Measurement in Circuits

A measurement symbol in a quantum circuit is an interface from a quantum input to a classical record and, only when the circuit declares one, a retained quantum output. Its meaning is not fixed by an outcome probability alone: the circuit must state which register is measured, in which basis, how outcomes are encoded, what state-update rule is used, and whether an output wire survives.

At terminal readout, an outcome distribution may be the only required output. If the measured register is reused, the record is forgotten, or several fine outcomes are merged, the instrument matters. In particular, classically relabeling fine measurement results need not implement the same quantum process as measuring a coarse projector directly.

The discussion below uses finite-dimensional projective measurements to connect circuit symbols with probabilities, retained quantum states and classical records. It also explains how basis rotations and repeated trials turn those probabilities into measured quantities.

Helpful background. The circuit model distinguishes quantum and classical wires. Measurement in a chosen basis explains basis projectors and the difference between the apparatus frame and the rotated-back output state.

Measurement Operations in a Circuit

Let $Q$ denote the measured quantum register, let $R$ collect every other retained quantum register, and let $M$ denote the classical outcome register. Declare the input tensor order as $Q\otimes R$. For computational reference-basis outcome $m$,

$$P_m = |m\rangle\langle m|_Q\otimes I_R, \qquad p_m = \operatorname{Tr}(P_m\rho).$$

An ideal projective branch is the subnormalized map

$$\mathcal I_m(\rho) = P_m\rho P_m, \qquad \rho_m = \frac{\mathcal I_m(\rho)}{p_m} \quad (p_m>0).$$

Here $\mathcal I_m$ retains $Q\otimes R$, and $\operatorname{Tr}_{QR}\mathcal I_m(\rho)=p_m$. The branch trace already contains its probability. Normalizing every block before adding it to a classical record would give the wrong weights. A general quantum instrument replaces the projectors by completely positive, trace-nonincreasing maps into a declared quantum output space whose sum is trace preserving. The associated effects fix the probabilities, but they do not by themselves fix the disturbance or the surviving output system. Projective measurement derives the Lüders update; quantum instruments describes more general outcome maps.

If the outcome is ignored but the quantum output is retained, the result is

$$\Phi_{QR}(\rho) = \sum_m\mathcal I_m(\rho).$$

If the classical record and the measured quantum register are both retained, their normalized classical-quantum state is

$$\omega_{MQR} = \sum_m |m\rangle\langle m|_M \otimes \mathcal I_m(\rho), \qquad \operatorname{Tr}\omega_{MQR}=1.$$

Each quantum block inside this sum is still subnormalized. If $Q$ is destroyed or discarded while $R$ survives, the corresponding cq state is instead

$$\omega_{MR} = \sum_m |m\rangle\langle m|_M \otimes \operatorname{Tr}_Q\!\left[\mathcal I_m(\rho)\right], \qquad \operatorname{Tr}\omega_{MR}=1.$$

A different destructive instrument may map into a replacement or flag space; that output space must be declared rather than inferred from a meter symbol. A terminal outcome distribution alone does not specify a post-measurement register that is unused or absent.

When a local measurement acts on $Q$ and $Q$ is then discarded, there is a useful simplification. For a given outcome effect $E_m$ on $Q$, the unnormalized remote state is

$$\sigma_{R,m}=\operatorname{Tr}_Q[(E_m\otimes I_R)\rho_{QR}].$$

It depends on the effect, not on which local instrument realizes that effect. This does not determine the state of a retained $Q$. Summing over outcomes gives the original $\rho_R$, so a local unread measurement cannot change the remote state.

Basis Choice and Pre-Measurement Rotations

Let $\{|e_a\rangle\}$ be the fixed apparatus basis and $\{|f_a\rangle\}$ the desired orthonormal basis. Define the column matrix $V$ by

$$V|e_a\rangle=|f_a\rangle.$$

The desired-basis measurement is implemented by applying $V^\dagger$ before fixed-reference readout, since

$$\langle e_a|V^\dagger|\psi\rangle = \langle f_a|\psi\rangle.$$

This is an active physical rotation, not merely a passive rewrite of coordinates. If $V^\dagger=AB$, then $B$ acts first in time. After a selected reference-basis outcome, a reusable output in the original frame requires applying $V$; a terminal probability-only measurement does not. Multiplying any $|f_a\rangle$ by an individual phase leaves its projector unchanged, but reversing noncommuting gates can change the implemented PVM.

