Quantum Foundations · Chapter 09
What form must reversible change take, and where does the Schrödinger equation come from?
Starting point: Chapter 8: density operators, pure states, and transition probabilities.
Anchor example: A spin in a uniform magnetic field.
Principle at work: Reversible transformations preserve transition probabilities and depend continuously on time.
Foil: Antiunitary evolution, which preserves probabilities but cannot be reached continuously from doing nothing.
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
The reconstruction has given us the states and measurements of every finite system: density operators on \(\mathbb C^N\), effects \(0\le E\le I\), and probabilities \(\tr(\rho E)\). The theorem also said that reversible transformations act as \(\rho\mapsto U\rho U^\dagger\) with \(U\) unitary, but we accepted that as part of a cited result.
Can we see why reversible change must be unitary, and how it unfolds in time? A reversible transformation must preserve the probabilities with which pure states pass each other’s tests. A theorem of Wigner then leaves two kinds of map, unitary and antiunitary, and continuity together with composition removes the second. Time evolution is a continuous family of unitaries, one for each duration, and such a family has a single generator. Differentiating it gives the equation of motion, so no dynamical law needs to be postulated.
The next chapter asks what it means to measure a quantity, such as the generator itself.
What property must every reversible transformation keep? We first find it: the transition probabilities between pure states. A theorem then lists every map that keeps them, and the list has two kinds. We then ask which kind can actually happen as a physical process.
What does a reversible transformation preserve? The transition probability between pure states, \(|\braket\phi\psi|^2\), is the probability that the test for \(\ket\phi\) passes on \(\ket\psi\). A reversible transformation carries the test for a pure state to the test for its image. So it preserves transition probabilities. Any map of pure states with this property is called a symmetry in Wigner’s sense.
Which maps are symmetries? Eugene Wigner answered in 1931.
Wigner’s theorem. Every bijection of the pure states of \(\mathbb C^N\) that preserves all transition probabilities is induced by a map \(\ket\psi\mapsto U\ket\psi\), where \(U\) is either unitary or antiunitary. \(U\) is unique up to a phase factor.
Valentine Bargmann gave a complete proof in 1964. Richard Kadison proved the matching statement for mixed states: every bijection of the density operators that respects mixing has the form \(\rho\mapsto U\rho U^\dagger\) or \(\rho\mapsto U\rho^TU^\dagger\), where \(\rho^T\) is the transpose and \(U\) is unitary. For a qubit the proof is geometric. A bijection of the Bloch ball that respects mixing is a linear map preserving the sphere. So it is a rotation, or a rotation combined with a reflection.
What is the second kind of map? An antiunitary map preserves the magnitude of inner products but conjugates them: \(\braket{U\phi}{U\psi}=\braket\phi\psi^*\). The basic example is complex conjugation of components in a fixed basis. On a qubit, conjugation sends \(\rho\) to its transpose, which changes the sign of the \(Y\) component: \((x,y,z)\mapsto(x,-y,z)\). It reflects the Bloch ball through a plane.
Can an antiunitary map be the result of continuous evolution? Two arguments say no. The first is topological. The identity is unitary, and a continuous path of transformations cannot jump between the two classes. So everything connected to the identity is unitary.
The second argument uses composition. A transformation is a recipe, so we can apply it to A while A is part of a pair, leaving B alone, and the result must still be a state of the pair. Apply the transpose to B alone: on a product \(E\otimes F\) it gives \(E\otimes F^T\), and it extends linearly. This is called the partial transpose. Take the correlated pure state \(\ket{\Phi}=(\ket{00}+\ket{11})/\sqrt2\) of two qubits, whose density matrix is
\[\ketbra\Phi\Phi=\tfrac12\big(\ketbra00\otimes\ketbra00+\ketbra01\otimes\ketbra01+\ketbra10\otimes\ketbra10+\ketbra11\otimes\ketbra11\big).\]
Transposing the second factor of each term swaps \(\ketbra01\) and \(\ketbra10\) there, giving
\[\tfrac12\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\]
in the basis \(\ket{00},\ket{01},\ket{10},\ket{11}\). The vector \(\ket{01}-\ket{10}\) is an eigenvector with eigenvalue \(-1/2\). A negative eigenvalue means some effect would get a negative probability, so the result is not a state. So transposition is not a physical transformation at all, though it remains a symmetry of single-system transition probabilities. Antiunitary operators enter physics as symmetries, such as time reversal, which map one possible history to another, not as processes that happen.
