Lecture Notes on Advanced Quantum Mechanics: Bell’s Theorem, Zirson Inequality, and Density Matrix
Introduction
This lecture concludes our exploration of quantum mechanics, focusing on advanced topics such as Bell’s Theorem, Zirson Inequality, and the Density Matrix. We will delve into the implications of quantum entanglement, the limits of classical descriptions of quantum phenomena, and the formalism necessary to describe quantum systems in realistic, noisy environments. The lecture will culminate in exercises applying the density matrix formalism to understand the effects of noise on quantum states.
Bell’s Theorem and Bell Inequalities
CHSH Inequality
To test local realism experimentally, Bell’s Theorem and specifically the Clauser-Horne-Shimony-Holt (CHSH) inequality are crucial. Consider Alice and Bob, each receiving one particle from an entangled pair. Alice chooses to measure property \(Q\) or \(R\), and Bob chooses to measure \(S\) or \(T\). Measurement outcomes are binary, \(\pm 1\).
Define the Bell quantity \(B\): \[B = QS + RS + RT - QT = Q(S-T) + R(S+T)\] Under local realism, for any given hidden variable state, the outcomes \(Q, R, S, T\) are predetermined. Since \(S, T \in \{\pm 1\}\), either \((S-T) = 0\) or \((S+T) = 0\), while the other term is \(\pm 2\). As \(Q, R \in \{\pm 1\}\), \(B\) is always \(\pm 2\).
Therefore, the expectation value of \(B\), \(\langle B \rangle\), under local realism, must satisfy the CHSH inequality:
Theorem 1 (Bell-CHSH Inequality). The Bell-CHSH inequality sets an upper bound on the expectation value of the Bell operator under the assumptions of local realism. \[|\langle B \rangle| \leq 2\label{eq:bell_inequality}\]
This inequality is a direct consequence of local realism and hidden variable assumptions.
Quantum Mechanical Violation
Quantum mechanics predicts violations of the Bell-CHSH inequality for certain entangled states and measurements, challenging local realism.
Consider the Bell state: \[\left| \psi \right\rangle = \frac{1}{\sqrt{2}} (\left| 01 \right\rangle - \left| 10 \right\rangle)\label{eq:bell_state}\] Let measurement operators be:
\(Q = \sigma_z^{(1)}\)
\(R = \sigma_x^{(1)}\)
\(S = -\sigma_z^{(2)}\)
\(T = \frac{-\sigma_z^{(2)} - \sigma_x^{(2)}}{\sqrt{2}}\)
Quantum mechanics predicts an expectation value \(|\langle B \rangle| > 2\) for these choices, violating the inequality [eq:bell_inequality]. For instance, the expectation value of \(QS = -\sigma_z^{(1)} \otimes \sigma_z^{(2)}\) in the state \(\left| \psi \right\rangle\) is: \[\begin{aligned}\langle QS \rangle &= \left\langle \psi \right| (-\sigma_z^{(1)} \otimes \sigma_z^{(2)}) \left| \psi \right\rangle \\&= \frac{1}{2} (\left\langle 01 \right| - \left\langle 10 \right|) (-\sigma_z^{(1)} \otimes \sigma_z^{(2)}) (\left| 01 \right\rangle - \left| 10 \right\rangle) \\&= 1\end{aligned}\] Similar calculations for \(\langle RS \rangle\), \(\langle RT \rangle\), and \(\langle QT \rangle\) (not explicitly computed here, but derivable analogously) would show that quantum mechanics predicts a value for \(\langle B \rangle\) that exceeds the bound of 2.
Experimental Validation and Implications
Numerous experiments have confirmed the violation of Bell inequalities, validating quantum mechanical predictions and refuting local realism. These results indicate that our universe operates in a way fundamentally incompatible with classical intuitions of locality and realism when describing quantum phenomena.
The experimental violations of Bell inequalities and the theoretical framework of quantum mechanics strongly suggest that quantum correlations are non-local. This non-locality, a cornerstone of quantum entanglement, is not merely an abstract concept but a measurable feature of quantum systems with profound implications for quantum technologies like quantum computing and quantum communication.
Note: The transcript’s mention of \(\langle QS \rangle\) being zero or related to \(1/\sqrt{2}\) appears to be inconsistent with standard quantum mechanical calculations and may represent an error or misinterpretation in the transcription. The key takeaway is the experimental and theoretical demonstration of Bell inequality violation, which is robustly established regardless of minor discrepancies in transcribed numerical examples.
Zirson Inequality and Maximum Quantum Correlation
Beyond Bell Inequalities: Maximum Quantum Correlations
While Bell inequalities, such as CHSH, define the limits of correlations under local realism, the Zirson inequality (Tsirelson’s bound) addresses a different question: What are the maximum correlations achievable within the framework of quantum mechanics itself, irrespective of local realism constraints? This bound explores the extremity of quantum correlations, pushing beyond the classical limits set by Bell inequalities.
