Measurement Backaction

Measurement backaction is the change caused by the interaction that measures a system. It can be a mechanical impulse, dephasing, loss, reset, or conditional filtering. An instrument specifies this change as well as the outcome probabilities; a POVM alone does not.

Helpful background. State update rules explains normalized conditional states and unread channels. The qubit examples below are finite dimensional; the position and oscillator examples state their additional assumptions locally.

Backaction as an Operation

Let a measurement have outcome-resolved operations $\{\mathcal I_m\}$. For input state $\rho$,

$$p(m) = \operatorname{Tr}\mathcal I_m(\rho).$$

If outcome $m$ is known and has positive probability, the conditional post-measurement state is

$$\rho_m = \frac{\mathcal I_m(\rho)} {\operatorname{Tr}\mathcal I_m(\rho)}.$$

If the outcome is ignored, the nonselective state is

$$\rho' = \sum_m\mathcal I_m(\rho).$$

The maps $\mathcal I_m$ and their sum encode the measurement backaction. Two devices can have the same outcome probabilities for every input state and still have different backaction.

In Kraus form,

$$\mathcal I_m(\rho) = \sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger.$$

The associated POVM effect is

$$E_m = \sum_\alpha K_{m\alpha}^\dagger K_{m\alpha}.$$

The effect $E_m$ determines $p(m)=\operatorname{Tr}(E_m\rho)$, but the operators $K_{m\alpha}$ determine the state left behind. That is where backaction enters.

Projective Backaction

For the ideal Lüders implementation of a measurement with projectors $\{P_a\}$, the selective operation is

$$\mathcal I_a(\rho)=P_a\rho P_a.$$

The conditional state for outcome $a$ is

$$\rho_a = \frac{P_a\rho P_a} {\operatorname{Tr}(P_a\rho)}.$$

The unread measurement is the channel

$$\mathcal M(\rho) = \sum_aP_a\rho P_a.$$

Writing the input in blocks,

$$\rho = \sum_{a,b}P_a\rho P_b,$$

the unread channel removes off-diagonal blocks:

$$\rho \mapsto \sum_aP_a\rho P_a.$$

Thus even when no one uses the outcome, the measurement interaction can disturb future interference experiments by dephasing the measured decomposition. This is the simplest form of measurement backaction.

For the degenerate ideal case and the meaning of minimal disturbance inside an eigenspace, see Lüders Rule.

Noncommuting Observables

Backaction is most visible when a later observable does not commute with the measured one. Suppose an unread measurement $\{P_a\}$ is followed by a projective measurement $\{Q_b\}$. The probability of $b$ after the unread first measurement is

$$p'(b) = \operatorname{Tr} \!\left[ Q_b\sum_aP_a\rho P_a \right].$$

Without the first measurement, the probability would be

$$p(b)=\operatorname{Tr}(Q_b\rho).$$

If every $Q_b$ commutes with every $P_a$, then $p'(b)=p(b)$ for all $\rho$. If they do not commute, the first measurement can change the statistics of the second.

For a spin-$1/2$ example, prepare

$$\lvert +x\rangle = \frac{\lvert +z\rangle+\lvert -z\rangle}{\sqrt2}.$$

If $\sigma_x$ is measured immediately, outcome $+x$ has probability one. If an unread $\sigma_z$ measurement is inserted first, the state becomes

$$\rho' = \frac{1}{2} \lvert+z\rangle\langle+z\rvert + \frac{1}{2} \lvert-z\rangle\langle-z\rvert = \frac{I}{2}.$$

The later probability for $+x$ is then

$$\operatorname{Tr} \!\left[ \lvert+x\rangle\langle+x\rvert \frac{I}{2} \right] = \frac{1}{2}.$$

The unread $\sigma_z$ measurement has disturbed the phase information needed to predict $\sigma_x$.

Pointer Correlations

Backaction is easiest to see in an explicit apparatus model. Suppose a system basis $\{\lvert a\rangle\}$ is correlated with pointer states:

$$\lvert a\rangle\lvert R\rangle \mapsto \lvert a\rangle\lvert M_a\rangle.$$

For an input

$$\lvert\psi\rangle = \sum_a c_a\lvert a\rangle,$$

the joint state becomes

$$\sum_a c_a \lvert a\rangle\lvert M_a\rangle.$$

If the pointer is ignored, the system state is

$$\rho_S' = \sum_{a,b} c_a c_b^* \langle M_b|M_a\rangle \lvert a\rangle\langle b\rvert.$$

The pointer overlaps $\langle M_b|M_a\rangle$ control the disturbance. Orthogonal pointer states carry fully distinguishable outcome information and remove coherences between different $a$. Overlaps close to unit magnitude cause little loss of coherence magnitude. Their complex phase can still produce a unitary phase shift, so little information gain does not exclude a substantial coherent disturbance.

