Quantum Foundations · Chapter 15
How should a system be described when it interacts with an environment we ignore?
Starting point: Chapters 9, 11, and 13: unitary dynamics, instruments, and reduced states.
Anchor example: A qubit coupled to a second qubit that is then discarded.
Principle at work: Purification: every state is the marginal of a pure state, unique up to reversible change of the purifying system.
Foil: Classical theory, in which mixed states have no purification.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Until now every system was either isolated, and evolved unitarily, or measured, and updated by the Lüders rule. We also found that a pure whole can have mixed parts, with all its information in correlations.
But real systems are never isolated. They interact with environments we do not track, and their evolution, viewed alone, is not reversible. How should we describe such a system? We start from the property that pure wholes can have mixed parts, and find that every mixed state is part of some pure whole, in an essentially unique way. That suggests describing every irreversible process as a reversible interaction with an unseen environment, followed by ignoring it. The resulting maps, channels, include every measurement as well. Finally we ask what they explain about everyday objects, which never show interference, and what they leave unexplained about single outcomes.
The next chapter leaves finite dimensions behind.
A correlated pure pair has a mixed part. Is every mixed state of this kind? We first show that it is, by constructing a pure extension. Then we find that the extension is unique up to a reversible change of the added system, and see a surprising consequence for what a distant measurement can do.
Can every mixed state be the part of a pure whole? Try to build one. Write a density operator in its eigenbasis, \(\rho=\sum_i\lambda_i\ketbra ii\). Introduce a second system B with orthonormal states \(\ket i\), and form
\[\ket\Psi=\sum_i\sqrt{\lambda_i}\,\ket i_A\ket i_B .\]
Its partial trace over B is \(\rho\). So every state is the reduced state of a pure state of a larger system, called a purification. In the Hilbert-space theory this is a theorem, a direct consequence of the partial trace.
Could we start from purification instead? Giulio Chiribella, Giacomo Mauro D’Ariano, and Paolo Perinotti did so in 2011. They took purification, with the uniqueness below, as a principle, and showed that together with five informational axioms it singles out quantum theory. The foil is classical theory, where no mixed state has a purification.
What does purification change about mixed states? A mixed state need not express ignorance of which pure state was prepared. It can instead express that the system is correlated with something else. Both readings give the same predictions for the system, just as different mixtures that give the same state cannot be told apart.
Is the purification unique? Suppose \(\ket\Psi\) and \(\ket{\Psi'}\) are two purifications of \(\rho\) on the same system B. Write both in Schmidt form, using the same eigenbasis of \(\rho\) on A. Both have Schmidt coefficients \(\sqrt{\lambda_i}\), so they differ only in their orthonormal vectors on B, and a unitary \(V\) on B relates those. So \(\ket{\Psi'}=(I\otimes V)\ket\Psi\). All purifications are equivalent, differing by a reversible change of the purifying system alone.
What can this do? Suppose \(\rho\) is written as a mixture of pure states, \(\rho=\sum_kp_k\ketbra{\psi_k}{\psi_k}\). A suitable measurement on B, with outcome \(k\) occurring with probability \(p_k\), leaves A in \(\ket{\psi_k}\). So any decomposition of \(\rho\) can be produced remotely. This is called steering. Erwin Schrödinger noticed it in 1935, and the general statement is due to Lane Hughston, Richard Jozsa, and William Wootters. Can steering send a signal? No: A’s overall state remains \(\rho\) until the outcome is communicated.
Problem 15.1. Purify \(\rho=\mathrm{diag}(3/4,1/4)\) and check the reduced state.
\(\ket\Psi=\tfrac{\sqrt3}2\ket{00}+\tfrac12\ket{11}\). The partial trace keeps \(\tfrac34\ketbra00+\tfrac14\ketbra11\) and removes the cross terms, giving \(\rho\).
If every mixed state is part of a pure whole, perhaps every irreversible process is part of a reversible one. We first compute what a reversible interaction with an environment does to a system viewed alone. Then we ask whether every irreversible process arises this way, and find that it does, by a theorem we sketch. Measurements turn out to be a special case.
