where R is an auxiliary reference system. This process highlights that any mixed state can be viewed as arising from a pure state by disregarding or \"tracing out\" the degrees of freedom of an environment (in this case, the reference system R). ## Spectral Decomposition of the Density Operator The foundation of purification lies in the spectral decomposition of density operators. Any density operator $\rho_{A}$ is, by definition, a Hermitian, positive semi-definite operator with trace equal to one. Consequently, $\rho_{A}$ can be diagonalized in an orthonormal basis $\{\lvert i \rangle_A\}$ of the Hilbert space $\mathcal{H}_A$. The spectral decomposition of $\rho_{A}$ is given by: $$\rho_{A}= \sum_i p_i \lvert i \rangle\langle i \rvert_A$$ where $p_i$ are the non-negative eigenvalues of $\rho_{A}$, representing the probabilities of the system being in the corresponding eigenstate $\lvert i \rangle_A$, and they satisfy $\sum_i p_i = 1$. ## Purification Procedure: Constructing a Pure State ::: definition **Definition 2** (Purification of a Mixed State). *To purify the mixed state $\rho_{A}$, we introduce a reference system R, which is chosen to have a Hilbert space $\mathcal{H}_R$ of at least the same dimension as the support of $\rho_{A}$. We then construct a pure state $\lvert \Psi \rangle_{AR}$ for the composite system AR in the Hilbert space $\mathcal{H}_{AR} = \mathcal{H}_A \otimes \mathcal{H}_R$ as follows: $$\lvert \Psi \rangle_{AR} = \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R$$ Here, $\{\lvert i \rangle_A\}$ are the eigenstates from the spectral decomposition of $\rho_{A}$, and $\{\lvert i \rangle_R\}$ is an orthonormal basis for the reference system R, chosen to be in one-to-one correspondence with the eigenstates $\{\lvert i \rangle_A\}$. The coefficients $\sqrt{p_i}$ are the square roots of the eigenvalues of $\rho_{A}$.* ::: ::: theorem **Theorem 3** (Verification of Purification). *The density matrix for this composite system AR in the pure state $\lvert \Psi \rangle_{AR}$ is given by $\rho_{AR} = \lvert \Psi \rangle\langle \Psi \rvert_{AR}$. To verify that $\lvert \Psi \rangle_{AR}$ is indeed a purification of $\rho_{A}$, we compute the reduced density matrix for system A by taking the partial trace of $\rho_{AR}$ over the reference system R: $$\begin{aligned}\rho'_A &= \text{Tr}_R [\rho_{AR}] \\&= \text{Tr}_R [\lvert \Psi \rangle\langle \Psi \rvert_{AR}] \\&= \text{Tr}_R \left[ \left( \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R \right) \left( \sum_j \sqrt{p_j} \langle j \rvert_A \otimes \langle j \rvert_R \right) \right] \\&= \text{Tr}_R \left[ \sum_{i,j} \sqrt{p_i p_j} (\lvert i \rangle\langle j \rvert_A \otimes \lvert i \rangle\langle j \rvert_R) \right] \\&= \sum_{i,j} \sqrt{p_i p_j} \lvert i \rangle\langle j \rvert_A \text{Tr}_R [\lvert i \rangle\langle j \rvert_R] \\&= \sum_{i,j} \sqrt{p_i p_j} \lvert i \rangle\langle j \rvert_A \langle j \rvert i \rangle_R\end{aligned}$$ Since $\{\lvert i \rangle_R\}$ is an orthonormal basis, $\langle j \rvert i \rangle_R = \delta_{ij}$. Thus, the expression simplifies to: $$\begin{aligned}\rho'_A &= \sum_{i,j} \sqrt{p_i p_j} \lvert i \rangle\langle j \rvert_A \delta_{ij} \\&= \sum_i \sqrt{p_i p_i} \lvert i \rangle\langle i \rvert_A \\&= \sum_i p_i \lvert i \rangle\langle i \rvert_A \\&= \rho_{A}\end{aligned}$$ Therefore, by tracing out the reference system R from the pure state $\lvert \Psi \rangle_{AR}$, we recover the original mixed state $\rho_{A}$. This confirms that $\lvert \Psi \rangle_{AR}$ is a purification of $\rho_{A}$. The purification procedure demonstrates that any mixed state can be regarded as part of a larger pure state, providing a deeper understanding of mixed states in quantum mechanics.