Mathematics for quantum mechanics, chapter 6: functional analysis

A particle can have infinitely many possible position outcomes. Finite matrices then give way to operators on spaces of functions. We need a notion of convergence, and we must specify which functions an operator can act on. This chapter develops those distinctions before constructing spectral probabilities and time evolution.

Convention. Our inner product is conjugate-linear in its first slot; the Alphabeta library uses the reverse convention. The exact translation is given at the end of this chapter.

Read chapter 2 for complex vectors and inner products, chapter 3 for exponentials, and chapter 4 for integration and convergence theorems. We use the convention \(\langle x,y\rangle\) conjugate-linear in \(x\), linear in \(y\). The probability interpretations belong to chapter 5.

Proof boundary. The Hilbert-space geometry, operator examples, trace calculations, tensor construction, and consequences of a spectral representation are proved below. The general existence and uniqueness of that representation uses the linked spectral-theorem proof and its dependencies. That substantial external proof is part of the reading route, not a theorem proved by our multiplication example. The converse direction of Stone’s theorem and the analytic construction of the differential Hamiltonian are further reading, outside the internal proofs here.

6.1. Convergence belongs to the space

Unless stated otherwise, the Hilbert spaces \(H\) and \(K\) in this chapter are nonzero complex separable spaces, as in the framework. Orthonormal-basis sums are finite or countable.

Definition 6.1. Norm, metric, completeness, and Hilbert space

A norm on a complex vector space \(X\) is a function \(\|x\|\ge0\), zero only at \(x=0\), satisfying \(\|cx\|=|c|\|x\|\) and \(\|x+y\|\le\|x\|+\|y\|\). It defines the metric \(d(x,y)=\|x-y\|\), a distance satisfying symmetry, separation, and the triangle inequality. A sequence converges to \(x\) when \(\|x_n-x\|\to0\). It is Cauchy when for every \(\varepsilon>0\) all sufficiently late pairs satisfy \(\|x_n-x_m\|<\varepsilon\). A space is complete if every Cauchy sequence converges in it; a complete normed space is a Banach space.

An inner product has the properties from chapter 2: conjugate symmetry, linearity in its second slot, and \(\langle x,x\rangle>0\) for \(x\ne0\). Its norm is \(\sqrt{\langle x,x\rangle}\). A complete inner-product space is a Hilbert space. A subset is dense if every vector is a limit of vectors in that subset. A space is separable if it has a countable dense subset. A subset is closed if it contains the limits of all its convergent sequences.

Alphabeta definition

Worked example for Definition 6.1

Three models. \(\mathbb C^n\), with \(\langle x,y\rangle=\sum_{j=1}^n\overline{x_j}y_j\), is complete because each coordinate of a Cauchy sequence converges in \(\mathbb C\). The space \(\ell^2\) consists of sequences \(x=(x_1,x_2,\ldots)\) for which \(\sum_j|x_j|^2<\infty\), with inner product \(\sum_j\overline{x_j}y_j\). This is \(L^2\) for counting measure on the positive integers. The space \(L^2(\mathbb R^3)\) consists of measurable functions with \(\int|f|^2<\infty\), identifying functions equal almost everywhere, and has inner product \(\int\overline f g\). Chapter 4 proves completeness of \(L^2\), including the counting-measure case. Both integrals and series defining these inner products converge absolutely by Cauchy–Schwarz.

An incomplete example. Let \(c_{00}\) be the finitely supported sequences with the \(\ell^2\) norm. The sequences \(x^{(N)}=(1,1/2,\ldots,1/N,0,\ldots)\) are Cauchy because \(\sum_{j>N}j^{-2}\le1/N\). Any norm limit would have coordinate \(1/j\) for every \(j\), since \(|x_j-y_j|\le\|x-y\|\); it therefore cannot belong to \(c_{00}\). Its completion is \(\ell^2\): truncations approximate every square-summable sequence.

Finite sequences with rational real and imaginary parts form a countable dense subset of \(\ell^2\), by truncation followed by rational approximation. Rational step functions on bounded rational boxes give a countable dense subset of \(L^2(\mathbb R^3)\), by the following measure approximation argument. Thus the spaces used here are separable.

Proof of separability of Euclidean L²

For a Lebesgue-measurable set \(A\) of finite measure and \(\varepsilon>0\), the box-cover definition of Lebesgue outer measure supplies countably many bounded boxes covering \(A\) with total volume less than \(\mu(A)+\varepsilon\). Enlarge their endpoints to rational numbers with extra total volume less than \(\varepsilon\), assigning an error budget \(\varepsilon2^{-j}\) to box \(j\). Let \(U\) be the union and \(F_N\) the first \(N\) boxes. Then \(\mu(U\setminus A)<2\varepsilon\), and \(\mu(A\setminus F_N)\) is bounded by the tail of the convergent sum of box volumes. For large \(N\), \(\mu(A\mathbin{\triangle}F_N)<3\varepsilon\), where the symmetric difference consists of points belonging to exactly one set. Thus \(\|\mathbf1_A-\mathbf1_{F_N}\|_2^2<3\varepsilon\). A finite union of rational boxes can be partitioned, up to boundary null sets, into finitely many disjoint rational boxes by sorting all its endpoints.

For \(f\in L^2\), first truncate to \(f\mathbf1_{\{|x|\le N,\ |f|\le N\}}\); dominated convergence makes the error tend to zero. On this bounded finite-measure region, round the real and imaginary values to a rational grid of mesh \(\delta\). The pointwise error is at most \(\sqrt2\delta\), so its squared integral tends to zero with \(\delta\). This produces a simple function with finitely many rational complex values on measurable sets of finite measure. Approximate each such set by a finite rational-box union as above. The triangle inequality bounds the resulting function error by the sum of its coefficient magnitudes times the square roots of the symmetric-difference measures. These can be made arbitrarily small. There are only countably many finite lists of rational boxes and rational coefficients, proving separability. The proof works in every finite Euclidean dimension.

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6.2. Orthogonal projections and continuous functionals

Theorem 6.2. Nearest vector in a closed subspace

For a closed linear subspace \(M\) of a Hilbert space \(H\), every \(x\in H\) has a unique decomposition \(x=m+r\), with \(m\in M\) and \(r\in M^\perp\). Here \(M^\perp=\{r:\langle v,r\rangle=0\text{ for all }v\in M\}\). The vector \(m=P_Mx\) uniquely minimizes \(\|x-v\|\) over \(v\in M\). The orthogonal projection \(P_M\) is linear, satisfies \(P_M^2=P_M\), and obeys \(\|P_Mx\|\le\|x\|\).

Worked example for Theorem 6.2

Why closedness matters. In \(\ell^2\), the vector \((1/j)_j\) has distance zero from \(c_{00}\), but has no nearest vector there. In \(L^2(\mathbb R)\), functions vanishing outside a measurable set \(B\) form a closed subspace, and \(P_Bf=\mathbf1_Bf\). Indeed \(\|\mathbf1_{B^c}f\|\le\|f\|\), so limits preserve the vanishing condition, and the two pieces \(\mathbf1_Bf\) and \(\mathbf1_{B^c}f\) are orthogonal.

Theorem 6.3. Riesz representation for a linear functional

A bounded linear functional is a linear map \(F:H\to\mathbb C\) with \(|F(x)|\le C\|x\|\) for some \(C\). There is a unique \(z\in H\) such that \(F(x)=\langle z,x\rangle\). Its functional norm \(\sup_{\|x\|=1}|F(x)|\) equals \(\|z\|\).

Worked example for Theorem 6.3

Example. For \(z\in\ell^2\), the functional \(F(x)=\sum_j\overline{z_j}x_j\) is bounded and has norm \(\|z\|\). This follows from Cauchy–Schwarz, with equality at \(x=z/\|z\|\) when \(z\ne0\). Riesz says every bounded linear functional on \(\ell^2\) has exactly this form. On \(L^2\), the corresponding expression is \(F(f)=\int\overline g f\) for a unique \(g\in L^2\).

6.3. Countable coordinates and Parseval’s identity

Theorem 6.4. Orthonormal basis in a separable Hilbert space

An orthonormal family satisfies \(\langle e_j,e_k\rangle=\delta_{jk}\). It is an orthonormal basis when its finite linear combinations are dense. Every separable Hilbert space has a finite or countable such basis. For this basis,

\[x=\sum_j e_j\langle e_j,x\rangle,\qquad \|x\|^2=\sum_j|\langle e_j,x\rangle|^2.\]

The series converges in norm. Its terms need not be finite in number; this differs from an algebraic basis.

