Quantum Foundations · Chapter 07
Which ball is the state space of the simplest system, and why does it need complex numbers?
Starting point: Chapter 6: the state space of a two-outcome system is a ball of some dimension.
Anchor example: Two qubits that interact reversibly and continuously.
Principle at work: Local tomography together with continuous reversible interaction between two bits.
Foil: Real quantum theory (the disc) and quaternionic quantum theory (the five-dimensional ball).
This lesson includes explanations, worked examples, and problems with solutions. Advanced results whose proofs are outside the chapter are identified explicitly.
Continuous reversibility has reduced the simplest non-classical system to a ball. Every point of its boundary sphere is a pure state, and opposite points are perfectly distinguishable. But the ball’s dimension \(d\) is still open, and every \(d\ge2\) gives a consistent single system.
A single system cannot decide between the balls, so we turn to pairs. Counting parameters under local tomography excludes most dimensions, and the demand that two bits can interact reversibly excludes all but one: \(d=3\). That ball is the bit of quantum theory. To connect it with the familiar formalism we then ask how to write its states as matrices that keep measurement probabilities visible. The answer forces complex numbers, and in that language the rule for probabilities turns out to be a theorem rather than a postulate.
The next chapter asks whether the same picture extends to systems with more than two distinguishable states.
Every ball is a consistent bit, so the dimension \(d\) must be fixed by something beyond one system. The composition rules give a first test: the parameter count of a pair must fit the power law \(K=N^r\). Counting leaves an infinite list. The demand that two bits can interact reversibly then cuts the list to one entry.
Which balls pass the parameter count? A ball of dimension \(d\) has \(K=d+1\): \(d\) coordinates plus normalization. The power law \(K=N^r\) with \(N=2\) requires
\[d+1=2^r .\]
For a pair, local tomography gives \(K_{AB}=(d+1)^2=4^r\), and the composite capacity gives \(N_{AB}=4\), so the pair fits the same law automatically. So \(d\) must be \(1,3,7,15,\dots\). The classical segment (\(d=1\)) is already excluded by continuity. The disc (\(d=2\), \(K=3\)) and the quaternionic ball (\(d=5\), \(K=6\)) fail the count, because \(3\) and \(6\) are not powers of \(2\).
What becomes of the disc, then? The disc is the bit of a consistent theory, quantum theory with real numbers. That theory keeps its bit and gives up local tomography instead. Its pairs need \(K_{AB}=10\) numbers, one more than the \(3\times3=9\) that local statistics supply. We construct the extra number explicitly once we can write states as matrices.
Counting leaves \(d=3,7,15,\dots\). What can exclude the larger balls? Continuous reversibility, applied to the pair. Two bits that never interact are of little use. The principle asks for reversible transformations that carry product pure states continuously to every pure state of the pair, including correlated ones. A transformation of this kind cannot be merely a product of separate transformations on each bit.
Theorem (Masanes, Müller, Augusiak, and Pérez-García, 2013). Suppose that
Then \(d=3\), and the theory of the pair is quantum theory.
The proof asks which continuous motions of the \((d+1)^2\)-dimensional space of pair states can keep the set of states inside itself. It lies beyond this course. For \(d=3\) such interactions exist, and we write one down when we study dynamics. For larger \(d\) none do.
So the bit of our theory is the three-dimensional ball, with \(K=4\), and the exponent in \(K=N^r\) is \(r=2\). From now on we call this system a qubit, and its state space the Bloch ball.
Problem 7.1. For a ball bit of dimension \(d=7\), compute \(K\) for one bit and for a pair under local tomography, and find the exponent \(r\). Why does counting alone not exclude \(d=7\)?
One bit has \(K=8=2^3\), and a pair has \(K_{AB}=64=4^3\), so \(r=3\). The count \(K=N^r\) holds with an integer exponent, so counting is satisfied. What excludes \(d=7\) is the theorem above: no continuous reversible transformation of a pair of such balls correlates them.
We now know the qubit is a ball, but quantum mechanics is written in vectors and matrices. Can the ball be written so that measurement probabilities stay visible and mixing stays simple? Trying to do so with real matrices reproduces the disc. The third direction of the ball forces complex entries, and the same matrices then show concretely why the disc fails for pairs.
