Problem 29 - A Modern Approach to Quantum Mechanics

Question

(a) Show that the time evolution of the density operator is given by $$ \hat{\rho}(t)=\hat{U}(t) \hat{\rho}(0) \hat{U}^{\dagger}(t) $$ where \(\hat{U}(t)\) is the time-evolution operator, namely $$ \hat{U}(t)|\psi(0)\rangle=|\psi(t)\rangle $$ (b) Suppose that an ensemble of particles is in a pure state at \(t=0\). Show the ensemble cannot evolve into a mixed state as long as time evolution is governed by the Schrödinger equation.

Step-by-Step Solution

Answer

(a) The evolution is given by \( \hat{\rho}(t)=\hat{U}(t) \hat{\rho}(0) \hat{U}^{\dagger}(t) \). (b) Pure states can't evolve into mixed states via unitary evolution.

Step 1: Understand the Time-Evolution Operator

The time-evolution operator \( \hat{U}(t) \) is defined such that it evolves the quantum state \( |\psi(0)\rangle \) at time \( t = 0 \) to \( |\psi(t)\rangle \) at time \( t \). Specifically, \( \hat{U}(t)|\psi(0)\rangle = |\psi(t)\rangle \). The operator \( \hat{U}(t) \) is unitary, meaning \( \hat{U}(t) \hat{U}^{\dagger}(t) = \hat{I} \), where \( \hat{I} \) is the identity operator.

Step 2: Derive the Density Operator at Time t

The density operator at \( t = 0 \) is \( \hat{\rho}(0) = |\psi(0)\rangle \langle \psi(0)| \). To find the density operator at time \( t \), apply the time-evolution operator: \( |\psi(t)\rangle \langle \psi(t)| = \hat{U}(t) |\psi(0)\rangle \langle \psi(0)| \hat{U}^{\dagger}(t) \). Thus, \( \hat{\rho}(t) = \hat{U}(t) \hat{\rho}(0) \hat{U}^{\dagger}(t) \).

Step 3: Confirm the Properties of Unitary Evolution

Since \( \hat{U}(t) \) is unitary, it preserves inner products: \( \langle \psi(t)|\psi(t)\rangle = \langle \psi(0)|\psi(0)\rangle \). This means the trace of the density operator \( \text{Tr}(\hat{\rho}(t)) = \text{Tr}(\hat{\rho}(0)) = 1 \), confirming it stays as a pure state.

Step 4: Analyze Evolution from Pure to Mixed State

A pure state at \( t=0 \) has a density operator \( \hat{\rho}(0) \) such that \( \hat{\rho}(0)^2 = \hat{\rho}(0) \). Since unitary operators preserve eigenvalues, the evolved density operator also satisfies \( \hat{\rho}(t)^2 = \hat{\rho}(t) \), proving it remains pure. The state remains unchanged in terms of being pure.

Key Concepts

Density Operator

Understanding the density operator is essential in quantum mechanics, as it provides a way to describe the statistical state of a quantum system. A density operator, denoted as \( \hat{\rho} \), can represent both pure and mixed states. For a pure state, the density operator is formulated as \( \hat{\rho} = |\psi\rangle \langle \psi| \) where \(|\psi\rangle\) is the wave function, meaning the quantum system is precisely known. For a mixed state, it is described as a statistical mixture of pure states: \[ \hat{\rho} = \sum_i p_i |\psi_i\rangle \langle \psi_i| \] Here, \( p_i \) are probabilities summing to 1, representing the system being in states \(|\psi_i\rangle\).The density operator is more versatile than wave functions and is especially useful in scenarios where information is incomplete, providing a broader understanding of the system's quantum state. With key properties like *trace* (\( \text{Tr}(\hat{\rho}) = 1 \)) and *hermiticity* (ensures real-valued probabilities), it is used extensively in quantum statistical mechanics.

Time Evolution

In quantum mechanics, the concept of time evolution describes how quantum states change over time. This is typically handled using the time-evolution operator \( \hat{U}(t) \). For any initial state \(|\psi(0)\rangle\), the state at a later time \( t \) is given by \( |\psi(t)\rangle = \hat{U}(t)|\psi(0)\rangle \).The time-evolution operator is unitary, meaning it preserves lengths and angles, maintained by the property: \( \hat{U}(t) \hat{U}^{\dagger}(t) = \hat{I} \), where \( \hat{I} \) is the identity operator. This invariance ensures probabilities remain consistent through time, a crucial aspect of physical realism.When applied to the density operator, time evolution maintains the system's statistical properties, evolving as: \[ \hat{\rho}(t) = \hat{U}(t) \hat{\rho}(0) \hat{U}^{\dagger}(t) \] Staying consistent with its statistical nature, the trace condition persists, showing the probabilities continue to add up to one. This guarantees any initial pure state cannot evolve into a mixed state purely by time evolution, maintaining its purity under unitary operations.

Schrödinger Equation

The Schrödinger Equation lies at the heart of quantum mechanics, elucidating how quantum states evolve over time. Known as a fundamental equation, it dictates the dynamics of a quantum system. The time-dependent Schrödinger equation for a quantum state \(|\psi(t)\rangle\) is given by: \[ i\hbar \frac{d}{dt} |\psi(t)\rangle = \hat{H}|\psi(t)\rangle \] where \( \hat{H} \) is the Hamiltonian operator signifying the total energy of the system, and \( i\hbar \) reflects the imaginary unit and reduced Planck constant, respectively.This equation isn't just a means to calculate the future state but reveals conservation properties intrinsic to quantum systems, like energy conservation. By solving it, one derives the time-evolution operator \( \hat{U}(t) \), showing the equation's foundational role in understanding time evolution in quantum mechanics.A key takeaway is that the Schrödinger equation preserves the probabilistic nature of quantum states. It guarantees that pure states remain pure, enabling us to predict the behavior of a quantum system accurately, as shown in the exercise where unitary evolution prevents a pure state from transforming into a mixed one. This underlines the deterministic yet probabilistic framework within which quantum systems operate.