Problem 9.1. Show that \(\rho\mapsto\rho^T\) maps the Bloch vector \((x,y,z)\) to \((x,-y,z)\).
\(\rho=\tfrac12\begin{pmatrix}1+z&x-iy\\x+iy&1-z\end{pmatrix}\), so \(\rho^T=\tfrac12\begin{pmatrix}1+z&x+iy\\x-iy&1-z\end{pmatrix}\), which is the matrix of \((x,-y,z)\). This is a reflection, of determinant \(-1\), so it is not a rotation.
A unitary map describes one reversible change. But time evolution is a family of changes, one for each duration. What must such a family look like if it is consistent and continuous? We find that it has a single generator, a Hermitian matrix, and then ask what that generator means physically.
Consider an isolated system whose dynamics does not change in time. Evolving for time \(s\) and then for time \(t\) is the same as evolving for \(t+s\). So the transformations satisfy \(T_{t+s}=T_tT_s\), with \(T_0\) the identity, and they depend continuously on \(t\), as we argued when motivating continuous reversibility. Each \(T_t\) is represented by a unitary \(U(t)\), fixed only up to a phase. Can the phases be chosen so that the unitaries compose in the same way?
Phase choice (quoted result, Bargmann 1954). For a continuous one-parameter family of this kind, the phases can be chosen so that \(U(t+s)=U(t)U(s)\) and \(U(t)\) is continuous in \(t\).
What can such a family be?
Theorem. Every continuous one-parameter group of \(N\times N\) unitary matrices has the form \(U(t)=e^{-iHt}\) for a unique Hermitian matrix \(H\).
Sketch. A standard result of matrix analysis says that a continuous homomorphism from the real line into a matrix group is automatically differentiable. Differentiating \(U(t+s)=U(t)U(s)\) at \(s=0\) gives \(U'(t)=U(t)A\) with \(A=U'(0)\), whose solution with \(U(0)=I\) is \(U(t)=e^{tA}\). Differentiating \(U(t)^\dagger U(t)=I\) at \(t=0\) gives \(A^\dagger+A=0\), so \(A=-iH\) with \(H\) Hermitian. \(\square\)
In infinite dimensions the same conclusion is Stone’s theorem, and \(H\) may be unbounded. We return to it when we leave finite dimensions.
The matrix \(H\) is the generator of the evolution. How much freedom is left in it? Adding a multiple of the identity changes \(U(t)\) only by a phase factor, which has no effect on states. So the generator is fixed up to an additive constant. To give it physical units we write \(U(t)=e^{-iHt/\hbar}\), where \(\hbar\) converts the energy unit of \(H\) into a frequency, and call \(H\) the Hamiltonian.
Why call the generator energy? That identification is a physical link, not something the principles supply. It is the same link as in classical mechanics, where the generator of time evolution is the energy and is conserved when the dynamics does not depend on time. Is \(H\) conserved here too? Write \(H=\sum_kE_kP_k\), with eigenvalues \(E_k\) and projections \(P_k\) onto its eigenspaces. Each \(P_k\) is an effect, and each commutes with every \(U(t)\). So \(\tr\big(U\rho U^\dagger P_k\big)=\tr(\rho P_k)\): the probabilities of the effects \(P_k\) never change.
What does a generator do in practice? A qubit carried by the spin of a particle in a magnetic field along \(z\) has \(H=\tfrac12\hbar\omega Z\). Then \(U(t)=\mathrm{diag}(e^{-i\omega t/2},e^{i\omega t/2})\), which rotates the Bloch ball about the \(z\)-axis at angular frequency \(\omega\). This is called precession.
Can a generator correlate two qubits, as reversible interaction required when we fixed the ball? Take \(H=\hbar JZ\otimes Z\), and start from the product state \(\ket+\otimes\ket+\), where \(\ket+=(\ket0+\ket1)/\sqrt2\) points along \(+x\). At time \(t=\pi/(4J)\) the evolution is \(U=(I-iZ\otimes Z)/\sqrt2\). The resulting pure state of the pair gives each qubit, on its own, the state \(I/2\): every local effect \(E\otimes I\) has the probability \(\tr(E)/2\). A pure whole whose parts are maximally mixed is as correlated as two qubits can be. So continuous reversible evolution has created the strongest correlation, as the theorem fixing the ball required.
Problem 9.2. A qubit starts in \(\ket+=(1,1)^T/\sqrt2\) and evolves with \(H=\tfrac12\hbar\omega Z\). Find the probability that a test for \(\ket+\) passes at time \(t\).