Write a state of the Bloch ball as a vector \(\mathbf r=(x,y,z)\) with \(|\mathbf r|\le1\). The effects \(e_{\pm z}(\mathbf r)=\tfrac12(1\pm z)\) form a measurement that distinguishes the two poles. Can we write each state as a \(2\times2\) matrix whose diagonal holds these two probabilities, and in which mixing states means averaging matrices?
Try real symmetric matrices first. Averaging is linear, so the matrix must depend linearly on \(\mathbf r\). The diagonal is fixed as \(\tfrac12(1+z)\) and \(\tfrac12(1-z)\). The only freedom left is the off-diagonal entry, which must be the same above and below the diagonal. Put \(x/2\) there. Writing
\[I=\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\]
the state becomes \(\tfrac12(I+xX+zZ)\). This uses up every real symmetric matrix with the given diagonal. So real symmetric matrices hold only two directions, \(x\) and \(z\). They describe the disc.
Where can the third direction go? It must not change the diagonal, so it is again an off-diagonal entry. For eigenvalues, and so probabilities, to stay real, the matrix must equal its conjugate transpose; such a matrix is called Hermitian. A Hermitian off-diagonal direction independent of \(X\) needs an imaginary entry:
\[Y=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}.\]
So we assign to \(\mathbf r\) the matrix
\[\rho=\tfrac12\big(I+xX+yY+zZ\big)=\tfrac12\begin{pmatrix}1+z&x-iy\\x+iy&1-z\end{pmatrix}.\]
Does every point of the ball, and only the ball, give a sensible matrix? The sum of the diagonal entries, the trace \(\tr\rho\), is \(1\). The eigenvalues are \((1\pm|\mathbf r|)/2\), which are nonnegative exactly when \(|\mathbf r|\le1\). So the Bloch ball is the set of \(2\times2\) Hermitian matrices with unit trace and nonnegative eigenvalues. These are the density matrices of a qubit. Hermitian \(2\times2\) matrices form a real vector space of dimension \(4\), matching \(K=4\). The matrices \(X\), \(Y\), \(Z\) are called the Pauli matrices.
What do pure states look like? For \(|\mathbf r|=1\) the eigenvalues are \(1\) and \(0\), so \(\rho\) is the projection onto a single direction in \(\mathbb C^2\). Such a projection can be written \(\rho=\psi\psi^\dagger\), where \(\psi\) is a unit column vector, fixed up to a phase factor, and \(\dagger\) is the conjugate transpose. We write \(\psi\) as \(\ket\psi\) and its conjugate transpose as \(\bra\psi\), so \(\rho=\ketbra\psi\psi\). The point \(\mathbf r=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)\) corresponds to \(\ket\psi=\big(\cos\tfrac\theta2,\ e^{i\phi}\sin\tfrac\theta2\big)^T\). State vectors are not postulated. They are a convenient way to write the pure points of the ball.
We have seen that the third direction of the ball needs an imaginary entry. So complex numbers enter because the bit is three-dimensional, and the bit is three-dimensional because of local tomography and reversible interaction. Can we see the failure of local tomography in the real theory directly?
Quantum theory with real numbers is defined by these rules. A state of a system with \(N\) distinguishable states is a real symmetric \(N\times N\) matrix with unit trace and nonnegative eigenvalues. An effect is a real symmetric matrix \(E\) with eigenvalues between \(0\) and \(1\), and its probability is \(\tr(\rho E)\); we derive the same rule for the complex ball in the next section. A pair of bits has \(N=4\), so its states are real symmetric \(4\times4\) matrices.
How many numbers does such a pair need? Real symmetric \(4\times4\) matrices form a ten-dimensional space. Local effects on one bit are combinations of \(I\), \(X\), \(Z\). The joint effects “\(A\) on the first bit and \(B\) on the second” are the Kronecker products \(A\otimes B\), and these span only \(3\times3=9\) dimensions. The tenth direction is \(Y\otimes Y\). It has entries \(-1,1,1,-1\) on the antidiagonal and zeros elsewhere, so it is real and symmetric, although \(Y\) itself is not real.
