Qubits and the Bloch sphere

In chapter 5 we worked hard on spin 1/2: a system with only two basic states, a two-dimensional complex state space, and measurements described by the Pauli operators. Back then it was “the smallest angular momentum”. From this chapter on, it takes the stage under a new identity — the smallest quantum carrier of information.

A concrete question: how much information fits in a two-level system?

A classical bit has exactly two states: 0 or 1. A coin, a switch, a voltage level. Now replace it with a quantum two-level system — say a spin-1/2 particle — and adopt the convention

$$\ket{0}\equiv\ket{\uparrow},\qquad \ket{1}\equiv\ket{\downarrow}\tag{9.1.1}$$

By the superposition principle of chapter 3, the legal states go far beyond those two:

$$\ket{\psi}=\alpha\ket{0}+\beta\ket{1},\qquad \alpha,\beta\in\mathbb{C},\quad |\alpha|^2+|\beta|^2=1\tag{9.1.2}$$

Such a system is called a quantum bit (qubit). And here comes the puzzle: $\alpha,\beta$ are two complex numbers, four real parameters — does a single qubit hold “infinitely much information”? The classical “0 or 1” picture breaks down completely here. We need a new map, one that draws every legal state at once.

Where the old tools fall short

Chapter 3 hands us two constraints: normalisation eats one real parameter, and the unobservable global phase eats another. Two real parameters remain — so the set of qubit pure states is a two-dimensional surface. Which surface?

From (α, β) to spherical coordinates (θ, φ)basic~6 minexpand

Step 1: rewrite in modulus and argument. Let $\alpha=r_0\ee^{\ii\gamma_0}$ and $\beta=r_1\ee^{\ii\gamma_1}$. Normalisation requires $r_0^2+r_1^2=1$, so we may set

$$r_0=\cos\frac{\theta}{2},\qquad r_1=\sin\frac{\theta}{2},\qquad \theta\in[0,\pi]\tag{9.1.3}$$

(Why $\theta/2$ rather than $\theta$? Hold that thought — step 3 will settle it.)

Step 2: throw away the global phase. Pull out the common factor $\ee^{\ii\gamma_0}$:

$$\ket{\psi}=\ee^{\ii\gamma_0}\left(\cos\frac{\theta}{2}\ket{0} +\ee^{\ii\varphi}\sin\frac{\theta}{2}\ket{1}\right), \qquad \varphi\equiv\gamma_1-\gamma_0\tag{9.1.4}$$

The global phase has no measurable consequence whatsoever (every probability and every expectation value is unchanged), so delete it. What remains is

$$\ket{\psi}=\cos\frac{\theta}{2}\ket{0}+\ee^{\ii\varphi}\sin\frac{\theta}{2}\ket{1}, \qquad \theta\in[0,\pi],\ \varphi\in[0,2\pi)\tag{9.1.5}$$

$(\theta,\varphi)$ are exactly spherical coordinates: the set of qubit pure states is a unit sphere, called the Bloch sphere.

Step 3: justify the θ/2 — compute the Pauli expectation values. Recall from chapter 5 that $\sigma_z\ket{0}=\ket{0}$, $\sigma_z\ket{1}=-\ket{1}$, together with the matrices of $\sigma_x,\sigma_y$. One by one:

$$\braket{\psi}{\sigma_z|\psi}=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}=\cos\theta\tag{9.1.6}$$$$\braket{\psi}{\sigma_x|\psi}=2\cos\frac{\theta}{2}\sin\frac{\theta}{2}\cos\varphi=\sin\theta\cos\varphi\tag{9.1.7}$$$$\braket{\psi}{\sigma_y|\psi}=\sin\theta\sin\varphi\tag{9.1.8}$$

The three expectation values assemble into precisely the Cartesian components of a unit vector:

$$\vec r\equiv(\langle\sigma_x\rangle,\langle\sigma_y\rangle,\langle\sigma_z\rangle) =(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta)\tag{9.1.9}$$

Had we written $\cos\theta$ instead of $\cos(\theta/2)$ in step 1, this would have come out as $\cos 2\theta$ and the coordinates would not match. The half-angle is the built-in price of mapping a complex two-dimensional space onto a real three-dimensional sphere — it is the very same fact as chapter 5’s “rotate a spin by 360° and the wavefunction flips sign; only 720° brings it back”.

