Pure vs Mixed States

So far you have described a quantum system by a single state vector $|\psi\rangle$ . That is the most complete description physics allows — but it is not the most general situation you can be in. Often you do not know exactly which state was prepared; you only know a list of possibilities and how likely each one is. Capturing that uncertainty is what mixed states are for.

Pure states: maximal knowledge

A pure state is one that can be written as a single normalized ket,

|\psi\rangle = \alpha|0\rangle + \beta|1\rangle, \qquad |\alpha|^2 + |\beta|^2 = 1.

A pure state encodes everything that can possibly be known about the system. The superposition above is not ignorance about whether the qubit is "really" $|0\rangle$ or "really" $|1\rangle$ : it is a definite state with definite phase relationships, and there exists a measurement (in the right basis) whose outcome is certain. For example, $\tfrac{1}{\sqrt2}(|0\rangle + |1\rangle) = |+\rangle$ gives the outcome $+$ with probability $1$ when measured along the $x$ -axis.

Mixed states: classical uncertainty about which pure state

A mixed state describes a situation where the system is in one of several pure states $|\psi_i\rangle$ with classical probabilities $p_i$ (with $p_i \ge 0$ and $\sum_i p_i = 1$ ). This is an ensemble

\{(p_i, |\psi_i\rangle)\}.

Crucially, the system is genuinely in one of the $|\psi_i\rangle$ — we simply do not know which. This is ordinary classical ignorance layered on top of quantum mechanics. A common laboratory example is an unpolarized beam: each photon is in a definite polarization state, but the source emits them with random, uniformly distributed polarizations.

Why a superposition is not a mixture

It is tempting to think the superposition $\tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ is "the same as" a 50/50 mixture of $|0\rangle$ and $|1\rangle$ . It is not. Measure both in the computational basis and you get $0$ or $1$ with probability $1/2$ either way — they look identical. But measure along the $x$ -axis:

The superposition $|+\rangle$ yields $+$ with certainty.
The 50/50 mixture yields $+$ or $-$ with probability $1/2$ each.

The superposition has a definite phase relationship between $|0\rangle$ and $|1\rangle$ ; the mixture has none. Interference distinguishes them. No single state vector can represent the mixture, because a state vector always carries definite phases.

The need for a new tool

Ensembles are awkward to manipulate directly: many different ensembles can be physically indistinguishable (a 50/50 mix of $|0\rangle,|1\rangle$ behaves exactly like a 50/50 mix of $|+\rangle,|-\rangle$ ). We need a single mathematical object that depends only on the physically observable content of the state and not on the particular list we wrote down. That object is the density operator $\rho$ , introduced in the next lesson. It represents pure and mixed states on equal footing, makes expectation values a one-line trace, and — as we will see — is the only way to describe a subsystem of an entangled pair.