\(U(t)\ket+=(e^{-i\omega t/2},e^{i\omega t/2})^T/\sqrt2\), and \(\braket{+}{U(t)+}=\cos(\omega t/2)\). The probability is \(\cos^2(\omega t/2)\), oscillating with period \(2\pi/\omega\) as the Bloch vector rotates around the \(z\)-axis.
We have the evolution over any finite time as an exponential of the generator. What law holds from one instant to the next? Differentiating gives the familiar equations of motion, first for states and then, equivalently, for effects. They are now consequences of continuity and reversibility rather than postulates.
How does a state change from instant to instant? For a pure state \(\ket{\psi(t)}=U(t)\ket{\psi(0)}\), differentiating \(U(t)=e^{-iHt/\hbar}\) gives
\[i\hbar\frac{d}{dt}\ket{\psi(t)}=H\ket{\psi(t)},\]
the Schrödinger equation. For a density operator, \(\rho(t)=U(t)\rho(0)U(t)^\dagger\) gives \(\frac{d\rho}{dt}=-\frac i\hbar[H,\rho]\), where \([A,B]=AB-BA\) is the commutator. What if \(H\) itself changes in time? The evolution over a short interval is still generated by the Hamiltonian at that time, and the equations keep the same form.
Which states do not change at all? An eigenvector of \(H\) with eigenvalue \(E\) only acquires the phase \(e^{-iEt/\hbar}\), which changes no prediction, so it is stationary. In a superposition of two eigenvectors, the relative phase turns at the angular frequency \((E_1-E_0)/\hbar\). This is why energy differences, not energies themselves, appear as observable frequencies.
Must it be the state that changes? Only probabilities \(\tr(\rho E)\) are observable. Since \(\tr\big(U\rho U^\dagger E\big)=\tr\big(\rho\,U^\dagger EU\big)\), we may leave the state fixed and evolve the effects instead: \(E(t)=U(t)^\dagger EU(t)\). This is the Heisenberg picture, in which
\[\frac{dE}{dt}=\frac i\hbar[H,E].\]
The two pictures make identical predictions. What does the second one show more clearly? Conservation laws. An effect that commutes with \(H\) does not change, so its probability is constant in every state. The eigenprojections of the energy are the first example.
Problem 9.3. For \(H=\tfrac12\hbar\omega Z\), find \(X(t)\) in the Heisenberg picture.
\(U^\dagger XU\) multiplies the off-diagonal entries of \(X\) by \(e^{\pm i\omega t}\), giving \(X(t)=\cos(\omega t)X-\sin(\omega t)Y\). The \(x\)-component of the spin rotates into the \(y\)-component at angular frequency \(\omega\), the Heisenberg description of precession.
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
For \(H=\tfrac12\hbar\omega Z\), what is the period of the Bloch-vector rotation? What is \(U(2\pi/\omega)\), and why does it not change the state?
\(U(t)=\mathrm{diag}(e^{-i\omega t/2},e^{i\omega t/2})\).
A common phase changes no density matrix.
The Bloch vector returns after \(2\pi/\omega\). Then \(U=\mathrm{diag}(e^{-i\pi},e^{i\pi})=-I\), a global phase, so \(U\rho U^\dagger=\rho\). Unitaries are fixed by rotations only up to sign.
Why does replacing \(H\) by \(H+cI\) change no prediction?
\(e^{-i(H+cI)t/\hbar}=e^{-ict/\hbar}e^{-iHt/\hbar}\).
Global phases cancel in \(U\rho U^\dagger\).
Because \(cI\) commutes with \(H\), the evolution acquires only the global phase \(e^{-ict/\hbar}\), which cancels in \(U\rho U^\dagger\). Only energy differences are observable.
Is the transpose map on a qubit in the connected component \(G_0\) of reversible transformations? Give a geometric reason.
Transposition flips the sign of \(y\).
Rotations have determinant \(+1\).
No. It is the reflection \((x,y,z)\mapsto(x,-y,z)\), of determinant \(-1\), and no continuous path of rotations, all of determinant \(+1\), reaches it.
A qubit starts in \(\ket0\) and evolves with \(H=\tfrac12\hbar\Omega X\). Find the probability of \(\ket1\) at time \(t\) and the time of the first complete transfer.
\(e^{-i\theta X/2}=\cos\tfrac\theta2\,I-i\sin\tfrac\theta2\,X\).
Here \(\theta=\Omega t\).