What happens to a system that interacts reversibly with an environment we then ignore? Let the environment E start in a pure state \(\ket{e_0}\), let the two interact through a unitary \(U\), and then discard E. The system’s state becomes
\[\Phi(\rho)=\tr_E\!\big[U(\rho\otimes\ketbra{e_0}{e_0})U^\dagger\big]=\sum_kK_k\rho K_k^\dagger,\qquad K_k=\bra kU\ket{e_0},\]
where \(\ket k\) is any basis of E. The Kraus operators \(K_k\) satisfy \(\sum_kK_k^\dagger K_k=I\), which makes \(\Phi\) preserve trace. The unitary on the larger system is called a dilation of \(\Phi\).
What kind of map is \(\Phi\)? Applied to part of a larger system, it still maps states to states, because it came from a unitary on a larger system. A map with this property is called completely positive, and a completely positive, trace-preserving map is a channel. Transposition, which we excluded as a physical process, maps states to states but is not completely positive. A simple example of a channel is dephasing: with \(K_0=\sqrt{1-p}\,I\) and \(K_1=\sqrt p\,Z\), the diagonal of \(\rho\) is unchanged and the off-diagonal entries are multiplied by \(1-2p\).
Does every channel have a dilation? William Stinespring proved in 1955 that it does. In the reconstruction of Chiribella, D’Ariano, and Perinotti the same conclusion follows from purification. For finite dimensions the argument is short.
Sketch. Let \(\Phi\) be a channel on \(\mathbb C^N\). Apply it to half of the unnormalized correlated state \(\sum_i\ket i\ket i\), giving \(J=\sum_{ij}\Phi(\ketbra ij)\otimes\ketbra ij\). Since \(\Phi\) is completely positive, \(J\) has no negative eigenvalues, so it is a sum \(J=\sum_k\ketbra{v_k}{v_k}\). Define \(K_k\) by \(K_k\ket i=(I\otimes\bra i)\ket{v_k}\). Comparing entries gives \(\Phi(\ketbra ij)=\sum_kK_k\ketbra ijK_k^\dagger\), and by linearity \(\Phi(\rho)=\sum_kK_k\rho K_k^\dagger\). Preserving the trace for every \(\rho\) forces \(\sum_kK_k^\dagger K_k=I\). Then \(V\ket\psi=\sum_kK_k\ket\psi\ket k\) preserves inner products, and any such map extends to a unitary \(U\) with \(U\ket\psi\ket{e_0}=V\ket\psi\). That \(U\) is a dilation of \(\Phi\). \(\square\)
Can measurements be dilated too? Couple the system to a probe by a unitary, then make a projective measurement on the probe. Suppose outcome \(k\) of the probe leaves the system in the unnormalized state \(K_k\rho K_k^\dagger\). Then the measured effect on the system is \(E_k=K_k^\dagger K_k\), and the instrument is \(\mathcal I_k(\rho)=K_k\rho K_k^\dagger\).
Can every general measurement be realized this way? Yes. Given effects \(\{E_k\}\), take \(K_k=\sqrt{E_k}\); then \(\sum_kK_k^\dagger K_k=I\), and the construction above gives the unitary. This is Naimark’s theorem: every POVM is a projective measurement on a larger system. The Lüders rule is the special case \(K_k=P_k\). The three-outcome trine measurement on a qubit is realized by coupling the qubit to a probe and measuring the probe in a basis of three states. Every instrument, more generally, can be realized by a unitary interaction with a probe followed by a projective measurement of the probe, as Masanao Ozawa showed in 1984. So nothing in the general theory of measurement goes beyond unitary interaction, projective tests, and discarding systems.
This lets us finish an argument left open: can any measurement, ideal or not, learn about nonorthogonal states without disturbing them? Realize it by a probe, and suppose it leaves two nonorthogonal pure states \(\ket\psi\) and \(\ket\phi\) undisturbed. After the interaction the system is still in a pure state, and a pure part of a pure whole means a product, by the Schmidt decomposition. So \(U\ket\psi\ket0=\ket\psi\ket a\) and \(U\ket\phi\ket0=\ket\phi\ket b\). Unitaries preserve inner products, so \(\braket\phi\psi=\braket\phi\psi\braket ba\). Since \(\braket\phi\psi\ne0\), \(\braket ba=1\). The probe states are identical, and the probe carries no information about which state was present.
Problem 15.2. Apply dephasing with \(p=0.1\) to \(\ketbra++\). What state results, and what is the probability that a test for \(\ket+\) passes?
