Lecture Notes on Density Matrices

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February 5, 2025

Introduction

This lecture introduces the density matrix, also known as the density operator, as a fundamental tool for describing quantum systems when complete knowledge of their state is unavailable. This situation is common in quantum mechanics, arising from various sources such as environmental noise, entanglement with unobserved systems, or when dealing with statistical ensembles of quantum systems, as in statistical quantum mechanics. In such scenarios, the density matrix provides a comprehensive probabilistic description, enabling the calculation of average values for physical observables. Unlike pure state descriptions using wave functions, the density matrix formalism is essential for handling mixed states, which represent statistical mixtures of pure quantum states.

Motivation and Definition of Density Matrix

Need for Density Matrix: Partial Knowledge

In quantum mechanics, the state of a system is typically described by a state vector $\ensuremath{\left|{\psi}\right\rangle}$ in a Hilbert space. This description, known as a pure state, assumes complete knowledge of the system. However, in many physical situations, our knowledge is incomplete, and the system might be in one of several possible states, each with a certain probability. Such states are called mixed states.

Consider a scenario where a quantum system is prepared to be in state $\ensuremath{\left|{0}\right\rangle}$ with a probability of $\frac{1}{2}$ and in state $\ensuremath{\left|{1}\right\rangle}$ with a probability of $\frac{1}{2}$. A common misconception is to describe this situation with a superposition state like $\frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle})$. However, $\frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle})$ represents a pure state, specifically the eigenstate of the $\sigma_x$ operator with eigenvalue $+1$. This pure state is fundamentally different from the statistical mixture of $\ensuremath{\left|{0}\right\rangle}$ and $\ensuremath{\left|{1}\right\rangle}$. The density matrix formalism is necessary to correctly describe such statistical mixtures of quantum states, where we have probabilistic knowledge rather than a definite pure state.

Statistical Definition of the Density Matrix

Definition 1 (Density Matrix for a Statistical Ensemble). For a quantum system that is in a state $\ensuremath{\left|{\psi_i}\right\rangle}$ with probability $p_i$, where $\sum_i p_i = 1$, the density matrix $\rho$ is defined as: \[\rho = \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|}\] Here, $\ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|}$ is the projection operator onto the state $\ensuremath{\left|{\psi_i}\right\rangle}$.

This definition mathematically formalizes the concept of a statistical ensemble of quantum states. Each term in the sum represents a pure state $\ensuremath{\left|{\psi_i}\right\rangle}$ weighted by its classical probability $p_i$. The density matrix $\rho$ is an operator that encapsulates all the statistical information about the ensemble.

Expectation Values using the Density Matrix

To calculate the average value of a quantum observable $A$ for a system described by a density matrix $\rho$, we consider the ensemble average of the expectation values in each possible pure state. If the system is in state $\ensuremath{\left|{\psi_i}\right\rangle}$ with probability $p_i$, the expectation value of $A$ in the state $\ensuremath{\left|{\psi_i}\right\rangle}$ is given by $\ensuremath{\left\langle{\psi_i}\right|} A \ensuremath{\left|{\psi_i}\right\rangle}$. The average expectation value over the entire ensemble is then: \[\langle A \rangle = \sum_i p_i \langle \psi_i | A | \psi_i \rangle\] This ensemble average can be conveniently expressed using the density matrix. By inserting the identity operator $I = \sum_k \ensuremath{\left|{k}\right\rangle}\ensuremath{\left\langle{k}\right|}$, which represents the completeness relation for any orthonormal basis $\{\ensuremath{\left|{k}\right\rangle}\}$, we can rewrite the expression as: \[\begin{aligned}\langle A \rangle &= \sum_i p_i \ensuremath{\left\langle{\psi_i}\right|} A \left( \sum_k \ensuremath{\left|{k}\right\rangle}\ensuremath{\left\langle{k}\right|} \right) \ensuremath{\left|{\psi_i}\right\rangle} \\&= \sum_i p_i \sum_k \ensuremath{\left\langle{\psi_i}\right|} A \ensuremath{\left|{k}\right\rangle} \ensuremath{\left\langle{k}\middle|{\psi_i}\right\rangle} \\&= \sum_k \sum_i p_i \ensuremath{\left\langle{k}\middle|{\psi_i}\right\rangle} \ensuremath{\left\langle{\psi_i}\right|} A \ensuremath{\left|{k}\right\rangle} \\&= \sum_k \ensuremath{\left\langle{k}\right|} \left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) A \ensuremath{\left|{k}\right\rangle} \\&= \sum_k \ensuremath{\left\langle{k}\right|} \rho A \ensuremath{\left|{k}\right\rangle}\end{aligned}\] The final expression $\sum_k \ensuremath{\left\langle{k}\right|} \rho A \ensuremath{\left|{k}\right\rangle}$ is precisely the trace of the operator product $\rho A$.

Theorem 1 (Expectation Value via Density Matrix). The average value of an observable $A$ for a system described by the density matrix $\rho$ is given by the trace of the operator product $\rho A$: \[\langle A \rangle = \text{Tr}(\rho A)\] where $\text{Tr}(\cdot)$ denotes the trace of an operator.

Proof. Proof. The derivation above shows that by inserting the completeness relation and rearranging the summation, the ensemble average of the expectation values $\sum_i p_i \langle \psi_i | A | \psi_i \rangle$ is transformed into $\sum_k \ensuremath{\left\langle{k}\right|} \rho A \ensuremath{\left|{k}\right\rangle}$, which is the definition of the trace of the operator $\rho A$. ◻

This theorem establishes the density matrix as the central object for calculating expectation values when dealing with statistical ensembles or situations of incomplete knowledge in quantum mechanics. It provides a powerful and concise way to handle mixed quantum states.

Properties of Density Matrix

The density matrix possesses several key properties that are essential for its interpretation and application in quantum mechanics. These properties arise directly from its definition and the probabilistic nature of mixed states.

Hermiticity

The density matrix $\rho$ is a Hermitian operator, meaning it is equal to its own Hermitian conjugate ($\rho = \rho^\dagger$). This property is crucial because it ensures that the eigenvalues of $\rho$ are real, which is necessary for their interpretation as probabilities.

Proof. Proof. Starting from the definition of the density matrix $\rho = \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|}$, we take the Hermitian conjugate: \[\begin{aligned}\rho^\dagger &= \left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right)^\dagger \\&= \sum_i p_i^* \left( \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right)^\dagger && \text{ (Linearity and conjugate transpose of scalar)} \\&= \sum_i p_i \left( \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right)^\dagger && \text{ (Probabilities } p_i \text{ are real, } p_i^* = p_i \text{)} \\&= \sum_i p_i \left( \ensuremath{\left\langle{\psi_i}\right|} \right)^\dagger \left( \ensuremath{\left|{\psi_i}\right\rangle} \right)^\dagger && \text{ (Definition of projector)} \\&= \sum_i p_i \ensuremath{\left|{\psi_i}\right\rangle} \ensuremath{\left\langle{\psi_i}\right|} && \text{ (Conjugate transpose of ket and bra)} \\&= \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} && \text{ (Definition of projector)} \\&= \rho && \text{ (Definition of } \rho \text{)}\end{aligned}\] Since $\rho^\dagger = \rho$, the density matrix $\rho$ is Hermitian. ◻

Trace Property

The trace of the density matrix is always equal to unity, i.e., $\text{Tr}(\rho) = 1$. This property is essential for the probabilistic interpretation of the density matrix, as it ensures that the probabilities sum up to one.