The Zirson Bound
The Zirson inequality quantifies the maximum possible violation of Bell-type inequalities allowed by quantum mechanics. For the CHSH scenario, it determines the upper limit for the expectation value of the Bell operator \(B = QS + RS + RT - QT\) when \(Q, R, S, T\) are quantum mechanical observables. The question shifts from "Can we exceed the classical bound of 2?" to "How far beyond 2 can we go within quantum mechanics?".
Mathematical Origin of the Bound
The derivation of the Zirson bound, as hinted in the transcript, involves examining the square of the Bell operator \(B\). By analyzing \(B^2 = (QS + RS + RT - QT)^2\) and leveraging properties of quantum operators, particularly commutators, one can find the quantum mechanical limit.
Specifically, the transcript mentions the relation to commutators like \([Q,R] = QR - RQ\) and \([S,T] = ST - TS\). A detailed derivation (beyond the scope of this lecture note but found in advanced texts on quantum information theory) reveals that the expectation value of \(B^2\) is bounded in quantum mechanics.
As indicated in the transcript, it can be shown that the expectation value of \(B^2\) is bounded by 8: \[\langle B^2 \rangle \leq 8\] From this bound on the square, we can deduce the bound on the expectation value of \(B\) itself:
Theorem 2 (Zirson (Tsirelson’s) Bound for CHSH). The Zirson bound defines the maximum quantum mechanical violation of the CHSH inequality. \[|\langle B \rangle| \leq 2\sqrt{2} \approx 2.828\label{eq:zirson_inequality}\]
This Zirson bound, \(2\sqrt{2}\), is the maximum quantum mechanical violation of the CHSH inequality. It is notably greater than the classical limit of 2, but still finite, indicating that quantum correlations, while non-classical, are not limitless.
Implications of the Zirson Bound
The Zirson inequality is fundamentally important because it delineates the boundary between classical and quantum correlations, and further, it sets a limit on quantum correlations themselves. It tells us that while quantum mechanics allows for correlations that violate Bell inequalities and thus cannot be explained by local realism, these correlations are still constrained. The value \(2\sqrt{2}\) represents the ultimate extent to which quantum mechanics can violate the CHSH inequality.
This bound is not just a theoretical curiosity; it has significant implications for quantum information science. It helps define the power of quantum entanglement as a resource and sets limits on what can be achieved with quantum correlations. For example, in device-independent quantum cryptography, the Zirson bound plays a role in determining the security limits achievable through the exploitation of Bell inequality violations.
In essence, the Zirson inequality provides a deeper understanding of the nature of quantum correlations, showing that they are non-classical yet bounded, and it highlights the unique characteristics of quantum mechanics that enable technologies beyond classical capabilities.
Density Matrix and Quantum Measurement
Beyond Pure States: The Need for Density Matrix
The wave function \(\left| \psi \right\rangle\) provides a complete description for pure states in quantum mechanics, representing systems in a definite quantum state. However, quantum systems are not always in pure states. Mixed states describe statistical ensembles or situations with incomplete knowledge about the system’s state. A mixed state is a probabilistic mixture of pure states, where there is a probability \(p_i\) of the system being in a pure state \(\left| \psi_i \right\rangle\). The wave function formalism is insufficient to directly represent mixed states.
Density Matrix: Definition and Properties
The density matrix \(\rho\) is a more general operator that can describe both pure and mixed quantum states. For a mixed state composed of pure states \(\left| \psi_i \right\rangle\) with probabilities \(p_i\), the density matrix is defined as:
Definition 1 (Density Matrix Definition). The density matrix is defined as a sum over pure states weighted by their probabilities in a mixed state ensemble. \[\rho = \sum_i p_i \left| \psi_i \middle\rangle \middle\langle \psi_i \right|\label{eq:density_matrix_definition}\]
For a pure state \(\left| \psi \right\rangle\), the density matrix simplifies to \(\rho = \left| \psi \middle\rangle \middle\langle \psi \right|\), which is a special case of [eq:density_matrix_definition] with a single probability \(p_1 = 1\) for \(\left| \psi_1 \right\rangle = \left| \psi \right\rangle\).
The density matrix \(\rho\) has the following key properties:
Hermitian: \(\rho^\dagger = \rho\)
Positive semi-definite: \(\left\langle \phi \right| \rho \left| \phi \right\rangle \geq 0\) for any state \(\left| \phi \right\rangle\)
Trace class: \(\text{Tr}(\rho) = \sum_i p_i = 1\)
The trace condition \(\text{Tr}(\rho) = 1\) ensures that the probabilities are normalized.
