In quantum statistical mechanics, the density matrix (or density operator) provides a complete description of a system when the exact quantum state is not known. It generalizes the concept of a wavefunction to describe statistical mixtures of quantum states.

The density matrix allows us to treat both pure states and mixed states in a unified formalism.

Pure and Mixed States

A quantum system is in a pure state if it is described by a single wavefunction

$ \psi\rangle $

$ \rho = |\psi\rangle \langle \psi| $

A mixed state occurs when the system is in state

$ |\psi_i\rangle $ with probability $ p_i $.

$ \rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|$

Pure state → complete information
Mixed state → statistical uncertainty

Properties of Density Matrix

Property	Mathematical Condition
Hermitian	$\rho^\dagger = \rho$
Unit Trace	$\text{Tr}(\rho) = 1$
Positive Definite	$\langle \psi\|\rho\|\psi\rangle \ge 0$
Pure State Condition	$\rho^2 = \rho$

Expectation Value of an Observable

The expectation value of an observable $A$ is given by:

$\langle A \rangle = \text{Tr}(\rho A)$

This formula replaces $ \langle \psi | A | \psi \rangle $ when dealing with mixed states.

Density Matrix in Energy Basis

In thermal equilibrium, the density operator is:

$ \rho = \dfrac{e^{-\beta H}}{Z}$

where:

$ H $ = Hamiltonian
$ \beta = \frac{1}{k_B T} $
$ Z = \text{Tr}(e^{-\beta H}) $ (partition function)

The canonical density operator connects quantum mechanics with statistical mechanics.

Time Evolution of Density Matrix

The time evolution is governed by the Liouville–von Neumann equation:

$i\hbar \frac{d\rho}{dt} = [H,\rho]$

For isolated systems → unitary evolution
For open systems → decoherence may occur

Reduced Density Matrix

For a composite system $A + B$, the reduced density matrix of subsystem A is:

$\rho_A = \text{Tr}_B (\rho_{AB})$

This formalism is essential in:

Quantum entanglement
Open quantum systems
Quantum information theory

Von Neumann Entropy

Quantum entropy is defined as:

$S = -k_B \text{Tr}(\rho \ln \rho)$

Pure state → $S = 0$
Mixed state → $S > 0$

Summary

✔ Density matrix describes quantum statistical states
✔ Works for both pure and mixed states
✔ Expectation values given by Tr(ρA)
✔ Connects quantum mechanics with thermodynamics
✔ Fundamental in quantum information theory

1. What is a Density Matrix?

In quantum statistical mechanics, the density matrix (or density operator) describes the statistical state of a quantum system. It generalizes the concept of a state vector to include mixed states.

2. Importance in Statistical Mechanics

The density matrix is crucial for describing quantum systems in equilibrium or interacting with an environment. It allows us to compute ensemble averages, account for thermal mixtures, and analyze decoherence.

3. Properties of the Density Matrix

Hermitian: $ \rho = \rho^\dagger $
Trace one: $ \mathrm{Tr}(\rho) = 1 $
Positive semi-definite: $ \langle \phi | \rho | \phi \rangle \geq 0 $

4. Application Examples

Example 1: Canonical Ensemble

In thermal equilibrium at temperature $ T $, the density matrix is: \[ \rho = \frac{1}{Z} e^{-\beta H}, \quad \text{where } Z = \mathrm{Tr}(e^{-\beta H}) \]

This allows computation of thermodynamic quantities like: \[ \langle A \rangle = \mathrm{Tr}(\rho A) \]

Example 2: Two-Level System

A two-state system (e.g., spin-1/2) in thermal equilibrium has: \[ \rho = \frac{1}{Z} \]

\begin{pmatrix} e^{-\beta \varepsilon_0} & 0 \\ 0 & e^{-\beta \varepsilon_1} \end{pmatrix},

\[ \quad Z = e^{-\beta \varepsilon_0} + e^{-\beta \varepsilon_1} \]