Proof. Proof. Using the definition of the density matrix and the linearity of the trace, we have: \[\begin{aligned}\text{Tr}(\rho) &= \text{Tr}\left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) \\&= \sum_i p_i \text{Tr}\left( \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) && \text{ (Linearity of trace)} \\&= \sum_i p_i \ensuremath{\left\langle{\psi_i}\middle|{\psi_i}\right\rangle} && \text{ (Property of trace for projector } \text{Tr}(\ensuremath{\left|{a}\middle\rangle\middle\langle{b}\right|}) = \ensuremath{\left\langle{b}\middle|{a}\right\rangle} \text{)}\end{aligned}\] Assuming that the states $\ensuremath{\left|{\psi_i}\right\rangle}$ are normalized, we have $\ensuremath{\left\langle{\psi_i}\middle|{\psi_i}\right\rangle} = 1$. Therefore, \[\text{Tr}(\rho) = \sum_i p_i \cdot 1 = \sum_i p_i\] By definition, the probabilities $p_i$ must sum to unity, $\sum_i p_i = 1$. Thus, \[\text{Tr}(\rho) = 1\] ◻

Eigenvalues and Probabilistic Interpretation

Because the density matrix $\rho$ is Hermitian, its eigenvalues are real, and its eigenvectors form an orthonormal basis. Let $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$ be the orthonormal eigenvectors of $\rho$, and $\{\lambda_j\}$ be the corresponding eigenvalues, such that $\rho \ensuremath{\left|{\phi_j}\right\rangle} = \lambda_j \ensuremath{\left|{\phi_j}\right\rangle}$. The density matrix can be diagonalized in this eigenbasis: \[\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}\] From the trace property, we know that $\text{Tr}(\rho) = \sum_j \lambda_j = 1$. Furthermore, the eigenvalues $\lambda_j$ are non-negative ($\lambda_j \geq 0$).

Proof. Non-negativity of Eigenvalues. Consider any state $\ensuremath{\left|{\psi}\right\rangle}$. Then, \[\begin{aligned}\ensuremath{\left\langle{\psi}\right|} \rho \ensuremath{\left|{\psi}\right\rangle} &= \ensuremath{\left\langle{\psi}\right|} \left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) \ensuremath{\left|{\psi}\right\rangle} \\&= \sum_i p_i \ensuremath{\left\langle{\psi}\right|} \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \ensuremath{\left|{\psi}\right\rangle} \\&= \sum_i p_i \ensuremath{\left\langle{\psi}\middle|{\psi_i}\right\rangle} \ensuremath{\left\langle{\psi_i}\middle|{\psi}\right\rangle} \\&= \sum_i p_i |\ensuremath{\left\langle{\psi_i}\middle|{\psi}\right\rangle}|^2\end{aligned}\] Since $p_i \geq 0$ and $|\ensuremath{\left\langle{\psi_i}\middle|{\psi}\right\rangle}|^2 \geq 0$, each term in the sum is non-negative. Therefore, $\ensuremath{\left\langle{\psi}\right|} \rho \ensuremath{\left|{\psi}\right\rangle} \geq 0$ for any state $\ensuremath{\left|{\psi}\right\rangle}$, which means $\rho$ is a positive semi-definite operator. Consequently, its eigenvalues must be non-negative, $\lambda_j \geq 0$.

For the eigenvector $\ensuremath{\left|{\phi_k}\right\rangle}$, we have $\ensuremath{\left\langle{\phi_k}\right|} \rho \ensuremath{\left|{\phi_k}\right\rangle} = \ensuremath{\left\langle{\phi_k}\right|} \lambda_k \ensuremath{\left|{\phi_k}\right\rangle} = \lambda_k \ensuremath{\left\langle{\phi_k}\middle|{\phi_k}\right\rangle} = \lambda_k$. Since $\ensuremath{\left\langle{\phi_k}\right|} \rho \ensuremath{\left|{\phi_k}\right\rangle} \geq 0$, it follows that $\lambda_k \geq 0$. ◻

Combining the properties $\lambda_j \geq 0$ and $\sum_j \lambda_j = 1$, we conclude that the eigenvalues $\lambda_j$ of the density matrix can be interpreted as probabilities. In the eigenbasis $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$, the density matrix $\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$ represents a statistical ensemble where the system is in the pure state $\ensuremath{\left|{\phi_j}\right\rangle}$ with probability $\lambda_j$. This provides a crucial link between the mathematical formalism of the density matrix and the probabilistic interpretation of quantum mechanics for mixed states. The condition $0 \leq \lambda_j \leq 1$ is automatically satisfied since $\lambda_j \geq 0$ and $\sum_j \lambda_j = 1$.

Pure and Mixed States

In quantum mechanics, the distinction between pure and mixed states is fundamental. The density matrix formalism provides a clear way to differentiate and characterize these two types of quantum states.

Density Matrix for Pure States

When a quantum system is known to be in a specific state $\ensuremath{\left|{\psi}\right\rangle}$, it is described as being in a pure state. In this case, the density matrix simplifies to the projector onto the state $\ensuremath{\left|{\psi}\right\rangle}$: \[\rho_{pure} = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|}\] For a pure state, there is no statistical uncertainty; the system is definitively in the state $\ensuremath{\left|{\psi}\right\rangle}$. The density matrix for a pure state is thus a rank-one projector.

Condition for Pure States: Idempotency

A key property that distinguishes pure states from mixed states in terms of their density matrices is idempotency. A density matrix $\rho$ represents a pure state if and only if it is idempotent, i.e., $\rho^2 = \rho$.

Theorem 2 (Pure State Condition). A density matrix $\rho$ describes a pure state if and only if $\rho^2 = \rho$.

Proof. Proof. First, we show that if $\rho$ represents a pure state, then $\rho^2 = \rho$. If $\rho = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|}$ is the density matrix of a pure state $\ensuremath{\left|{\psi}\right\rangle}$, then: \[\rho^2 = \rho \cdot \rho = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|} \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|} = \ensuremath{\left|{\psi}\right\rangle} \ensuremath{\left\langle{\psi}\middle|{\psi}\right\rangle} \ensuremath{\left\langle{\psi}\right|}\] Assuming the state $\ensuremath{\left|{\psi}\right\rangle}$ is normalized, $\ensuremath{\left\langle{\psi}\middle|{\psi}\right\rangle} = 1$. Thus, \[\rho^2 = \ensuremath{\left|{\psi}\right\rangle} \cdot 1 \cdot \ensuremath{\left\langle{\psi}\right|} = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|} = \rho\] Therefore, for a pure state, $\rho^2 = \rho$.