Worked example for Theorem 6.4

Coordinates are amplitudes. In \(\ell^2\), \(e_j\) is the sequence with a single 1 in slot \(j\). A normalized vector has \(\sum_j|x_j|^2=1\). Chapter 5 explains why a declared measurement in this basis has probabilities \(|x_j|^2\). The norm identity alone does not choose a physical measurement.

6.4. Bounded operators and their adjoints

Definition 6.5. Bounded operator and operator norm

A linear map \(T:H\to K\) is bounded if \(\|Tx\|\le C\|x\|\) for every \(x\). Its operator norm is \(\|T\|=\sup_{\|x\|=1}\|Tx\|\). The notation \(\mathcal B(H,K)\) denotes these maps; \(\mathcal B(H)=\mathcal B(H,H)\). A bounded map is continuous because \(\|Tx-Ty\|\le\|T\|\|x-y\|\). Conversely continuity at zero gives a \(\delta>0\) such that \(\|x\|<\delta\) implies \(\|Tx\|<1\); applying this to \(\delta x/(2\|x\|)\) gives \(\|Tx\|\le2\|x\|/\delta\). Thus boundedness and continuity are equivalent for an everywhere-defined linear map.

Worked example for Definition 6.5

The bounded operators form a Banach space in this norm. Indeed an operator-norm Cauchy sequence \(T_n\) gives a Cauchy sequence \(T_nx\) for every \(x\); define \(Tx=\lim_nT_nx\). Linearity passes to the limit, a common bound for \(\|T_n\|\) bounds \(T\), and \(\|(T_n-T)x\|\le\varepsilon\|x\|\) follows by taking the pointwise limit in the eventual Cauchy bound. Taking a supremum proves norm convergence.

Proposition 6.6. The adjoint

Every bounded \(T:H\to K\) has a unique bounded map \(T^*:K\to H\) satisfying \(\langle y,Tx\rangle_K=\langle T^*y,x\rangle_H\). It satisfies \(\|T^*\|=\|T\|\), \((ST)^*=T^*S^*\), and \((T^*)^*=T\). A bounded operator on \(H\) is self-adjoint if \(T=T^*\), positive if it is self-adjoint and \(\langle x,Tx\rangle\ge0\), and unitary if \(T^*T=TT^*=I\).

Worked example for Proposition 6.6

A one-sided inverse is insufficient. On \(\ell^2\), the right shift \(S(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)\) preserves norms. Its adjoint is \(S^*(x_1,x_2,\ldots)=(x_2,x_3,\ldots)\), as direct summation shows. Then \(S^*S=I\), but \(SS^*\) removes the first coordinate. A linear map preserving norms is called an isometry. Thus an isometry need not be unitary.

6.5. Unbounded operators require domains

Definition 6.7. Domain, unbounded adjoint, and self-adjointness

An operator \(A\) on \(H\) consists of a linear subspace \(D(A)\subset H\), its domain, and a linear rule \(A:D(A)\to H\). Equality of operators means equality of both domains and rules. It is densely defined if \(D(A)\) is dense. For such an operator,

\[D(A^*)=\{y\in H:\text{there is }z\in H\text{ with }\langle y,Ax\rangle=\langle z,x\rangle\text{ for all }x\in D(A)\},\quad A^*y=z.\]

Density makes \(z\) unique: the difference of two choices is orthogonal to a dense subspace and hence zero. Equivalently \(y\in D(A^*)\) when \(x\mapsto\langle y,Ax\rangle\) is bounded with respect to \(\|x\|\), since a bounded functional on a dense subspace extends continuously to \(H\) and Riesz applies.

The operator is symmetric if \(D(A)\subset D(A^*)\) and \(A^*x=Ax\) on \(D(A)\). It is self-adjoint if \(A=A^*\), including domain equality. It is closed if whenever \(x_n\to x\) and \(Ax_n\to y\), with \(x_n\in D(A)\), one has \(x\in D(A)\) and \(Ax=y\).

Adjoint definition · Symmetry and self-adjointness

Worked example for Definition 6.7

Counterexample: A symmetric rule on a dense domain need not be self-adjoint

Every adjoint is closed. To see this, if \(y_n\to y\) and \(A^*y_n\to z\), then for each \(x\in D(A)\), \(\langle y,Ax\rangle=\lim_n\langle y_n,Ax\rangle=\lim_n\langle A^*y_n,x\rangle=\langle z,x\rangle\). This is exactly \(y\in D(A^*)\), \(A^*y=z\). Self-adjoint operators are therefore closed.

Same formula, different operators. On \(\ell^2\), define \(Ax=(jx_j)_j\) on \(c_{00}\). Finite sums show symmetry. Testing the adjoint identity on each \(e_j\) forces \((A^*y)_j=jy_j\). This represents a vector in \(\ell^2\) exactly when \(\sum_j j^2|y_j|^2<\infty\). Conversely that condition makes the adjoint identity valid by Cauchy–Schwarz. Thus \(A\) is symmetric but not self-adjoint: \(y_j=j^{-2}\) lies in \(D(A^*)\), not in \(c_{00}\). The maximal operator \(Tx=(jx_j)_j\), with domain \(\{x:\sum_jj^2|x_j|^2<\infty\}\), is self-adjoint by the same coordinate test applied to its adjoint. It is unbounded since \(\|Te_j\|=j\). The graph of an operator is the set of pairs \((x,Ax)\) in \(H\times H\). Its closure, when the closure of that set is still the graph of an operator, is called the operator’s closure. Truncation approximates both \(x\) and \(Tx\); hence the closed graph of \(T\) is exactly the closure of the graph of \(A\). Thus the closure of \(A\) is \(T\). A symmetric operator whose closure is self-adjoint is called essentially self-adjoint.

For two operators, \(D(AB)=\{x\in D(B):Bx\in D(A)\}\). A commutator \([A,B]=AB-BA\) is defined on \(D(AB)\cap D(BA)\). Writing a formula without its domain can therefore conceal a different mathematical assertion.

6.6. Spectrum without eigenvectors

Definition 6.8. Resolvent and spectrum

For a closed densely defined \(A\), a complex number \(z\) belongs to the resolvent set when \(A-zI:D(A)\to H\) is bijective and its inverse is a bounded everywhere-defined operator \(H\to H\). The spectrum \(\sigma(A)\) is the complement of that set. The point spectrum consists of eigenvalues: numbers \(z\) with \(Ax=zx\) for a nonzero \(x\in D(A)\). A number is in the continuous spectrum when \(A-zI\) is injective with dense range, but is not onto. These definitions suffice for the self-adjoint examples below.

Worked example for Definition 6.8

Proposition 6.9. Position multiplication on the line

On \(H=L^2(\mathbb R,dx)\), set \((Qf)(x)=xf(x)\) and \(D(Q)=\{f\in L^2:xf\in L^2\}\). Then \(Q\) is self-adjoint, \(\sigma(Q)=\mathbb R\), and every real number lies in its continuous spectrum. It has no eigenvalues.

Worked example for Proposition 6.9

Proof: Proposition 6.9. Position multiplication on the line

Proof.

Truncations \(\mathbf1_{[-N,N]}f\) lie in \(D(Q)\) and converge to \(f\), so the domain is dense. Symmetry follows from the integral identity for multiplication by a real function. If \(g\in D(Q^*)\) and \(h=Q^*g\), then \(\langle g,Qf\rangle=\langle h,f\rangle\) for all \(f\in D(Q)\). On a bounded interval, take \(f=\mathbf1_{[-N,N]}(xg-h)\); this is an admissible \(L^2\) function and the identity gives \(\int_{[-N,N]}|xg-h|^2=0\). Hence \(xg=h\) almost everywhere and belongs to \(L^2\). This proves the maximal domain equals its adjoint domain.

If \(\operatorname{Im}z\ne0\), multiplication by \((x-z)^{-1}\) has norm at most \(1/|\operatorname{Im}z|\). It sends \(L^2\) into \(D(Q)\), since \(x/(x-z)=1+z/(x-z)\) is bounded, and is the inverse of \(Q-zI\). Thus nonreal numbers are in the resolvent.