Now compare the two states
\[\rho_\pm=\tfrac14\big(I\otimes I\pm Y\otimes Y\big).\]
The matrix \(Y\otimes Y\) has eigenvalues \(1,1,-1,-1\), so each \(\rho_\pm\) has eigenvalues \(0,0,\tfrac12,\tfrac12\) and is a valid state. What does a local effect \(A\otimes B\) give? Using \(\tr(M\otimes M')=\tr M\,\tr M'\),
\[\tr\big(\rho_\pm(A\otimes B)\big)=\tfrac14\big(\tr A\,\tr B\pm\tr(AY)\,\tr(BY)\big).\]
For \(A\) and \(B\) among \(I\), \(X\), \(Z\), the traces \(\tr(AY)\) are all zero. So both states give the same probability for every local effect and every correlation between local effects. But the effect \(P=\tfrac12(I\otimes I+Y\otimes Y)\) is real and symmetric, and gives \(\tr(\rho_+P)=1\) and \(\tr(\rho_-P)=0\). The two states are perfectly distinguishable by a joint measurement and completely indistinguishable by local ones.
What are the reversible transformations of the complex ball? Its rotations. For a \(2\times2\) matrix \(U\) with \(U^\dagger U=I\), called unitary, the map \(\rho\mapsto U\rho U^\dagger\) keeps \(\rho\) Hermitian and keeps its trace and eigenvalues. So it maps density matrices onto density matrices linearly, and is a reversible transformation. Take \(U=\mathrm{diag}(e^{-i\alpha/2},e^{i\alpha/2})\). It leaves the diagonal of \(\rho\) unchanged and multiplies the lower off-diagonal entry \((x+iy)/2\) by \(e^{i\alpha}\). That is a rotation of the ball by angle \(\alpha\) about the \(z\)-axis. A standard result of group theory says that every rotation of the ball arises in this way, from a unitary \(U\) of determinant \(1\), with \(U\) and \(-U\) giving the same rotation. So unitary matrices appear as the way rotations of the Bloch ball are written. Reflections of the ball are not continuously connected to the identity, and we return to them when we study symmetries.
Problem 7.2. Find the density matrix and a state vector for the pure state \(\mathbf r=(0,1,0)\).
\(\rho=\tfrac12(I+Y)=\tfrac12\begin{pmatrix}1&-i\\ i&1\end{pmatrix}\). With \(\theta=\pi/2\), \(\phi=\pi/2\), \(\ket\psi=(1,i)^T/\sqrt2\), and indeed \(\ketbra\psi\psi=\tfrac12\begin{pmatrix}1&-i\\ i&1\end{pmatrix}\). No real vector gives this matrix.
The Born rule, which turns states into probabilities, is usually stated as a postulate. We already know every effect on the ball as an affine function. So the question is what those effects look like in matrix language. They become matrices too, and the probability turns out to be a trace, which for pure states is the squared magnitude of an inner product.
What are the effects of the Bloch ball? An effect is an affine functional \(e(\mathbf r)=a+\mathbf b\cdot\mathbf r\) with values between \(0\) and \(1\) on the ball. Its extreme values on the ball are \(a\pm|\mathbf b|\), so the condition is
\[0\le a-|\mathbf b|,\qquad a+|\mathbf b|\le1 .\]
Are all of them measurable? We assume so here. This is the principle of all measurements allowed, which we state in general when we study larger systems. Now write each effect as a matrix, as we did for states: \(E=aI+b_xX+b_yY+b_zZ\). Its eigenvalues are \(a\pm|\mathbf b|\). So the conditions say exactly that \(E\) has eigenvalues between \(0\) and \(1\), written \(0\le E\le I\). The unit effect is \(I\), and a measurement is a list of such matrices summing to \(I\).
How do we compute the probability \(e(\mathbf r)\) from the two matrices \(\rho\) and \(E\)? Each Pauli matrix squares to \(I\), and products of two different Pauli matrices have zero trace. Hence
\[\tr(\rho E)=\tfrac12\tr\big((I+\mathbf r\cdot\vec\sigma)(aI+\mathbf b\cdot\vec\sigma)\big)=\tfrac12(2a+2\,\mathbf r\cdot\mathbf b)=a+\mathbf b\cdot\mathbf r=e(\mathbf r),\]
where \(\vec\sigma=(X,Y,Z)\). So the probability of an outcome is \(\tr(\rho E)\).
What does this give for pure states and sharp effects? Take \(\rho=\ketbra\psi\psi\) and \(E=\ketbra\phi\phi\), a sharp effect: it is \(1\) on \(\ket\phi\) and \(0\) on the opposite point of the sphere. Then \(\tr(\rho E)=|\braket\phi\psi|^2\), the squared magnitude of an inner product. We have derived the Born rule for a qubit from the shape of its state space and the linearity of probabilities.
