Infinite Dimensions

16.1 Why Finite Dimensions Are Not Enough

Where exactly does the finite-dimensional derivation stop? We locate the failure in one line of algebra, the trace of a commutator. Then we meet the new kind of operator that infinite dimensions require, and the care its domain needs.

16.1.1 The trace argument

Can finite matrices satisfy the relation \(xp-px=i\hbar I\) that physics uses for position and momentum? For finite matrices, \(\tr(AB)=\tr(BA)\), so every commutator has trace zero. The relation \([x,p]=i\hbar I\) has trace \(i\hbar N\) on the right, so it has no finite-dimensional solution. Does infinite dimension alone rescue it? Not with bounded operators, those with a finite largest stretching factor \(\|A\|\): Aurel Wintner and Helmut Wielandt proved in the 1940s that the commutator of two bounded operators is never a nonzero multiple of the identity.

Why should the relation hold at all? In the usual presentation it is postulated, by analogy with classical mechanics. We will instead derive it from the requirement that position measurements shift correctly when the particle is moved. Whatever its source, the relation forces position or momentum to be unbounded, so we need operators of that kind.

16.1.2 Unbounded operators

What does an unbounded operator look like? Position and momentum have unbounded sets of possible values, so their operators are unbounded. We take the standard representation now, and justify it from symmetry once we have the tools. On the space \(L^2(\mathbb R)\) of square-integrable wave functions, position acts as \((x\psi)(x)=x\,\psi(x)\) and momentum as \((p\psi)(x)=-i\hbar\,\psi'(x)\). Neither is defined on every square-integrable function: \(x\psi\) may fail to be square-integrable, and \(\psi\) may fail to be differentiable. Each is defined only on a subspace, its domain. The domain is dense: every square-integrable function can be approximated as closely as we like by functions in it.

Do domains matter? They do. Call an operator symmetric if \(\braket\phi{A\psi}=\braket{A\phi}\psi\) for all \(\phi\), \(\psi\) in its domain. In finite dimensions this is the same as being Hermitian. In infinite dimensions it is weaker. The adjoint \(A^\dagger\) is defined on every \(\phi\) for which \(\psi\mapsto\braket\phi{A\psi}\) is continuous, and the operator is self-adjoint when \(A^\dagger=A\), domains included. Only self-adjoint operators have a spectral decomposition and generate unitary groups. The momentum operator on a half-line, for instance, is symmetric but has no self-adjoint version, reflecting that a particle confined to a half-line cannot be translated freely. In finite dimensions the distinction does not arise, which is one reason we kept to finite dimensions until now.

16.2 Spectra and Groups

Which finite-dimensional results survive? Two that we relied on need upgrading. Observables were finite sums of projections, and continuous one-parameter groups had Hermitian generators. We first replace eigenvectors by projections onto intervals, then find that generators still exist, and that one of them is momentum.

16.2.1 Continuous spectra

Can position be written as a sum over eigenvalues? No. Position has no eigenvectors in \(L^2(\mathbb R)\): a function concentrated at a single point is not square-integrable. Instead of projections onto eigenvectors, position comes with projections onto intervals: \(P(\Delta)\) multiplies \(\psi\) by the indicator function of the interval \(\Delta\). The probability that a position measurement finds the particle in \(\Delta\) is \(\tr(\rho P(\Delta))\), and for a pure state \(\int_\Delta|\psi(x)|^2dx\).

Does every observable have such projections? The spectral theorem for self-adjoint operators says that every self-adjoint \(A\), bounded or not, has a unique projection-valued measure \(P\) with \(A=\int a\,dP(a)\). This replaces the finite sum \(\sum_ka_kP_k\) that defined an observable. The Born rule takes the same form, \(\tr(\rho P(\Delta))\), and Gleason’s theorem extends to separable infinite-dimensional spaces for countably additive probability assignments. What fails is finiteness itself: there is no finite fiducial set, and \(K\) and \(N\) are infinite.