Fix real \(\lambda\). An eigenfunction would vanish outside the singleton \(\{\lambda\}\), a null set, so is the zero vector in \(L^2\). The range of \(Q-\lambda I\) is dense: for any \(h\in L^2\), truncate it to \(|x|\le N\), \(|x-\lambda|\ge1/N\). The truncated \(h_N\) equals \((Q-\lambda I)f_N\), where \(f_N=h_N/(x-\lambda)\in D(Q)\), and \(h_N\to h\) by dominated convergence. It is not onto: \(h=\mathbf1_{[\lambda-1,\lambda+1]}\) would require \(f=h/(x-\lambda)\), whose squared integral diverges at \(\lambda\). Indeed on \((\lambda+2^{-k-1},\lambda+2^{-k}]\), its squared magnitude is at least \(2^{2k}\); multiplying by the interval length \(2^{-k-1}\) gives an integral contribution at least \(2^{k-1}\), for every \(k\ge0\). Thus \(\lambda\) is in the continuous spectrum. As an independent check, normalized indicators of \([\lambda-\varepsilon,\lambda+\varepsilon]\) obey \(\|(Q-\lambda I)f_\varepsilon\|\le\varepsilon\). A bounded inverse would force \(1\le C\varepsilon\), which fails as \(\varepsilon\to0\).

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A “position eigenvector” is not a normalizable state. A Dirac point mass is a measure, not a nonzero \(L^2\) function supported at one point. The calculation above is the precise obstruction. Continuous-spectrum probabilities come from intervals and spectral projections, not from a list of normalizable position eigenvectors.

6.7. Spectral measures and functional calculus

A projection-valued measure (PVM) \(E\) on \(\mathbb R\) assigns an orthogonal projection \(E(B)\) to each Borel set \(B\). It satisfies \(E(\mathbb R)=I\), \(E(\varnothing)=0\), \(E(B)E(C)=E(B\cap C)\), and for disjoint \(B_j\), \(E(\bigcup_jB_j)x=\sum_jE(B_j)x\), with norm convergence for every \(x\). This last requirement is strong countable additivity. Chapter 5 proves that \(\mu_x(B)=\langle x,E(B)x\rangle=\|E(B)x\|^2\) is a finite positive measure, of total mass \(\|x\|^2\).

The multiplication PVM. For \(Q\), put \(E_Q(B)f=\mathbf1_B f\). All projection and intersection identities follow by multiplying indicators. For disjoint sets, the squared norm of the remainder after \(N\) terms is the integral of \(|f|^2\) over \(\bigcup_{j>N}B_j\), tending to zero by countable additivity of the integral. Here \(\mu_f(B)=\int_B|f(x)|^2dx\), an ordinary measure with a density.

Proposition 6.10. Integration against a PVM

For a bounded Borel function \(b\), one can define a bounded operator \(b(E)=\int b(\lambda)\,dE(\lambda)\). It satisfies

\[\|b(E)x\|^2=\int|b|^2\,d\mu_x,\quad \|b(E)\|\le\sup|b|,\quad b(E)^*=\overline b(E),\quad b(E)c(E)=(bc)(E).\]

For a finite-valued, possibly unbounded Borel function \(a\), define

\[D(a(E))=\left\{x:\int|a|^2d\mu_x<\infty\right\},\qquad a(E)x=\lim_{n\to\infty}(a\mathbf1_{\{|a|\le n\}})(E)x.\]

The domain is dense, the operator is closed, its adjoint is \(\overline a(E)\), and it is self-adjoint when \(a\) is real.

Worked example for Proposition 6.10

Proof of the unbounded domain and self-adjointness claims

Put \(P_n=E(\{|a|\le n\})\), \(a_n=a\mathbf1_{\{|a|\le n\}}\), and \(A_n=a_n(E)\). Strong countable additivity gives \(P_nx\to x\). Moreover \(\mu_{P_nx}(B)=\|E(B)P_nx\|^2=\mu_x(B\cap\{|a|\le n\})\), so \(\int|a|^2d\mu_{P_nx}\le n^2\|x\|^2\). Thus \(P_nx\in D(a(E))\), proving density. If the squared integral is finite, the bounded norm identity gives \(\|(A_m-A_n)x\|^2=\int_{n<|a|\le m}|a|^2d\mu_x\to0\); thus the definition has a limit. Monotone convergence gives \(\|a(E)x\|^2=\int|a|^2d\mu_x\). It also shows the domain is a subspace: the truncated norm inequality \(\|A_n(x+y)\|\le\|A_nx\|+\|A_ny\|\) bounds the integrals uniformly. Linearity follows by taking limits.

Write \(A=a(E)\). Bounded multiplication identities and limits give \(P_nAx=A_nx=AP_nx\) for \(x\in D(A)\). If \(x_k\to x\) and \(Ax_k\to y\), then \(A_nx=\lim_k A_nx_k=\lim_kP_nAx_k=P_ny\). Thus \(\int_{|a|\le n}|a|^2d\mu_x=\|P_ny\|^2\le\|y\|^2\). Monotone convergence puts \(x\) in \(D(A)\), and \(Ax=\lim_nP_ny=y\); this proves closedness.

The bounded conjugation identity and limits show \(\overline a(E)\subset A^*\), with the common squared-integral domain. Conversely let \(y\in D(A^*)\), \(z=A^*y\). Testing its identity on \(P_nx\), for arbitrary \(x\in H\), gives \(\langle y,A_nx\rangle=\langle z,P_nx\rangle\); hence \(\overline{a_n}(E)y=P_nz\). Therefore \(\int_{|a|\le n}|a|^2d\mu_y\le\|z\|^2\). Monotone convergence shows \(y\in D(\overline a(E))\), and taking limits identifies its value with \(z\). This proves exact adjoint equality, including domains.

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Theorem 6.11. General spectral representation: external proof dependency

For each self-adjoint operator \(A\) on a complex Hilbert space, there is a unique PVM \(E_A\) on \(\mathbb R\) such that

\[D(A)=\left\{x:\int\lambda^2d\mu_x^A(\lambda)<\infty\right\},\qquad Ax=\int\lambda\,dE_A(\lambda)x,\qquad \mu_x^A(B)=\|E_A(B)x\|^2.\]

Conversely integration of the real function \(\lambda\) against a PVM, with this domain, is self-adjoint, as Proposition 6.10 proves. The forward existence and uniqueness proof is supplied at Alphabeta’s spectral theorem for unbounded self-adjoint operators, including its stated mathematical hypotheses and linked intermediate proofs. The page was checked on October 15, 2026 and contains an explicit proof, rather than only a theorem statement.

Worked example for Theorem 6.11

The external proof constructs the bounded unitary Cayley transform \(C=(A-iI)(A+iI)^{-1}\), obtains its bounded normal spectral measure (an operator \(C\) is normal when \(C^*C=CC^*\)), and transports that measure from the unit circle to the real line using \(\lambda\mapsto(\lambda-i)/(\lambda+i)\). Existence of the bounded spectral measure is a separate theorem in that chain. This description locates the missing proof; it does not replace it.

Once that representation is supplied, we write \(b(A)=b(E_A)\). For normalized \(x\), the adopted Born rule assigns probability \(\mu_x^A(B)\) to a result in \(B\). If \(x\in D(A)\), then \(\langle x,Ax\rangle=\int\lambda\,d\mu_x^A\): first prove this for truncated \(\lambda\), then use convergence of the vectors and of the scalar integrals. The latter follows from Cauchy–Schwarz and finiteness of \(\int\lambda^2d\mu_x^A\). A finite absolute first moment may exist even when \(x\notin D(A)\); in that case the scalar mean is defined, but the expression \(Ax\) is not.

6.8. Why the exponential gives dynamics

Proposition 6.12. Evolution from a self-adjoint generator

Let \(H\) be a self-adjoint energy operator, and let \(\hbar>0\) be the action constant. The operators

\[U(t)=\int e^{-it\lambda/\hbar}\,dE_H(\lambda),\qquad t\in\mathbb R,\]

form a strongly continuous one-parameter unitary group: each is unitary, \(U(0)=I\), \(U(t+s)=U(t)U(s)\), and \(\|U(t)x-U(t_0)x\|\to0\) as \(t\to t_0\) for each vector \(x\). They preserve \(D(H)\). For \(x\in D(H)\), the norm derivative exists and

\[i\hbar\frac d{dt}U(t)x=HU(t)x=U(t)Hx.\]

The derivative at zero exists precisely for \(x\in D(H)\).