Conversely, we prove that if $\rho^2 = \rho$, then $\rho$ represents a pure state. Consider the eigenvalue decomposition of the Hermitian operator $\rho$: \[\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}\] where $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$ are orthonormal eigenvectors and $\{\lambda_j\}$ are the corresponding eigenvalues. Then, \[\rho^2 = \rho \cdot \rho = \left( \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} \right) \left( \sum_k \lambda_k \ensuremath{\left|{\phi_k}\middle\rangle\middle\langle{\phi_k}\right|} \right) = \sum_{j,k} \lambda_j \lambda_k \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} \ensuremath{\left|{\phi_k}\middle\rangle\middle\langle{\phi_k}\right|}\] Using the orthogonality of eigenvectors $\ensuremath{\left\langle{\phi_j}\middle|{\phi_k}\right\rangle} = \delta_{jk}$, we have $\ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} \ensuremath{\left|{\phi_k}\middle\rangle\middle\langle{\phi_k}\right|} = \delta_{jk} \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_k}\right|} = \delta_{jk} \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$. Thus, \[\rho^2 = \sum_j \lambda_j^2 \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}\] If $\rho^2 = \rho$, then we must have: \[\sum_j \lambda_j^2 \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}\] For this equality to hold, the coefficients of each projector $\ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$ must be equal, i.e., $\lambda_j^2 = \lambda_j$ for all $j$. This condition is satisfied if and only if each eigenvalue $\lambda_j$ is either 0 or 1. Since we also know that $\text{Tr}(\rho) = \sum_j \lambda_j = 1$, there must be exactly one eigenvalue equal to 1 and all other eigenvalues must be 0. Let’s assume $\lambda_1 = 1$ and $\lambda_j = 0$ for all $j > 1$. Then, the density matrix becomes: \[\rho = \lambda_1 \ensuremath{\left|{\phi_1}\middle\rangle\middle\langle{\phi_1}\right|} + \sum_{j>1} \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} = 1 \cdot \ensuremath{\left|{\phi_1}\middle\rangle\middle\langle{\phi_1}\right|} + \sum_{j>1} 0 \cdot \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} = \ensuremath{\left|{\phi_1}\middle\rangle\middle\langle{\phi_1}\right|}\] This is the density matrix for the pure state $\ensuremath{\left|{\phi_1}\right\rangle}$. Therefore, if $\rho^2 = \rho$, then $\rho$ represents a pure state. ◻

Mixed States and Interpretation as Ignorance

If a density matrix $\rho$ does not satisfy the condition $\rho^2 = \rho$, i.e., $\rho^2 \neq \rho$, it represents a mixed state. Mixed states arise when we have incomplete statistical knowledge about the quantum system. They are probabilistic mixtures of different pure states. The eigenvalues $\{\lambda_j\}$ in the diagonal representation $\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$ of a mixed state, where more than one $\lambda_j$ is non-zero, can be interpreted as the probabilities of the system being in the corresponding eigenstates $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$. In essence, a mixed state density matrix describes our "ignorance" about the system’s true pure state, representing an ensemble of possible pure states with associated probabilities.

Examples of Density Matrices

To illustrate the concepts of pure and mixed states and their density matrix representations, we consider several examples. These examples will clarify how different statistical mixtures lead to different density matrices and how to distinguish between pure and mixed states.

Mixed State: Statistical Mixture in the Computational Basis

Example 1 (Mixed State: Statistical Mixture in the Computational Basis). Consider a qubit system that is prepared such that it is in the state $\ensuremath{\left|{0}\right\rangle}$ with a probability of $p_0 = \frac{1}{2}$ and in the state $\ensuremath{\left|{1}\right\rangle}$ with a probability of $p_1 = \frac{1}{2}$. These states are defined with respect to the computational basis $\{\ensuremath{\left|{0}\right\rangle}, \ensuremath{\left|{1}\right\rangle}\}$. The density matrix for this statistical ensemble is given by: \[\begin{aligned}\rho &= p_0 \ensuremath{\left|{0}\middle\rangle\middle\langle{0}\right|} + p_1 \ensuremath{\left|{1}\middle\rangle\middle\langle{1}\right|} \\&= \frac{1}{2} \ensuremath{\left|{0}\middle\rangle\middle\langle{0}\right|} + \frac{1}{2} \ensuremath{\left|{1}\middle\rangle\middle\langle{1}\right|} \\&= \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\&= \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} = \frac{1}{2} I\end{aligned}\] where $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix. To determine if this density matrix represents a pure or mixed state, we compute $\rho^2$: \[\rho^2 = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} = \begin{pmatrix} 1/4 & 0 \\ 0 & 1/4 \end{pmatrix} = \frac{1}{4} I\] Since $\rho^2 = \frac{1}{4} I \neq \rho = \frac{1}{2} I$, the condition $\rho^2 = \rho$ is not satisfied. Therefore, this density matrix represents a mixed state. This mixed state arises from our probabilistic uncertainty about whether the system is in state $\ensuremath{\left|{0}\right\rangle}$ or $\ensuremath{\left|{1}\right\rangle}$.

Mixed State: Statistical Mixture in the X-Basis

Example 2 (Mixed State: Statistical Mixture in the X-Basis). Now consider a different statistical mixture. Suppose a qubit system is in the state $\ensuremath{\left|{+}\right\rangle} = \frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle})$ with probability $p_+ = \frac{1}{2}$ and in the state $\ensuremath{\left|{-}\right\rangle} = \frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} - \ensuremath{\left|{1}\right\rangle})$ with probability $p_- = \frac{1}{2}$. Here, $\ensuremath{\left|{+}\right\rangle}$ and $\ensuremath{\left|{-}\right\rangle}$ are eigenstates of the $\sigma_x$ operator, forming the X-basis. The density matrix for this ensemble is: \[\begin{aligned}\rho &= p_+ \ensuremath{\left|{+}\middle\rangle\middle\langle{+}\right|} + p_- \ensuremath{\left|{-}\middle\rangle\middle\langle{-}\right|} \\&= \frac{1}{2} \ensuremath{\left|{+}\middle\rangle\middle\langle{+}\right|} + \frac{1}{2} \ensuremath{\left|{-}\middle\rangle\middle\langle{-}\right|} \\&= \frac{1}{2} \left( \frac{\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle}}{\sqrt{2}} \frac{\ensuremath{\left\langle{0}\right|} + \ensuremath{\left\langle{1}\right|}}{\sqrt{2}} \right) + \frac{1}{2} \left( \frac{\ensuremath{\left|{0}\right\rangle} - \ensuremath{\left|{1}\right\rangle}}{\sqrt{2}} \frac{\ensuremath{\left\langle{0}\right|} - \ensuremath{\left\langle{1}\right|}}{\sqrt{2}} \right) \\&= \frac{1}{4} (\ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{0}\right|} + \ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{1}\right|} + \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{0}\right|} + \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{1}\right|}) + \frac{1}{4} (\ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{0}\right|} - \ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{1}\right|} - \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{0}\right|} + \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{1}\right|}) \\&= \frac{1}{4} (2\ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{0}\right|} + 2\ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{1}\right|}) \\&= \frac{1}{2} \ensuremath{\left|{0}\middle\rangle\middle\langle{0}\right|} + \frac{1}{2} \ensuremath{\left|{1}\middle\rangle\middle\langle{1}\right|} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} = \frac{1}{2} I\end{aligned}\] Remarkably, this density matrix is identical to the one obtained in the previous example, even though the statistical mixtures are defined in terms of different bases ($\{\ensuremath{\left|{0}\right\rangle}, \ensuremath{\left|{1}\right\rangle}\}$ vs. $\{\ensuremath{\left|{+}\right\rangle}, \ensuremath{\left|{-}\right\rangle}\}$). This illustrates a crucial point: different statistical preparations can lead to the same density matrix. In this case, the density matrix $\rho = \frac{1}{2} I$ represents a state of maximal mixture or maximal ignorance. It indicates complete uncertainty about the state of the qubit, regardless of whether we consider measurements in the Z-basis or the X-basis.

