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10.2

Mixed states and the shrinking Bloch sphere

Every state of a qubit lives inside one unit ball: pure states on the surface, mixed states in the interior, and "complete ignorance" at the centre. Unitary evolution is rotation; the environment's damage is contraction.

Recommended first

After this section you should be able to

  • Write any qubit's ρ in Bloch-vector form, and recover the Bloch vector from the matrix
  • Compute purity from the length of the Bloch vector, and explain what the surface, the interior, and the centre each correspond to
  • Write down the probability formula P = (1 + r·n)/2 for measurement along an arbitrary direction
  • Describe the geometric pictures of unitary evolution and decoherence on the Bloch sphere — rotation vs contraction

The conclusion of the last section: a qubit in the lab must be described by a density matrix, and its “going bad” is a process of falling purity. But a 2×22\times2 complex matrix is not exactly intuitive — four complex numbers, three constraints; it is hard to see at a glance “how mixed” a state is.

Fortunately the qubit enjoys a luxury: its entire state space can be drawn in three-dimensional space.

The phenomenon first: three numbers are enough

Section 9.1 already placed the pure states on the Bloch sphere: ∣ψ⟩=cos⁡θ2∣0⟩+eiϕsin⁡θ2∣1⟩\ket{\psi}=\cos\frac\theta2\ket{0}+\ee^{\ii\phi}\sin\frac\theta2\ket{1} corresponds to the unit vector in direction (θ,ϕ)(\theta,\phi). But the sphere is only a skin. Where do the mixed states live?

Count degrees of freedom. A 2×22\times2 matrix has 4 complex numbers, 8 real parameters; Hermiticity cuts 4, Tr⁡ρ^=1\operatorname{Tr}\hat\rho=1 cuts one more, leaving 3 real parameters. Three real numbers are exactly a three-dimensional vector — and that is no coincidence, it is the following expansion:

From 2×2 matrix to Bloch vectorbasic~6 min

Step 1: choose a basis. The Pauli matrices plus the identity, {1^,σ^x,σ^y,σ^z}\{\hat 1,\hat\sigma_x,\hat\sigma_y,\hat\sigma_z\}, form a complete basis for 2×22\times2 Hermitian matrices (4 linearly independent Hermitian matrices — exactly enough to span). Any Hermitian ρ^\hat\rho can be expanded as

ρ^=a01^+axσ^x+ayσ^y+azσ^z,aμ∈R(10.2.1)\hat\rho=a_0\hat 1+a_x\hat\sigma_x+a_y\hat\sigma_y+a_z\hat\sigma_z, \qquad a_\mu\in\mathbb{R}\tag{10.2.1}

Step 2: fix the coefficients with traces. The Pauli matrices are traceless and Tr⁡(σ^iσ^j)=2δij\operatorname{Tr}(\hat\sigma_i\hat\sigma_j)=2\delta_{ij}, so

Tr⁡ρ^=2a0=1  ⇒  a0=12,ai=12Tr⁡(ρ^ σ^i)=12⟨σi⟩(10.2.2)\operatorname{Tr}\hat\rho=2a_0=1\;\Rightarrow\;a_0=\tfrac12, \qquad a_i=\tfrac12\operatorname{Tr}(\hat\rho\,\hat\sigma_i)=\tfrac12\langle\sigma_i\rangle\tag{10.2.2}

The second equality used last section’s ledger formula ⟨A⟩=Tr⁡(ρ^A^)\langle A\rangle=\operatorname{Tr}(\hat\rho\hat A). Define the Bloch vector r⃗≡(⟨σx⟩,⟨σy⟩,⟨σz⟩)\vec r\equiv(\langle\sigma_x\rangle,\langle\sigma_y\rangle,\langle\sigma_z\rangle); then

ρ^=1^+r⃗⋅σ⃗^2=12(1+rzrx−iryrx+iry1−rz)(10.2.3)\hat\rho=\frac{\hat 1+\vec r\cdot\hat{\vec\sigma}}{2} =\frac12\begin{pmatrix}1+r_z&r_x-\ii r_y\\ r_x+\ii r_y&1-r_z\end{pmatrix}\tag{10.2.3}

The Bloch vector is simply the three spin expectation values — not an abstract parameter; every component is directly measurable.