Measurement in Density Matrix Formalism
In standard quantum mechanics, projective measurements are described by a set of measurement operators \(\{M_m\}\) that satisfy the completeness relation \(\sum_m M_m^\dagger M_m = I\). For a system in a state \(\left| \psi \right\rangle\), upon measurement, the state collapses to \(\left| \psi' \right\rangle = \frac{M_m \left| \psi \right\rangle}{\sqrt{\left\langle \psi \right| M_m^\dagger M_m \left| \psi \right\rangle}}\) with probability \(P(m) = \left\langle \psi \right| M_m^\dagger M_m \left| \psi \right\rangle\).
In the density matrix formalism, if we perform a measurement described by \(M_m\) on a system with density matrix \(\rho\), and we obtain outcome \(m\), the density matrix of the system after measurement becomes:
\[\rho'_m = \frac{M_m \rho M_m^\dagger}{\text{Tr}(M_m \rho M_m^\dagger)}\label{eq:density_matrix_measurement_outcome}\]
The probability of obtaining the outcome \(m\) is given by:
\[P(m) = \text{Tr}(M_m \rho M_m^\dagger)\label{eq:probability_measurement_outcome}\]
Equation [eq:density_matrix_measurement_outcome] describes the state of the system conditioned on obtaining a specific measurement outcome \(m\), normalized such that the trace remains unity, representing a valid density matrix.
Density Matrix for Open Quantum Systems and Noise
The density matrix formalism is particularly powerful for describing open quantum systems, which interact with their environment. Interactions with the environment lead to phenomena like decoherence and dissipation, effectively introducing noise into the system. The evolution of the density matrix under noise can be described using quantum channels, often represented using Kraus operators or in Lindblad form.
In the context of noise, the evolution of a density matrix is often given in terms of a sum of operations, reflecting different possible interactions with the environment. For example, a simple noise model might describe the density matrix evolving as a probabilistic mixture of the original state and a transformed state, as seen in the exercise section. This approach contrasts with the unitary evolution of closed quantum systems described by the Schrödinger equation, and provides a framework for understanding and mitigating noise in quantum systems, which is crucial for developing practical quantum technologies.
Exercises: Density Matrix and Noise
Example 1: Density Matrix Evolution under Noise
Problem Statement: Bit-Flip Noise Channel
Consider a bit-flip noise channel acting on a single qubit. This channel is described by the evolution: \[\rho' = (1-P) \rho + P \sigma_x \rho \sigma_x\label{eq:bit_flip_channel}\] where \(\rho\) is the initial density matrix, \(\rho'\) is the density matrix after noise, \(P\) is the probability of a bit-flip error (noise probability), and \(\sigma_x\) is the Pauli-X operator representing the bit-flip.
Our task is to analyze the effect of this noise channel on a qubit initially in a pure state \(\left| \psi \right\rangle = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle\). We will determine the density matrix of the qubit after passing through the noise channel and investigate whether the state remains pure or becomes mixed.
Step 1: Initial Density Matrix
For a qubit in the pure state \(\left| \psi \right\rangle = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle\), the initial density matrix is: \[\rho_0 = \left| \psi \middle\rangle \middle\langle \psi \right| = (\alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle)(\alpha^* \left\langle 0 \right| + \beta^* \left\langle 1 \right|) = |\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\label{eq:initial_rho_noise_ex}\]
Step 2: Applying the Noise Channel