* ::: # The Einstein-Podolsky-Rosen Paradox and Bell's Theorem The Einstein-Podolsky-Rosen (EPR) paradox, introduced in their seminal 1935 paper, and Bell's Theorem, developed by John Stewart Bell in 1964, are pivotal in understanding the conceptual challenges posed by quantum mechanics, particularly concerning entanglement and the nature of reality. These concepts highlight the profound departure of quantum mechanics from classical physics. ## The EPR Paradox: Challenging Completeness of Quantum Mechanics The EPR paradox emerges from considering the implications of quantum entanglement for spatially separated particles. EPR's thought experiment involved entangled particles, though their original paper focused on position and momentum. A more illustrative example, and closer to the transcript's discussion, involves spin-entangled particles. Consider a source thatemits pairs of particles in the spin singlet state: $$\lvert \Psi^- \rangle = \frac{1}{\sqrt{2}} (\lvert 0 \rangle_A \lvert 1 \rangle_B - \lvert 1 \rangle_A \lvert 0 \rangle_B)$$ where particle A is sent to Alice and particle B is sent to Bob, who are assumed to be far apart, such that no physical signal can travel between them faster than light. ## EPR's Criterion of Physical Reality Central to the EPR paradox is their definition of an \"element of reality.\" They posited: \"If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity.\" Imagine Alice measures the spin of particle A along the z-axis ($\sigma_z^A$). If she obtains spin-up (outcome $\lvert 0 \rangle_A$), due to the entangled nature of the singlet state, she can predict with certainty that a measurement of the spin of particle B along the z-axis ($\sigma_z^B$) will yield spin-down (outcome $\lvert 1 \rangle_B$). Crucially, this prediction is made without performing any measurement on particle B, thus, according to EPR, without disturbing it. Applying their criterion, EPR concluded that the spin of particle B along the z-axis must be an \"element of reality\" even before Bob performs any measurement. ## The Argument for Incompleteness and Hidden Variables EPR extended this argument by considering that Alice could have chosen to measure the spin of particle A along the x-axis ($\sigma_x^A$) instead. Depending on her outcome, she could similarly predict with certainty the outcome of a $\sigma_x^B$ measurement by Bob. This led them to argue that both the spin along the z-axis and the spin along the x-axis (and indeed, along any axis) for particle B are simultaneously \"elements of reality.\" However, quantum mechanics dictates that spin components along different axes are incompatible observables; they cannot have simultaneously definite values. This apparent contradiction led EPR to conclude that quantum mechanics provides an incomplete description of physical reality. They hypothesized that there must be some \"hidden variables\" not accounted for in quantum mechanics that predetermine the outcomes of all possible measurements, thus restoring determinism and locality. These hidden variables would ensure that particles possess definite properties independently of measurement, consistent with classical intuition. ## Bell's Theorem: Testing Local Realism ::: theorem **Theorem 4** (Bell's Theorem). *Bell's Theorem, formulated by John Stewart Bell, provides a way to experimentally test the validity of local hidden variable theories against the predictions of quantum mechanics. Bell formalized the concept of \"local realism,\" which underpins the EPR argument. Local realism combines two key assumptions:* - ***Realism**: Physical properties of systems have definite values at all times, independent of observation.* - ***Locality**: The outcome of a measurement on one particle is independent of measurement choices made on a spatially separated particle. There is no instantaneous influence between distant events.* *Bell derived inequalities, known as Bell's inequalities, that must be satisfied by any theory adhering to local realism. He considered scenarios where Alice and Bob independently choose to measure spin along different directions on entangled particles. Quantum mechanics, however, predicts correlations between measurement outcomes for certain entangled states that violate Bell's inequalities.