Part I — The Need for Density Matrices
In quantum mechanics, a pure state is described by a single state vector $|\psi\rangle$ — it represents maximal information about the system. But in practice, we often have incomplete information. A mixed state arises when we have a statistical ensemble of pure states: we know the system is in one of several states $|\psi_i\rangle$, each with classical probability $p_i$, but we don't know which one.
There are two fundamentally distinct reasons a system may be in a mixed state:
- Classical ignorance: imperfect preparation — the lab equipment doesn't reliably put every electron into the same state.
- Quantum entanglement: the system is entangled with another system (say, Bob's) whose state we don't control or measure.
The density matrix $\rho$ captures both types of ignorance in a single unified formalism. For a pure state $|\psi\rangle$, the density matrix is defined as the outer product:
This is a rank-1 matrix — a projector onto the state $|\psi\rangle$. For a classical mixture with probabilities $p_i$ and pure states $|\psi_i\rangle$, the density matrix is the convex combination:
Part II — Properties of the Density Matrix
The density matrix satisfies three fundamental properties that follow directly from its definition:
| Property | Statement | Why? |
|---|---|---|
| Hermitian | $\rho = \rho^\dagger$ | Each $|\psi_i\rangle\langle\psi_i|$ is Hermitian, sum of Hermitians is Hermitian |
| Positive semi-definite | $\langle\phi|\rho|\phi\rangle \geq 0$ for all $|\phi\rangle$ | Follows from $p_i \geq 0$ and $|\langle\phi|\psi_i\rangle|^2 \geq 0$ |
| Unit trace | $\mathrm{Tr}(\rho) = 1$ | Probabilities sum to 1 |
Beyond these, the density matrix carries a powerful diagnostic for purity. For a pure state, $\rho^2 = \rho$ (it is idempotent), so $\mathrm{Tr}(\rho^2) = 1$. For any mixed state, $\mathrm{Tr}(\rho^2) < 1$. The maximally mixed state for an $N$-dimensional system is $\rho = I/N$ — it encodes zero information about the state.
The expectation value of any observable $A$ is computed simply as:
Part III — Partial Trace and Reduced Density Matrix
Consider a composite system $AB$ with joint state $\rho_{AB}$. Alice has access only to subsystem $A$. Her reduced density matrix is obtained by tracing out Bob's degrees of freedom:
The partial trace is defined on basis elements by:
This is the only operation that correctly describes Alice's local measurements when she has no access to Bob's system. Consider the singlet state $|\mathrm{sing}\rangle = \tfrac{1}{\sqrt{2}}(|ud\rangle - |du\rangle)$. Its density matrix is:
Taking the partial trace over $B$:
Alice's reduced state is the maximally mixed state — even though the composite state $\rho_{AB}$ is pure! This is one of the most striking features of quantum entanglement: a subsystem of a pure entangled state can have maximal uncertainty.
Part IV — Entanglement Entropy
How much entanglement does a bipartite pure state contain? The answer is quantified by the von Neumann entropy:
where $\lambda_i$ are the eigenvalues of $\rho$ and we use the convention $0 \cdot \log 0 = 0$. Key benchmark values:
- Pure state: eigenvalues are $\{1, 0, 0, \ldots\}$, so $S = -1\cdot\log 1 = 0$.
- Maximally mixed state: $\rho = I/N$ has all eigenvalues $1/N$, giving $S(I/N) = \log N$.
For a bipartite pure state $|\psi\rangle_{AB}$, the entanglement entropy of subsystem $A$ is:
For the singlet: $\rho_A = I/2$ has eigenvalues $1/2, 1/2$, giving:
For a product state $|\psi\rangle = |\alpha\rangle_A \otimes |\beta\rangle_B$: $\rho_A$ is a pure state, so $S_E = 0$ — no entanglement.
Part V — The Measurement Chain (Von Neumann)
Quantum mechanics raises a profound question: where does the collapse happen? Von Neumann analyzed the measurement process as a chain of interactions:
At each stage, the quantum system entangles with the next object in the chain. From the perspective of an outside observer, the entire chain is in a superposition — like Schrödinger's cat. The remarkable result is that the physical predictions are independent of where you "cut" the chain. The reduced density matrix of the measured system, at any cut point, becomes diagonal in the eigenbasis of the measured observable — the off-diagonal coherences vanish.
This is decoherence: from the perspective of a local observer, quantum interference terms (off-diagonal elements) disappear because they are encoded in correlations with an inaccessible environment. The system appears to be in a classical mixture, even though the global state remains pure.
Part VI — Pure States From Entanglement
We encounter an apparent paradox: the joint state of Alice and Bob is pure (zero entropy), yet Alice's reduced state is maximally mixed (maximum entropy). How is this possible?
This is only possible through entanglement. Classical correlations cannot increase local entropy while keeping global entropy zero. The key theorem is:
Theorem: For a pure bipartite state $|\psi\rangle_{AB}$, the subsystems have equal entanglement entropy: $S(\rho_A) = S(\rho_B)$.
This is why the singlet is called "maximally entangled": Alice and Bob each have maximum uncertainty about their own subsystem (their local states are $I/2$), yet together they are in a perfectly definite pure state. All the information is encoded in the correlations between the subsystems, not in either subsystem alone.
Part VII — Distinguishing Pure from Mixed States
Three equivalent methods to determine whether a given $\rho$ is pure or mixed:
Method 1 — Purity: Compute $\mathrm{Tr}(\rho^2)$. Pure $\Leftrightarrow \mathrm{Tr}(\rho^2) = 1$. Mixed $\Leftrightarrow \mathrm{Tr}(\rho^2) < 1$.
Method 2 — Eigenvalues: All eigenvalues $\lambda_i \geq 0$ sum to 1. Pure $\Leftrightarrow$ one eigenvalue equals 1, the rest 0. Mixed $\Leftrightarrow$ more than one nonzero eigenvalue.
Method 3 — Entropy: $S(\rho) = 0$ if and only if $\rho$ is pure.
A concrete comparison illustrating a crucial physical difference:
| State | Matrix | $\mathrm{Tr}(\rho^2)$ | Type | $\rho_{ud}$ |
|---|---|---|---|---|
| $|\psi\rangle = \tfrac{1}{\sqrt{2}}(|u\rangle+|d\rangle)$ | $\tfrac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}$ | $1$ | Pure | $1/2 \neq 0$ |
| $\rho = I/2$ | $\tfrac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}$ | $1/2$ | Mixed | $0$ |
These two states give identical probabilities for $\sigma_z$ measurements (both give 50/50), but the pure state has nonzero off-diagonal coherences — it will give different predictions for $\sigma_x$ measurements. They are physically distinguishable.