Pure State Condition and Bloch Vector Parameterization

A general density matrix for a qubit can be parameterized using the Bloch vector $\mathbf{R} = (X, Y, Z)$, where $|\mathbf{R}| \leq 1$, as: \[\rho = \frac{1}{2} (I + X\sigma_x + Y\sigma_y + Z\sigma_z) = \frac{1}{2} \begin{pmatrix} 1+Z & X-iY \\ X+iY & 1-Z \end{pmatrix}\] Here, $\sigma_x, \sigma_y, \sigma_z$ are the Pauli matrices, and $I$ is the identity matrix. The purity of the state, i.e., whether it is pure or mixed, can be determined by examining the condition $\rho^2 = \rho$ or, equivalently, by considering the magnitude of the Bloch vector $\mathbf{R}$. Let $R^2 = X^2 + Y^2 + Z^2 = |\mathbf{R}|^2$. We compute $\rho^2$: \[\begin{aligned}\rho^2 &= \frac{1}{4} (I + X\sigma_x + Y\sigma_y + Z\sigma_z)^2 \\&= \frac{1}{4} \left[ I^2 + (X\sigma_x)^2 + (Y\sigma_y)^2 + (Z\sigma_z)^2 + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) \right. \\& \qquad \left. + XY(\sigma_x\sigma_y + \sigma_y\sigma_x) + XZ(\sigma_x\sigma_z + \sigma_z\sigma_x) + YZ(\sigma_y\sigma_z + \sigma_z\sigma_y) \right]\end{aligned}\] Using the properties of Pauli matrices: $\sigma_i^2 = I$ and $\sigma_i\sigma_j + \sigma_j\sigma_i = 2\delta_{ij}I$, we simplify the expression: \[\begin{aligned}\rho^2 &= \frac{1}{4} \left[ I + X^2\sigma_x^2 + Y^2\sigma_y^2 + Z^2\sigma_z^2 + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) + 0 + 0 + 0 \right] \\&= \frac{1}{4} \left[ I + X^2I + Y^2I + Z^2I + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) \right] \\&= \frac{1}{4} \left[ (1 + X^2 + Y^2 + Z^2)I + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) \right] \\&= \frac{1}{2} \left[ \frac{1+R^2}{2} I + (X\sigma_x + Y\sigma_y + Z\sigma_z) \right]\end{aligned}\] For $\rho^2 = \rho$, we require: \[\frac{1}{2} \left[ \frac{1+R^2}{2} I + (X\sigma_x + Y\sigma_y + Z\sigma_z) \right] = \frac{1}{2} \left[ I + (X\sigma_x + Y\sigma_y + Z\sigma_z) \right]\] This equality holds if and only if $\frac{1+R^2}{2} = 1$, which simplifies to $1+R^2 = 2$, or $R^2 = 1$. Thus, $\rho^2 = \rho$ if and only if $R^2 = X^2 + Y^2 + Z^2 = 1$.

Remark. Remark 1 (Pure and Mixed States via Bloch Vector). The density matrix $\rho = \frac{1}{2} (I + X\sigma_x + Y\sigma_y + Z\sigma_z)$ represents a pure state if and only if the Bloch vector $\mathbf{R} = (X, Y, Z)$ has a magnitude $|\mathbf{R}| = \sqrt{X^2 + Y^2 + Z^2} = 1$. In this case, the Bloch vector lies on the surface of the Bloch sphere. If $|\mathbf{R}| < 1$, then $\rho$ represents a mixed state, and the Bloch vector is inside the Bloch sphere. The case $|\mathbf{R}| = 0$, corresponding to $\rho = \frac{1}{2}I$, represents the maximally mixed state, located at the center of the Bloch sphere, as seen in [subsection:mixed-state-computational-basis,subsection:mixed-state-x-basis].

Conclusion

This lecture has introduced the density matrix as a crucial tool for describing quantum states, particularly in scenarios where complete state information is unavailable. We have established the statistical definition of the density matrix, explored its fundamental properties—Hermiticity, unit trace, and probabilistic eigenvalues—and delineated the distinction between pure and mixed states through the idempotency condition $\rho^2 = \rho$. Illustrative examples demonstrated density matrices for mixed states in different bases and the pure state condition within the Bloch vector parameterization.

Key insights from this lecture are:

Thedensity matrix is indispensable for representing statistical ensembles of quantum states and situations characterized by partial information.
Expectation values of observables for systems described by a density matrix are calculated using the trace formula: $\langle A \rangle = \text{Tr}(\rho A)$.
Density matrices are Hermitian operators with a trace of one and non-negative eigenvalues summing to unity, interpretable as probabilities.
The condition $\rho^2 = \rho$ serves as a definitive criterion for distinguishing pure states from mixed states.
A single density matrix can arise from multiple distinct statistical mixtures, reflecting an inherent description of our lack of complete knowledge about the system’s preparation.

Further study in this area should consider:

The density matrix provides a robust framework for advancing into more complex areas of quantum mechanics where statistical descriptions are essential.

