Measurement Effect

quantum-computing quantum-mechanics

Definition

Measurement Effect

Let $\mathcal{H}$ be a complex Hilbert space. A measurement effect is a positive semidefinite operator $E:\mathcal{H}\to \mathcal{H}$ satisfying

$$0\le E\le I.$$

Equivalently,

$$0\le ⟨\psi |E|\psi ⟩\le 1\forall |\psi ⟩\in \mathcal{H}\text{ with }⟨\psi |\psi ⟩=1.$$

The number $⟨\psi |E|\psi ⟩$ is the Born-rule probability assigned to the measurement outcome represented by $E$.

Effect as One Outcome

A positive operator-valued measure is a family of effects

$$\{E_j{\}}_{j\in J}$$

whose total effect is the identity operator:

$$\sum _{j\in J}{E}_j=I.$$

Each $E_j$ represents one possible measurement outcome. For a pure state $|\psi ⟩$,

$$\mathrm{Pr}(j\mid \psi )=⟨\psi |E_j|\psi ⟩.$$

For a mixed state represented by a density operator $\rho$,

$$\mathrm{Pr}(j\mid \rho )=\mathrm{tr}(\rho E_j).$$

Why $0\le E\le I$

The condition

$$E\ge 0$$

ensures

$$⟨\psi |E|\psi ⟩\ge 0.$$

The condition

$$E\le I$$

means

$$I-E\ge 0,$$

which ensures

$$⟨\psi |E|\psi ⟩\le ⟨\psi |I|\psi ⟩=1$$

for normalised $|\psi ⟩$.

Thus an effect is exactly an operator whose expectation value is always a valid probability.

Projectors as Sharp Effects

Every orthogonal projection $P$ is a measurement effect:

$$P=P^{†},P^2=P,0\le P\le I.$$

Projection effects are sharp: for states inside the projected subspace, the outcome has probability $1$; for states orthogonal to it, the outcome has probability $0$.

General effects need not be projections. They can represent noisy or unsharp outcomes.

Qubit Example

For $0\le \epsilon \le 1$, define

$$E=(1-\epsilon )|0⟩⟨0|+\epsilon |1⟩⟨1|.$$

The eigenvalues of $E$ are $1-\epsilon$ and $\epsilon$, so

$$0\le E\le I.$$

For

$$|\psi ⟩=\alpha |0⟩+\beta |1⟩,$$

the effect assigns probability

$$⟨\psi |E|\psi ⟩=(1-\epsilon )|\alpha {|}^2+\epsilon |\beta {|}^2.$$

If $\epsilon =0$, then $E=|0⟩⟨0|$ is a sharp projection. If $0<\epsilon <1$, then $E$ is a noisy effect rather than a projection.