* ::: In essence, Bell's theorem demonstrates that if nature is governed by local realism, the correlations between measurements on entangled particles must be weaker than what quantum mechanics predicts. Numerous experimental tests of Bell's inequalities, starting with Alain Aspect's experiments in the 1980s and continuing with increasingly sophisticated experiments, have consistently shown violations of Bell's inequalities. These experimental violations strongly support quantum mechanics and rule out local realistic hidden variable theories as a complete description of nature. The implication is profound: at least one of the core assumptions of local realism---realism or locality---must be abandoned in the quantum realm. While interpretations vary, the most widely accepted conclusion is that locality, in the classical sense, is violated by quantum mechanics, suggesting the existence of non-local correlations between entangled particles. The transcript's mention of measurements along directions A and B and the expectation value of $\sigma_1 \cdot A \otimes \sigma_2 \cdot B$ alludes to the mathematical framework used to derive and test Bell's inequalities, which quantify these correlations and demonstrate the conflict with local realism. # Conclusion This lecture explored fundamental concepts related to composite quantum systems and the intriguing phenomenon of entanglement. We introduced the reduced density matrix as a crucial tool for describing the state of subsystems within a larger composite system, and illustrated its application in the context of quantum teleportation, demonstrating how it describes the state of Bob's qubit before classical communication from Alice. We then delved into the nature of entanglement, using Schmidt decomposition as a method to quantify and detect entanglement in pure bipartite states, providing illustrative examples. The concept of purification revealed how mixed states can be understood as arising from pure states of larger systems by tracing out environmental degrees of freedom. Finally, we discussed the profound implications of entanglement through the Einstein-Podolsky-Rosen paradox and Bell's Theorem, underscoring the fundamental conflict between quantum mechanics and classical notions of local realism and highlighting the non-local nature of quantum correlations. Key takeaways from this lecture are: - The **reduced density matrix** is an indispensable tool for describing the quantum state of subsystems and for making predictions about local measurements on composite systems. - **Entanglement** is a uniquely quantum correlation that defies classical explanation and represents a fundamental resource in quantum information processing. - **Schmidt decomposition** provides a practical method to determine and quantify entanglement in pure bipartite states by analyzing the rank of the coefficient matrix. - **Purification** offers a conceptual framework to understand mixed states as partial descriptions of larger pure states, linking mixed states to entanglement with an environment. - **Bell's Theorem** and its experimental verification present a compelling challenge to local realism, demonstrating the non-local nature of quantum mechanics and the profound implications of entanglement for our understanding of physical reality. Further study should delve into the experimental verifications of Bell's inequalities, explore the diverse interpretations of quantum mechanics in light of non-locality, and investigate the burgeoning field of quantum technologies that harness entanglement for applications such as quantum computation, quantum communication, and quantum cryptography. # Algorithm for Schmidt Decomposition ::: algorithm **Algorithm 1** (Schmidt Decomposition Algorithm). ***Input:** A pure bipartite state $\lvert \psi \rangle_{AB} = \sum_{j,k} C_{jk} \lvert j \rangle_A \otimes \lvert k \rangle_B$ in a product basis $\{\lvert j \rangle_A \otimes \lvert k \rangle_B\}$.