Unsharp Measurements: Smearing and Moment Noise

An unsharp observable has at least one effect that is not a projection. Its outcome statistics may arise from classical blurring of a sharp observable, but not every unsharp POVM has that form. The canonical Unsharp Measurements article develops measurement models; this page analyzes smearing, moment noise, and the separation between outcome resolution and state disturbance.

Helpful background. POVMs supplies operator integrals; Naimark dilation supplies the compression used in the noise inequality.

Smearing a sharp observable

Let $P$ be a PVM on an outcome space $X$ and let $k(x,\Delta)$ be a Markov kernel into a reported outcome space $Y$. Then

$$E(\Delta)=\int_Xk(x,\Delta)\,P(dx)$$

defines a POVM. The kernel is a classical conditional probability: after the sharp value $x$, the detector reports an event $\Delta$ with probability $k(x,\Delta)$.

Normalization and positivity follow from $0\leq k(x,\Delta)\leq1$ and $k(x,Y)=1$. Countable additivity follows by monotone convergence in each vector spectral measure. Thus smearing provides a rigorous way to introduce detector uncertainty without using formal pointwise eigenstates.

All effects of this POVM commute, since they are functions of the same PVM. Therefore a POVM with noncommuting effects, such as the qubit trine measurement, cannot be obtained by classical smearing of a single sharp observable on the original Hilbert space. It still admits a projective dilation on a larger space.

Intrinsic noise from the second moment

For a real bounded outcome function $f$, write

$$E[f]=\int f\,dE,\qquad N_f=E[f^2]-E[f]^2.$$

The operator $N_f$ is positive. To prove this, use $E(\Delta)=V^*P(\Delta)V$ and let $M=\int f\,dP$. Since $V$ is an isometry,

$$N_f=V^*M(I-VV^*)MV\geq0.$$

The positivity follows because $I-VV^*$ is an orthogonal projection. This calculation shows exactly how compression can destroy multiplicativity of the sharp functional calculus.

In a state $\rho$, the variance of the reported value separates as

$$\operatorname{Var}_{p_\rho}(f) =\operatorname{Var}_\rho(E[f]) +\operatorname{Tr}(\rho N_f).$$

Both terms are nonnegative. The first is the state variance of the first-moment operator; the second is additional variance intrinsic to the POVM for this outcome labeling. A PVM has $N_f=0$ for every bounded $f$. Vanishing for one particular $f$ alone does not establish sharpness of the entire observable.

Noisy binary spin

For outcomes $s=\pm1$ and effects

$$E_s=\tfrac12(I+s\eta\,n\cdot\sigma), \qquad0\leq\eta\leq1,$$

the first moment is $\eta n\cdot\sigma$ and the second is $I$. Thus

$$N=I-(\eta n\cdot\sigma)^2 =(1-\eta^2)I.$$

At $\eta=1$ the measurement is sharp; at $\eta=0$ it is an uninformative fair coin. Intermediate values describe a state-dependent signal with added measurement variance.

These effects arise by measuring the sharp spin and flipping its reported sign with probability $(1-\eta)/2$. This gives a classical smearing interpretation for this particular binary model. Compatibility of differently oriented noisy spins is a separate question answered by the threshold in Joint Measurability.

Gaussian position resolution

For position $Q$ on $L^2(\mathbb R)$, take a centered Gaussian response $g_s$ of standard deviation $s>0$:

$$k(x,\Delta)=\int_\Delta \frac{\exp[-(y-x)^2/(2s^2)]}{\sqrt{2\pi s^2}}\,dy.$$

The outcome law is the position distribution convolved with the Gaussian. Its first moment agrees with position, and its second moment contains the added detector variance. Write $\mu_\psi^E(\Delta)=\langle\psi,E(\Delta)\psi\rangle$. For vectors in $D(Q)$, the quadratic-form identity is

$$\int y^2\,d\mu_\psi^E(y) =\|Q\psi\|^2+s^2\|\psi\|^2.$$

For states with finite second position moment,

$$\operatorname{Var}(Y)=\operatorname{Var}_\rho(Q)+s^2.$$

The domain restriction matters. If the input position variance is infinite, adding finite Gaussian noise does not make it finite. The bounded-$f$ operator proof above cannot be applied to $f(y)=y$ without justifying the unbounded moments; the convolution and quadratic-form calculation provide that justification in this example.

Resolution does not determine disturbance

The POVM fixes outcome probabilities and the noise quantities just computed. It does not fix the post-measurement state. Composing every outcome operation with a chosen output unitary leaves its effect unchanged while changing later measurements.

Thus a small resolution parameter or a small intrinsic-noise operator does not by itself imply small disturbance of every observable. Conversely, a state-independent coin-toss POVM can be implemented while leaving the state untouched or while replacing it completely. The instrument supplies the missing dynamical information.

Exercises

1. For a binary measurement with outcomes $0,1$ and effect $E$ for outcome one, compute its intrinsic-noise operator and determine when it vanishes.

Solution

Both the first and second moments are $E$, so $N=E-E^2\geq0$. It vanishes exactly when $E$ is a projection. This is the sharpness criterion of the effect algebra.

2. Can a noncommuting trine POVM be implemented by adding only classical reporting noise after one sharp system observable?

Solution

No. Every smeared effect would be a bounded function of the same PVM, so all effects would commute. A trine measurement needs a more general implementation, such as a Naimark dilation; this does not contradict the universality of enlarged projective measurements.

References

• Busch, Paul, Pekka Lahti, Juha-Pekka Pellonpää, and Kari Ylinen, Quantum Measurement, Springer, 2016 — smearing, moment operators, and approximate measurement. Publisher.
• Heinosaari, Teiko, and Mário Ziman, The Mathematical Language of Quantum Theory: From Uncertainty to Entanglement, Cambridge University Press, 2012 — unsharp observables and measurement models. Publisher.