16.2.2 Stone’s theorem

Does a continuous family of unitaries still have a generator? Call a family \(U(t)\) strongly continuous if \(U(t)\ket\psi\) changes continuously with \(t\) for each fixed \(\ket\psi\).

Stone’s theorem (1932). Every strongly continuous one-parameter group of unitary operators on a Hilbert space has the form \(U(t)=e^{-iHt}\) for a unique self-adjoint operator \(H\), possibly unbounded. Conversely, every self-adjoint operator generates such a group.

This is the infinite-dimensional version of the theorem that gave the Hamiltonian, and its proof requires the spectral theorem. What does it say about moving a particle? Applied to spatial translations, \((T(a)\psi)(x)=\psi(x-a)\), it gives a generator, and a Taylor expansion identifies it: \(T(a)=e^{-iap/\hbar}\) with \(p=-i\hbar\,d/dx\). Momentum is the generator of translations, just as the Hamiltonian is the generator of time translations. The connection is structural, not an analogy with classical mechanics.

16.3 Position and Momentum

With these tools, can we derive position and momentum instead of taking them as given? Symmetry does it. A specific group, the symmetries of space, takes over the role continuous reversibility played. We first require that position shift correctly under translations, which gives the commutation relation. Then a theorem shows that the result is unique.

16.3.1 Translations, boosts, and commutation

What must a position observable do when we move the particle? A position observable for a particle on a line is a projection-valued measure \(P(\Delta)\). Translating the particle by \(a\) should translate its position distribution: \(T(a)^\dagger P(\Delta)T(a)=P(\Delta-a)\). A pair \((P,T)\) with this property is called covariant. Call it irreducible if no proper subspace is left invariant by all the \(P(\Delta)\) and \(T(a)\). Two descriptions are unitarily equivalent if one unitary carries every operator of the first to the corresponding operator of the second, so that they make the same predictions.

Imprimitivity theorem (Mackey, 1949). Every irreducible covariant pair is unitarily equivalent to the standard one on \(L^2(\mathbb R)\), in which positions multiply and translations shift.

So the standard representation we took earlier is forced by symmetry. With \(V(b)=e^{ibx/\hbar}\), which boosts momentum by \(b\), the covariance relation becomes the Weyl relation

\[T(a)\,V(b)=e^{-iab/\hbar}\,V(b)\,T(a),\]

and differentiating in \(a\) and \(b\) at zero gives \([x,p]=i\hbar\) on a suitable domain. The commutation relation is the infinitesimal form of the statement that translations shift positions.

What about the rest of the symmetries of space? Valentine Bargmann analysed the full group of translations, rotations, and boosts in 1954. A projective representation of a group is one in which the unitaries compose correctly only up to a phase, as we allowed for time evolution. Bargmann showed that mass appears as a parameter of these phases, and that states of different total mass cannot be superposed. This is another superselection rule, like the one that separates symmetric and antisymmetric states of identical particles.

16.3.2 The Stone–von Neumann theorem

Is the answer unique for several particles too? Yes, for finitely many.

Stone–von Neumann theorem (1930s). Every irreducible representation of the Weyl relations for finitely many degrees of freedom by strongly continuous unitary groups is unitarily equivalent to the standard representation on \(L^2(\mathbb R^n)\).

So there is essentially one quantum theory of a finite number of particles, once the symmetry of space is imposed: the Schrödinger representation. For infinitely many degrees of freedom, as in quantum field theory, the theorem fails and inequivalent representations exist, which is the source of phenomena such as superselection sectors and spontaneous symmetry breaking.

Where does this leave the derivation from principles? The reconstruction programmes we followed assume finiteness, and no equally complete principle-based derivation of infinite-dimensional quantum theory from operational principles alone exists. The usual route is the one described here: accept the finite-dimensional structure, then fix the infinite-dimensional representation by symmetry. Another is to treat infinite-dimensional systems as limits of finite ones, which works for many purposes but requires care with domains and limits.