--- title: "Lecture Notes on Density Matrices" 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 density matrix, also known as the density operator, as a fundamental tool for describing quantum systems when complete knowledge of their state is unavailable. This situation is common in quantum mechanics, arising from various sources such as environmental noise, entanglement with unobserved systems, or when dealing with statistical ensembles of quantum systems, as in statistical quantum mechanics. In such scenarios, the density matrix provides a comprehensive probabilistic description, enabling the calculation of average values for physical observables. Unlike pure state descriptions using wave functions, the density matrix formalism is essential for handling mixed states, which represent statistical mixtures of pure quantum states. # Motivation and Definition of Density Matrix ## Need for Density Matrix: Partial Knowledge In quantum mechanics, the state of a system is typically described by a state vector $\ensuremath{\left|{\psi}\right\rangle}$ in a Hilbert space. This description, known as a pure state, assumes complete knowledge of the system. However, in many physical situations, our knowledge is incomplete, and the system might be in one of several possible states, each with a certain probability. Such states are called mixed states. Consider a scenario where a quantum system is prepared to be in state $\ensuremath{\left|{0}\right\rangle}$ with a probability of $\frac{1}{2}$ and in state $\ensuremath{\left|{1}\right\rangle}$ with a probability of $\frac{1}{2}$. A common misconception is to describe this situation with a superposition state like $\frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle})$. However, $\frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle})$ represents a pure state, specifically the eigenstate of the $\sigma_x$ operator with eigenvalue $+1$. This pure state is fundamentally different from the statistical mixture of $\ensuremath{\left|{0}\right\rangle}$ and $\ensuremath{\left|{1}\right\rangle}$. The density matrix formalism is necessary to correctly describe such statistical mixtures of quantum states, where we have probabilistic knowledge rather than a definite pure state. ## Statistical Definition of the Density Matrix :::: tcolorbox ::: definition **Definition 1** (Density Matrix for a Statistical Ensemble). *For a quantum system that is in a state $\ensuremath{\left|{\psi_i}\right\rangle}$ with probability $p_i$, where $\sum_i p_i = 1$, the density matrix $\rho$ is defined as: $$\rho = \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|}$$ Here, $\ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|}$ is the projection operator onto the state $\ensuremath{\left|{\psi_i}\right\rangle}$.* ::: :::: This definition mathematically formalizes the concept of a statistical ensemble of quantum states. Each term in the sum represents a pure state $\ensuremath{\left|{\psi_i}\right\rangle}$ weighted by its classical probability $p_i$. The density matrix $\rho$ is an operator that encapsulates all the statistical information about the ensemble. ## Expectation Values using the Density Matrix To calculate the average value of a quantum observable $A$ for a system described by a density matrix $\rho$, we consider the ensemble average of the expectation values in each possible pure state. If the system is in state $\ensuremath{\left|{\psi_i}\right\rangle}$ with probability $p_i$, the expectation value of $A$ in the state $\ensuremath{\left|{\psi_i}\right\rangle}$ is given by $\ensuremath{\left\langle{\psi_i}\right|} A \ensuremath{\left|{\psi_i}\right\rangle}$. The average expectation value over the entire ensemble is then: $$\langle A \rangle = \sum_i p_i \langle \psi_i | A | \psi_i \rangle$$ This ensemble average can be conveniently expressed using the density matrix. By inserting the identity operator $I = \sum_k \ensuremath{\left|{k}\right\rangle}\ensuremath{\left\langle{k}\right|}$, which represents the completeness relation for any orthonormal basis $\{\ensuremath{\left|{k}\right\rangle}\}$, we can rewrite the expression as: $$\begin{aligned}\langle A \rangle &= \sum_i p_i \ensuremath{\left\langle{\psi_i}\right|} A \left( \sum_k \ensuremath{\left|{k}\right\rangle}\ensuremath{\left\langle{k}\right|} \right) \ensuremath{\left|{\psi_i}\right\rangle} \\&= \sum_i p_i \sum_k \ensuremath{\left\langle{\psi_i}\right|} A \ensuremath{\left|{k}\right\rangle} \ensuremath{\left\langle{k}\middle|{\psi_i}\right\rangle} \\&= \sum_k \sum_i p_i \ensuremath{\left\langle{k}\middle|{\psi_i}\right\rangle} \ensuremath{\left\langle{\psi_i}\right|} A \ensuremath{\left|{k}\right\rangle} \\&= \sum_k \ensuremath{\left\langle{k}\right|} \left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) A \ensuremath{\left|{k}\right\rangle} \\&= \sum_k \ensuremath{\left\langle{k}\right|} \rho A \ensuremath{\left|{k}\right\rangle}\end{aligned}$$ The final expression $\sum_k \ensuremath{\left\langle{k}\right|} \rho A \ensuremath{\left|{k}\right\rangle}$ is precisely the trace of the operator product $\rho A$. :::: tcolorbox ::: theorem **Theorem 1** (Expectation Value via Density Matrix). *The average value of an observable $A$ for a system described by the density matrix $\rho$ is given by the trace of the operator product $\rho A$: $$\langle A \rangle = \text{Tr}(\rho A)$$ where $\text{Tr}(\cdot)$ denotes the trace of an operator.* ::: :::: ::: proof *Proof.* The derivation above shows that by inserting the completeness relation and rearranging the summation, the ensemble average of the expectation values $\sum_i p_i \langle \psi_i | A | \psi_i \rangle$ is transformed into $\sum_k \ensuremath{\left\langle{k}\right|} \rho A \ensuremath{\left|{k}\right\rangle}$, which is the definition of the trace of the operator $\rho A$. ◻ ::: This theorem establishes the density matrix as the central object for calculating expectation values when dealing with statistical ensembles or situations of incomplete knowledge in quantum mechanics. It provides a powerful and concise way to handle mixed quantum states. # Properties of Density Matrix The density matrix possesses several key properties that are essential for its interpretation and application in quantum mechanics. These properties arise directly from its definition and the probabilistic nature of mixed states. ## Hermiticity The density matrix $\rho$ is a Hermitian operator, meaning it is equal to its own Hermitian conjugate ($\rho = \rho^\dagger$). This property is crucial because it ensures that the eigenvalues of $\rho$ are real, which is necessary for their interpretation as probabilities. ::: proof *Proof.