Apply the bit-flip noise channel [eq:bit_flip_channel] to the initial density matrix \(\rho_0\): \[\begin{aligned}\rho' &= (1-P) \rho_0 + P \sigma_x \rho_0 \sigma_x \\&= (1-P) \left(|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\right) \\&+ P \sigma_x \left(|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\right) \sigma_x\end{aligned}\] Using the properties of Pauli-X operator: \(\sigma_x \left| 0 \right\rangle = \left| 1 \right\rangle\), \(\sigma_x \left| 1 \right\rangle = \left| 0 \right\rangle\), \(\sigma_x \left\langle 0 \right| = \left\langle 1 \right|\), \(\sigma_x \left\langle 1 \right| = \left\langle 0 \right|\), we have: \[\begin{aligned}\sigma_x \left| 0 \middle\rangle \middle\langle 0 \right| \sigma_x &= \left| 1 \middle\rangle \middle\langle 1 \right| \\\sigma_x \left| 1 \middle\rangle \middle\langle 1 \right| \sigma_x &= \left| 0 \middle\rangle \middle\langle 0 \right| \\\sigma_x \left| 0 \middle\rangle \middle\langle 1 \right| \sigma_x &= \sigma_x \left| 0 \right\rangle \left\langle 1 \right| \sigma_x = \left| 1 \right\rangle \left\langle 0 \right| = \left| 1 \middle\rangle \middle\langle 0 \right| \\\sigma_x \left| 1 \middle\rangle \middle\langle 0 \right| \sigma_x &= \sigma_x \left| 1 \right\rangle \left\langle 0 \right| \sigma_x = \left| 0 \right\rangle \left\langle 1 \right| = \left| 0 \middle\rangle \middle\langle 1 \right|\end{aligned}\] Substituting these back into the equation for \(\rho'\): \[\begin{aligned}\rho' &= (1-P) \left(|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\right) \\&+ P \left(|\alpha|^2 \left| 1 \middle\rangle \middle\langle 1 \right| + \alpha \beta^* \left| 1 \middle\rangle \middle\langle 0 \right| + \beta \alpha^* \left| 0 \middle\rangle \middle\langle 1 \right| + |\beta|^2 \left| 0 \middle\rangle \middle\langle 0 \right|\right) \\&= (1-P)|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + (1-P)|\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right| + P|\alpha|^2 \left| 1 \middle\rangle \middle\langle 1 \right| + P|\beta|^2 \left| 0 \middle\rangle \middle\langle 0 \right| \\&+ (1-P)\alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + P\beta \alpha^* \left| 0 \middle\rangle \middle\langle 1 \right| + (1-P)\beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + P\alpha \beta^* \left| 1 \middle\rangle \middle\langle 0 \right| \\&= \left((1-P)|\alpha|^2 + P|\beta|^2\right) \left| 0 \middle\rangle \middle\langle 0 \right| + \left((1-P)|\beta|^2 + P|\alpha|^2\right) \left| 1 \middle\rangle \middle\langle 1 \right| \\&+ \left((1-P)\alpha \beta^* + P\beta \alpha^*\right) \left| 0 \middle\rangle \middle\langle 1 \right| + \left((1-P)\beta \alpha^* + P\alpha \beta^*\right) \left| 1 \middle\rangle \middle\langle 0 \right|\end{aligned}\]
Step 3: Matrix Representation
In matrix form, the evolved density matrix \(\rho'\) is: \[\rho' = \begin{pmatrix} (1-P)|\alpha|^2 + P|\beta|^2 & (1-P)\alpha \beta^* + P\beta \alpha^* \\ (1-P)\beta \alpha^* + P\alpha \beta^* & (1-P)|\beta|^2 + P|\alpha|^2 \end{pmatrix}\label{eq:rho_prime_noise_ex}\]
Step 4: Purity Analysis
To determine if the state remains pure after noise, we calculate the purity \(\text{Tr}(\rho'^2)\). For a pure state, \(\text{Tr}(\rho'^2) = 1\), and for a mixed state, \(\text{Tr}(\rho'^2) < 1\). \[\begin{aligned}\text{Tr}(\rho'^2) &= \text{Tr}\left( \begin{pmatrix} (1-P)|\alpha|^2 + P|\beta|^2 & (1-P)\alpha \beta^* + P\beta \alpha^* \\ (1-P)\beta \alpha^* + P\alpha \beta^* & (1-P)|\beta|^2 + P|\alpha|^2 \end{pmatrix}^2 \right) \\&= \left((1-P)|\alpha|^2 + P|\beta|^2\right)^2 + \left((1-P)|\beta|^2 + P|\alpha|^2\right)^2 \\&+ 2 \left|(1-P)\alpha \beta^* + P\beta \alpha^*\right|^2\end{aligned}\] For a general state, it is clear that \(\text{Tr}(\rho'^2) \leq 1\). To check for purity, we need to see if \(\text{Tr}(\rho'^2) = 1\). This condition is met if and only if \(P=0\) (no noise) or if the initial state is \(\left| 0 \right\rangle\) or \(\left| 1 \right\rangle\) (i.e., \(|\alpha|^2 |\beta|^2 = 0\)). In general, for \(P>0\) and a superposition state (i.e., \(|\alpha|^2 |\beta|^2 \neq 0\)), the purity \(\text{Tr}(\rho'^2) < 1\), indicating that the noise channel transforms a pure state into a mixed state.
Exercise 2: Tensor Product Operators and Hamiltonian
Problem Statement: Two-Qubit Hamiltonian
Consider a two-qubit Hamiltonian given by: \[H = \frac{1}{2} (\sigma_x \otimes \sigma_x + \sigma_y \otimes \sigma_y + \sigma_z \otimes \sigma_z + I \otimes I)\label{eq:two_qubit_hamiltonian}\] where \(\sigma_x, \sigma_y, \sigma_z\) are Pauli matrices and \(I\) is the identity matrix.
