Quantum Foundations · Chapter 11
How does a state change when a measurement reveals an outcome?
Starting point: Chapter 10: projective measurements, compatibility, and commutation.
Anchor example: Three Stern–Gerlach devices in sequence: z, then x, then z.
Principle at work: Repeatable measurements that disturb the state as little as their outcome requires.
Foil: Measurement as passive reading of pre-existing values.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
We can now say what is measured: an observable is a labelled projective measurement, and two observables can be measured together exactly when they commute. Every prediction so far concerns the outcome of one measurement on a freshly prepared state.
But a system often survives a measurement, and can be measured again. What state should we assign to it afterwards? Textbooks postulate the answer, the projection rule. We first find that outcome probabilities alone cannot fix it, because devices that measure the same observable can leave different states behind. So we must say which measurements the rule describes. Two conditions on an ideal measurement, that it confirms its own result and disturbs nothing it need not, turn out to force the rule. With an update rule in hand, sequences of measurements become calculable, and they show that learning about a state has a cost.
The next chapter asks whether the outcomes of measurements could have been fixed in advance after all.
To ask what state follows a measurement, we need a description of a measurement that records more than its effects. We bundle each outcome’s probability with the state that follows it. Then we ask whether the effects alone fix that state, and find that they do not.
What must a complete description of a measurement contain, if the system survives it? For each outcome \(k\), both its probability and the state that follows. Bundle these into one map \(\mathcal I_k\). It takes the initial state \(\rho\) to an unnormalized state \(\mathcal I_k(\rho)\), whose trace is the probability of outcome \(k\) and whose normalized form is the state after outcome \(k\). The collection \(\{\mathcal I_k\}\) is called an instrument.
What do we know about these maps? Each \(\mathcal I_k\) respects mixing, so it is linear. The sum \(\sum_k\mathcal I_k\) describes measuring and forgetting the result, so it preserves the trace.
Does an instrument determine its measurement? Yes. The effect \(E_k\) is the unique operator with \(\tr(\rho E_k)=\tr\mathcal I_k(\rho)\) for all \(\rho\). Does the measurement determine the instrument?
Take two devices that both measure \(Z\) on a qubit, with projections \(P_0=\ketbra00\) and \(P_1=\ketbra11\). The first leaves the system in \(\ket0\) after outcome \(0\) and in \(\ket1\) after outcome \(1\). The second does the same, then flips the qubit with \(X\). Both have identical outcome probabilities for every input, yet they leave different states. A third device could even discard the system and prepare a fresh \(\ket+\) whatever the outcome.
So the update is not a property of the observable. It depends on how the measurement is physically carried out, which the effects do not record. A general update rule must therefore refer to a special class of measurements. Which class? The natural candidate is the measurements that disturb the system least.
Problem 11.1. For the second device above (measure \(Z\), then apply \(X\)), what is the probability that an immediate repetition of the same device gives the same outcome?
After outcome \(0\) the qubit is in \(\ket1\), so a repetition gives outcome \(1\) with certainty, and vice versa. The probability of the same outcome is \(0\): the device is not repeatable, although it measures the same observable as the first device.
What should “disturbs least” mean? We split it into two conditions on a measurement of a projective observable. The first, repeatability, fixes the update only when each outcome has one state. The second, minimal disturbance, fixes it in general, and the two together force a unique rule.
What should it mean that a measurement “found the value \(a_k\)”? A natural answer is that testing again confirms it. Call a measurement repeatable if an immediate second application gives the same outcome with certainty.
What does this say about the state after the measurement? Take a projective measurement \(\{P_k\}\), and let \(\rho_k\) be the state after outcome \(k\). Repeatability says \(\tr(\rho_kP_k)=1\). A state that gives probability \(1\) to a projection lies in its range, so \(P_k\rho_kP_k=\rho_k\). If the range of \(P_k\) is one-dimensional, spanned by \(\ket k\), this fixes the answer: \(\rho_k=\ketbra kk\), whatever the initial state.