Step 3: positivity gives the ball. The eigenvalues of ρ^\hat\rho can be computed outright: ρ^−121^=12r⃗⋅σ⃗^\hat\rho-\frac12\hat 1=\frac12\vec r\cdot\hat{\vec\sigma}, and (r⃗⋅σ⃗^)2=r21^(\vec r\cdot\hat{\vec\sigma})^2=r^2\hat 1 (the signature algebra of the Pauli matrices), so the eigenvalues of r⃗⋅σ⃗^\vec r\cdot\hat{\vec\sigma} are ±r\pm r and those of ρ^\hat\rho are

λ±=1±r2,r≡∣r⃗∣(10.2.4)\lambda_\pm=\frac{1\pm r}{2},\qquad r\equiv|\vec r|\tag{10.2.4}

Positive semi-definiteness requires λ−≥0\lambda_-\ge0, i.e.

r≤1(10.2.5)r\le1\tag{10.2.5}

The complete set of legitimate qubit states = the solid unit ball. On the surface (r=1r=1), λ−=0\lambda_-=0 and only one eigenstate survives — a pure state; in the interior (r<1r<1) both eigenvalues are non-zero — a mixed state.

Step 4: purity. Using Tr⁡(ρ^2)=λ+2+λ−2\operatorname{Tr}(\hat\rho^2)=\lambda_+^2+\lambda_-^2:

Tr⁡(ρ^2)=(1+r)2+(1−r)24=1+r22(10.2.6)\operatorname{Tr}(\hat\rho^2)=\frac{(1+r)^2+(1-r)^2}{4}=\frac{1+r^2}{2}\tag{10.2.6}

r=1r=1 gives purity 1; r=0r=0 gives 12\tfrac12 — precisely the maximally mixed state 121^\frac12\hat 1 of section 3.10, which lives at the centre of the ball.

The picture

A map of states

  • Surface: pure states. You know as much about the system as quantum mechanics permits.
  • Interior: mixed states. The farther from the surface, the more mixed.
  • Centre: the maximally mixed state 121^\frac12\hat 1. Every direction measures 50/50 — a coin, through and through.

The direction of the vector tells you “which way to measure with most confidence”; its length tells you “how much confidence that is”. rr is the state’s “health index”, the whole quantum state compressed into one number.

The mathematics

The matching formulas

ρ^=1^+r⃗⋅σ⃗^2,0≤r≤1(10.2.7)\hat\rho=\frac{\hat 1+\vec r\cdot\hat{\vec\sigma}}{2},\qquad 0\le r\le1\tag{10.2.7}Tr⁡(ρ^2)=1+r22(10.2.8)\operatorname{Tr}(\hat\rho^2)=\frac{1+r^2}{2}\tag{10.2.8}

Measuring the spin along the unit direction n⃗\vec n, the probability of getting +1+1:

P(+n⃗)=Tr⁡ ⁣(ρ^ 1^+n⃗⋅σ⃗^2)=1+r⃗⋅n⃗2(10.2.9)P(+\vec n)=\operatorname{Tr}\!\Big(\hat\rho\,\frac{\hat 1+\vec n\cdot\hat{\vec\sigma}}{2}\Big) =\frac{1+\vec r\cdot\vec n}{2}\tag{10.2.9}

(The second step used Tr⁡(σiσj)=2δij\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}.) The outcome depends only on the projection of r⃗\vec r onto n⃗\vec n.

Rotation versus contraction: the watershed between two kinds of evolution

Now for dynamics. How do states move inside the ball?

Unitary evolution = rigid rotationadvanced~5 min

What does unitary evolution ρ^→U^ρ^U^†\hat\rho\to\hat U\hat\rho\hat U^\dagger do to the Bloch vector? Take rotation about the zz axis, U^=e−iϕσ^z/2\hat U=\ee^{-\ii\phi\hat\sigma_z/2}, as the example. Using the Pauli algebra U^†σ^xU^=cos⁡ϕ σ^x−sin⁡ϕ σ^y\hat U^\dagger\hat\sigma_x\hat U=\cos\phi\,\hat\sigma_x-\sin\phi\,\hat\sigma_y (verifiable by term-by-term expansion),

rx→rxcos⁡ϕ+rysin⁡ϕ,ry→−rxsin⁡ϕ+rycos⁡ϕ,rz→rz(10.2.10)r_x\to r_x\cos\phi+r_y\sin\phi,\quad r_y\to -r_x\sin\phi+r_y\cos\phi,\quad r_z\to r_z\tag{10.2.10}

which is exactly a three-dimensional rotation by angle ϕ\phi about the zz axis. The general result: any 2×22\times2 unitary (up to an unobservable global phase) corresponds to a rigid rotation of the Bloch sphere, with axis and angle given by the n⃗\vec n and θ\theta in U^=e−iθ n⃗⋅σ⃗^/2\hat U=\ee^{-\ii\theta\,\vec n\cdot\hat{\vec\sigma}/2}.