* Starting from the definition of the density matrix $\rho = \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|}$, we take the Hermitian conjugate: $$\begin{aligned}\rho^\dagger &= \left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right)^\dagger \\&= \sum_i p_i^* \left( \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right)^\dagger && \text{ (Linearity and conjugate transpose of scalar)} \\&= \sum_i p_i \left( \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right)^\dagger && \text{ (Probabilities } p_i \text{ are real, } p_i^* = p_i \text{)} \\&= \sum_i p_i \left( \ensuremath{\left\langle{\psi_i}\right|} \right)^\dagger \left( \ensuremath{\left|{\psi_i}\right\rangle} \right)^\dagger && \text{ (Definition of projector)} \\&= \sum_i p_i \ensuremath{\left|{\psi_i}\right\rangle} \ensuremath{\left\langle{\psi_i}\right|} && \text{ (Conjugate transpose of ket and bra)} \\&= \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} && \text{ (Definition of projector)} \\&= \rho && \text{ (Definition of } \rho \text{)}\end{aligned}$$ Since $\rho^\dagger = \rho$, the density matrix $\rho$ is Hermitian. ◻ ::: ## Trace Property The trace of the density matrix is always equal to unity, i.e., $\text{Tr}(\rho) = 1$. This property is essential for the probabilistic interpretation of the density matrix, as it ensures that the probabilities sum up to one. ::: proof *Proof.* Using the definition of the density matrix and the linearity of the trace, we have: $$\begin{aligned}\text{Tr}(\rho) &= \text{Tr}\left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) \\&= \sum_i p_i \text{Tr}\left( \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) && \text{ (Linearity of trace)} \\&= \sum_i p_i \ensuremath{\left\langle{\psi_i}\middle|{\psi_i}\right\rangle} && \text{ (Property of trace for projector } \text{Tr}(\ensuremath{\left|{a}\middle\rangle\middle\langle{b}\right|}) = \ensuremath{\left\langle{b}\middle|{a}\right\rangle} \text{)}\end{aligned}$$ Assuming that the states $\ensuremath{\left|{\psi_i}\right\rangle}$ are normalized, we have $\ensuremath{\left\langle{\psi_i}\middle|{\psi_i}\right\rangle} = 1$. Therefore, $$\text{Tr}(\rho) = \sum_i p_i \cdot 1 = \sum_i p_i$$ By definition, the probabilities $p_i$ must sum to unity, $\sum_i p_i = 1$. Thus, $$\text{Tr}(\rho) = 1$$ ◻ ::: ## Eigenvalues and Probabilistic Interpretation Because the density matrix $\rho$ is Hermitian, its eigenvalues are real, and its eigenvectors form an orthonormal basis. Let $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$ be the orthonormal eigenvectors of $\rho$, and $\{\lambda_j\}$ be the corresponding eigenvalues, such that $\rho \ensuremath{\left|{\phi_j}\right\rangle} = \lambda_j \ensuremath{\left|{\phi_j}\right\rangle}$. The density matrix can be diagonalized in this eigenbasis: $$\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$$ From the trace property, we know that $\text{Tr}(\rho) = \sum_j \lambda_j = 1$. Furthermore, the eigenvalues $\lambda_j$ are non-negative ($\lambda_j \geq 0$). ::: proof *Non-negativity of Eigenvalues.* Consider any state $\ensuremath{\left|{\psi}\right\rangle}$. Then, $$\begin{aligned}\ensuremath{\left\langle{\psi}\right|} \rho \ensuremath{\left|{\psi}\right\rangle} &= \ensuremath{\left\langle{\psi}\right|} \left( \sum_i p_i \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \right) \ensuremath{\left|{\psi}\right\rangle} \\&= \sum_i p_i \ensuremath{\left\langle{\psi}\right|} \ensuremath{\left|{\psi_i}\middle\rangle\middle\langle{\psi_i}\right|} \ensuremath{\left|{\psi}\right\rangle} \\&= \sum_i p_i \ensuremath{\left\langle{\psi}\middle|{\psi_i}\right\rangle} \ensuremath{\left\langle{\psi_i}\middle|{\psi}\right\rangle} \\&= \sum_i p_i |\ensuremath{\left\langle{\psi_i}\middle|{\psi}\right\rangle}|^2\end{aligned}$$ Since $p_i \geq 0$ and $|\ensuremath{\left\langle{\psi_i}\middle|{\psi}\right\rangle}|^2 \geq 0$, each term in the sum is non-negative. Therefore, $\ensuremath{\left\langle{\psi}\right|} \rho \ensuremath{\left|{\psi}\right\rangle} \geq 0$ for any state $\ensuremath{\left|{\psi}\right\rangle}$, which means $\rho$ is a positive semi-definite operator. Consequently, its eigenvalues must be non-negative, $\lambda_j \geq 0$. For the eigenvector $\ensuremath{\left|{\phi_k}\right\rangle}$, we have $\ensuremath{\left\langle{\phi_k}\right|} \rho \ensuremath{\left|{\phi_k}\right\rangle} = \ensuremath{\left\langle{\phi_k}\right|} \lambda_k \ensuremath{\left|{\phi_k}\right\rangle} = \lambda_k \ensuremath{\left\langle{\phi_k}\middle|{\phi_k}\right\rangle} = \lambda_k$. Since $\ensuremath{\left\langle{\phi_k}\right|} \rho \ensuremath{\left|{\phi_k}\right\rangle} \geq 0$, it follows that $\lambda_k \geq 0$. ◻ ::: Combining the properties $\lambda_j \geq 0$ and $\sum_j \lambda_j = 1$, we conclude that the eigenvalues $\lambda_j$ of the density matrix can be interpreted as probabilities. In the eigenbasis $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$, the density matrix $\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$ represents a statistical ensemble where the system is in the pure state $\ensuremath{\left|{\phi_j}\right\rangle}$ with probability $\lambda_j$. This provides a crucial link between the mathematical formalism of the density matrix and the probabilistic interpretation of quantum mechanics for mixed states. The condition $0 \leq \lambda_j \leq 1$ is automatically satisfied since $\lambda_j \geq 0$ and $\sum_j \lambda_j = 1$. # Pure and Mixed States In quantum mechanics, the distinction between pure and mixed states is fundamental. The density matrix formalism provides a clear way to differentiate and characterize these two types of quantum states. ## Density Matrix for Pure States When a quantum system is known to be in a specific state $\ensuremath{\left|{\psi}\right\rangle}$, it is described as being in a pure state. In this case, the density matrix simplifies to the projector onto the state $\ensuremath{\left|{\psi}\right\rangle}$: $$\rho_{pure} = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|}$$ For a pure state, there is no statistical uncertainty; the system is definitively in the state $\ensuremath{\left|{\psi}\right\rangle}$. The density matrix for a pure state is thus a rank-one projector. ## Condition for Pure States: Idempotency A key property that distinguishes pure states from mixed states in terms of their density matrices is idempotency. A density matrix $\rho$ represents a pure state if and only if it is idempotent, i.e., $\rho^2 = \rho$. :::: tcolorbox ::: theorem **Theorem 2** (Pure State Condition). *A density matrix $\rho$ describes a pure state if and only if $\rho^2 = \rho$.* ::: :::: ::: proof *Proof.