Pure vs Mixed States (Comparison & Summary)

Overview

By now you have seen two complementary ways to describe quantum states:

• Pure states using kets $|\psi\rangle$.
• Mixed states using density matrices $\rho$.

This page gives a calm, side-by-side comparison of what each description means, when you need it, and how to interpret it physically and probabilistically.

Intuition

A pure state is “maximal information” about a quantum system within the theory: it is not uncertainty-free (measurement can still be random), but it is not a classical mixture of possibilities.

A mixed state represents additional uncertainty beyond quantum randomness:

• either because the preparation involved classical randomness,
• or because you are looking at a subsystem and ignoring the rest.

Geometrically:

• Pure single-qubit states live on the surface of the Bloch sphere.
• Mixed single-qubit states live inside it (the Bloch ball).

The closer a mixed state is to the center, the less “directional information” it contains.

Formal Description

Pure states

A pure state is represented by a ket $|\psi\rangle$ and its density matrix

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

For a single qubit, $|\psi\rangle = \alpha|0\rangle+\beta|1\rangle$ with $|\alpha|^2+|\beta|^2=1$.

Mixed states

A mixed state is represented directly by a density matrix

$$\rho = \sum_k p_k|\psi_k\rangle\langle\psi_k|,$$

where:

• each $p_k\ge 0$ is a classical probability,
• and $\sum_k p_k=1$.

Measurement rule (works for both)

For a projective measurement with outcome projector $\Pi_k$, the probability is

$$P(k)=\mathrm{Tr}(\rho\Pi_k).$$

This formula reduces to the usual Born rule when $\rho=|\psi\rangle\langle\psi|$.

Summary table

Topic	Pure state	Mixed state
Representation	$	\psi\rangle$(or$\rho=
Meaning	Single state (not classical uncertainty)	Classical uncertainty or ignored subsystem
Single-qubit geometry	Point on Bloch sphere surface	Point inside Bloch ball
Can be written as a single ket?	Yes	Not in general
Measurement probabilities	From overlaps (Born rule)	$P(k)=\mathrm{Tr}(\rho\Pi_k)$

Worked Example

Compare two preparations that look the same in the computational basis.

Case 1: Pure state

$$|+\rangle=\tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle).$$

Measuring in the computational basis gives $P(0)=P(1)=1/2$.

Case 2: Mixed state

A 50/50 mixture of $|0\rangle$ and $|1\rangle$ has density matrix

$$\rho=\tfrac{1}{2}|0\rangle\langle 0|+\tfrac{1}{2}|1\rangle\langle 1|.$$

Measuring in the computational basis also gives $P(0)=P(1)=1/2$.

Now measure in the $\{|+\rangle,|-\rangle\}$ basis.

• In the pure case $|+\rangle$, you get “$+$” with probability 1.
• In the mixed case, you get “$+$” with probability $1/2$.

So the difference is operational: it shows up when you change what you measure.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

You now have the full conceptual and mathematical maturity needed for the next step: describing controlled state changes (unitary transformations) and composing them into computations.

In the Gates & Circuits track, we will introduce transformations as precise state updates—without changing the measurement rules you’ve learned here.