Rotations preserve length: rr is unchanged, and the purity 1+r22\frac{1+r^2}{2} is unchanged. This is the geometric version of “unitary evolution preserves purity” from section 3.10 — an isolated system’s state can only circle around inside the ball, and can never reach the centre.

The corollary follows immediately: since qubits in the lab do slide from the surface into the interior, what they undergo cannot be unitary evolution. What does a map that shortens the vector look like? That is exactly the protagonist of the next section.

First, a concrete preview. Suppose the environment’s action is “with probability pp, scramble the state completely into 121^\frac12\hat 1; with probability 1−p1-p, do nothing”:

ρ^→(1−p) ρ^+p 1^2⟺r⃗→(1−p) r⃗(10.2.11)\hat\rho\to(1-p)\,\hat\rho+p\,\frac{\hat 1}{2} \qquad\Longleftrightarrow\qquad \vec r\to(1-p)\,\vec r\tag{10.2.11}

The whole Bloch ball shrinks uniformly into a ball of radius 1−p1-p — this is called the depolarising channel. Other environments squash only the transverse components (losing phase but not energy), and others drag the whole ball towards the north pole (an atom spontaneously emitting, falling to the ground state). The next section works out all three “deformations”; on the map they are, respectively, uniform shrinking, squashing into an olive, and “pulling towards a point”.

Key formulas

  • Bloch representation

    ρ^=1^+r⃗⋅σ⃗^2,ri=⟨σi⟩\hat\rho=\frac{\hat 1+\vec r\cdot\hat{\vec\sigma}}{2},\qquad r_i=\langle\sigma_i\rangle

    Bloch vector = the three spin expectation values, each component measurable

  • Eigenvalues and the legal domain

    λ±=1±r2,r≤1\lambda_\pm=\frac{1\pm r}{2},\qquad r\le1

    Positive semi-definite ⇔ vector inside the unit ball; r = 1 pure, r < 1 mixed

  • Purity

    Tr⁡(ρ^2)=1+r22\operatorname{Tr}(\hat\rho^2)=\frac{1+r^2}{2}

    1 on the surface, 1/2 at the centre (maximally mixed state)

  • Measurement along any direction

    P(+n⃗)=1+r⃗⋅n⃗2P(+\vec n)=\frac{1+\vec r\cdot\vec n}{2}

    Depends only on the projection of the Bloch vector onto the measurement direction

  • Two kinds of evolution

    U^ρ^U^† ⇔ rotation;r⃗→(1−p)r⃗ ⇔ contraction\hat U\hat\rho\hat U^\dagger\ \Leftrightarrow\ \text{rotation};\qquad \vec r\to(1-p)\vec r\ \Leftrightarrow\ \text{contraction}

    Unitaries preserve length; open-system evolution shortens the vector

Self-check4 questions

  1. 1.

    A qubit has Bloch vector of length r = 0. This state is:

  2. 2.

    A qubit has Bloch vector of length r = 0.6. What is its purity Tr(ρ²)?

    1% relative tolerance
  3. 3.

    Which statements about the action of unitary evolution on the Bloch sphere are correct? (Select all that apply.)

    Select all that apply

  4. 4.

    On the Bloch sphere, the state √0.7|↑⟩ + √0.3|↓⟩ and the ensemble "70% |↑⟩, 30% |↓⟩" sit, respectively:

What comes next

The map is drawn: unitary evolution circles around inside the ball, while real qubits fall inward. What kind of mathematical object can send states from the surface into the interior without violating the three iron laws — a density matrix must stay Hermitian, unit-trace, and positive semi-definite?

The answer is called a quantum channel. Starting from “system and environment evolve unitarily together, then look only at the system”, the next section constructs the most general legitimate evolution — the CPTP map — step by step.

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