Worked example for Proposition 6.12

Counterexample: Strong continuity does not imply operator-norm continuity

Continuity is weaker than differentiability. For the diagonal energy \((Hx)_j=jx_j\) on its maximal domain, \((U(t)x)_j=e^{-itj/\hbar}x_j\). The vector \(x_j=1/j\) is in \(\ell^2\), but outside \(D(H)\), since \(\sum_jj^2|x_j|^2=\sum_j1\). Its evolution is continuous in norm, with no norm derivative at zero. Also this group is not continuous in operator norm at zero: at \(t_n=\pi\hbar/n\), its action on \(e_n\) changes sign, giving \(\|U(t_n)-I\|=2\) although \(t_n\to0\).

The converse construction, starting from a strongly continuous unitary group, is the subject of Stone’s theorem and its proof. The framework’s autonomous-dynamics postulate already supplies the self-adjoint energy operator, so the forward construction just proved is the direction required here. We do not use the converse as an unproved step.

6.9. Trace-class states beyond matrices

Definition 6.13. Rank-one maps and trace class

For \(a,b\in H\), define \(|a\rangle\langle b|x=a\langle b,x\rangle\). This is a bounded rank-one map when \(a,b\ne0\), meaning its range has dimension one. Cauchy–Schwarz gives its norm \(\|a\|\|b\|\), with equality at \(x=b/\|b\|\). We use this characterization of a trace-class operator:

\[T=\sum_{j=1}^\infty|a_j\rangle\langle b_j|,\qquad \sum_j\|a_j\|\|b_j\|<\infty.\]

The series converges in operator norm. This is the nuclear-series definition. Its equivalence with other standard trace-class definitions and trace-norm identification are proved in Alphabeta’s nuclear-series characterization. None of the trace computations below requires that equivalence as an extra step. The trace is \(\operatorname{tr}T=\sum_k\langle e_k,Te_k\rangle\) for an orthonormal basis. Its convergence and independence of that basis are proved next.

Worked example for Definition 6.13

Proposition 6.14. Basis independence and products

For the preceding series, \(\operatorname{tr}T=\sum_j\langle b_j,a_j\rangle\), absolutely. The trace is independent of the orthonormal basis and of the representation of \(T\). If \(B\) is bounded, then \(BT\) and \(TB\) are trace-class and \(\operatorname{tr}(BT)=\operatorname{tr}(TB)\).

Worked example for Proposition 6.14

Proof of absolute convergence, basis independence, and cyclicity

For any orthonormal basis \((e_k)\), Cauchy–Schwarz for the coefficient sequences and Parseval give

\[\sum_k|\langle e_k,a_j\rangle\langle b_j,e_k\rangle|\le\|a_j\|\|b_j\|.\]

The double series over \(j,k\) is therefore absolutely convergent. For each \(k\), operator-norm convergence permits evaluation of \(Te_k\). Interchanging absolutely convergent sums then gives

\[\sum_k\langle e_k,Te_k\rangle=\sum_j\sum_k\langle e_k,a_j\rangle\langle b_j,e_k\rangle=\sum_j\langle b_j,a_j\rangle.\]

The last equality is the norm-convergent basis expansion of \(a_j\), paired with \(b_j\). The right side contains no basis. Representation independence follows because the left side depends only on the operator. Products have series \(BT=\sum_j|Ba_j\rangle\langle b_j|\) and \(TB=\sum_j|a_j\rangle\langle B^*b_j|\), with total norm products bounded by \(\|B\|\sum_j\|a_j\|\|b_j\|\). Their traces agree because \(\langle B^*b_j,a_j\rangle=\langle b_j,Ba_j\rangle\). This also proves trace linearity by combining two absolutely convergent series.

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Definition 6.14a. Trace norm

For a trace-class operator, its trace norm is

\[\|T\|_1=\inf\left\{\sum_j\|a_j\|\|b_j\|:T=\sum_j|a_j\rangle\langle b_j|\right\}.\]

The infimum ranges over all its nuclear-series representations. It is a norm, with \(\|T\|\le\|T\|_1\) and \(|\operatorname{tr}T|\le\|T\|_1\). Trace-class operators are complete in this norm. For positive \(T\), \(\|T\|_1=\operatorname{tr}T\).

Worked example for Definition 6.14a

Proof of the trace-norm properties and completeness

Every series representation bounds the operator norm by its sum of norm products and bounds the absolute trace by the same sum, by Proposition 6.14. Taking the infimum proves both inequalities. Nondegeneracy follows from the operator-norm bound. Scaling representations and scaling back proves absolute homogeneity; concatenating two representations and choosing sums arbitrarily close to their infima proves the triangle inequality.

Let \((T_n)\) be trace-norm Cauchy. Choose a subsequence \(T_{n_k}\) with \(\|T_{n_{k+1}}-T_{n_k}\|_1<2^{-k}\). For each difference choose a nuclear series with total norm products less than \(2^{1-k}\), and choose one series for \(T_{n_1}\). Concatenating all these series defines a trace-class operator \(T\), because their total sum of norm products is finite. The tail representation gives \(\|T-T_{n_k}\|_1\le\sum_{j\ge k}2^{1-j}\to0\). The Cauchy property then gives convergence of the original sequence to \(T\). Thus the space is complete. For positive \(T\), the diagonalization proof in Theorem 6.15, applied without requiring trace 1, gives a nuclear representation with positive eigenvalue sum \(\operatorname{tr}T\). Hence \(\|T\|_1\le\operatorname{tr}T\), and the opposite inequality follows from the trace bound. This last conclusion depends on that later proof.

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Theorem 6.15. Positive density operators are countable mixtures

A density operator is a positive trace-class operator \(\rho\) with \(\operatorname{tr}\rho=1\). On a separable Hilbert space it can be written

\[\rho=\sum_jr_j|u_j\rangle\langle u_j|,\qquad r_j>0,\quad \langle u_j,u_k\rangle=\delta_{jk},\quad \sum_jr_j=1.\]

The family may be finite. Vectors orthogonal to all these \(u_j\) are in the kernel of \(\rho\). Conversely every such series is a density operator. For every bounded \(B\),

\[\operatorname{tr}(\rho B)=\sum_jr_j\langle u_j,Bu_j\rangle.\]

In particular \(B=E_A(C)\) gives a mixture of the scalar spectral probability measures, and \(B\) a finite-outcome effect gives its Born probability. This formula applies to bounded \(B\); an unbounded observable requires separate moment and domain conditions.

Worked example for Theorem 6.15

Proof of positive compact diagonalization and the density formula

Call a bounded operator compact when the image of every bounded sequence has a convergent subsequence. An operator is finite-rank when its range is finite-dimensional. Each finite-rank operator is compact: in its finite-dimensional range, bounded sequences have convergent subsequences by coordinate bisection. An operator-norm limit \(T\) of finite-rank maps is compact. Indeed successively extract subsequences on which each approximating map converges; on the diagonal subsequence, use \(\|Tx_n-Tx_m\|\le\|(T-T_N)(x_n-x_m)\|+\|T_Nx_n-T_Nx_m\|\), choosing first \(N\) large and then \(n,m\) large, to make it Cauchy. Completeness gives its limit. A trace-class series has finite-rank partial sums converging in operator norm, so it is compact.

We prove diagonalization for a positive compact \(R\). The positive sesquilinear form (conjugate-linear in its first slot and linear in its second) \(q(x,y)=\langle x,Ry\rangle\) satisfies \(|q(x,y)|^2\le q(x,x)q(y,y)\), by minimizing \(q(x+ty,x+ty)\ge0\) over complex \(t\). If \(q(y,y)=0\), the same quadratic argument forces \(q(x,y)=0\). Set \(a=\sup_{\|x\|=1}q(x,x)\). Since \(\|v\|=\sup_{\|y\|=1}|\langle y,v\rangle|\), the form inequality gives \(\|Rx\|^2\le a q(x,x)\le a^2\|x\|^2\). Hence \(\|R\|\le a\), while \(q(x,x)\le\|R\|\|x\|^2\) gives equality.

If \(a=0\), the norm identity gives \(R=0\). If \(a>0\), choose unit \(x_n\) with \(q(x_n,x_n)\to a\). A sequence in the unit ball of a separable Hilbert space has a subsequence converging weakly to some \(x\) of norm at most 1. Here weak convergence means \(\langle y,x_n\rangle\to\langle y,x\rangle\) for every \(y\). To prove this fact, diagonalize subsequences so each orthonormal-basis coordinate converges; Bessel’s inequality bounds the sum of the limiting coordinate squares, producing \(x\) by Theorem 6.4. For any \(y\), finite-coordinate pairings converge, and its remaining tail has arbitrarily small norm, uniformly controlling the pairing error by Cauchy–Schwarz. This proves weak convergence.

