Pure vs Mixed (Preview)

Every state we have worked with so far — $|0\rangle$, $|1\rangle$, $|+\rangle$ and their rotations — sits on the surface of the Bloch sphere. That surface is reserved for pure states: states that are fully described by a single ket $|\psi\rangle$ and carry the maximum amount of quantum information a single qubit can hold.

What makes a state pure?

A pure state has two features. First, it satisfies $|\alpha|^2 + |\beta|^2 = 1$ with a definite pair of amplitudes; no randomness in the state specification itself. Second, its Bloch vector has unit length — it lands exactly on the sphere's surface. The north pole $|0\rangle$, the south pole $|1\rangle$, and every equatorial superposition like $|+\rangle$ are all pure.

Formally, the density operator of a pure state is $\rho = |\psi\rangle\langle\psi|$, a rank-one projector. One useful test is $\operatorname{Tr}(\rho^2) = 1$: squaring a rank-one projector gives back itself, so the trace of the square is one.

Interior points: mixed states

The Bloch sphere's interior is populated by mixed states. A mixed state arises when there is genuine classical uncertainty about which pure state the qubit is in — for example, a preparation device that flips a coin and produces either $|0\rangle$ or $|1\rangle$ with equal probability.

Such a situation cannot be captured by any single ket. Instead it is described by a weighted mixture of pure-state projectors:

For the equal-coin flip, $p_0 = p_1 = \tfrac{1}{2}$, giving

This is the maximally mixed state — its Bloch vector has zero length and sits at the origin. Measuring it in any basis still gives $0$ or $1$ with probability $\tfrac{1}{2}$, but that is now a result of classical ignorance rather than quantum superposition. The purity test confirms: $\operatorname{Tr}(\rho^2) = \operatorname{Tr}(\tfrac{I}{4}) = \tfrac{1}{2} < 1$.

Where the Bloch vector lands

For any state described by a $2\times 2$ density operator $\rho$, the Bloch vector $(r_x, r_y, r_z)$ is defined by

where $X$, $Y$, $Z$ are the Pauli matrices. The Bloch vector has length $r = \sqrt{r_x^2 + r_y^2 + r_z^2}$. Physical states require $r \le 1$:

• $r = 1$: pure state, lives on the surface.
• $0 < r < 1$: partially mixed state, lives strictly inside the sphere.
• $r = 0$: maximally mixed state $I/2$, sits at the center.

This geometric picture is why the Bloch sphere is so powerful: the formal purity $\operatorname{Tr}(\rho^2) = (1+r^2)/2$ increases monotonically with the Bloch vector length $r$, so a greater distance from the center always means a less mixed, more pure state — with $r=1$ (surface) corresponding to purity 1 and $r=0$ (center) corresponding to purity $\tfrac{1}{2}$.

A preview of why this matters

In the beginner track we work almost exclusively with pure states, because they are the ones our gates and state-vector simulator act on directly. Mixed states appear naturally when a qubit is entangled with another system and we look at only one half of the pair, or whenever the qubit interacts with an environment it cannot control. Intermediate lessons will return to density operators and quantum channels in full detail; for now, keep in mind that the Bloch sphere's surface and interior are two genuinely different physical regimes.

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