\(\ketbra++\) has all entries \(\tfrac12\). Dephasing keeps the diagonal and multiplies the off-diagonal by \(0.8\), giving \(\begin{pmatrix}0.5&0.4\\0.4&0.5\end{pmatrix}\). The test passes with probability \(\tr(\rho\ketbra++)=0.5+0.4=0.9\).
Why does a dust grain never show interference? Channels answer this: an environment that records which state a system is in removes the coherence between those states. We first compute the effect and how fast it grows with the size of the environment. Then we ask whether the same mechanism explains why a measurement has a single outcome, and find that it does not.
What happens to a superposition when the environment records it? Take a qubit in \(a\ket0+b\ket1\), and let it interact with an environment so that the environment ends in \(\ket{e_0}\) or \(\ket{e_1}\) depending on the qubit’s basis state: \(a\ket0\ket{e_0}+b\ket1\ket{e_1}\). The qubit’s reduced state keeps its diagonal \(|a|^2,|b|^2\), and its off-diagonal entries are multiplied by \(\braket{e_1}{e_0}\). If the environment states are nearly orthogonal, the coherence is nearly gone, and in this basis the qubit behaves like a classical mixture. This is decoherence.
Why is decoherence fast for large systems? Because environments have many parts. Suppose each of \(n\) environmental particles is left in states with overlap \(\cos\theta\). Then the total overlap is \(\cos^n\theta\). For \(\theta=0.5\) and \(n=10\) this is about \(0.27\); for \(n=100\) it is about \(2\times10^{-6}\). A dust grain scattering air molecules loses its coherence between positions almost instantly. The basis singled out by the interaction, here \(\ket0,\ket1\), is called the pointer basis.
Does decoherence explain why a measurement has one outcome? Treat a measuring apparatus as a quantum system, and let it interact unitarily with a qubit in a superposition. The result is an entangled superposition of “qubit 0, pointer reads 0” and “qubit 1, pointer reads 1.” Decoherence makes the two branches unable to interfere. It does not select one of them. Yet a measurement, as our framework assumes, produces a single classical record. The tension between universal unitary dynamics and definite outcomes is the measurement problem.
Does the reconstruction resolve it? No. The framework takes outcomes as primitive, and the principles then fix the theory that predicts their statistics. Interpretations answer the remaining question in different ways.
A principles-first derivation narrows the question, but it does not answer it.
Problem 15.3. The decoherence argument shows that the qubit’s reduced state is \(\mathrm{diag}(|a|^2,|b|^2)\). Does this mean the qubit is in \(\ket0\) or \(\ket1\), with those probabilities?
Not by itself. The joint state is still a pure entangled superposition; the reduced state is the same as that of a proper mixture, so no measurement on the qubit alone can tell the difference, but the global state includes both branches. Reading the reduced state as ignorance of a definite outcome is precisely the step the measurement problem questions.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Give a purification of \(I/2\).
Use eigenvalues \(\tfrac12,\tfrac12\).
\(\sum_i\sqrt{\lambda_i}\ket i\ket i\).
\((\ket{00}+\ket{11})/\sqrt2\), or any state related to it by a unitary on the second system.
Check that the dephasing Kraus operators \(K_0=\sqrt{1-p}\,I\), \(K_1=\sqrt p\,Z\) satisfy \(\sum K_k^\dagger K_k=I\).
\(Z^\dagger Z=I\).
Add.
\((1-p)I+pZ^\dagger Z=(1-p)I+pI=I\).
For the trine effects \(E_k=\tfrac23\ketbra{\psi_k}{\psi_k}\), what are the Kraus operators \(K_k=\sqrt{E_k}\), and what is the state after outcome \(k\)?
The square root of a rank-one projection times \(\tfrac23\) is \(\sqrt{2/3}\) times the projection.
\(K_k\rho K_k^\dagger\) is proportional to \(\ketbra{\psi_k}{\psi_k}\).
\(K_k=\sqrt{2/3}\,\ketbra{\psi_k}{\psi_k}\), and after outcome \(k\) the state is \(\ket{\psi_k}\), whatever the input.
Amplitude damping has \(K_0=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix}\) and \(K_1=\begin{pmatrix}0&\sqrt\gamma\\0&0\end{pmatrix}\). Verify completeness and apply the channel to \(\ketbra11\).
Compute \(K_0^\dagger K_0+K_1^\dagger K_1\).
\(K_1\) maps \(\ket1\) to \(\sqrt\gamma\ket0\).