5. Theorem: Expectation Value from Density Matrix

Theorem: For any observable $ A $, the ensemble average is given by:

\[ \langle A \rangle = \mathrm{Tr}(\rho A) \]

Proof:

If the system is in state $ |\psi_i\rangle $ with probability $ p_i $: \[ \langle A \rangle = \sum_i p_i \langle \psi_i | A | \psi_i \rangle \] But: \[ \mathrm{Tr}(\rho A) = \mathrm{Tr}\left(\sum_i p_i |\psi_i\rangle\langle\psi_i| A\right) = \sum_i p_i \langle \psi_i | A | \psi_i \rangle = \langle A \rangle \]

6. Summary

The density matrix is a powerful tool in quantum statistical mechanics. It unifies treatment of pure and mixed states, encodes thermal properties, and facilitates calculation of measurable quantities through trace operations.

6. Theorem: Density Matrix Commutes with Hamiltonian in Equilibrium

Theorem: In thermal equilibrium, the density matrix $ \rho $ commutes with the Hamiltonian $ H $:

\[ [\rho, H] = 0 \]

Proof:

In the canonical ensemble, \[ \rho = \frac{1}{Z} e^{-\beta H} \] Since $ e^{-\beta H} $ is a function of $ H $, and any operator commutes with functions of itself: \[ [\rho, H] = \left[\frac{1}{Z} e^{-\beta H}, H\right] = 0 \] Therefore, $ \rho $ and $ H $ share the same eigenbasis in equilibrium.

7. Summary

The density matrix is a powerful tool in quantum statistical mechanics. It unifies treatment of pure and mixed states, encodes thermal properties, facilitates calculation of measurable quantities, and commutes with the Hamiltonian at equilibrium.

Show that density matrix satisfies a quantum Mechanical analogue of Liouville’s theorem

Density matrices satisfy the von Neumann Equation

\[ i\hbar \frac{\partial }{\partial t}\rho =\left[ H,\rho \right] \] since

\[\begin{align} \rho & =\sum\limits_{j=1}^{n}{{{P}_{J}}}\left\langle {{\psi }_{j}} \right|\left. {{\psi }_{j}} \right\rangle \\ & \Rightarrow \frac{d\rho }{dt}=\sum\limits_{j=1}^{n}{{{P}_{j}}}\left\langle \frac{d{{\psi }_{j}}}{dt} \right|\left. {{\psi }_{j}} \right\rangle +{{P}_{j}}\left\langle {{\psi }_{j}} \right|\left. \frac{d{{\psi }_{j}}}{dt} \right\rangle \\ & =\frac{1}{i\hbar }\sum\limits_{j=1}^{n}{{{P}_{j}}}\left\langle \frac{i\hbar d{{\psi }_{j}}}{dt} \right|\left. {{\psi }_{j}} \right\rangle +{{P}_{j}}\left\langle {{\psi }_{j}} \right|\left. \frac{i\hbar d{{\psi }_{j}}}{dt} \right\rangle \\ & =\frac{1}{i\hbar }\sum\limits_{j=1}^{n}{{{P}_{j}}}\left( H\left\langle {{\psi }_{j}} \right|\left. {{\psi }_{j}} \right\rangle \right)+{{P}_{j}}\left( \left\langle {{\psi }_{j}} \right|\left. {{\psi }_{j}} \right\rangle {{H}^{*}} \right) \\ & =\frac{1}{i\hbar }\sum\limits_{j=1}^{n}{H}\left( {{P}_{J}}\left\langle {{\psi }_{j}} \right|\left. {{\psi }_{j}} \right\rangle \right)-H\left( {{P}_{j}}\left\langle {{\psi }_{j}} \right|\left. {{\psi }_{j}} \right\rangle \right) \\ & =\frac{1}{i\hbar }\sum\limits_{j=1}^{n}{H}\rho -\rho H \\ & =\frac{1}{i\hbar }\left[ H,\rho \right] \\ \end{align}\]

Express Density Matrix of Canonical

\[\rho =\frac{{{e}^{-\beta H}}}{Tr\left( {{e}^{-\beta H}} \right)}\]

If the time derivative of density matrix is zero then the density matrix commutes with Hamiltonian:

Proof : Since When time derivative of density matrix is zero. The value of \[ [H,\rho]=0 \] Thus $$ H\rho-\rho H=0 $$ , Hence $\rho $ commutes with Hamiltonian.