Compactness allows a further subsequence with \(Rx_n\) convergent in norm. Its limit is \(Rx\), since \(\langle y,Rx_n\rangle=\langle Ry,x_n\rangle\to\langle Ry,x\rangle\). Consequently \(q(x_n,x_n)\to q(x,x)\): replace \(Rx_n\) by \(Rx\) with a norm-small error, then use weak convergence. Thus \(q(x,x)=a\), and \(q(x,x)\le a\|x\|^2\) forces \(\|x\|=1\). The operator \(aI-R\) is positive. Since \(\langle x,(aI-R)x\rangle=0\), its positive-form Cauchy–Schwarz inequality implies \(\langle y,(aI-R)x\rangle=0\) for every \(y\). Therefore \(Rx=ax\): the norm is a positive eigenvalue.

Restrict \(R\) to \(x^\perp\), which is invariant because \(R\) is self-adjoint, and repeat. Either the remainder becomes zero after finitely many steps or we obtain orthonormal eigenvectors \(u_j\) with positive decreasing eigenvalues \(r_j\). They must tend to zero; otherwise \(Ru_j=r_ju_j\) has no convergent subsequence, since distinct such images have squared distance \(r_j^2+r_k^2\) bounded below, contradicting compactness. After \(N\) steps the norm of the remaining restriction is \(r_{N+1}\), by its construction as a maximal eigenvalue. Thus the finite spectral sums converge to \(R\) in operator norm. The orthogonal complement of all \(u_j\) is its kernel. Complete the \(u_j\) to an orthonormal basis by choosing a basis of this kernel; it is separable because projecting a countable dense set onto a closed subspace gives a countable dense set there.

Apply the already proved basis-independent trace formula to \(R=\rho\), in this eigenbasis. Its diagonal entries are \(r_j\) and zeros, so \(\sum_jr_j=\operatorname{tr}\rho=1\). This makes the spectral series a trace-class representation with total norm products 1. The trace-product formula gives \(\operatorname{tr}(\rho B)=\sum_jr_j\langle u_j,Bu_j\rangle\), absolutely because each term is bounded by \(r_j\|B\|\). Conversely, a series with these positive summable coefficients has nonnegative quadratic form, is self-adjoint by termwise adjoints and operator-norm convergence, and has trace 1. It is therefore a density operator.

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A normalized vector gives a pure density operator. If \(\|u\|=1\), then \(\rho_u=|u\rangle\langle u|\) is positive, \(\operatorname{tr}\rho_u=1\), and \(\rho_u^2=\rho_u\). Multiplying \(u\) by a complex phase changes neither \(\rho_u\) nor its probabilities. For \(\ell^2\), \(\rho=\sum_{j\ge1}2^{-j}|e_j\rangle\langle e_j|\) is a mixed state with countably many nonzero weights. The identity on \(\ell^2\) is bounded and positive but cannot be trace-class: its diagonal trace would be \(\sum_j1=\infty\).

The density operators form a convex set: for density operators \(\sigma,\tau\) and \(0\le a\le1\), the operator \(a\sigma+(1-a)\tau\) is positive, trace-class, and has trace 1, by positivity of each quadratic form and linearity of trace. They form a closed set in trace norm: if \(\rho_n\to T\) in that norm, then operator-norm convergence passes self-adjointness and nonnegative quadratic forms to \(T\), while \(|\operatorname{tr}(T-\rho_n)|\le\|T-\rho_n\|_1\) gives trace 1. Completeness ensures that trace-norm Cauchy sequences have a trace-class limit.

A density operator is called pure when it has rank one, and mixed otherwise. An extreme point of a convex set is one that cannot be written as \(a\sigma+(1-a)\tau\) with distinct members and \(0<a<1\). The pure states are precisely the extreme density operators, and precisely those with \(\operatorname{tr}(\rho^2)=1\).

Proof of the infinite-dimensional purity and extreme-point criteria

Using Theorem 6.15, \(\operatorname{tr}(\rho^2)=\sum_jr_j^2\). Since the positive \(r_j\) sum to 1, this equals 1 precisely when exactly one weight is nonzero: \(1-\sum_jr_j^2=2\sum_{j<k}r_jr_k\), an identity obtained by multiplying the absolutely convergent series. In that case \(\rho=|u\rangle\langle u|\). If at least two eigenvalues are positive, choose \(0<\varepsilon<\min(r_1,r_2)\) and define \(\rho_\pm=\rho\pm\varepsilon(|u_1\rangle\langle u_1|-|u_2\rangle\langle u_2|)\). Both remain positive trace-class with trace 1, are distinct, and average to \(\rho\). Thus a mixed state is not extreme.

Conversely suppose \(|u\rangle\langle u|=a\sigma+(1-a)\tau\), with \(0<a<1\). For any \(v\perp u\), its quadratic form on the left is zero. Both quadratic forms on the right are nonnegative, so \(\langle v,\sigma v\rangle=\langle v,\tau v\rangle=0\). The positive-form Cauchy–Schwarz argument in Theorem 6.15 gives \(\sigma v=\tau v=0\). Self-adjointness then implies \(\sigma u\) and \(\tau u\) belong to the span of \(u\). Each operator is a nonnegative multiple of \(|u\rangle\langle u|\), and trace 1 makes each multiple 1. They coincide with the pure state, proving extremality.

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Under a unitary \(U\), \(\rho\) evolves to \(U\rho U^*\). The displayed mixture becomes \(\sum_jr_j|Uu_j\rangle\langle Uu_j|\); its weights remain positive and sum to 1, proving preservation of positivity and trace without differentiating an unbounded Hamiltonian.

6.10. Composite Hilbert spaces

Definition 6.16. Completed tensor product

Start with the vector space of finite formal complex linear combinations of symbols \([x,y]\), one for every pair \((x,y)\in H\times K\). Let \(R\) be the subspace spanned by \[[x+x',y]-[x,y]-[x',y],\quad [x,y+y']-[x,y]-[x,y'],\quad [cx,y]-c[x,y],\quad [x,cy]-c[x,y].\] Two formal sums represent the same element when their difference belongs to \(R\). Their equivalence classes form the algebraic tensor product; addition and scalar multiplication are well defined because \(R\) is a linear subspace. Denote the class of \([x,y]\) by \(x\otimes y\). Thus every element is a finite tensor sum and the prescribed bilinear relations hold. This construction extends the finite-dimensional tensor product from chapter 2. On finite sums, prescribe

\[\left\langle\sum_jx_j\otimes y_j,\sum_kv_k\otimes w_k\right\rangle=\sum_{j,k}\langle x_j,v_k\rangle_H\langle y_j,w_k\rangle_K.\]

The Hilbert tensor product \(H\widehat\otimes K\) is its completion in this norm. In particular \(\|x\otimes y\|=\|x\|\|y\|\). If \((e_j)\) and \((f_k)\) are orthonormal bases, then \((e_j\otimes f_k)_{j,k}\) is an orthonormal basis, so a vector is specified by square-summable double coefficients.

Worked example for Definition 6.16

Proof that the inner product and completion are well defined

The bilinear relations preserve the displayed expression because each factor inner product is linear in the appropriate slot and conjugate-linear in the first. For any finite tensor sum, its first factors lie in a finite-dimensional subspace of \(H\) and its second factors in one of \(K\). Choose finite orthonormal bases there. The bilinear relations express the sum as \(\sum_{j,k}c_{jk}e_j\otimes f_k\). For each \(j,k\), the scalar map \((x,y)\mapsto\langle e_j,x\rangle\langle f_k,y\rangle\) is bilinear, so its linear extension to formal sums vanishes on every generator of \(R\) and therefore on \(R\). It descends to the quotient and evaluates this tensor sum to \(c_{jk}\). Hence the coefficients are unique, including when the same element is represented using other vectors. The displayed squared norm equals \(\sum_{j,k}|c_{jk}|^2\). Hence it is positive definite and independent of presentation.

A completion can be constructed as follows. Take Cauchy sequences in the algebraic space; identify two when their difference tends to zero. Add and scale sequences termwise, define their norm by the limit of the norms, and their inner product by the limit of the inner products. These limits exist: Cauchy sequences are bounded, and Cauchy–Schwarz bounds differences of pairings. The zero-norm classes are precisely the identified zero sequences. Constant sequences embed the original space densely. This space is complete: for a Cauchy sequence of classes, choose a subsequence with successive distances at most \(2^{-n}\), and for each subsequence term choose an original-space point within \(2^{-n}\). These chosen points are Cauchy, hence define a class to which the subsequence converges. The original Cauchy sequence converges to that same class by the triangle inequality.