This is the operational core of the information-disturbance tradeoff. The disturbance is not added after the measurement; it is the reduced-system effect of correlating the system with degrees of freedom that store information.

See von Neumann Measurement Model for the unitary measurement model behind this calculation.

Extra disturbance with unchanged readout

Replacing any outcome operator $M_m$ by $U_mM_m$, where $U_m$ is unitary, preserves the effect $M_m^\dagger M_m$ and all its probabilities. It rotates the conditional output. The reset example demonstrates this explicitly. Thus the information gained does not impose an upper bound on extra disturbance that an apparatus might add.

Weak Measurement

A weak measurement interaction can acquire little information and produce a small state change per shot. Weak information alone does not guarantee weak disturbance; here we explicitly choose operators close to scalar multiples of the identity. A schematic two-outcome weak measurement of a qubit in the $\sigma_z$ basis may use diagonal measurement operators such as

$$M_\pm = \sqrt{\frac{1}{2}} \left( \sqrt{1\pm\epsilon}\,P_+ + \sqrt{1\mp\epsilon}\,P_- \right), \qquad 0\le\epsilon\ll1.$$

The associated effects are

$$E_\pm = \frac{1}{2} \left( I\pm\epsilon\sigma_z \right).$$

For small $\epsilon$, each outcome only slightly biases the state toward one $\sigma_z$ eigenspace. Repeating many weak measurements can build up strong information and substantial cumulative backaction.

For these operators, the unread off-diagonal entry is multiplied by $\sqrt{1-\epsilon^2}=1-\epsilon^2/2+O(\epsilon^4)$, while the populations are unchanged. The effect accumulates over repeated shots. Continuous Monitoring is the limit in which many weak updates are taken over small time steps and the state is conditioned on a time-dependent record.

Position Measurement and Momentum Disturbance

A finite-resolution position measurement can be modeled by Gaussian measurement operators

$$M_y = \frac{1}{(2\pi\sigma^2)^{1/4}} \exp\!\left[ -\frac{(y-X)^2}{4\sigma^2} \right],$$

where $X$ is position on the line, $y$ is the continuous reported position, and $\sigma>0$ is the detector standard deviation. These are operator densities: $\int dy\,M_y^\dagger M_y=I$. The outcome density is

$$p(y)=\operatorname{Tr}(M_y^\dagger M_y\rho).$$

The nonselective map suppresses spatial coherence:

$$\rho(x,x') \mapsto \exp\!\left[ -\frac{(x-x')^2}{8\sigma^2} \right] \rho(x,x').$$

The same unread channel is an average of momentum translations $e^{i\xi X/\hbar}$, where the added momentum $\xi$ is an independent zero-mean Gaussian. Matching its characteristic function $e^{-\operatorname{Var}(\xi)(x-x')^2/(2\hbar^2)}$ gives the added variance

$$\Delta p_{\text{ba}}^2 = \frac{\hbar^2}{4\sigma^2}.$$

For an input with finite momentum variance, the output variance is the input variance plus this amount. A sharper position readout has smaller $\sigma$ and produces stronger suppression of long-range spatial coherence, along with larger momentum disturbance. This is a measurement-model statement, not merely a slogan about uncertainty.

Photon Counting

Photon counting gives an example where backaction includes loss and conditioning. For ideal monitoring of a cavity decay channel with rate $\kappa$ and no thermal input, a detected photon during $dt$ is represented to first order by the jump operation

$$\mathcal I_{\text{click}}(\rho) = \kappa\,dt\,a\rho a^\dagger.$$

Conditioned on a click, the state becomes

$$\rho_{\text{click}} = \frac{a\rho a^\dagger} {\operatorname{Tr}(a^\dagger a\rho)}.$$

For a number state $|n\rangle$ with $n>0$, a click leaves $|n-1\rangle$. But a click does not always lower the conditional mean relative to the original ensemble: it preferentially selects preparations with larger occupation. Writing $n=a^\dagger a$,

$$\langle n\rangle_{\rm click} =\frac{\langle n(n-1)\rangle}{\langle n\rangle} =\langle n\rangle-1+\frac{\operatorname{Var}(n)}{\langle n\rangle}.$$

This assumes positive finite mean and finite second moment. A coherent state remains the same normalized coherent state under $a$; its Poisson variance cancels the subtraction of one. The absence of a click also carries information. To first order in $dt$, a no-click Kraus operator has the form

$$K_0 \approx I-\frac{\kappa\,dt}{2}a^\dagger a,$$

apart from Hamiltonian evolution. The one-click approximation requires $\kappa n\,dt\ll1$ over occupations with appreciable weight; it is not uniform on the entire unbounded oscillator Hilbert space. Thus even “nothing happened” in the detector can update the conditional state because it changes what is inferred about the photon number.