Does this square conflict with the linearity of probabilities? No. The rule is quadratic in the vectors \(\ket\psi\) but linear in the state \(\rho\), which is all that linearity required. Consider the sharp effect that tests for the pure state \(\mathbf m\), \(E=\tfrac12(I+\mathbf m\cdot\vec\sigma)\), with \(a=\tfrac12\) and \(\mathbf b=\mathbf m/2\). On the pure state \(\mathbf n\) at angle \(\theta\) from \(\mathbf m\) it gives \(\tfrac12(1+\cos\theta)=\cos^2(\theta/2)\). So the two-device statistics with which we began are a consequence of the principles, not their foundation.
Problem 7.3. Compute \(|\braket\phi\psi|^2\) for \(\ket\psi=(1,0)^T\) and \(\ket\phi=(\cos\tfrac\theta2,\sin\tfrac\theta2)^T\), and check that it agrees with \(\tfrac12(1+\mathbf n\cdot\mathbf m)\) for the corresponding Bloch vectors.
\(\braket\phi\psi=\cos\tfrac\theta2\), so the probability is \(\cos^2\tfrac\theta2\). The Bloch vectors are \((0,0,1)\) and \((\sin\theta,0,\cos\theta)\), whose dot product is \(\cos\theta\), and \(\tfrac12(1+\cos\theta)=\cos^2\tfrac\theta2\).
These exercises draw on the whole chapter. Attempt each question before opening a hint, explain your reasoning, then compare it with the worked solution.
Find the density matrix of the state \(\mathbf r=(0.6,0,0.8)\). Is it pure? Give a state vector.
\(\rho=\tfrac12(I+xX+yY+zZ)\).
Pure states have \(|\mathbf r|=1\).
\(\rho=\begin{pmatrix}0.9&0.3\\0.3&0.1\end{pmatrix}\). Since \(|\mathbf r|=1\) it is pure, and \(\ket\psi=(3,1)^T/\sqrt{10}\), because \(\ketbra\psi\psi=\tfrac1{10}\begin{pmatrix}9&3\\3&1\end{pmatrix}\).
What are the eigenvalues of the density matrix of a state with \(|\mathbf r|=0.5\)?
The eigenvalues are \((1\pm|\mathbf r|)/2\).
They must be nonnegative and sum to one.
\(0.75\) and \(0.25\). The state is mixed, a quarter of the way from the center to the surface in terms of eigenvalues.
Is \(E=0.5I+0.3Z\) an effect? What probability does it give on \(\ket0\) and on \(\ket1\)?
Its eigenvalues are \(a\pm|\mathbf b|\).
Use \(\tr(\rho E)\).
Its eigenvalues are \(0.8\) and \(0.2\), both in \([0,1]\), so it is an effect, a noisy test for \(\ket0\). It gives \(0.8\) on \(\ket0\) and \(0.2\) on \(\ket1\).
For \(\rho_\pm=\tfrac14(I\otimes I\pm Y\otimes Y)\): (a) confirm both are valid states; (b) show that they give the same expectation for every product of observables from \(\{I,X,Z\}\); (c) explain why, in complex quantum theory, local tomography still distinguishes them.
\(Y\otimes Y\) is real, with eigenvalues \(\pm1\).
In complex quantum theory, \(Y\) is a local observable.
(a) \(Y\otimes Y\) has eigenvalues \(\pm1\), each twice, so \(\rho_\pm\) have eigenvalues \(0,0,\tfrac12,\tfrac12\): valid. (b) Distinct Pauli products are orthogonal under the trace, so every product from \(\{I,X,Z\}\), other than \(I\otimes I\), has zero expectation in both. (c) In complex quantum theory each party can measure \(Y\), and the correlation \(\langle Y\otimes Y\rangle=\pm1\) is a product of local outcomes. The states differ in a local correlation, as local tomography requires.
Show that \(U=\mathrm{diag}(e^{-i\alpha/2},e^{i\alpha/2})\) rotates the Bloch vector by \(\alpha\) about the \(z\)-axis.
\(U\rho U^\dagger\) multiplies the off-diagonal entry \(\tfrac12(x-iy)\) by a phase.