Does repeatability fix the state when the range is larger? No. After outcome \(0\) of a measurement of \(Z\otimes I\) on two qubits, any state in the range of \(\ketbra00\otimes I\) passes the repeated test. That includes states in which the second qubit has been scrambled entirely.
What extra condition stops the measurement from scrambling the second qubit? It should disturb nothing it does not have to. We make this precise.
Minimal disturbance. Every effect that is compatible with the measured observable, and is certain before the measurement, remains certain after it.
Does this, with repeatability, fix the update? Take a pure state \(\ket\psi\), and let \(\ket{\psi_j}=P_j\ket\psi/\|P_j\ket\psi\|\) for each \(j\) with \(P_j\ket\psi\ne0\). These vectors lie in different eigenspaces, so they are orthogonal. Let \(F\) be the projection onto their span. Then \(F\) commutes with every \(P_j\), so it is compatible with the measured observable. And \(\ket\psi=\sum_jP_j\ket\psi\) lies in the range of \(F\), so \(F\) is certain before the measurement.
What happens after outcome \(k\)? By minimal disturbance the new state still gives \(F\) probability \(1\), so it lies in the range of \(F\). By repeatability it also lies in the range of \(P_k\). A vector \(\sum_jc_j\ket{\psi_j}\) lies in the range of \(P_k\) only if \(c_j=0\) for \(j\ne k\). So the only state left is \(\ket{\psi_k}\):
\[\ket\psi\ \longmapsto\ \frac{P_k\ket\psi}{\|P_k\ket\psi\|}\qquad\text{after outcome }k.\]
What about mixed states? The unnormalized map \(\mathcal I_k\) sends \(\ketbra\psi\psi\) to a multiple of \(P_k\ketbra\psi\psi P_k\). Its trace must be the probability \(\braket\psi{P_k\psi}\), which is also the trace of \(P_k\ketbra\psi\psi P_k\), so the multiple is \(1\). Every density operator is a mixture of pure states, and \(\mathcal I_k\) is linear, so \(\mathcal I_k(\rho)=P_k\rho P_k\) for every \(\rho\). Normalizing gives
\[\rho\ \longmapsto\ \frac{P_k\rho P_k}{\tr(\rho P_k)}\qquad\text{after outcome }k.\]
This is the Lüders rule. Gerhard Lüders proposed it in 1951, and Fedor Herbut showed in 1969 that minimal requirements of this kind characterize it in general. The rule keeps every coherence within the eigenspace of the outcome and removes those between eigenspaces. In the two-qubit example it leaves the second qubit exactly as it was.
Does every measurement obey it? No. The Lüders rule describes ideal measurements, not every measurement. Real devices realize other instruments, and we find the general form of an instrument when we study open systems. Nor is the rule a dynamical process inside the theory. It updates the description given the recorded outcome. Whether something physical “collapses” is the measurement problem, which the principles do not settle.
Problem 11.2. A measurement of \(Z\otimes I\) is applied to \((\ket{00}+\ket{01}+\ket{10}-\ket{11})/2\). Find the probability of outcome \(0\) and the Lüders post-measurement state.
Projecting onto first-qubit \(0\) gives \((\ket{00}+\ket{01})/2\), of squared norm \(1/2\). The post-measurement state is \(\ket0\otimes(\ket0+\ket1)/\sqrt2\). The second qubit is in \(\ket+\), the coherence within the outcome-\(0\) subspace having been kept.
With an update rule, we can compute what a sequence of measurements does. We first work out the three-device sequence and see that the order of measurements matters. Then we ask whether any measurement can learn about a state without changing it, and find that it cannot.
What does the update rule predict for the three devices? After the first device the qubit is in \(\ket0\). The \(x\)-device keeps the outcome \(\ket+\) with probability \(|\braket+0|^2=1/2\), and by the Lüders rule the state becomes \(\ket+\). The third device then finds \(\ket1\) with probability \(|\braket1+|^2=1/2\). Overall, a quarter of the particles leaving the first device exit the lower port of the third. Without the middle device none do.