Our tasks are to:
Construct the \(4 \times 4\) matrix representation of \(H\).
Calculate \(H^2\).
Find the eigenvalues and eigenvectors of \(H\).
Determine the time evolution operator \(U(t) = e^{-iHt}\).
Step 1: Matrix Representation of H
Using the Kronecker product and the matrix representations of Pauli matrices: \(\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), \(\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\), \(\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\), \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).
\[\begin{aligned}\sigma_x \otimes \sigma_x &= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}, \quad\sigma_y \otimes \sigma_y = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \\\sigma_z \otimes \sigma_z &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \quadI \otimes I = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\end{aligned}\] Adding these matrices and multiplying by \(\frac{1}{2}\): \[H = \frac{1}{2} \left(\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} +\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\right) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\label{eq:matrix_H_ex2}\]
Step 2: Calculating \(H^2\)
\[H^2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = I\label{eq:H_squared_ex2}\] Thus, \(H^2 = I\), the identity matrix.
Step 3: Eigenvalues and Eigenvectors
From the matrix form of \(H\) [eq:matrix_H_ex2], we can find the eigenvalues. The characteristic polynomial is given by: \[\begin{aligned}\det(H - \lambda I) &= \det \begin{pmatrix} 1-\lambda & 0 & 0 & 0 \\ 0 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 0 \\ 0 & 0 & 0 & 1-\lambda \end{pmatrix} \\&= (1-\lambda) \det \begin{pmatrix} -\lambda & 1 & 0 \\ 1 & -\lambda & 0 \\ 0 & 0 & 1-\lambda \end{pmatrix} \\&= (1-\lambda)^2 \det \begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} \\&= (1-\lambda)^2 (\lambda^2 - 1) = (1-\lambda)^3 (1+\lambda)\end{aligned}\] The eigenvalues are \(\lambda_1 = 1\) (with multiplicity 3) and \(\lambda_2 = -1\) (with multiplicity 1).
Eigenvectors:
For \(\lambda = 1\): We need to solve \((H - I)v = 0\). \(\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}\). This gives \(-v_2 + v_3 = 0 \Rightarrow v_2 = v_3\). Eigenvectors are of the form \(\begin{pmatrix} v_1 \\ v_2 \\ v_2 \\ v_4 \end{pmatrix}\). We can choose an orthonormal basis: \(\left| v_{1} \right\rangle = \left| 00 \right\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}\), \(\left| v_{2} \right\rangle = \frac{1}{\sqrt{2}}(\left| 01 \right\rangle + \left| 10 \right\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}\), \(\left| v_{3} \right\rangle = \left| 11 \right\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\).
For \(\lambda = -1\): We need to solve \((H + I)v = 0\). \(\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}\). This gives \(v_1 = 0\), \(v_4 = 0\), \(v_2 + v_3 = 0 \Rightarrow v_3 = -v_2\). Eigenvector is of the form \(\begin{pmatrix} 0 \\ v_2 \\ -v_2 \\ 0 \end{pmatrix}\). We can choose \(\left| v_{4} \right\rangle = \frac{1}{\sqrt{2}}(\left| 01 \right\rangle - \left| 10 \right\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix}\).
Step 4: Time Evolution Operator \(U(t) = e^{-iHt}\)
Since \(H^2 = I\), we use the Taylor expansion of \(e^{-iHt} = \sum_{n=0}^\infty \frac{(-iHt)^n}{n!} = \cos(t) I - i \sin(t) H\). \[U(t) = e^{-iHt} = \cos(t) I - i \sin(t) H = \begin{pmatrix} \cos(t) - i \sin(t) & 0 & 0 & 0 \\ 0 & \cos(t) & -i \sin(t) & 0 \\ 0 & -i \sin(t) & \cos(t) & 0 \\ 0 & 0 & 0 & \cos(t) - i \sin(t) \end{pmatrix}\label{eq:time_evo_op_ex2}\] Which simplifies to: \[U(t) = \begin{pmatrix} e^{-it} & 0 & 0 & 0 \\ 0 & \cos(t) & -i \sin(t) & 0 \\ 0 & -i \sin(t) & \cos(t) & 0 \\ 0 & 0 & 0 & e^{-it} \end{pmatrix}\label{eq:time_evo_op_simplified_ex2}\] This time evolution operator describes howthe two-qubit system evolves under the Hamiltonian \(H\).
Conclusion
This lecture explored advanced concepts in quantum mechanics, starting with Bell’s Theorem and the profound implications of Bell inequality violations for local realism. We then discussed the Zirson inequality and the maximum quantum correlations allowed by quantum mechanics. Finally, we introduced the density matrix formalism as a powerful tool for describing quantum states, especially mixed states and systems subject to noise. We worked through examples of density matrix evolution under noise and exercises involving tensor product operators and Hamiltonian evolution.