Then the density matrix of this 3 level quantum states is represented by the 3X3 matrix which is hermitian

\[\begin{align} \rho &=\left\langle \psi \right|\left. \psi \right\rangle \\ & =\left( \begin{matrix} {{a}_{1}} \\ {{a}_{2}} \\ {{a}_{3}} \\ \end{matrix} \right)\left( \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ \end{matrix} \right) \\ & =\left( \begin{matrix} a_{1}^{2} & {{a}_{1}}{{a}_{2}} & {{a}_{1}}{{a}_{3}} \\ {{a}_{2}}{{a}_{1}} & a_{2}^{2} & {{a}_{2}}{{a}_{3}} \\ {{a}_{3}}{{a}_{1}} & {{a}_{3}}{{a}_{2}} & a_{3}^{2} \\ \end{matrix} \right) \\ \end{align}\]

Suppose we have a system with half the spins pointing up and half pointing down. Then represent the system as 2x2 density matrix

\[\begin{align} {{\rho }_{mix}} & =\sum\limits_{i}{{{p}_{i}}}\left\langle {{\psi }_{i}} \right.><\left. {{\psi }_{i}} \right\rangle \\ & =\frac{1}{2}\left| +><+ \right\rangle +\frac{1}{2}\left| -><- \right\rangle \\ & =\frac{1}{2}\left( \begin{matrix} 1 \\ 0 \\ \end{matrix} \right)\left( \begin{matrix} 1 & 0 \\ \end{matrix} \right)+\frac{1}{2}\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right)\left( \begin{matrix} 0 & 1 \\ \end{matrix} \right) \\ & =\frac{1}{2}\left( \begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right)+\frac{1}{2}\left( \begin{matrix} 0 & 0 \\ 0 & 1 \\ \end{matrix} \right) \\ & =\frac{1}{2}\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right) \\ & =\frac{1}{2}I \\ \end{align}\]

Find the density matrix of the state

\[\psi \left( \theta \right)=\frac{1}{\sqrt{2}}\left[ \left| \uparrow \right\rangle +{{e}^{i\theta }}\left| \downarrow \right\rangle \right]\] Answer:

\[\begin{align} \rho \left( \theta \right) &=\left| \psi \left( \theta \right) \right\rangle \left| \psi \left( \theta \right) \right\rangle \\ & =\frac{1}{\sqrt{2}}\left[ \left( \begin{matrix} 1 \\ 0 \\ \end{matrix} \right)+{{e}^{i\theta }}\left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right) \right]\frac{1}{\sqrt{2}}\left[ \left( \begin{matrix} 1 & 0 \\ \end{matrix} \right)+{{e}^{-i\theta }}\left( \begin{matrix} 0 & 1 \\ \end{matrix} \right) \right] \\ & =\frac{1}{\sqrt{2}}\left[ \begin{matrix} 1 \\ {{e}^{i\theta }} \\ \end{matrix} \right]\frac{1}{\sqrt{2}}\left[ \begin{matrix} 1 & {{e}^{-i\theta }} \\ \end{matrix} \right] \\ & =\frac{1}{2}\left[ \begin{matrix} 1 & {{e}^{-i\theta }} \\ {{e}^{-i\theta }} & 1 \\ \end{matrix} \right] \\ \end{align}\]

Property	Mathematical Condition
Hermitian	\(\rho^\dagger = \rho\)
Unit Trace	\(\text{Tr}(\rho) = 1\)
Positive Definite	\(\langle \psi\|\rho\|\psi\rangle \ge 0\)
Pure State Condition	\(\rho^2 = \rho\)