Bell States in the Bloch Picture

After building the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)$ , you may have noticed something puzzling in the Bloch-sphere display: each qubit's arrow has collapsed to the origin. The two qubits are in a perfectly pure joint state — yet individually each looks maximally uncertain. This apparent contradiction sits at the heart of quantum entanglement.

The Bloch sphere represents a single qubit

The Bloch sphere is a picture of one qubit's state. A pure single-qubit state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ corresponds to a point on the surface of the sphere. Its position is determined by three expectation values:

r_x = \langle X \rangle, \quad r_y = \langle Y \rangle, \quad r_z = \langle Z \rangle,

with $r_x^2 + r_y^2 + r_z^2 = 1$ for a pure state. The north pole is $|0\rangle$ , the south pole is $|1\rangle$ , and equatorial points like $|+\rangle$ and $|-\rangle$ are superpositions.

A mixed state — one about which we have less than complete information — maps to a point inside the ball, with $r_x^2 + r_y^2 + r_z^2 < 1$ . The origin $(0, 0, 0)$ represents the maximally mixed state: every outcome is equally likely along every axis.

Why entanglement forces the arrow to the origin

When two qubits are entangled, you cannot describe qubit 0 in isolation by any pure state. The correct single-qubit description is a reduced density matrix, obtained by tracing out the other qubit:

\rho_0 = \mathrm{Tr}_1\!\bigl(|\Phi^+\rangle\langle\Phi^+|\bigr).

For $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr)$ , this trace gives

\rho_0 = \frac{1}{2}\bigl(|0\rangle\langle 0| + |1\rangle\langle 1|\bigr) = \frac{I}{2}.

This is precisely the maximally mixed state. Its Bloch vector has all three components equal to zero:

r_x = \mathrm{Tr}(X\,\rho_0) = 0, \quad r_y = \mathrm{Tr}(Y\,\rho_0) = 0, \quad r_z = \mathrm{Tr}(Z\,\rho_0) = 0.

The same calculation holds for qubit 1 by symmetry. Every Bell state produces the maximally mixed reduced state on each qubit, so every Bell state places both Bloch arrows at the origin.

All four Bell states behave the same way

There are four Bell states, often labeled $|\Phi^\pm\rangle$ and $|\Psi^\pm\rangle$ :

|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle + |11\rangle\bigr), \qquad |\Phi^-\rangle = \tfrac{1}{\sqrt{2}}\bigl(|00\rangle - |11\rangle\bigr),

|\Psi^+\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle + |10\rangle\bigr), \qquad |\Psi^-\rangle = \tfrac{1}{\sqrt{2}}\bigl(|01\rangle - |10\rangle\bigr).

For any one of these, partial-tracing over either qubit yields $I/2$ . The overall phase or the relative sign between the two terms does not survive the trace — it is encoded purely in the joint correlations. This is why all four Bell states look identical on the single-qubit Bloch sphere: the sphere literally cannot display entanglement.

A practical consequence

This property is not just geometrical curiosity. When Alice and Bob share a Bell pair and Bob measures his qubit, Bob's local statistics are completely random ( $P(0) = P(1) = \tfrac{1}{2}$ ) regardless of what basis Alice chose to measure in. No information travels faster than light, even though the outcomes are perfectly correlated. The Bloch-origin picture makes this explicit: before any measurement, Bob's qubit alone carries zero information.