* ***Output:** Schmidt decomposition $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$, Schmidt coefficients $\lambda_i$, and Schmidt rank $r$.* ***Steps:*** 1. *Construct the coefficient matrix $C$ with entries $C_{jk}$.* 2. *Compute the Singular Value Decomposition (SVD) of $C = U S V^\dagger$, where $U$ and $V$ are unitary matrices, and $S$ is a diagonal matrix with singular values $s_1, s_2, \dots, s_r, 0, \dots, 0$ on the diagonal.* 3. *The Schmidt coefficients are $\lambda_i = s_i^2$.* 4. *The Schmidt rank $r$ is the number of non-zero singular values (or Schmidt coefficients).* 5. *The orthonormal basis $\{\lvert i \rangle_A\}$ for subsystem A is given by the columns of $U$.* 6. *The orthonormal basis $\{\lvert i \rangle_B\}$ for subsystem B is given by the columns of $V$.* 7. *The Schmidt decomposition is then $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} s_i \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$, where $\lvert u_i \rangle_A$ are columns of $U$ and $\lvert v_i \rangle_B$ are columns of $V$. Using $\sqrt{\lambda_i} = s_i$, we get $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$.* ***Note:** In the main text, for simplicity of notation, we used the same index for bases in both subsystems in the Schmidt decomposition $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$. In general, the bases can be different, as shown in the algorithm output using $U$ and $V$ matrices from SVD. For square coefficient matrices and when using SVD directly to find Schmidt decomposition, using different bases $\{\lvert u_i \rangle_A\}$ and $\{\lvert v_i \rangle_B\}$ from SVD is more accurate. For the purpose of entanglement detection via Schmidt rank, the rank of the coefficient matrix $C$ is sufficient.* ::: # Mathematical Statements in Tcolorboxes {#mathematical-statements-in-tcolorboxes .unnumbered} ## Definitions {#definitions .unnumbered} ::: tcolorbox **Definition:** The reduced density matrix for subsystem A, denoted as $\rho_{A}$, is obtained by taking the partial trace of the density matrix $\rho_{AB}$ of the composite system over subsystem B: $$\rho_{A}= \text{Tr}_B [\rho_{AB}]$$ Similarly, the reduced density matrix for subsystem B is obtained by tracing over subsystem A: $$\rho_{B}= \text{Tr}_A [\rho_{AB}]$$ The reduced density matrix $\rho_{A}$ provides a complete description of subsystem A when subsystem B is not considered or when we are only interested in measurements performed on subsystem A. All expectation values of observables acting solely on subsystem A can be calculated using $\rho_{A}$. ::: ::: tcolorbox **Definition:** To purify the mixed state $\rho_{A}$, we introduce a reference system R, which is chosen to have a Hilbert space $\mathcal{H}_R$ of at least the same dimension as the support of $\rho_{A}$. We then construct a pure state $\lvert \Psi \rangle_{AR}$ for the composite system AR in the Hilbert space $\mathcal{H}_{AR} = \mathcal{H}_A \otimes \mathcal{H}_R$ as follows: $$\lvert \Psi \rangle_{AR} = \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R$$ Here, $\{\lvert i \rangle_A\}$ are the eigenstates from the spectral decomposition of $\rho_{A}$, and $\{\lvert i \rangle_R\}$ is an orthonormal basis for the reference system R, chosen to be in one-to-one correspondence with the eigenstates $\{\lvert i \rangle_A\}$. The coefficients $\sqrt{p_i}$ are the square roots of the eigenvalues of $\rho_{A}$. ::: ## Theorems {#theorems .unnumbered} ::: tcolorbox **Theorem:** In the quantum teleportation protocol described, the reduced density matrix $\rho_{B}$ of Bob's qubit, before Alice sends classical information, is the maximally mixed state: $$\rho_{B}= \text{Tr}_{\text{in}A} [\lvert \Psi_2 \rangle\langle \Psi_2 \rvert] = \frac{1}{2} (\lvert 0 \rangle\langle 0 \rvert + \lvert 1 \rangle\langle 1 \rvert) = \frac{1}{2} \mathbb{I}= \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}$$ This indicates that Bob's qubit is in a completely random state until he receives classical information from Alice. ::: ::: tcolorbox **Theorem:** **(Schmidt Decomposition)** For any pure state $\lvert \psi \rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$, there exist orthonormal bases $\{\lvert i \rangle_A\}$ for $\mathcal{H}_A$ and $\{\lvert i \rangle_B\}$ for $\mathcal{H}_B$ such that $\lvert \psi \rangle_{AB}$ can be written as: $$\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$$ where $\lambda_i$ are positive real numbers called Schmidt coefficients, which satisfy $\sum_{i=1}^{r} \lambda_i = 1$. The number of non-zero Schmidt coefficients, $r$, is called the Schmidt rank of the state. The square roots $\sqrt{\lambda_i}$ are the singular values of the coefficient matrix when the state $\lvert \psi \rangle_{AB}$ is expanded in any product basis. ::: ::: tcolorbox **Theorem:** **(Bell's Theorem)** Bell's Theorem demonstrates the incompatibility between quantum mechanics and local realism. It states that any theory adhering to local realism must satisfy certain inequalities (Bell's inequalities). Quantum mechanics predicts violations of these inequalities for entangled states, which has been experimentally confirmed, thus challenging the assumptions of local realism. Local realism combines two key assumptions: - **Realism**: Physical properties of systems have definite values at all times, independent of observation. - **Locality**: The outcome of a measurement on one particle is independent of measurement choices made on a spatially separated particle. There is no instantaneous influence between distant events. Bell derived inequalities that must be satisfied by local realistic theories, and showed that quantum mechanics violates them. ::: ::: tcolorbox **Theorem:** **(Verification of Purification)** Given a mixed state $\rho_{A}$ and its purification $\lvert \Psi \rangle_{AR} = \sum_i \sqrt{p_i} \lvert i \rangle_A \otimes \lvert i \rangle_R$, tracing out the reference system R from the pure state density matrix $\rho_{AR} = \lvert \Psi \rangle\langle \Psi \rvert_{AR}$ recovers the original mixed state $\rho_{A}$: $$\text{Tr}_R [\rho_{AR}] = \rho_{A}$$ This confirms that $\lvert \Psi \rangle_{AR}$ is indeed a valid purification of $\rho_{A}$. ::: ## Examples {#examples .unnumbered} ::: tcolorbox **Example:** Consider the Bell state $\lvert \Phi^+ \rangle = \frac{1}{\sqrt{2}} (\lvert 00 \rangle + \lvert 11 \rangle)$. In the computational basis $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ for both subsystems, the coefficient matrix $C$ is: $$C = \begin{pmatrix} 1/\sqrt{2} & 0 \\ 0 & 1/\sqrt{2} \end{pmatrix}$$ The rank of the matrix $C$ is 2, as there are two linearly independent rows (or columns). Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the Bell state $\lvert \Phi^+ \rangle$ is entangled. ::: ::: tcolorbox **Example:** Consider the state $\lvert \psi \rangle = \frac{1}{2} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle + \lvert 11 \rangle)$. The coefficient matrix $C$ in the computational basis is: $$C = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$$ The rank of the matrix $C$ is 1, because the rows (and columns) are linearly dependent (the second row is identical to the first). Thus, the Schmidt rank is $r=1$. Since $r = 1$, this state is separable, meaning it is not entangled. Indeed, this state can be written as a product state: $$\lvert \psi \rangle = \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right) \otimes \left( \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} \right)$$ ::: ::: tcolorbox **Example:** Consider the state $\lvert \phi \rangle = \frac{1}{\sqrt{3}} (\lvert 00 \rangle + \lvert 01 \rangle + \lvert 10 \rangle)$. The coefficient matrix $C$ is: $$C = \begin{pmatrix} 1/\sqrt{3} & 1/\sqrt{3} \\ 1/\sqrt{3} & 0 \end{pmatrix}$$ The rank of the matrix $C$ is 2, as the rows are linearly