Does the order of measurements matter? Measuring \(Z\) and then \(X\) gives a different joint distribution of outcomes from measuring \(X\) and then \(Z\), for most initial states. So the failure of observables to commute shows up here as the dependence of sequential statistics on order.
Can a measurement learn something about a state without changing it? Suppose an ideal measurement \(\{P_k\}\), obeying the Lüders rule, leaves a pure state \(\ket\psi\) undisturbed whatever the outcome. Then each nonzero \(P_k\ket\psi\) is a multiple of \(\ket\psi\). Since \(\ket\psi=\sum_kP_k\ket\psi\), and the nonzero terms lie in different eigenspaces, only one term can be nonzero. So \(\ket\psi\) lies in a single eigenspace, and gives one outcome with certainty.
Now suppose the measurement leaves two nonorthogonal pure states \(\ket\psi\) and \(\ket\phi\) undisturbed. Each lies in a single eigenspace. Different eigenspaces are orthogonal, and \(\braket\phi\psi\ne0\), so both lie in the same eigenspace. Both then give the same outcome with certainty, and the measurement reveals nothing about which was present.
So gaining any information about nonorthogonal states requires disturbing at least one of them. The same holds for every instrument, not only ideal ones; the proof describes the instrument as an interaction with a probe, and we give it when we study open systems. This limit is the basis of quantum key distribution, in which two parties build a shared secret key and detect an eavesdropper by the disturbance she must cause. It is a consequence of the derived structure, not an extra postulate. Classical theory has a counterpart only for mixed states: a classical pure state can be read without any disturbance at all.
Problem 11.3. In the three-device sequence, replace the middle \(x\)-device by one along a direction at angle \(\theta\) from \(z\) in the \(xz\)-plane, again keeping its upper port. What fraction leaves the lower port of the final \(z\)-device?
The middle device passes \(\cos^2(\theta/2)\) of the particles and leaves them in the state at angle \(\theta\). The final device sends a fraction \(\sin^2(\theta/2)\) of those down. The total is \(\cos^2(\theta/2)\sin^2(\theta/2)=\tfrac14\sin^2\theta\), largest, \(1/4\), at \(\theta=\pi/2\).
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
A measurement of \(Z\) is applied to \(\ket+\). Give the probability of outcome \(0\) and the Lüders post-measurement state.
\(|\braket0+|^2\).
Project and normalize.
Probability \(\tfrac12\); the state becomes \(\ket0\).
A device measures \(Z\) and then prepares \(\ket+\) regardless of the outcome. Does it measure the same observable as an ideal \(Z\) measurement? Is it repeatable?
Compare outcome probabilities.
What does a second application give?
It has the same outcome probabilities, so it measures \(Z\). It is not repeatable: after it, \(\ket+\) gives each outcome with probability \(\tfrac12\). The observable does not determine the update.
In the three-device experiment, what fraction leaves the lower port of the final \(z\)-device if the middle \(x\)-device is removed?
The state after the first device is \(\ket0\).
\(\braket10=0\).
None. The first device leaves \(\ket0\), which the final device passes upward with certainty.
A two-qubit parity measurement has projections \(P_{\text{even}}=\ketbra{00}{00}+\ketbra{11}{11}\) and \(P_{\text{odd}}=I-P_{\text{even}}\). Apply it to \((\ket{00}+\ket{01}+\ket{11})/\sqrt3\). Find the probabilities and the Lüders states.
Project onto each subspace.
Keep coherence within the subspace.
Even: \((\ket{00}+\ket{11})/\sqrt3\), probability \(\tfrac23\), state \((\ket{00}+\ket{11})/\sqrt2\). Odd: \(\ket{01}/\sqrt3\), probability \(\tfrac13\), state \(\ket{01}\). The even outcome preserves the superposition of \(\ket{00}\) and \(\ket{11}\), producing an entangled state: minimal disturbance does not mean no change.
On \(\ket0\), compare the joint outcome distributions of (a) measuring \(Z\) then \(X\), and (b) measuring \(X\) then \(Z\), with Lüders updates.