For sign-sensitive Y-basis readout, this page fixes

$$Y= \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}, \qquad |+y\rangle = \frac{|0\rangle+i|1\rangle}{\sqrt2}, \qquad |-y\rangle = \frac{|0\rangle-i|1\rangle}{\sqrt2}.$$

With $S=\operatorname{diag}(1,i)$,

$$V_Y=SH, \qquad V_Y^\dagger=HS^\dagger,$$

so the physical pre-readout order is $S^\dagger$ first and $H$ second. For

$$\rho = \frac{I+\mathbf r\cdot\boldsymbol{\sigma}}{2},$$

the same convention gives

$$p(\pm y) = \frac{1\pm r_y}{2}, \qquad r_y = \operatorname{Tr}(\rho Y) = -2\operatorname{Im}\rho_{01}.$$

For the derivation with arbitrary basis phases and output-frame choices, see Measurement in a Chosen Basis.

Outcomes, Classical Registers, and Bitstring Order

A drawn meter does not determine a software record convention. For an $n$-qubit register, state whether a displayed string is $q_{n-1}\cdots q_0$ or $q_0\cdots q_{n-1}$, which bit is most significant, and whether the register list uses the same order as the string. Changing a displayed label is classical re-encoding; permuting quantum wires is a physical operation.

Outcome bits also need an interpretation. For computational measurement of $Z$, a common convention is

$$b=0\longleftrightarrow z=+1, \qquad b=1\longleftrightarrow z=-1, \qquad z=(-1)^b.$$

This map must be stated rather than assumed. A qutrit stored in two classical bits likewise needs a code and an explicit invalid word. An invalid recorded code must be accounted for, for example as a detector error or a rejected trial; silently discarding it changes the sampled ensemble.

If only part of a record is reported, its probability is a marginal. For displayed order $q_1q_0$,

$$p(q_1=0) = p(00)+p(01).$$

Probabilities of exclusive recorded outcomes are added; amplitudes are not. One shot produces one outcome. The full distribution is an ensemble statement estimated from repeated, comparably prepared trials.

Conditional States, Unread Measurements, and Discard

For $p_m>0$, the selected state is

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

When $p_m=0$, the subnormalized branch is the zero operator and no normalized posterior state exists. Postselection therefore requires both the selected state and its success probability; it is not a deterministic transformation.

Three different objects answer three different questions:

• the family $\{\rho_m,p_m\}$ predicts a state after a known result;
• $\Phi_{QR}(\rho)=\sum_m\mathcal I_m(\rho)$ predicts the retained state after the result is forgotten;
• $\omega_{MQR}$ retains the classical result together with $Q\otimes R$, while $\omega_{MR}$ uses $\operatorname{Tr}_Q$ when $Q$ is discarded.

Tracing the classical record out of $\omega_{MQR}$ gives $\Phi_{QR}(\rho)$, not the unmeasured input in general. For example, unread computational Lüders measurement removes computational-basis coherences. This is a physical dephasing channel induced by the measurement interaction and discarded record, not a unitary on the measured qubit alone.

The State Update Rule explains the general selective/nonselective distinction. The compact Born Rule formula card is useful for probability lookup, but it does not determine the instrument or the output space.

Classical Postprocessing and Quantum Coarse Graining

Suppose a fine outcome $m$ is converted to a reported label $y$ by a classical stochastic map satisfying

$$q(y|m)\ge 0, \qquad \sum_y q(y|m)=1.$$

The resulting outcome operation is

$$\mathcal J_y = \sum_m q(y|m)\mathcal I_m.$$

For deterministic relabeling $y=f(m)$, this sums the subnormalized branches whose recorded labels share the same reported value. It describes measuring the fine alternatives first and then hiding their labels.

A direct coarse Lüders measurement is generally different. If

$$Q_y = \sum_{m\in y}P_m,$$

then

$$Q_y\rho Q_y = \sum_{m,n\in y}P_m\rho P_n$$

retains cross terms within the coarse subspace, whereas

$$\sum_{m\in y}P_m\rho P_m$$

removes them. The two procedures can agree on the distribution of $y$ while making different predictions for later interference. For degenerate observables, see Degenerate Measurements and Lüders Rule.