Remark. Remark 1 (Important Remarks and Key Takeaways).
Bell’s Theorem and Non-Locality: Bell’s Theorem and experimental violations of Bell inequalities demonstrate that quantum mechanics is incompatible with local realism. This implies that quantum correlations are non-local, a concept central to understanding entanglement.
Zirson Inequality and Quantum Limits: The Zirson inequality defines the maximum quantum correlation, setting the limits of how much quantum mechanics can violate classical bounds. This is crucial for understanding the power and limitations of quantum systems.
Density Matrix for MixedStates and Noise: The density matrix is essential for describing mixed states, which are probabilistic mixtures of pure states. It is also a fundamental tool for analyzing open quantum systems and the effects of noise on quantum states, crucial for practical quantum technologies. Its ability to represent noisy and probabilistic quantum states makes it indispensable for realistic quantum system analysis and design.
Tensor Product Operators and Multi-Qubit Systems: Understanding tensor product operators is vital for dealing with multi-qubit systems and their Hamiltonians, as seen in the exercise. This is fundamental for quantum computing and quantum simulations.
Remark. Remark 2 (Follow-up Questions and Topics for Further Study).
How does decoherence, as modeled by noise channels, affect quantum entanglement and quantum computation?
What are other types of noise channels and how do they impact qubit states differently?
How can we use error correction techniques to mitigate the effects of noise in quantum systems?
Explore further applications of density matrices in quantum statistical mechanics and quantum information theory.
Investigate the experimental setups used to test Bell inequalities and the challenges in performing such experiments.
This lecture provides a foundation for understanding advanced topics in quantum mechanics and their relevance to modern quantum technologies. Further exploration into these areas will deepen your understanding of the quantum world and its potential applications. These concepts are not only pivotal for theoretical advancements but also for the ongoing development of quantum technologies that promise to revolutionize computation, communication, and sensing. Mastering these principles is crucial for anyone seeking to contribute to or understand the future landscape of quantum information science and technology.
Examples
Example 1 (Example 1: Density Matrix Evolution under Noise).
Problem Statement: Bit-Flip Noise Channel
Consider a bit-flip noise channel acting on a single qubit. This channel is described by the evolution: \[\rho' = (1-P) \rho + P \sigma_x \rho \sigma_x\] where \(\rho\) is the initial density matrix, \(\rho'\) is the density matrix after noise, \(P\) is the probability of a bit-flip error (noise probability), and \(\sigma_x\) is the Pauli-X operator representing the bit-flip.
Our task is to analyze the effect of this noise channel on a qubit initially in a pure state \(\left| \psi \right\rangle = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle\). We will determine the density matrix of the qubit after passing through the noise channel and investigate whether the state remains pure or becomes mixed.
Step 1: Initial Density Matrix
For a qubit in the pure state \(\left| \psi \right\rangle = \alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle\), the initial density matrix is: \[\rho_0 = \left| \psi \middle\rangle \middle\langle \psi \right| = (\alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle)(\alpha^* \left\langle 0 \right| + \beta^* \left\langle 1 \right|) = |\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\]
Step 2: Applying the Noise Channel