* First, we show that if $\rho$ represents a pure state, then $\rho^2 = \rho$. If $\rho = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|}$ is the density matrix of a pure state $\ensuremath{\left|{\psi}\right\rangle}$, then: $$\rho^2 = \rho \cdot \rho = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|} \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|} = \ensuremath{\left|{\psi}\right\rangle} \ensuremath{\left\langle{\psi}\middle|{\psi}\right\rangle} \ensuremath{\left\langle{\psi}\right|}$$ Assuming the state $\ensuremath{\left|{\psi}\right\rangle}$ is normalized, $\ensuremath{\left\langle{\psi}\middle|{\psi}\right\rangle} = 1$. Thus, $$\rho^2 = \ensuremath{\left|{\psi}\right\rangle} \cdot 1 \cdot \ensuremath{\left\langle{\psi}\right|} = \ensuremath{\left|{\psi}\middle\rangle\middle\langle{\psi}\right|} = \rho$$ Therefore, for a pure state, $\rho^2 = \rho$. Conversely, we prove that if $\rho^2 = \rho$, then $\rho$ represents a pure state. Consider the eigenvalue decomposition of the Hermitian operator $\rho$: $$\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$$ where $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$ are orthonormal eigenvectors and $\{\lambda_j\}$ are the corresponding eigenvalues. Then, $$\rho^2 = \rho \cdot \rho = \left( \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} \right) \left( \sum_k \lambda_k \ensuremath{\left|{\phi_k}\middle\rangle\middle\langle{\phi_k}\right|} \right) = \sum_{j,k} \lambda_j \lambda_k \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} \ensuremath{\left|{\phi_k}\middle\rangle\middle\langle{\phi_k}\right|}$$ Using the orthogonality of eigenvectors $\ensuremath{\left\langle{\phi_j}\middle|{\phi_k}\right\rangle} = \delta_{jk}$, we have $\ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} \ensuremath{\left|{\phi_k}\middle\rangle\middle\langle{\phi_k}\right|} = \delta_{jk} \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_k}\right|} = \delta_{jk} \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$. Thus, $$\rho^2 = \sum_j \lambda_j^2 \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$$ If $\rho^2 = \rho$, then we must have: $$\sum_j \lambda_j^2 \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$$ For this equality to hold, the coefficients of each projector $\ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$ must be equal, i.e., $\lambda_j^2 = \lambda_j$ for all $j$. This condition is satisfied if and only if each eigenvalue $\lambda_j$ is either 0 or 1. Since we also know that $\text{Tr}(\rho) = \sum_j \lambda_j = 1$, there must be exactly one eigenvalue equal to 1 and all other eigenvalues must be 0. Let's assume $\lambda_1 = 1$ and $\lambda_j = 0$ for all $j > 1$. Then, the density matrix becomes: $$\rho = \lambda_1 \ensuremath{\left|{\phi_1}\middle\rangle\middle\langle{\phi_1}\right|} + \sum_{j>1} \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} = 1 \cdot \ensuremath{\left|{\phi_1}\middle\rangle\middle\langle{\phi_1}\right|} + \sum_{j>1} 0 \cdot \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|} = \ensuremath{\left|{\phi_1}\middle\rangle\middle\langle{\phi_1}\right|}$$ This is the density matrix for the pure state $\ensuremath{\left|{\phi_1}\right\rangle}$. Therefore, if $\rho^2 = \rho$, then $\rho$ represents a pure state. ◻ ::: ## Mixed States and Interpretation as Ignorance If a density matrix $\rho$ does not satisfy the condition $\rho^2 = \rho$, i.e., $\rho^2 \neq \rho$, it represents a mixed state. Mixed states arise when we have incomplete statistical knowledge about the quantum system. They are probabilistic mixtures of different pure states. The eigenvalues $\{\lambda_j\}$ in the diagonal representation $\rho = \sum_j \lambda_j \ensuremath{\left|{\phi_j}\middle\rangle\middle\langle{\phi_j}\right|}$ of a mixed state, where more than one $\lambda_j$ is non-zero, can be interpreted as the probabilities of the system being in the corresponding eigenstates $\{\ensuremath{\left|{\phi_j}\right\rangle}\}$. In essence, a mixed state density matrix describes our \"ignorance\" about the system's true pure state, representing an ensemble of possible pure states with associated probabilities. # Examples of Density Matrices To illustrate the concepts of pure and mixed states and their density matrix representations, we consider several examples. These examples will clarify how different statistical mixtures lead to different density matrices and how to distinguish between pure and mixed states. ## Mixed State: Statistical Mixture in the Computational Basis {#subsection:mixed-state-computational-basis} :::: tcolorbox ::: example **Example 1** (Mixed State: Statistical Mixture in the Computational Basis). *Consider a qubit system that is prepared such that it is in the state $\ensuremath{\left|{0}\right\rangle}$ with a probability of $p_0 = \frac{1}{2}$ and in the state $\ensuremath{\left|{1}\right\rangle}$ with a probability of $p_1 = \frac{1}{2}$. These states are defined with respect to the computational basis $\{\ensuremath{\left|{0}\right\rangle}, \ensuremath{\left|{1}\right\rangle}\}$. The density matrix for this statistical ensemble is given by: $$\begin{aligned}\rho &= p_0 \ensuremath{\left|{0}\middle\rangle\middle\langle{0}\right|} + p_1 \ensuremath{\left|{1}\middle\rangle\middle\langle{1}\right|} \\&= \frac{1}{2} \ensuremath{\left|{0}\middle\rangle\middle\langle{0}\right|} + \frac{1}{2} \ensuremath{\left|{1}\middle\rangle\middle\langle{1}\right|} \\&= \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\&= \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} = \frac{1}{2} I\end{aligned}$$ where $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix. To determine if this density matrix represents a pure or mixed state, we compute $\rho^2$: $$\rho^2 = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} = \begin{pmatrix} 1/4 & 0 \\ 0 & 1/4 \end{pmatrix} = \frac{1}{4} I$$ Since $\rho^2 = \frac{1}{4} I \neq \rho = \frac{1}{2} I$, the condition $\rho^2 = \rho$ is not satisfied. Therefore, this density matrix represents a mixed state. This mixed state arises from our probabilistic uncertainty about whether the system is in state $\ensuremath{\left|{0}\right\rangle}$ or $\ensuremath{\left|{1}\right\rangle}$.