Backaction Evasion

Backaction cannot be made to disappear from arbitrary quantum measurement. It can, however, be directed into variables that are not the target of the measurement.

A quantum nondemolition-style measurement aims to monitor an observable $A$ whose future values remain predictable. A simplified sufficient condition is that the measured observable commute with the free Hamiltonian and with the interaction in a way that does not randomize $A$:

$$[A,H_S]=0, \qquad [A,H_I]=0.$$

Then the measurement may dephase superpositions of different $A$ values while preserving the value of $A$ itself. The backaction is pushed into conjugate observables or phases rather than into the repeated readout variable.

Backaction evasion is therefore not “measurement without disturbance” in an absolute sense. It is measurement designed so that the disturbance avoids the variable one wants to keep predicting.

Measurement-Disturbance Relations

The textbook slogan that measuring position disturbs momentum captures an important idea, but precise measurement-disturbance relations are subtler than the preparation uncertainty relation

$$\Delta X\,\Delta P\ge\frac{\hbar}{2}.$$

Preparation uncertainty concerns spreads in a single quantum state. Measurement disturbance concerns how a measurement apparatus changes later statistics. Different formal definitions of measurement error and disturbance lead to different inequalities.

For this volume, the practical rule is:

Do not infer the backaction from the POVM alone. Specify the instrument or the physical measurement model.

Ozawa (2003) and Busch, Lahti, and Werner (2014) develop distinct operational error–disturbance formulations. An inequality must be read with its definitions of error, disturbance, and state dependence; preparation uncertainty alone does not specify detector performance.

Exercises

Unread spin measurement

A qubit is prepared in $\lvert+x\rangle$. An unread $\sigma_z$ measurement is performed, followed by a $\sigma_x$ measurement. What is the probability of obtaining $+x$ at the end?

Solution

The unread $\sigma_z$ measurement maps

$$\lvert+x\rangle\langle+x\rvert \mapsto \frac{1}{2} \lvert+z\rangle\langle+z\rvert + \frac{1}{2} \lvert-z\rangle\langle-z\rvert = \frac{I}{2}.$$

Therefore

$$p(+x) = \operatorname{Tr} \!\left[ \lvert+x\rangle\langle+x\rvert\frac{I}{2} \right] = \frac{1}{2}.$$

The unread $\sigma_z$ measurement has removed the phase coherence that made the original state a definite $\sigma_x$ eigenstate.

Mean photon number after a click

A thermal oscillator state has mean occupation $\bar n>0$ and variance $\bar n(\bar n+1)$. Find its conditional mean after an ideal photon-counting click. Explain why this does not mean the click adds energy.

Solution

The formula above gives $\langle n\rangle_{\rm click}=\bar n-1+(\bar n+1)=2\bar n$. The click preferentially selects the high-occupation part of the initial ensemble. Each selected number-state component still loses one quantum: the conditional weighting and the loss are different effects. The unconditional decay channel, including no-click runs, lowers the ensemble mean.

Pointer overlap

In the pointer model, what happens to the coherence $\lvert a\rangle\langle b\rvert$ when the pointer states have overlap $\langle M_b|M_a\rangle=\gamma$?

Solution

After tracing out the pointer, the coefficient of $\lvert a\rangle\langle b\rvert$ is multiplied by $\gamma$:

$$\rho_{ab} \mapsto \gamma\rho_{ab}.$$

If $\gamma=1$, the pointer carries no distinguishing information between $a$ and $b$, and that coherence is preserved. If $\gamma=0$, the pointer states are orthogonal, and the coherence is fully removed in the reduced system state.

References

• V. B. Braginsky and F. Y. Khalili, Quantum Measurement, Cambridge University Press (1992).
• P. Busch, P. Lahti, and R. F. Werner, “Colloquium: Quantum root-mean-square error and measurement uncertainty relations,” Reviews of Modern Physics 86, 1261–1281 (2014). DOI.
• M. Ozawa, “Universally valid reformulation of the Heisenberg uncertainty principle on noise and disturbance in measurement,” Physical Review A 67, 042105 (2003). DOI.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). Publisher.