For two-qubit computational parity,

$$Q_0 = \frac{I+Z\otimes Z}{2} = P_{00}+P_{11},$$ $$Q_1 = \frac{I-Z\otimes Z}{2} = P_{01}+P_{10}.$$

An even/odd result is not a full Bell-state label. Direct parity primitives and ancilla constructions are described in Multi-Qubit Gates; the four states are introduced in Bell States.

What survives a measurement?

Terminal record-only use. If no quantum output is consumed later, a declared probability distribution and record encoding may be sufficient. It does not imply that a measured system survives.

Destructive measurement. A destructive detector has an output space that omits the measured subsystem or replaces it by a declared flag state. Destructiveness is an instrument property, not a consequence of drawing the meter at the end of a diagram.

Retained quantum output. If later gates act on the measured register, the circuit must specify the selected or unread state, the frame in which it is returned, and any reset. Identical first-outcome probabilities do not guarantee identical later behavior. Reuse alone also does not imply a quantum nondemolition measurement: repeatability of the measured observable is an additional property.

Measured classical control. A recorded result may choose a later operation, but the intermediate state differs from a coherent controlled unitary. Under suitable conditions, replacing early measurement and feedforward by coherent control followed by a later measurement preserves the final classical–quantum output; Exercise 7 proves a simple case. It does not preserve every possible intermediate interference experiment.

Mid-circuit measurement and feedforward follows the causal use of outcomes, reset and later branches. Detector timing, assignment accuracy and repeatability are discussed in control, readout and calibration.

Shot counts and sampling uncertainty

For $N$ stable independent repetitions with $K$ outcome categories,

$$(n_1,\ldots,n_K) \sim \operatorname{Multinomial}(N;p_1,\ldots,p_K), \qquad \widehat p_m = \frac{n_m}{N}.$$

The estimator covariance is

$$\operatorname{Cov}(\widehat p_m,\widehat p_n) = \frac{p_m\delta_{mn}-p_mp_n}{N}.$$

The negative off-diagonal covariance expresses normalization: category frequencies are not independent. A model standard error substitutes the declared $p_m$; a plug-in standard error substitutes $\widehat p_m$. Neither is automatically a confidence interval. Hoeffding’s bounded-variable inequality can provide a finite-sample tail bound, but it still assumes a stable sampling model.

Finite-shot variation is only one source of uncertainty. Preparation error, detector response and drift can change the sampled probabilities; correlated shots can also invalidate the multinomial model. A stationary noisy detector may still give IID outcomes, but then $p_m$ denotes its actual recorded distribution, not the ideal Born probabilities. Sampling formulas alone do not bound the bias between those distributions.

For example, the empirical discrepancy from an ideal model can be summarized by total variation distance,

$$D_{\mathrm{TV}}(\widehat p,p)=\frac12\sum_m|\widehat p_m-p_m|.$$

A nonzero value can arise from finite sampling even when the ideal model is correct. Conversely, agreement on one distribution cannot establish the full measurement instrument. Quantum measurement as estimation develops likelihoods and confidence regions. SPAM errors separates preparation and readout effects, and measurement-error mitigation explains inference with a calibrated detector-response model.

Worked example: Y-basis readout of a phase state

Prepare

$$|\psi\rangle=\frac{|0\rangle+e^{i\pi/3}|1\rangle}{\sqrt2}.$$

To measure $Y$, apply $S^\dagger$, then $H$, then computational-basis readout. Label zero as the $+1$ eigenvalue of $Y$ and one as $-1$. Since $r_y=\sin(\pi/3)=\sqrt3/2$,

$$\begin{aligned} p_0&=\frac{2+\sqrt3}{4}\simeq0.933013,\\ p_1&=\frac{2-\sqrt3}{4}\simeq0.066987. \end{aligned}$$

If the measured qubit survives a reference-basis Lüders readout, its selected state is $|0\rangle$ or $|1\rangle$ in the rotated frame. Applying $V=SH$ afterward gives $|+y\rangle$ or $|-y\rangle$ in the original frame. Ignoring the result yields a mixture with eigenvalues $p_0,p_1$ and purity $p_0^2+p_1^2=7/8$.