\(U(t)\ket0=\cos\tfrac{\Omega t}2\ket0-i\sin\tfrac{\Omega t}2\ket1\), so the probability is \(\sin^2(\Omega t/2)\), reaching \(1\) first at \(t=\pi/\Omega\). The Bloch vector rotates about \(x\) from the north to the south pole: a Rabi oscillation.
Given \(U(t)=\begin{pmatrix}\cos t&-i\sin t\\-i\sin t&\cos t\end{pmatrix}\), check that it is a one-parameter group of unitaries and find its generator \(H\) with \(U(t)=e^{-iHt}\).
Differentiate at \(t=0\).
\(U'(0)=-iH\).
\(U(t)=\cos t\,I-i\sin t\,X=e^{-iXt}\), and \(e^{-iXt}e^{-iXs}=e^{-iX(t+s)}\), with \(U^\dagger U=I\). Differentiating at zero gives \(U'(0)=-iX\), so \(H=X\).
Two qubits start in \(\ket+\ket+\) and evolve with \(H=\hbar JZ\otimes Z\). Find the state at \(t=\pi/(4J)\) and the reduced state of the first qubit.
\(Z\otimes Z\) is diagonal with entries \(1,-1,-1,1\).
Compute \(\rho_A\) from the \(2\times2\) array of coefficients \(c\) as \(cc^\dagger\).
The coefficients become \(\tfrac12(e^{-i\pi/4},e^{i\pi/4},e^{i\pi/4},e^{-i\pi/4})\). With \(c\) the \(2\times2\) array, \(cc^\dagger\) has diagonal \(\tfrac12,\tfrac12\) and off-diagonal \(\tfrac14(e^{-i\pi/4}e^{-i\pi/4}+e^{i\pi/4}e^{i\pi/4})=\tfrac14(-i+i)=0\). So \(\rho_A=I/2\): continuous reversible evolution has produced a maximally entangled state.
For \(H=Z\otimes Z\), show in the Heisenberg picture that \(Z\otimes I\) is conserved and \(X\otimes I\) is not.
A quantity is conserved if it commutes with \(H\).
\(XZ=-ZX\).
\([Z\otimes I,Z\otimes Z]=0\), so \(Z\otimes I\) is constant. \([X\otimes I,Z\otimes Z]=(XZ-ZX)\otimes Z=-2iY\otimes Z\ne0\), so \(X\otimes I\) changes; its evolution correlates with the second qubit’s \(Z\).
A student says: “The Schrödinger equation is a postulate justified by its experimental success.” What does the derivation of the dynamics prove, and what does it still import?
Which principle gives continuous one-parameter groups?
Where does energy enter?
Continuous reversibility and time-homogeneity give a continuous one-parameter group of unitaries, and the theorem on one-parameter groups gives \(U(t)=e^{-iHt/\hbar}\), from which the Schrödinger equation follows by differentiation. What is imported is the identification of the generator \(H\) with the energy of the system, a physical link shared with classical mechanics. The form of the equation is derived; which \(H\) describes a given system is physics.
Compute the eigenvalues of the partial transpose of \(\ketbra{\Phi^+}{\Phi^+}\), with \(\ket{\Phi^+}=(\ket{00}+\ket{11})/\sqrt2\), and explain why this excludes transposition as a physical transformation.
The partial transpose swaps the entries \(\ket{00}\bra{11}\) and \(\ket{01}\bra{10}\).
The result is half the swap operator.
The partial transpose is \(\tfrac12S\), with \(S\) the swap, whose eigenvalues are \(1,1,1,-1\). So it has eigenvalue \(-\tfrac12\), not a state. A transformation of one system must remain valid when that system is part of a larger one; transposition fails this, so it is a symmetry of single-system probabilities, not a process.
A qubit evolves with \(H_1=X\) for time \(\pi/4\), then \(H_2=Z\) for time \(\pi/4\) (units with \(\hbar=1\)). Is the result \(e^{-i(X+Z)\pi/4}\)? What does this show about time-dependent Hamiltonians?
The total evolution is \(e^{-iZ\pi/4}e^{-iX\pi/4}\).
\(X\) and \(Z\) do not commute.
No. The product \(e^{-iZ\pi/4}e^{-iX\pi/4}\) differs from \(e^{-i(X+Z)\pi/4}\) because \(X\) and \(Z\) do not commute, as a direct calculation confirms. For a time-dependent Hamiltonian the evolution is an ordered product of short steps, each generated by the Hamiltonian at that time, not the exponential of the integrated Hamiltonian.