For density of the claimed tensor basis, approximate \(x\) and \(y\) by finite sums in their bases, using

\[\|x\otimes y-x_N\otimes y_N\|\le\|x-x_N\|\|y\|+\|x_N\|\|y-y_N\|.\]

Finite elementary sums are dense by the construction of the completion. Thus finite sums of \(e_j\otimes f_k\) are dense, and Theorem 6.4 applies to give Parseval and square-summable coordinates.

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If \(A\) and \(B\) are bounded, \((A\otimes B)(x\otimes y)=Ax\otimes By\) extends to the completed space and has norm at most \(\|A\|\|B\|\). To check the bound, express a finite tensor as \(\sum_kx_k\otimes f_k\) with orthonormal \(f_k\); its squared norm is \(\sum_k\|x_k\|^2\). Applying \(A\otimes I\) bounds this by \(\|A\|^2\sum_k\|x_k\|^2\). The same argument for \(I\otimes B\) and composition gives the product bound. A bounded map on a dense subspace extends by limits uniquely, which proves the assertion. Pairing elementary tensors proves \((A\otimes B)^*=A^*\otimes B^*\).

Internal spin components. \(L^2(\mathbb R^3)\widehat\otimes\mathbb C^d\) is \(L^2(\mathbb R^3;\mathbb C^d)\): write a vector as \(\sum_{k=1}^d\psi_k\otimes e_k\) and identify it with \((\psi_1(x),\ldots,\psi_d(x))\). The squared norm is \(\sum_k\int|\psi_k|^2=\int\sum_k|\psi_k|^2\), and every tuple of \(L^2\) functions is obtained this way. A two-system vector such as \((e_1\otimes f_1+e_2\otimes f_2)/\sqrt2\) is normalized but is not a single elementary tensor: its coefficient matrix has rank two, whereas an elementary tensor has coefficients \(x_jy_k\), a matrix of rank at most one. Such a vector is called entangled.

6.11. Decoding the massive-particle notation

The framework’s one-particle space is \(L^2(\mathbb R^3;\mathbb C^{2s+1})\), where \(s\) is its adopted nonnegative integer or half-integer spin label. The previous example explains this notation. Position component \(Q_j\) means multiplication by \(x_j\), with domain \(\{\psi:x_j\psi\in L^2\}\). The same bounded-region test as in Proposition 6.9 proves that this maximal real multiplier is self-adjoint.

Definition 6.17. Weak derivatives, Sobolev space, and Schwartz space

A function is locally integrable if its absolute integral on every compact subset of its open domain is finite. For such \(u\), a locally integrable \(v\) is its weak derivative \(\partial_j u\) when

\[\int u\,\partial_j\phi=-\int v\phi\qquad\text{for every }\phi\in C_c^\infty.\]

Here smooth means that all orders of partial derivatives exist and are continuous. The support of a function is the closure of the set where it is nonzero. The notation \(C_c^\infty\) denotes smooth test functions with compact support. A multi-index \(\alpha=(\alpha_1,\ldots,\alpha_n)\) has nonnegative integer entries; \(|\alpha|=\sum_j\alpha_j\), \(x^\alpha=\prod_jx_j^{\alpha_j}\), and \(\partial^\alpha=\partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}\). The Sobolev space \(H^k(\mathbb R^n)\), for nonnegative integer \(k\), consists of \(L^2\) functions whose weak derivatives through order \(k\) belong to \(L^2\); its squared norm is \(\sum_{|\alpha|\le k}\|\partial^\alpha u\|_2^2\). On an arbitrary open set \(\Omega\), the same definition gives \(H^k(\Omega)\), with test functions supported in \(\Omega\) and all integrals over that set. Uniqueness of weak derivatives up to equality almost everywhere is needed to make this norm well defined; its external proof is weak derivatives are unique almost everywhere. Components are treated separately for vector-valued functions.

The Schwartz space \(\mathcal S(\mathbb R^n)\) consists of smooth functions for which \(\sup_x|x^\alpha\partial^\beta f(x)|<\infty\) for every pair of multi-indices. It provides functions on which multiplication, differentiation, and integration by parts can be checked directly. A core for a closed operator \(A\) is a subspace \(C\subset D(A)\) dense in its domain for the graph norm \((\|x\|^2+\|Ax\|^2)^{1/2}\). Density in \(H\) alone does not establish the core property.

Sobolev definition · Schwartz definition

Worked example for Definition 6.17

A corner can have a weak derivative. On \((-1,1)\), \(u(x)=|x|\) has weak derivative \(-1\) on \((-1,0)\) and \(1\) on \((0,1)\). Split \(\int|x|\phi'(x)dx\) at zero and integrate by parts on each side. The boundary terms at zero vanish because \(|0|=0\); the outer terms vanish because \(\phi\) has compact support. The result is \(-\int\operatorname{sgn}(x)\phi(x)dx\). Both functions are in \(L^2(-1,1)\), so \(|x|\in H^1(-1,1)\), though its classical derivative fails at zero. The almost-everywhere uniqueness used by the Sobolev definition is supplied by the external proof linked above.

On Schwartz functions the momentum rule is \(P_j=-i\hbar\partial_j\). Direct differentiation gives

\[[Q_j,P_k]\psi=x_j(-i\hbar\partial_k\psi)+i\hbar\partial_k(x_j\psi)=i\hbar\delta_{jk}\psi.\]

The products are defined on this common subspace: derivatives and multiplication preserve the Schwartz bounds by the product rule, and those bounds imply square integrability by choosing a decay power larger than the dimension. More explicitly, the Schwartz bounds control every monomial in the expansion of \( (1+|x_1|+\cdots+|x_n|)^N \), so \( |f(x)|\le C_N(1+|x|)^{-N} \); squaring and integrating over unit-width shells gives a convergent series when \(2N>n\). The same argument applies to every derivative. This computation proves the commutator there; it does not prove self-adjointness of a differential expression on an arbitrary domain.

The maximal momentum domain consists of \(\psi\in L^2\) whose weak \(\partial_j\psi\) belongs to \(L^2\). The framework supplies the full self-adjointness proof in its momentum-operator item. That proof uses the unitary Fourier transform, a further analytic construction. Its initial integral convention is

\[\widehat f(k)=(2\pi)^{-n/2}\int e^{-ik\cdot x}f(x)\,dx\qquad(f\in\mathcal S).\]

Integration by parts, with the vanishing boundary terms supplied by rapid decay, gives \(\widehat{\partial_j f}(k)=ik_j\widehat f(k)\). The Plancherel theorem, whose proof belongs to the linked Fourier theory rather than this chapter, extends this transform to a unitary map on \(L^2\). Under that map momentum becomes the real multiplier \(\hbar k_j\). This explains the domain choice and the route to a self-adjointness proof; the integral formula need not exist as an ordinary integral for every \(L^2\) vector.

Likewise \(\Delta=\sum_{j=1}^3\partial_j^2\) is the Laplacian, and the bounded-real-potential Hamiltonian in the framework is \(-\hbar^2\Delta/(2m)+V\), on \(H^2(\mathbb R^3;\mathbb C^{2s+1})\), with \(m>0\). Its self-adjointness and the required Fourier and Sobolev facts are proved at real multipliers and free bounded-potential operators. We have explained its symbols; proving those additional analytic constructions requires following that external proof route. Arbitrary singular potentials or other boundary conditions require other hypotheses and are not covered by this statement.

6.12. Return to the four physics pages

Framework and quantities now has precise meanings for a separable Hilbert space, density operator, completed tensor product, self-adjoint observable, and exponential of an energy operator. Its assignment of these objects to preparations and apparatus is a physical postulate. The mathematical definitions do not prove that assignment.

State geometry uses normalized vectors modulo phase and rank-one projections. Observables and Born probabilities combines the PVM with a vector or density operator. Autonomous dynamics uses the functional calculus and domain-sensitive derivative proved in Proposition 6.12. The distinction between a continuous orbit and a differentiable orbit is essential when the energy is unbounded.

Notation translation. The linked library uses an inner product linear in the first slot, whereas this series uses one linear in the second. If its pairing is denoted \((u,v)_{\mathrm{AB}}\), then \((u,v)_{\mathrm{AB}}=\langle v,u\rangle\) in our convention. Our \(|a\rangle\langle b|x=a\langle b,x\rangle\) therefore reads \((x,b)_{\mathrm{AB}}a\) there. Norms and real self-adjoint expectations agree after translation; off-diagonal pairings must have their arguments exchanged.