Compare \((x-iy)e^{-i\alpha}\) with a rotated vector.
The diagonal of \(\rho\) is unchanged, so \(z\) is fixed. The entry \(\tfrac12(x-iy)\) becomes \(\tfrac12(x-iy)e^{-i\alpha}\). Writing \(x'-iy'=(x-iy)(\cos\alpha-i\sin\alpha)\) gives \(x'=x\cos\alpha-y\sin\alpha\) and \(y'=x\sin\alpha+y\cos\alpha\): a rotation by \(\alpha\) about \(z\).
Compute the probability that \(\ket\psi=(\cos\tfrac\pi8,\sin\tfrac\pi8)^T\) passes a test for \(\ket+\) in two ways: from \(|\braket+\psi|^2\), and from the Bloch vectors.
\(\braket+\psi=(\cos\tfrac\pi8+\sin\tfrac\pi8)/\sqrt2\).
\(\ket\psi\) has Bloch angle \(\theta=\pi/4\) in the \(xz\)-plane.
\(|\braket+\psi|^2=\tfrac12(\cos\tfrac\pi8+\sin\tfrac\pi8)^2=\tfrac12(1+\sin\tfrac\pi4)\approx0.854\). The Bloch vectors \((\sin\tfrac\pi4,0,\cos\tfrac\pi4)\) and \((1,0,0)\) are at angle \(\pi/4\), giving \(\tfrac12(1+\cos\tfrac\pi4)\approx0.854\). The Born rule and the geometry of the ball agree, as the derivation of the qubit’s probability rule proved.
Counting alone allows a bit that is a ball of dimension \(7\). (a) What are \(K\) for one such bit and for a pair? (b) Do these satisfy \(K=N^r\)? (c) Which result excludes \(d=7\)?
\(K=d+1\).
\(8=2^3\).
(a) \(K=8\) for one bit and \(64\) for a pair. (b) Yes, with \(r=3\): \(8=2^3\) and \(64=4^3\). (c) The theorem of Masanes, Müller, Augusiak, and Pérez-García: with local tomography, balls of dimension \(d\ne3\) admit no continuous reversible interaction between two bits. Counting is not enough; reversible interaction selects \(d=3\).
A student says: “Complex numbers in quantum mechanics are just convenient bookkeeping; real amplitudes would give the same predictions.” What is wrong?
What state space does a real qubit have?
Recall the network experiments that separate real from complex quantum theory.
Real \(2\times2\) density matrices describe only the disc, so a real qubit lacks the states with \(y\ne0\). One can simulate a complex qubit with an extra real qubit, but then composite systems behave differently, and Renou and collaborators showed that real quantum theory with independent sources cannot reproduce certain network correlations of complex quantum theory, which experiments observed in 2022. Complex numbers are physically significant, and the course traces them to local tomography.
A \(2\times2\) Hermitian matrix with quaternion entries has real diagonal entries and an off-diagonal quaternion. (a) Count its real parameters and find the dimension of the corresponding ball of states. (b) What does this give for \(K\)?
A quaternion has four real components.
Subtract one for unit trace.
(a) Two real diagonal entries and four real components off the diagonal: \(6\) parameters. Unit trace leaves \(5\), so the state space is a \(5\)-dimensional ball. (b) \(K=6\), not a power of \(2\), so counting excludes it; its composites also lack a well-defined tensor product over the quaternions.
Show that \(\ket\psi\) and \(e^{i\phi}\ket\psi\) have the same density matrix, while \((1,e^{i\phi})^T/\sqrt2\) has Bloch vector \((\cos\phi,\sin\phi,0)\). Interpret global and relative phase geometrically.
\(\ketbra\psi\psi\) is unchanged by a common phase.
Read off \(x\) and \(y\) from the off-diagonal entry.
\((e^{i\phi}\ket\psi)(e^{i\phi}\ket\psi)^\dagger=\ketbra\psi\psi\), so a global phase is not a property of the state. For \((1,e^{i\phi})^T/\sqrt2\) the off-diagonal entry \(\rho_{10}=e^{i\phi}/2=\tfrac12(x+iy)\), so \(x=\cos\phi\), \(y=\sin\phi\), \(z=0\). A relative phase is the azimuthal angle on the Bloch sphere: a physical property, measurable by interference.