Apply the bit-flip noise channel to the initial density matrix \(\rho_0\): \[\begin{aligned}\rho' &= (1-P) \rho_0 + P \sigma_x \rho_0 \sigma_x \\&= (1-P) \left(|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\right) \\&+ P \sigma_x \left(|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\right) \sigma_x\end{aligned}\] Using the properties of Pauli-X operator: \(\sigma_x \left| 0 \right\rangle = \left| 1 \right\rangle\), \(\sigma_x \left| 1 \right\rangle = \left| 0 \right\rangle\), \(\sigma_x \left\langle 0 \right| = \left\langle 1 \right|\), \(\sigma_x \left\langle 1 \right| = \left\langle 0 \right|\), we have: \[\begin{aligned}\sigma_x \left| 0 \middle\rangle \middle\langle 0 \right| \sigma_x &= \left| 1 \middle\rangle \middle\langle 1 \right| \\\sigma_x \left| 1 \middle\rangle \middle\langle 1 \right| \sigma_x &= \left| 0 \middle\rangle \middle\langle 0 \right| \\\sigma_x \left| 0 \middle\rangle \middle\langle 1 \right| \sigma_x &= \sigma_x \left| 0 \right\rangle \left\langle 1 \right| \sigma_x = \left| 1 \right\rangle \left\langle 0 \right| = \left| 1 \middle\rangle \middle\langle 0 \right| \\\sigma_x \left| 1 \middle\rangle \middle\langle 0 \right| \sigma_x &= \sigma_x \left| 1 \right\rangle \left\langle 0 \right| \sigma_x = \left| 0 \right\rangle \left\langle 1 \right| = \left| 0 \middle\rangle \middle\langle 1 \right|\end{aligned}\] Substituting these back into the equation for \(\rho'\): \[\begin{aligned}\rho' &= (1-P) \left(|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + \alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + \beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + |\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right|\right) \\&+ P \left(|\alpha|^2 \left| 1 \middle\rangle \middle\langle 1 \right| + \alpha \beta^* \left| 1 \middle\rangle \middle\langle 0 \right| + \beta \alpha^* \left| 0 \middle\rangle \middle\langle 1 \right| + |\beta|^2 \left| 0 \middle\rangle \middle\langle 0 \right|\right) \\&= (1-P)|\alpha|^2 \left| 0 \middle\rangle \middle\langle 0 \right| + (1-P)|\beta|^2 \left| 1 \middle\rangle \middle\langle 1 \right| + P|\alpha|^2 \left| 1 \middle\rangle \middle\langle 1 \right| + P|\beta|^2 \left| 0 \middle\rangle \middle\langle 0 \right| \\&+ (1-P)\alpha \beta^* \left| 0 \middle\rangle \middle\langle 1 \right| + P\beta \alpha^* \left| 0 \middle\rangle \middle\langle 1 \right| + (1-P)\beta \alpha^* \left| 1 \middle\rangle \middle\langle 0 \right| + P\alpha \beta^* \left| 1 \middle\rangle \middle\langle 0 \right| \\&= \left((1-P)|\alpha|^2 + P|\beta|^2\right) \left| 0 \middle\rangle \middle\langle 0 \right| + \left((1-P)|\beta|^2 + P|\alpha|^2\right) \left| 1 \middle\rangle \middle\langle 1 \right| \\&+ \left((1-P)\alpha \beta^* + P\beta \alpha^*\right) \left| 0 \middle\rangle \middle\langle 1 \right| + \left((1-P)\beta \alpha^* + P\alpha \beta^*\right) \left| 1 \middle\rangle \middle\langle 0 \right|\end{aligned}\]
Step 3: Matrix Representation
In matrix form, the evolved density matrix \(\rho'\) is: \[\rho' = \begin{pmatrix} (1-P)|\alpha|^2 + P|\beta|^2 & (1-P)\alpha \beta^* + P\beta \alpha^* \\ (1-P)\beta \alpha^* + P\alpha \beta^* & (1-P)|\beta|^2 + P|\alpha|^2 \end{pmatrix}\]
Step 4: Purity Analysis
To determine if the state remains pure after noise, we calculate the purity \(\text{Tr}(\rho'^2)\). For a pure state, \(\text{Tr}(\rho'^2) = 1\), and for a mixed state, \(\text{Tr}(\rho'^2) < 1\). \[\begin{aligned}\text{Tr}(\rho'^2) &= \text{Tr}\left( \begin{pmatrix} (1-P)|\alpha|^2 + P|\beta|^2 & (1-P)\alpha \beta^* + P\beta \alpha^* \\ (1-P)\beta \alpha^* + P\alpha \beta^* & (1-P)|\beta|^2 + P|\alpha|^2 \end{pmatrix}^2 \right) \\&= \left((1-P)|\alpha|^2 + P|\beta|^2\right)^2 + \left((1-P)|\beta|^2 + P|\alpha|^2\right)^2 \\&+ 2 \left|(1-P)\alpha \beta^* + P\beta \alpha^*\right|^2\end{aligned}\] For a general state, it is clear that \(\text{Tr}(\rho'^2) \leq 1\). To check for purity, we need to see if \(\text{Tr}(\rho'^2) = 1\). This condition is met if and only if \(P=0\) (no noise) or if the initial state is \(\left| 0 \right\rangle\) or \(\left| 1 \right\rangle\) (i.e., \(|\alpha|^2 |\beta|^2 = 0\)). In general, for \(P>0\) and a superposition state (i.e., \(|\alpha|^2 |\beta|^2 \neq 0\)), the purity \(\text{Tr}(\rho'^2) < 1\), indicating that the noise channel transforms a pure state into a mixed state.
Exercises
Example 2 (Exercise 2: Tensor Product Operators and Hamiltonian).
Problem Statement: Two-Qubit Hamiltonian
Consider a two-qubit Hamiltonian given by: \[H = \frac{1}{2} (\sigma_x \otimes \sigma_x + \sigma_y \otimes \sigma_y + \sigma_z \otimes \sigma_z + I \otimes I)\] where \(\sigma_x, \sigma_y, \sigma_z\) are Pauli matrices and \(I\) is the identity matrix.