Worked examples

Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.

Worked example 1. Coordinates, projection, and a represented functional in ℓ²

Applies to: Definition 6.1, Theorem 6.2, Theorem 6.3, Theorem 6.4.

Let \(H=\ell^2(\mathbb N)\), with \(\langle x,y\rangle=\sum\bar x_ny_n\), and standard coordinate vectors \(e_n\). For \(x_n=2^{-n}\), \(\|x\|^2=\sum4^{-n}=1/3\). Its truncations \(x^{(N)}\) have error squared \(4^{-N}/3\), so are Cauchy and converge in norm to \(x\). Finite rational-complex coordinate sequences form a countable dense subset, proving separability; \(\{e_n\}\) is an orthonormal basis since every square-summable tail tends to zero. Completeness is the counting-measure case of Theorem 4.18.

The subspace \(M=\{x:x_1=0\}\) is closed because \(|x_1-y_1|\le\|x-y\|\). Its projection deletes the first coordinate. For every \(v\in M\), \(\|x-v\|^2=|x_1|^2+\sum_{n\ge2}|x_n-v_n|^2\), uniquely minimized by \(v=P_Mx\); also \(P_M^2=P_M\) and \(\|P_Mx\|\le\|x\|\). The functional \(F(x)=x_1+ix_2\) is represented by \(z=e_1-ie_2\), since \(\langle z,x\rangle=x_1+ix_2\). Its norm is \(\sqrt2\), attained at \(x=z/\sqrt2\), and its representing vector is unique by testing every \(e_n\). This verifies Riesz with the second-slot-linear convention.

Further proof details

A direct completeness proof for the particular Hilbert space also avoids relying only on a theorem name. If \(x^{(j)}\) is Cauchy in \(\ell^2\), each coordinate is Cauchy since \(|x_n^{(j)}-x_n^{(k)}|\le\|x^{(j)}-x^{(k)}\|\); let its limit be \(x_n\). The norms of the sequence are bounded, say by \(C\). Passing to the limit in each finite sum gives \(\sum_{n\le N}|x_n|^2\le C^2\); taking the supremum gives \(x\in\ell^2\). Given \(\varepsilon>0\), choose \(J\) so late pair differences have squared norm at most \(\varepsilon^2\). For fixed \(j\ge J\), first pass \(k\to\infty\) in \(\sum_{n\le N}|x_n^{(j)}-x_n^{(k)}|^2\le\varepsilon^2\), then take the supremum over \(N\). The result is \(\|x^{(j)}-x\|\le\varepsilon\). This proves completeness and, with the explicit coordinate inner product and countable dense subset, all the Hilbert-space hypotheses used in the example.

Worked example 2. The unilateral shift and its adjoint

Applies to: Definition 6.5, Proposition 6.6.

On \(\ell^2\) define \(S(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)\). It is linear and \(\|Sx\|=\|x\|\), so bounded and continuous with norm 1. The inner-product identity gives \(S^*(y_1,y_2,\ldots)=(y_2,y_3,\ldots)\), also of norm 1. Then \(S^*S=I\), but \(SS^*=I-P_{e_1}\), so \(S\) is not unitary. The latter is a positive self-adjoint projection: its quadratic form is \(\sum_{n\ge2}|x_n|^2\). Taking adjoints verifies \((SS^*)^*=SS^*\) and \((S^*)^*=S\) directly, illustrating the product and norm assertions while distinguishing isometries from unitary operators.

Worked example 3. An unbounded diagonal operator with a checked domain

Applies to: Definition 6.7, Definition 6.8.

On \(\ell^2\) put \((Ax)_n=nx_n\), \(D(A)=\{x:\sum n^2|x_n|^2<\infty\}\). Finite sequences lie in the domain and are dense. If \(y\in D(A^*)\), testing the adjoint identity on \(e_n\) forces \((A^*y)_n=ny_n\), so square summability forces \(y\in D(A)\). Conversely this condition makes the identity valid by Cauchy–Schwarz. Thus the adjoint domains and rules agree. If \(x^{(j)}\to x\) and \(Ax^{(j)}\to z\), coordinate convergence gives \(z_n=nx_n\); since \(z\in\ell^2\), \(x\in D(A)\), proving closedness. The norm \(\|Ae_n\|=n\) proves unboundedness.

Its spectrum is \(\{1,2,3,\ldots\}\). These are eigenvalues with eigenvectors \(e_n\). For any other complex \(z\), the distance \(d=\inf_n|n-z|\) is positive, since the integer sequence has no finite accumulation point. The inverse \(y_n\mapsto y_n/(n-z)\) is bounded by \(1/d\), and maps into \(D(A)\) because \(n/(n-z)=1+z/(n-z)\) is bounded. It is an everywhere-defined inverse of \(A-z\). This checks the full resolvent definition including the range’s domain condition.

Worked example 4. Position has continuous spectrum, not point eigenvectors

Applies to: Proposition 6.9.

For the chapter’s maximal position operator \(Q\) on \(L^2(\mathbb R)\), take \(z=i\). Multiplication by \((x-i)^{-1}\) has norm at most 1 and maps into \(D(Q)\), since \(|x/(x-i)|\le1\); hence it is a bounded inverse of \(Q-i\). At the real value zero, \(Qf=0\) forces \(f=0\) a.e., since the singleton zero has measure zero. The range is dense because it contains all \(L^2\) functions supported in bounded regions away from zero, which approximate every \(L^2\) vector by truncation. It is not onto: \(g=\mathbf1_{[-1,1]}\) would require \(f=g/x\), whose squared integral diverges near zero. Translations give the same argument at any real \(a\). Thus the real spectrum is continuous and has no eigenvectors, with all defining conditions exhibited.

Further proof details

The maximal position domain’s self-adjointness can be checked directly. It is dense because bounded-region truncations of \(L^2\) functions belong to it and converge in norm. If \(y\in D(Q^*)\), some \(z\in L^2\) satisfies \(\int\bar y\,xf=\int\bar z f\) for all \(f\in D(Q)\). On any bounded interval, both \(x\bar y\) and \(\bar z\) are integrable by Cauchy–Schwarz. Writing their difference as \(w\), test with the bounded supported function \(f=\bar w/|w|\) where \(w\ne0\), zero elsewhere. This test lies in \(D(Q)\), and gives \(\int|w|=0\), so \(xy=z\) a.e. on every bounded interval. Therefore \(xy\in L^2\), so \(y\in D(Q)\). Conversely, for such \(y\), Cauchy–Schwarz makes the same adjoint identity valid with \(z=xy\). Thus the domains and rules of \(Q\) and \(Q^*\) agree.

Worked example 5. Spectral calculus and a domain-sensitive time derivative

Applies to: Proposition 6.10, Theorem 6.11, Proposition 6.12.

For the diagonal \(A\) above, define \(E(B)x=(\mathbf1_B(n)x_n)_n\). Coordinate projections satisfy the PVM laws; strong countable additivity follows because the remaining squared-coordinate tail tends to zero. Then \(a(E)x=(a(n)x_n)\) with domain \(\sum|a(n)|^2|x_n|^2<\infty\). Bounded \(b\) gives \(\|b(E)x\|^2=\sum|b(n)|^2|x_n|^2\), and multiplication and adjoints are pointwise. Taking \(a(\lambda)=\lambda\) recovers exactly \(A\) and its domain, a concrete spectral representation. The PVM is fixed on each eigenvector by its eigenvalue, so this representation’s spectral projections are unique.

With \(\hbar=1\), \((U(t)x)_n=e^{-int}x_n\), proving the unitary group law and domain preservation. Strong continuity follows by bounding finitely many coordinate changes and then the tail by \(4\sum_{n>N}|x_n|^2\). If \(x\in D(A)\), the difference quotients converge in \(\ell^2\) to \((-inx_n)\), using \(|(e^{-int}-1)/t|\le n\) and a summable \(n^2|x_n|^2\) bound. For \(x_n=1/n\), the vector lies in \(\ell^2\) but not \(D(A)\); a norm derivative would have coordinates \(-i\) and could not lie in \(\ell^2\). This explicitly illustrates continuous evolution without a Schrödinger derivative on every vector.