Reversing the pre-rotation order would apply $S^\dagger H$. Its effective observable is $X$, giving probabilities $3/4,1/4$ on this input. The eigenstate test in Exercise 1 is a quick way to detect this mistake.

Worked example: identical parity, different coherence

Take $|\Phi^+\rangle=(|00\rangle+|11\rangle)/\sqrt2$. Fine computational readout yields $00$ or $11$ with probability $1/2$ each. Both have even parity. If the detailed outcome is hidden, the retained data state is

$$\rho_{\rm fine}=\frac12\bigl(|00\rangle\langle00|+|11\rangle\langle11|\bigr).$$

A direct even-parity Lüders measurement instead leaves $|\Phi^+\rangle$ unchanged. Both instruments predict $\langle Z\otimes Z\rangle=1$, but their later predictions differ:

Quantity	Fine readout, detail hidden	Direct parity Lüders readout
purity	$1/2$	$1$
squared fidelity with the initial Bell state	$1/2$	$1$
trace distance from the initial Bell state	$1/2$	$0$
$\langle X\otimes X\rangle$	$0$	$1$

The $X\otimes X$ measurement detects coherence that the parity distribution alone cannot see. Exercise 4 derives the entries.

Exercises

1. Basis-Rotation Direction and Time Order

Let $|f_a\rangle=V|e_a\rangle$. Prove which operation must precede fixed reference-basis readout. Apply the result to X and Y measurements, and use $|+y\rangle$ to diagnose the error made by applying $V$ instead.

Solution

The amplitude for reference outcome $a$ after applying $V^\dagger$ is

$$\langle e_a|V^\dagger|\psi\rangle = \langle f_a|\psi\rangle,$$

which is exactly the desired-basis amplitude. For X readout, $V=H=V^\dagger$. For Y readout,

$$V=SH, \qquad V^\dagger=HS^\dagger.$$

The rightmost operator acts first, so the chronological sequence is $S^\dagger$, then $H$, then reference readout. With

$$|+y\rangle = \frac{|0\rangle+i|1\rangle}{\sqrt2},$$

we have

$$S^\dagger|+y\rangle = |+\rangle, \qquad HS^\dagger|+y\rangle = H|+\rangle = |0\rangle.$$

Thus the $+y$ state produces outcome zero with certainty. If $V=SH$ is applied instead, direct multiplication gives

$$SH|+y\rangle = \frac{1+i}{2} \left( |0\rangle+|1\rangle \right),$$

which has computational probabilities $1/2,1/2$. The wrong operation fails the eigenstate test.

2. A Reset Instrument and Its cq Record

Consider

$$\rho= \begin{pmatrix} 3/4&1/4\\ 1/4&1/4 \end{pmatrix}, \qquad K_0=|0\rangle\langle0|, \qquad K_1=|0\rangle\langle1|.$$

Treat $K_m:\mathcal H_{Q_{\mathrm{in}}}\to \mathcal H_{Q_{\mathrm{out}}}$ as a same-dimensional reset instrument. The output label denotes the reset register after the interface; the map by itself does not claim that the physical input subsystem was destroyed.

Find the POVM, outcome branches and probabilities, selected outputs, unread output, and normalized classical-quantum state. Compare the result with the computational Lüders instrument.

Solution

The effects are

$$K_0^\dagger K_0 = |0\rangle\langle0| = P_0, \qquad K_1^\dagger K_1 = |1\rangle\langle1| = P_1,$$

and $P_0+P_1=I$. Thus the POVM is the computational PVM. The subnormalized branches are

$$\mathcal I_0(\rho) = K_0\rho K_0^\dagger = \frac34|0\rangle\langle0|,$$ $$\mathcal I_1(\rho) = K_1\rho K_1^\dagger = \frac14|0\rangle\langle0|.$$

Their traces give $p_0=3/4$ and $p_1=1/4$. Both normalized conditional outputs equal $|0\rangle\langle0|$, and the unread output is therefore

$$\mathcal I_0(\rho)+\mathcal I_1(\rho) = |0\rangle\langle0|.$$

The normalized cq state is

$$\omega_{M Q_{\mathrm{out}}} = \frac34 |0\rangle\langle0|_M \otimes |0\rangle\langle0|_{Q_{\mathrm{out}}} +\frac14 |1\rangle\langle1|_M \otimes |0\rangle\langle0|_{Q_{\mathrm{out}}}.$$

A computational Lüders instrument has the same effects and probabilities, but its selected outputs are $|0\rangle\langle0|$ and $|1\rangle\langle1|$, with unread output $\operatorname{diag}(3/4,1/4)$. Equal POVMs therefore do not imply equal instruments.