After \(Z\), the state is \(\ket0\).
After \(X\), the state is \(\ket\pm\).
(a) \(Z\) gives \(0\) with certainty, then \(X\) gives \(\pm\) with probability \(\tfrac12\) each: outcomes \((0,+)\) and \((0,-)\), each \(\tfrac12\). (b) \(X\) gives \(\pm\) with probability \(\tfrac12\); from \(\ket\pm\), \(Z\) gives each value with probability \(\tfrac12\): four outcomes, each \(\tfrac14\). The order matters, reflecting that \(X\) and \(Z\) do not commute.
A gentle measurement of \(Z\) has Kraus operators \(K_\pm=\sqrt{\tfrac12(I\pm\eta Z)}\). Apply it to \(\ket+\). For outcome \(+\), find the post-measurement Bloch vector and interpret the trade-off between information and disturbance.
\(K_+=\mathrm{diag}\big(\sqrt{(1+\eta)/2},\sqrt{(1-\eta)/2}\big)\).
Normalize \(K_+\ket+\) and compute \(x\) and \(z\).
\(K_+\ket+\propto(\sqrt{1+\eta},\sqrt{1-\eta})\), so the post-measurement state is pure with \(z=\eta\) and \(x=\sqrt{1-\eta^2}\). For \(\eta=0\) nothing is learned and nothing is disturbed; for \(\eta=1\) the outcome is sharp and the \(x\)-component is destroyed. The more the measurement reveals about \(Z\), the more it disturbs the state’s coherence.
Between a \(z\)-device keeping the upper port and a final device along \(-z\), insert \(n-1\) devices at angles \(k\pi/n\), \(k=1,\dots,n-1\), each keeping its upper port. Find the fraction passing all of them for \(n=10\) and \(n=100\).
Each step passes \(\cos^2(\pi/2n)\).
There are \(n\) steps.
Each step passes \(\cos^2(\pi/2n)\), so the total is \(\cos^{2n}(\pi/2n)\): about \(0.78\) for \(n=10\) and \(0.98\) for \(n=100\). Frequent measurements can drag a state from \(\ket0\) to \(\ket1\) almost without loss, a version of the quantum Zeno effect, made possible by the Lüders update.
A student says: “The Lüders rule proves that measurement physically collapses the wave function.” What does the rule describe, and what does it leave open?
From what assumptions was the rule derived?
Recall the measurement problem.
The rule describes how the state assigned to a system should be updated after an ideal measurement with a recorded outcome, derived from repeatability and minimal disturbance. It is not a dynamical law within the theory. Whether something physical collapses, or the update reflects new information, or branches persist, is the measurement problem, which the principles do not settle.
Give two different instruments for the observable \(Z\otimes I\) that are both repeatable but differ in their effect on the second qubit, and identify the one satisfying minimal disturbance.
After outcome \(a\), any state in the range of \(\ketbra aa\otimes I\) is repeatable.
Try applying a unitary to the second qubit.
Instrument 1: \(\rho\mapsto(P_a\otimes I)\rho(P_a\otimes I)\). Instrument 2: the same followed by \(X\) on the second qubit. Both leave the first qubit in \(\ket a\), so both are repeatable. Instrument 2 changes an initially certain property of the second qubit, such as \(Z\otimes I\) and \(I\otimes Z\) both definite in \(\ket{00}\), while instrument 1 does not. Minimal disturbance selects instrument 1, the Lüders rule.
A probe interacts unitarily with a qubit prepared in \(\ket0\) or \(\ket+\), leaving the qubit undisturbed and the probe in \(\ket a\) or \(\ket b\). Show that \(\braket ba=1\), and conclude that the probe gains no information.
Unitaries preserve inner products.
\(\braket0+=1/\sqrt2\ne0\).
Inner products before and after must agree: \(\braket0+=\braket0+\braket ba\). Dividing by \(1/\sqrt2\) gives \(\braket ba=1\), so the probe states are identical and carry no information. To learn anything about which of two nonorthogonal states was present, at least one must be disturbed.