Further proof details

For the uniqueness assertion, suppose a PVM \(E'\) also represents this diagonal operator with its spectral-calculus domain. Its scalar law at \(e_n\) satisfies \(\int|\lambda-n|^2d\mu_{e_n}'=\|(A-n)e_n\|^2=0\). For each positive integer \(j\), the set \(|\lambda-n|\ge1/j\) must have measure zero, since its integrand is at least \(1/j^2\). Their countable union is \(\mathbb R\setminus\{n\}\), so the law is the unit mass at \(n\). Hence \(\|E'(B)e_n\|^2\) equals 0 or 1 according as \(n\notin B\) or \(n\in B\). In the latter case \(\|(I-E'(B))e_n\|^2=0\), giving \(E'(B)e_n=e_n\); in the former it is zero. Thus \(E'\) agrees with the displayed coordinate projections on every basis vector, then on finite sums, and on all vectors by boundedness and norm approximation. This proves uniqueness in this example.

Worked example 6. A countable mixture and its trace norm

Applies to: Definition 6.13, Proposition 6.14, Definition 6.14a, Theorem 6.15.

On \(\ell^2\), let \(\rho=\sum_{n\ge1}2^{-n}|e_n\rangle\langle e_n|\). Taking \(a_n=2^{-n}e_n,b_n=e_n\) gives a nuclear series with \(\sum\|a_n\|\|b_n\|=1\), converging in operator norm with tail norm \(2^{-(N+1)}\). It is positive, and its trace is \(\sum2^{-n}=1\), so it is a density operator. The trace inequalities give \(1=|\operatorname{tr}\rho|\le\|\rho\|_1\le1\), hence \(\|\rho\|_1=1\), while \(\|\rho\|=1/2\). Its rank-one summands have the asserted norm and trace, each 1 before weighting.

For any bounded \(B\), the absolutely convergent sum is \(\operatorname{tr}(\rho B)=\sum2^{-n}\langle e_n,Be_n\rangle\), with absolute bound \(\|B\|\). In particular the projection onto even coordinates has probability \(\sum_{j\ge1}2^{-2j}=1/3\), and cyclicity gives the same trace for \(B\rho\). The rotated basis \((e_{2j-1}\pm e_{2j})/\sqrt2\) has diagonal contributions \((2^{-(2j-1)}+2^{-2j})/2\) for each sign, so summing gives the same trace 1. The finite partial mixtures satisfy \(\|\rho-\rho_N\|_1=\sum_{n>N}2^{-n}=2^{-N}\), illustrating trace-norm convergence and a countable, rather than only finite, density mixture.

Worked example 7. A vector requiring tensor completion

Applies to: Definition 6.16.

In \(\ell^2\widehat\otimes\ell^2\), the partial sums \(\Psi_N=\sqrt3\sum_{n=1}^N2^{-n}e_n\otimes e_n\) have orthogonal terms and are Cauchy, since their tail squared norms sum to \(3\sum_{n>N}4^{-n}=4^{-N}\). Their limit \(\Psi\) has norm one. It is not a finite algebraic tensor sum: its infinite diagonal coefficient matrix has arbitrarily large finite rank, whereas a sum of \(r\) elementary tensors has matrix rank at most \(r\). Thus completion adds a concrete vector. With \(P=P_{e_1}\), \((P\otimes I)\Psi=(\sqrt3/2)e_1\otimes e_1\), of norm \(\sqrt3/2\le1\); its adjoint is itself. This checks extension and boundedness of a product operator.

For an internal two-component space, \(f\otimes e_1+g\otimes e_2\) corresponds to the tuple \((f(x),g(x))\), with squared norm \(\|f\|_2^2+\|g\|_2^2\). Taking \(f=\mathbf1_{[0,1]}/\sqrt2,g=i\mathbf1_{[1,2]}/\sqrt2\) gives a normalized two-component function and checks the unboxed spin-component identification without assigning it a classical trajectory.

Worked example 8. A weak derivative, a Schwartz vector, and a core

Applies to: Definition 6.17.

On \((-1,1)\), \(u(x)=|x|\) has weak derivative \(\operatorname{sgn}x\): split \(\int|x|\phi'\) at zero and integrate by parts; the compactly supported test function removes the outer endpoints and the zero factor \(|0|\) removes the inner ones. Both functions are in \(L^2\), and \(\|u\|_{H^1}^2=2/3+2=8/3\). This is a Sobolev function without a classical derivative at zero.

On \(\mathbb R\), \(f=e^{-x^2}\) is Schwartz because every derivative is a polynomial times \(e^{-x^2}\), and every polynomial times that Gaussian is bounded. Thus all the products in \(Qf=xf\), \(Pf=-i\hbar f'=2i\hbar xf\) and their commutator are square integrable. Direct differentiation gives \((QP-PQ)f=i\hbar f\). Integration by parts against \(e^{-ikx}\) gives \(\widehat{f'}=ik\widehat f\), with no boundary term because of Gaussian decay, checking the unboxed Fourier derivative rule on an admissible vector.

For the closed diagonal operator \(A\) used above, finite sequences are a core: truncating \(x\in D(A)\) gives graph-norm error squared \(\sum_{n>N}(1+n^2)|x_n|^2\to0\). This checks the stronger core condition explicitly, rather than inferring it from ordinary density alone.

Further proof details

The Gaussian decay assertion follows from an explicit bound. Inductively, if \(f^{(m)}=p_m(x)e^{-x^2}\), then \(f^{(m+1)}=(p'_m-2xp_m)e^{-x^2}\), again a polynomial times the Gaussian. For any degree \(d\), choose an integer \(k\) with \(2k>d\). The nonnegative exponential series gives \(e^{x^2}\ge x^{2k}/k!\), so \(|x|^de^{-x^2}\le k!|x|^{d-2k}\to0\) as \(|x|\to\infty\); continuity bounds it on the remaining compact interval. This proves all Schwartz seminorm bounds. It also bounds polynomial-times-Gaussian tails by an integrable negative power, proving square integrability of the derivatives and products used. Compact support removes the boundary terms in the weak-derivative example; Gaussian decay removes them in the Fourier integration by parts. Thus each displayed operator product is defined on the stated vector.

Counterexamples and limits of the hypotheses

Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.

Counterexample 1. A symmetric rule on a dense domain need not be self-adjoint

Related statements: Definition 6.7.

Claim refuted. A densely defined symmetric operator is self-adjoint, regardless of its domain.

Proof.

On \(\ell^2(\mathbb N)\) let \(A_0x=(nx_n)\) have domain \(c_{00}\), the finite sequences. The domain is dense by truncation, and finite sums give \(\langle y,A_0x\rangle=\langle A_0y,x\rangle\) for \(x,y\in c_{00}\), so it is symmetric. If \(y\in D(A_0^*)\) has representing vector \(z\), testing on \(e_n\) forces \(z_n=ny_n\). Therefore \(D(A_0^*)\subset\{y:\sum n^2|y_n|^2<\infty\}\). Conversely any such \(y\) has representing vector \((ny_n)\), since the identity holds on every finite sequence. This proves equality of these domains and the adjoint rule.

The sequence \(y_n=1/n^2\) belongs to the adjoint domain because \(\sum n^2/n^4=\sum n^{-2}<\infty\), but is not finitely supported. Thus \(D(A_0)\ne D(A_0^*)\), so \(A_0\) is not self-adjoint. Its finite truncations converge to \(y\) while their images converge to \((1/n)\), showing also that \(A_0\) is not closed. The same multiplication expression on its maximal weighted-square-summable domain is the self-adjoint operator proved in the worked examples; the rules alone do not determine the operator.

Counterexample 2. Strong continuity does not imply operator-norm continuity

Related statements: Proposition 6.12.

Claim refuted. A strongly continuous unitary group is continuous in operator norm.

Proof.

On \(\ell^2\) take \((U(t)x)_n=e^{-int}x_n\). Each operator is unitary and the group law follows from multiplication of the scalar phases. For fixed \(x\), choose \(N\) with \(\sum_{n>N}|x_n|^2\) small. The finite sum \(\sum_{n\le N}|e^{-int}-1|^2|x_n|^2\) tends to zero with \(t\), and the tail is at most \(4\sum_{n>N}|x_n|^2\). This proves strong continuity. But at \(t_j=\pi/j\), the unit vector \(e_j\) obeys \(\|(U(t_j)-I)e_j\|=|-1-1|=2\). The operator-norm difference is at least 2 and at most \(\|U(t_j)\|+\|I\|=2\), hence exactly 2, although \(t_j\to0\). Strong continuity controls each fixed vector; the operator norm permits a different test vector at each time.