Line-by-line match with spin 1/2

The qubit is not new physics — it is chapter 5’s spin 1/2 in new notation. The dictionary:

Spin 1/2 (chapter 5)	Qubit (this chapter)
$\ket{\uparrow},\ket{\downarrow}$ (eigenstates of $S_z$)	Computational basis $\ket{0},\ket{1}$
$S_x,S_y,S_z=\frac{\hbar}{2}\sigma_{x,y,z}$	Pauli gates $X,Y,Z$ (next section)
Precession in a magnetic field $\ee^{-\ii\omega t\sigma_z/2}$	Phase gates / single-qubit rotations
Stern–Gerlach measurement along $\hat n$	Projective measurement along $\hat n$
720° to return to itself	The half-angle $\theta/2$

A real qubit need not literally be a spin: superconducting circuits use their two lowest levels, ion traps use two hyperfine levels, photons use two polarisation directions. As long as it is “a two-dimensional complex space + unitary evolution + projective measurement”, the mathematics is identical to spin 1/2 — which is precisely the power of chapter 3’s formalism: different physical carriers, isomorphic Hilbert spaces.

To spin this sphere with your own hands, head to Lab module 05: spin and the Bloch sphere — drag the point on the sphere and watch $\ket{\psi}$ update live in the readout, complete with precession and measurement collapse.

Mixed states live inside the ball

Pure states tile the surface — so what is the interior? It is the mixed states introduced in section 3.10. Any qubit density operator can be written

$$\rho=\frac{1}{2}\left(I+\vec r\cdot\vec\sigma\right),\qquad |\vec r|\le 1\tag{9.1.12}$$

$|\vec r|=1$ is a pure state (the surface), $|\vec r|<1$ is a mixed state (the interior), and the centre $\vec r=0$ is the maximally mixed state $\rho=I/2$ — fifty-fifty along every axis, telling you nothing. This picture returns again and again in section 9.3 (reduced density matrices measuring entanglement) and in chapter 10 (decoherence “shrinking” the surface toward the centre). For now, remember: the closer to the centre, the less pure the state.

∎Key formulas

• Qubit pure state

  $$\ket{\psi}=\cos\frac{\theta}{2}\ket{0}+\ee^{\ii\varphi}\sin\frac{\theta}{2}\ket{1}$$

  θ∈[0,π] latitude, φ∈[0,2π) longitude; the global phase has been dropped

• Bloch vector

  $$\vec r=(\langle\sigma_x\rangle,\langle\sigma_y\rangle,\langle\sigma_z\rangle)$$

  Pure states have |r| = 1; orthogonal states are antipodal

• Measurement probability

  $$P_\pm=\frac{1\pm\hat n\cdot\vec r}{2}$$

  Probability of ±1 when measuring n̂·σ along n̂; at angle Θ, P₊=cos²(Θ/2)

• Density operator

  $$\rho=\frac{1}{2}(I+\vec r\cdot\vec\sigma),\quad |\vec r|\le 1$$

  Surface = pure, interior = mixed, centre = I/2

?Self-check4 questions

1. 1.

   Where does the Bloch vector of the state ∣ψ⟩ = (∣0⟩ + i∣1⟩)/√2 point?

   • The +y axis
   • The +x axis
   • The +z axis
   • In the x-y plane, 45° from the x axis

   Submit
2. 2.

   Which of the following statements about the Bloch sphere are correct? (Select all that apply.)

   Select all that apply

   • Two states orthogonal in Hilbert space are antipodal points on the sphere
   • Two states 90° apart on the sphere have zero inner product
   • Points inside the ball represent mixed states
   • Two states differing only by a global phase are the same point on the sphere

   Submit
3. 3.

   A qubit is in ∣0⟩. Measure n̂·σ along a direction n̂ at 60° to the z axis. Find the probability of getting +1.

   2% relative tolerance

   Submit
4. 4.

   How much classical information can a single measurement extract from one unknown qubit state, at most?

   • 1 bit: the outcome is + or −, and afterwards the state has collapsed
   • 2 real numbers: both θ and φ can be read out
   • Infinitely much: the sphere has a continuum of points
   • Zero: a single outcome is completely random and carries no information

   Submit

What comes next

The map is drawn; now we learn to move on it. Classical computers process bits with AND, OR and NOT gates — what does a quantum computer process qubits with? The answer hides in chapter 3’s time-evolution postulate: every operation must be unitary — and on the Bloch sphere, that means rotations. In the next section we lay out the most-used rotations as a catalogue of “gates”, and answer a deeper question: how many kinds of gate does it take to assemble every possible operation?