3. Bitstring Order, Marginals, and Re-encoding

For the following distribution displayed in $q_2q_1q_0$ order:

• $000:1/8$, $001:1/8$, $010:1/4$, $011:0$;
• $100:0$, $101:1/4$, $110:1/8$, $111:1/8$.

Rewrite only the displayed labels in $q_0q_1q_2$ order, compute all one-bit marginals, and find $p(q_2\oplus q_0=1)$. Explain why relabeling is not a quantum wire permutation.

Solution

Reversing each displayed label, while leaving its probability attached to the same physical outcome, gives:

• $000:1/8$, $100:1/8$, $010:1/4$, $110:0$;
• $001:0$, $101:1/4$, $011:1/8$, $111:1/8$.

In the original order,

$$p(q_0=1) = p(001)+p(011)+p(101)+p(111) = \frac12,$$ $$p(q_1=1) = p(010)+p(011)+p(110)+p(111) = \frac12,$$ $$p(q_2=1) = p(100)+p(101)+p(110)+p(111) = \frac12.$$

Odd $q_2\oplus q_0$ occurs for $001,011,100,110$. Only $001$ and $110$ carry nonzero probability here, so

$$p(q_2\oplus q_0=1) = \frac18+\frac18 = \frac14.$$

The operation changed notation in the classical record only. A physical wire permutation would apply a SWAP network to quantum subsystems and could affect later gates.

4. Fine Readout versus Coarse Parity Measurement

For $|\Phi^+\rangle$, compare fine computational measurement followed by hiding the fine record with a direct even-parity Lüders measurement. Compute purity, squared fidelity, trace distance, and the $X\otimes X$ and $Z\otimes Z$ expectations.

Solution

The input density operator is

$$\rho_\Phi = \frac12 \left( P_{00}+P_{11} +|00\rangle\langle11| +|11\rangle\langle00| \right).$$

Fine computational measurement removes the two cross terms:

$$\rho_{\mathrm{fine}} = \frac12(P_{00}+P_{11}).$$

By contrast, $Q_0|\Phi^+\rangle=|\Phi^+\rangle$, so direct even-parity Lüders measurement leaves $\rho_\Phi$ unchanged. Since

$$\rho_{\mathrm{fine}}^2 = \frac14(P_{00}+P_{11}),$$

its purity is $1/2$. Under the squared-fidelity convention,

$$\langle\Phi^+| \rho_{\mathrm{fine}} |\Phi^+\rangle = \frac12.$$

The difference

$$\rho_\Phi-\rho_{\mathrm{fine}} = \frac12 \left( |00\rangle\langle11| +|11\rangle\langle00| \right)$$

has eigenvalues $+1/2$ and $-1/2$ in the even subspace. Its trace norm is one, so the trace distance is $1/2$. Because $X\otimes X$ exchanges $|00\rangle$ and $|11\rangle$,

$$\langle X\otimes X\rangle_{\mathrm{fine}}=0, \qquad \langle X\otimes X\rangle_{\Phi}=1.$$

Both states lie entirely in the even computational-parity subspace, so $\langle Z\otimes Z\rangle=1$ for both. Their parity statistics agree even though their retained coherence does not.

5. Shot Counts and Estimator Uncertainty

Let $K\sim\operatorname{Binomial}(400,3/4)$ and $\widehat p=K/400$. Compute the mean, variance, and standard deviation of the count and estimator, then state what these quantities exclude.

Solution

For a binomial count,

$$\mathbb E K = Np = 400\frac34 = 300,$$ $$\operatorname{Var}K = Np(1-p) = 400\frac34\frac14 = 75.$$

Therefore

$$\sigma_K = \sqrt{75} = 8.660254.$$

Dividing the count by $400$ gives

$$\operatorname{Var}\widehat p = \frac{\operatorname{Var}K}{400^2} = \frac3{6400},$$ $$\sigma_{\widehat p} = \frac{\sqrt3}{80} = 0.0216506.$$

These are sampling standard deviations under stable independent Bernoulli trials with recorded probability $p=3/4$. They are not confidence intervals or bounds on preparation/readout bias. Drift and shot correlations may invalidate this binomial model; the outcome statistics also do not determine a state-update rule.