\(K_0^\dagger K_0=\mathrm{diag}(1,1-\gamma)\) and \(K_1^\dagger K_1=\mathrm{diag}(0,\gamma)\), summing to \(I\). Applied to \(\ketbra11\): \((1-\gamma)\ketbra11+\gamma\ketbra00\). The excited state decays to the ground state with probability \(\gamma\).
Alice and Bob share \((\ket{00}+\ket{11})/\sqrt2\). Show that Bob measuring in the \(Z\) basis leaves Alice in \(\ket0\) or \(\ket1\), and in the \(X\) basis leaves her in \(\ket+\) or \(\ket-\), each with probability \(\tfrac12\). Why can Alice not tell which basis Bob used?
\((\ket{00}+\ket{11})/\sqrt2=(\ket{++}+\ket{--})/\sqrt2\).
Both ensembles average to \(I/2\).
Rewriting the state as \((\ket{++}+\ket{--})/\sqrt2\) shows that an \(X\) outcome \(\pm\) leaves Alice in \(\ket\pm\), while a \(Z\) outcome leaves her in \(\ket0\) or \(\ket1\). Both ensembles average to \(I/2\), so without Bob’s message Alice’s statistics are identical. Bob steers which decomposition of \(I/2\) Alice holds, without signaling.
A qubit decoheres through \(n\) environmental particles, each leaving states with overlap \(0.99\). Compute the remaining coherence factor for \(n=50\), and find how many particles reduce it below \(0.01\).
The factor is \(0.99^n\).
Solve \(0.99^n<0.01\).
\(0.99^{50}\approx0.61\). We need \(n\ln0.99<\ln0.01\), so \(n>\ln0.01/\ln0.99\approx458\). Even nearly negligible individual records accumulate into effective decoherence.
Realize dephasing with probability \(p\) as a unitary dilation: the qubit controls a rotation \(R_y(\phi)\) of an environment qubit starting in \(\ket0\). Find \(\phi\) in terms of \(p\).
The coherence factor is \(\braket{e_1}{e_0}=\bra0R_y(\phi)\ket0=\cos(\phi/2)\).
Dephasing multiplies coherences by \(1-2p\).
The environment ends in \(\ket{e_0}=\ket0\) or \(\ket{e_1}=R_y(\phi)\ket0\), with overlap \(\cos(\phi/2)\). Setting \(\cos(\phi/2)=1-2p\) gives \(\phi=2\arccos(1-2p)\). For \(p=\tfrac12\), \(\phi=\pi\) and the environment records the qubit perfectly, destroying all coherence.
A student says: “Decoherence solves the measurement problem, because the reduced state becomes a classical mixture.” What is missing?
Is the global state still a superposition?
Recall the difference between a reduced state and a proper mixture.
Decoherence makes the reduced state of the system look like a mixture in the pointer basis and prevents interference between branches. The global state remains an entangled superposition of all outcomes, and nothing in the unitary dynamics selects one. Decoherence explains why interference is unobservable locally, not why a single outcome is recorded.
The states \(\tfrac{\sqrt3}2\ket{00}+\tfrac12\ket{11}\) and \(\tfrac{\sqrt3}2\ket{01}+\tfrac12\ket{10}\) both purify \(\mathrm{diag}(\tfrac34,\tfrac14)\). Find the unitary on the second system relating them.
The first system’s basis states are paired with \(\ket0,\ket1\) in one and \(\ket1,\ket0\) in the other.
Try \(X\).
\((I\otimes X)\) maps \(\ket{00}\mapsto\ket{01}\) and \(\ket{11}\mapsto\ket{10}\), carrying the first purification to the second. All purifications on the same system are related by a unitary on the purifying system, as the uniqueness of purification states.
Explain, using a unitary dilation, why every channel obtained from a unitary on a larger system is completely positive, and why transposition is not of this form.
Apply the dilation to half of an entangled state.
Transposition on half of \(\ket{\Phi^+}\) gives a negative eigenvalue.
If the channel acts on A while A is entangled with R, the joint result is \(\tr_E[(U\otimes I_R)(\rho_{AR}\otimes\ketbra{e_0}{e_0})(U\otimes I_R)^\dagger]\), a partial trace of a state, hence a state. So such channels remain valid on parts of larger systems. Transposition applied to half of \(\ket{\Phi^+}\) gives eigenvalue \(-\tfrac12\), so it cannot come from any dilation.