Our tasks are to:
Construct the \(4 \times 4\) matrix representation of \(H\).
Calculate \(H^2\).
Find the eigenvalues and eigenvectors of \(H\).
Determine the time evolution operator \(U(t) = e^{-iHt}\).
Step 1: Matrix Representation of H
Using the Kronecker product and the matrix representations of Pauli matrices: \(\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), \(\sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\), \(\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\), \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).
\[\begin{aligned}\sigma_x \otimes \sigma_x &= \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}, \quad\sigma_y \otimes \sigma_y = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \\\sigma_z \otimes \sigma_z &= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}, \quadI \otimes I = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\end{aligned}\] Adding these matrices and multiplying by \(\frac{1}{2}\): \[H = \frac{1}{2} \left(\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} +\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} +\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\right) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\]
Step 2: Calculating \(H^2\)
\[H^2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = I\] Thus, \(H^2 = I\), the identity matrix.
Step 3: Eigenvalues and Eigenvectors
From the matrix form of \(H\), we can find the eigenvalues. The characteristic polynomial is given by: \[\begin{aligned}\det(H - \lambda I) &= \det \begin{pmatrix} 1-\lambda & 0 & 0 & 0 \\ 0 & -\lambda & 1 & 0 \\ 0 & 1 & -\lambda & 0 \\ 0 & 0 & 0 & 1-\lambda \end{pmatrix} \\&= (1-\lambda) \det \begin{pmatrix} -\lambda & 1 & 0 \\ 1 & -\lambda & 0 \\ 0 & 0 & 1-\lambda \end{pmatrix} \\&= (1-\lambda)^2 \det \begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} \\&= (1-\lambda)^2 (\lambda^2 - 1) = (1-\lambda)^3 (1+\lambda)\end{aligned}\] The eigenvalues are \(\lambda_1 = 1\) (with multiplicity 3) and \(\lambda_2 = -1\) (with multiplicity 1).
Eigenvectors:
For \(\lambda = 1\): We need to solve \((H - I)v = 0\). \(\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}\). This gives \(-v_2 + v_3 = 0 \Rightarrow v_2 = v_3\). Eigenvectors are of the form \(\begin{pmatrix} v_1 \\ v_2 \\ v_2 \\ v_4 \end{pmatrix}\). We can choose an orthonormal basis: \(\left| v_{1} \right\rangle = \left| 00 \right\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}\), \(\left| v_{2} \right\rangle = \frac{1}{\sqrt{2}}(\left| 01 \right\rangle + \left| 10 \right\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}\), \(\left| v_{3} \right\rangle = \left| 11 \right\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}\).
For \(\lambda = -1\): We need to solve \((H + I)v = 0\). \(\begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}\). This gives \(v_1 = 0\), \(v_4 = 0\), \(v_2 + v_3 = 0 \Rightarrow v_3 = -v_2\). Eigenvector is of the form \(\begin{pmatrix} 0 \\ v_2 \\ -v_2 \\ 0 \end{pmatrix}\). We can choose \(\left| v_{4} \right\rangle = \frac{1}{\sqrt{2}}(\left| 01 \right\rangle - \left| 10 \right\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix}\).
Step 4: Time Evolution Operator \(U(t) = e^{-iHt}\)
Since \(H^2 = I\), we use the Taylor expansion of \(e^{-iHt} = \sum_{n=0}^\infty \frac{(-iHt)^n}{n!} = \cos(t) I - i \sin(t) H\). \[U(t) = e^{-iHt} = \cos(t) I - i \sin(t) H = \begin{pmatrix} \cos(t) - i \sin(t) & 0 & 0 & 0 \\ 0 & \cos(t) & -i \sin(t) & 0 \\ 0 & -i \sin(t) & \cos(t) & 0 \\ 0 & 0 & 0 & \cos(t) - i \sin(t) \end{pmatrix}\] Which simplifies to: \[U(t) = \begin{pmatrix} e^{-it} & 0 & 0 & 0 \\ 0 & \cos(t) & -i \sin(t) & 0 \\ 0 & -i \sin(t) & \cos(t) & 0 \\ 0 & 0 & 0 & e^{-it} \end{pmatrix}\] This time evolution operator describes how the two-qubit system evolves under the Hamiltonian \(H\).
Remark. Remark 3 (Note:). The transcript’s mention of \(\langle QS \rangle\) being zero or related to \(1/\sqrt{2}\) appears to be inconsistent with standard quantum mechanical calculations and may represent an error or misinterpretation in the transcription. The key takeaway is the experimental and theoretical demonstration of Bell inequality violation, which is robustly established regardless of minor discrepancies in transcribed numerical examples.