* ::: :::: ## Mixed State: Statistical Mixture in the X-Basis {#subsection:mixed-state-x-basis} :::: tcolorbox ::: example **Example 2** (Mixed State: Statistical Mixture in the X-Basis). *Now consider a different statistical mixture. Suppose a qubit system is in the state $\ensuremath{\left|{+}\right\rangle} = \frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle})$ with probability $p_+ = \frac{1}{2}$ and in the state $\ensuremath{\left|{-}\right\rangle} = \frac{1}{\sqrt{2}}(\ensuremath{\left|{0}\right\rangle} - \ensuremath{\left|{1}\right\rangle})$ with probability $p_- = \frac{1}{2}$. Here, $\ensuremath{\left|{+}\right\rangle}$ and $\ensuremath{\left|{-}\right\rangle}$ are eigenstates of the $\sigma_x$ operator, forming the X-basis. The density matrix for this ensemble is: $$\begin{aligned}\rho &= p_+ \ensuremath{\left|{+}\middle\rangle\middle\langle{+}\right|} + p_- \ensuremath{\left|{-}\middle\rangle\middle\langle{-}\right|} \\&= \frac{1}{2} \ensuremath{\left|{+}\middle\rangle\middle\langle{+}\right|} + \frac{1}{2} \ensuremath{\left|{-}\middle\rangle\middle\langle{-}\right|} \\&= \frac{1}{2} \left( \frac{\ensuremath{\left|{0}\right\rangle} + \ensuremath{\left|{1}\right\rangle}}{\sqrt{2}} \frac{\ensuremath{\left\langle{0}\right|} + \ensuremath{\left\langle{1}\right|}}{\sqrt{2}} \right) + \frac{1}{2} \left( \frac{\ensuremath{\left|{0}\right\rangle} - \ensuremath{\left|{1}\right\rangle}}{\sqrt{2}} \frac{\ensuremath{\left\langle{0}\right|} - \ensuremath{\left\langle{1}\right|}}{\sqrt{2}} \right) \\&= \frac{1}{4} (\ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{0}\right|} + \ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{1}\right|} + \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{0}\right|} + \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{1}\right|}) + \frac{1}{4} (\ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{0}\right|} - \ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{1}\right|} - \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{0}\right|} + \ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{1}\right|}) \\&= \frac{1}{4} (2\ensuremath{\left|{0}\right\rangle}\ensuremath{\left\langle{0}\right|} + 2\ensuremath{\left|{1}\right\rangle}\ensuremath{\left\langle{1}\right|}) \\&= \frac{1}{2} \ensuremath{\left|{0}\middle\rangle\middle\langle{0}\right|} + \frac{1}{2} \ensuremath{\left|{1}\middle\rangle\middle\langle{1}\right|} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} = \frac{1}{2} I\end{aligned}$$ Remarkably, this density matrix is identical to the one obtained in the previous example, even though the statistical mixtures are defined in terms of different bases ($\{\ensuremath{\left|{0}\right\rangle}, \ensuremath{\left|{1}\right\rangle}\}$ vs. $\{\ensuremath{\left|{+}\right\rangle}, \ensuremath{\left|{-}\right\rangle}\}$). This illustrates a crucial point: different statistical preparations can lead to the same density matrix. In this case, the density matrix $\rho = \frac{1}{2} I$ represents a state of maximal mixture or maximal ignorance. It indicates complete uncertainty about the state of the qubit, regardless of whether we consider measurements in the Z-basis or the X-basis.* ::: :::: ## Pure State Condition and Bloch Vector Parameterization A general density matrix for a qubit can be parameterized using the Bloch vector $\mathbf{R} = (X, Y, Z)$, where $|\mathbf{R}| \leq 1$, as: $$\rho = \frac{1}{2} (I + X\sigma_x + Y\sigma_y + Z\sigma_z) = \frac{1}{2} \begin{pmatrix} 1+Z & X-iY \\ X+iY & 1-Z \end{pmatrix}$$ Here, $\sigma_x, \sigma_y, \sigma_z$ are the Pauli matrices, and $I$ is the identity matrix. The purity of the state, i.e., whether it is pure or mixed, can be determined by examining the condition $\rho^2 = \rho$ or, equivalently, by considering the magnitude of the Bloch vector $\mathbf{R}$. Let $R^2 = X^2 + Y^2 + Z^2 = |\mathbf{R}|^2$. We compute $\rho^2$: $$\begin{aligned}\rho^2 &= \frac{1}{4} (I + X\sigma_x + Y\sigma_y + Z\sigma_z)^2 \\&= \frac{1}{4} \left[ I^2 + (X\sigma_x)^2 + (Y\sigma_y)^2 + (Z\sigma_z)^2 + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) \right. \\& \qquad \left. + XY(\sigma_x\sigma_y + \sigma_y\sigma_x) + XZ(\sigma_x\sigma_z + \sigma_z\sigma_x) + YZ(\sigma_y\sigma_z + \sigma_z\sigma_y) \right]\end{aligned}$$ Using the properties of Pauli matrices: $\sigma_i^2 = I$ and $\sigma_i\sigma_j + \sigma_j\sigma_i = 2\delta_{ij}I$, we simplify the expression: $$\begin{aligned}\rho^2 &= \frac{1}{4} \left[ I + X^2\sigma_x^2 + Y^2\sigma_y^2 + Z^2\sigma_z^2 + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) + 0 + 0 + 0 \right] \\&= \frac{1}{4} \left[ I + X^2I + Y^2I + Z^2I + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) \right] \\&= \frac{1}{4} \left[ (1 + X^2 + Y^2 + Z^2)I + 2(X\sigma_x + Y\sigma_y + Z\sigma_z) \right] \\&= \frac{1}{2} \left[ \frac{1+R^2}{2} I + (X\sigma_x + Y\sigma_y + Z\sigma_z) \right]\end{aligned}$$ For $\rho^2 = \rho$, we require: $$\frac{1}{2} \left[ \frac{1+R^2}{2} I + (X\sigma_x + Y\sigma_y + Z\sigma_z) \right] = \frac{1}{2} \left[ I + (X\sigma_x + Y\sigma_y + Z\sigma_z) \right]$$ This equality holds if and only if $\frac{1+R^2}{2} = 1$, which simplifies to $1+R^2 = 2$, or $R^2 = 1$. Thus, $\rho^2 = \rho$ if and only if $R^2 = X^2 + Y^2 + Z^2 = 1$. :::: tcolorbox ::: {.remark:bloch-vector-purity .remark} **Remark 1** (Pure and Mixed States via Bloch Vector). *The density matrix $\rho = \frac{1}{2} (I + X\sigma_x + Y\sigma_y + Z\sigma_z)$ represents a pure state if and only if the Bloch vector $\mathbf{R} = (X, Y, Z)$ has a magnitude $|\mathbf{R}| = \sqrt{X^2 + Y^2 + Z^2} = 1$. In this case, the Bloch vector lies on the surface of the Bloch sphere. If $|\mathbf{R}| < 1$, then $\rho$ represents a mixed state, and the Bloch vector is inside the Bloch sphere. The case $|\mathbf{R}| = 0$, corresponding to $\rho = \frac{1}{2}I$, represents the maximally mixed state, located at the center of the Bloch sphere, as seen in [\[subsection:mixed-state-computational-basis,subsection:mixed-state-x-basis\]](#subsection:mixed-state-computational-basis,subsection:mixed-state-x-basis){reference-type="ref+label" reference="subsection:mixed-state-computational-basis,subsection:mixed-state-x-basis"}.* ::: :::: # Conclusion This lecture has introduced the density matrix as a crucial tool for describing quantum states, particularly in scenarios where complete state information is unavailable. We have established the statistical definition of the density matrix, explored its fundamental properties---Hermiticity, unit trace, and probabilistic eigenvalues---and delineated the distinction between pure and mixed states through the idempotency condition $\rho^2 = \rho$. Illustrative examples demonstrated density matrices for mixed states in different bases and the pure state condition within the Bloch vector parameterization. Key insights from this lecture are: - Thedensity matrix is indispensable for representing statistical ensembles of quantum states and situations characterized by partial information. - Expectation values of observables for systems described by a density matrix are calculated using the trace formula: $\langle A \rangle = \text{Tr}(\rho A)$. - Density matrices are Hermitian operators with a trace of one and non-negative eigenvalues summing to unity, interpretable as probabilities. - The condition $\rho^2 = \rho$ serves as a definitive criterion for distinguishing pure states from mixed states. - A single density matrix can arise from multiple distinct statistical mixtures, reflecting an inherent description of our lack of complete knowledge about the system's preparation. Further study in this area should consider: The density matrix provides a robust framework for advancing into more complex areas of quantum mechanics where statistical descriptions are essential.