6. Unread Measurement Is Not a Unitary

Start from $\rho_+=|+\rangle\langle+|$. Compute the unread computational Lüders output, its purity, its squared fidelity and trace distance from the input, and explain why no qubit-only unitary implements the map.

Solution

In the computational basis,

$$\rho_+ = \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.$$

The unread Lüders output is

$$P_0\rho_+P_0+P_1\rho_+P_1 = \frac12 \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} = \frac I2.$$

Its purity is

$$\operatorname{Tr} \left( \frac I2 \right)^2 = \frac12.$$

Because the input is pure, the squared fidelity is

$$\langle+| \frac I2 |+\rangle = \frac12.$$

Also,

$$\rho_+-\frac I2 = \frac X2,$$

whose eigenvalues are $\pm1/2$. Hence the trace distance is $\tfrac12\|\rho_+-I/2\|_1=1/2$. A qubit-only unitary preserves the spectrum and purity, so it cannot map a pure state to $I/2$. The lost coherence is carried by a measurement record or environment that has been ignored.

7. Terminal Readout versus Deferred Measurement

Let $P_m=|m\rangle\langle m|_c$. Compare measurement of $c$ followed by classically applying $X^m$ to target $t$ with a CNOT followed by later computational measurement of $c$. State the exact equivalence and its limitations.

Solution

The measured-and-conditioned branch for outcome $m$ is

$$(P_m\otimes X^m) \rho (P_m\otimes X^m).$$

The coherent operation is

$$U_{\mathrm{CNOT}} = \sum_mP_m\otimes X^m.$$

Projecting its control output gives

$$(P_m\otimes I)U_{\mathrm{CNOT}} = P_m\otimes X^m.$$

Therefore applying CNOT and eventually producing the same computational control record yields exactly the same subnormalized branch and hence the same final cq state as measuring first and applying the classically selected $X^m$.

Before that final measurement, however, CNOT retains off-diagonal control blocks

$$(P_m\otimes X^m) \rho (P_n\otimes X^n), \qquad m\ne n.$$

The early measured process removes them. Later interference on the control can therefore distinguish the processes. Detector destructiveness, reset, controller latency, noisy classically selected gates, and logical or physical resource accounting also prevent an unrestricted equivalence claim.

8. Qutrit Fourier readout

Let $\omega=e^{2\pi i/3}$ and

$$|f_a\rangle=\frac1{\sqrt3}\sum_{j=0}^2\omega^{ja}|j\rangle, \qquad V=F_3.$$

Measure $|0\rangle$ in this basis, encoding outcomes $0,1,2$ as $00,01,10$. Find the pre-rotation, probabilities, expected counts and frequency covariance for $N=900$ independent trials. Which classical word is unused, and what does the probability calculation tell us about a quantum output?

Solution

Apply $F_3^\dagger$ before computational readout. Since $\langle f_a|0\rangle=1/\sqrt3$, each valid outcome has probability $1/3$, and word $11$ has zero probability in this ideal model. Each expected count is $300$, with

$$\operatorname{Var}(n_a)=900\frac13\frac23=200, \qquad \sigma_{n_a}=10\sqrt2.$$

Each frequency has standard deviation

$$\sigma_{\widehat p_a}=\frac{\sqrt{200}}{900} =\frac{\sqrt2}{90}\simeq0.0157135.$$

For distinct outcomes,

$$\operatorname{Cov}(\widehat p_a,\widehat p_b) =-\frac{(1/3)^2}{900}=-\frac1{8100}.$$

The negative covariance is required because the three frequencies sum to one. These probabilities do not specify whether the measured qutrit survives. A terminal destructive instrument can have maps $\mathcal I_a(\rho)=\operatorname{Tr}(P_a\rho)\tau$ into a fixed one-dimensional output; a Lüders instrument $P_a\rho P_a$ instead retains the selected qutrit state. Both have the same effects $P_a=|f_a\rangle\langle f_a|$.

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