independent. Therefore, the Schmidt rank is $r=2$. Since $r > 1$, the state $\lvert \phi \rangle$ is entangled. ::: ## Remarks {#remarks .unnumbered} ::: tcolorbox **Remark:** The reduced density matrix for Bob's qubit, $\rho_{B}= \frac{1}{2} \mathbb{I}$, is a mixed state. This implies that before Alice communicates her measurement outcome, Bob's qubit is in a completely random, unpolarized state. Bob cannot gain any information about the original state $\lvert \psi_{\text{in}} \rangle$ by performing measurements on his qubit alone until he receives classical information from Alice about her measurement result. ::: ::: tcolorbox **Remark:** To confirm that $\rho_{B}$ is a mixed state and not a pure state, we can check if $\rho_{B}^2 = \rho_{B}$. $$\rho_{B}^2 = \left(\frac{1}{2} \mathbb{I}\right)^2 = \frac{1}{4} \mathbb{I}^2 = \frac{1}{4} \mathbb{I}\neq \rho_{B}$$ Since $\rho_{B}^2 \neq \rho_{B}$, $\rho_{B}$ is indeed a mixed state. This result is consistent with the no-communication theorem of quantum mechanics, which states that entanglement cannot be used to send classical information faster than light. Bob's state is random until he learns Alice's measurement outcome, at which point he can perform a correction operation to retrieve the teleported state. ::: ::: tcolorbox **Remark:** As demonstrated by the examples, the rank of the coefficient matrix $C$ provides a direct method to determine if a pure bipartite state is entangled. For a pure bipartite state of qubits, the criterion simplifies to: - $\text{Rank}(C) = 1$: The state is separable (not entangled). - $\text{Rank}(C) = 2$: The state is entangled. For qubit systems, a Schmidt rank greater than 1 is always equal to 2 because the dimension of the subsystems is 2, leading to at most a $2 \times 2$ coefficient matrix for a bipartite state. ::: ## Algorithms {#algorithms .unnumbered} ::: tcolorbox **Algorithm:** Schmidt Decomposition Algorithm **Input:** A pure bipartite state $\lvert \psi \rangle_{AB} = \sum_{j,k} C_{jk} \lvert j \rangle_A \otimes \lvert k \rangle_B$ in a product basis $\{\lvert j \rangle_A \otimes \lvert k \rangle_B\}$. **Output:** Schmidt decomposition $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$, Schmidt coefficients $\lambda_i$, and Schmidt rank $r$. **Steps:** 1. Construct the coefficient matrix $C$ with entries $C_{jk}$. 2. Compute the Singular Value Decomposition (SVD) of $C = U S V^\dagger$, where $U$ and $V$ are unitary matrices, and $S$ is a diagonal matrix with singular values $s_1, s_2, \dots, s_r, 0, \dots, 0$ on the diagonal. 3. The Schmidt coefficients are $\lambda_i = s_i^2$. 4. The Schmidt rank $r$ is the number of non-zero singular values (or Schmidt coefficients). 5. The orthonormal basis $\{\lvert i \rangle_A\}$ for subsystem A is given by the columns of $U$. 6. The orthonormal basis $\{\lvert i \rangle_B\}$ for subsystem B is given by the columns of $V$. 7. The Schmidt decomposition is then $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} s_i \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$, where $\lvert u_i \rangle_A$ are columns of $U$ and $\lvert v_i \rangle_B$ are columns of $V$. Using $\sqrt{\lambda_i} = s_i$, we get $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert u_i \rangle_A \otimes \lvert v_i \rangle_B$. **Note:** In the main text, for simplicity of notation, we used the same index for bases in both subsystems in the Schmidt decomposition $\lvert \psi \rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i} \lvert i \rangle_A \otimes \lvert i \rangle_B$. In general, the bases can be different, as shown in the algorithm output using $U$ and $V$ matrices from SVD. For square coefficient matrices and when using SVD directly to find Schmidt decomposition, using different bases $\{\lvert u_i \rangle_A\}$ and $\{\lvert v_i \rangle_B\}$ from SVD is more accurate. For the purpose of entanglement detection via Schmidt rank, the rank of the coefficient matrix $C$ is sufficient. :::