According to the Picard–Lindelof Theorem in the theory of ordinary differential ¨ equations, any complete vector field has a unique flow. In fact, the uniqueness part of this theorem implies that (3.19) is a consequence of (3.20) with (3.18), but it is convenient to state (3.19) separately, so as to make the point that the flow of a complete vector field ξ on *X* is a smooth R-action on *X*, as defined by conditions (3.18) - (3.19), whose orbits integrate ξ . In particular, each ϕ*t* : *X* → *X* is invertible, with inverse ϕ−1 *t* = ϕ−*t*. In particular, *X* is a disjoint union of the integral curves of ξ , which can never cross each other because of the uniqueness of the solution of the initial-value problem (3.13) with *c*(0) = *x*0).

If ξ is not complete, we do the best we can by defining the set

$$D\_{\xi} = \{(t, \mathbf{x}) \in \mathbb{R} \times X \mid \exists c : I \to X, c(\mathbf{0}) = \mathbf{x}, t \in I\} \subset \mathbb{R} \times X,\tag{3.25}$$

where it is understood that *c* satisfies (3.13). Obviously {0} × *X* ⊂ *D*ξ , and (less trivially) it turns out that *D*ξ is open. Then a flow of ξ is a map ϕ : *D*ξ → *X* that satisfies (3.18) for all *x*, eq. (3.21) for (*t*, *x*) ∈ *D*ξ , as well as (3.19) whenever defined.

#### 3.2 Poisson brackets and Hamiltonian vector fields

To obtain flows, classical mechanics requires more than a manifold structure:

Definition 3.6. *A* Poisson bracket *on a manifold X is a Lie bracket* {−,−} *on (the real vector space) C*∞(*X*)*, such that for each h* ∈ *C*∞(*X*) *the map*

$$
\mathfrak{S}\_h: f \mapsto \{h, f\} \tag{3.26}
$$

*is a vector field on X (or, equivalently, a derivation of C*∞(*X*,R) *with respect to its structure of a commutative algebra under pointwise multiplication). A manifold X equipped with a Poisson bracket is called a* Poisson manifold*,* (*C*∞(*X*),{, }) *is called a* Poisson algebra*, and* ξ*h is called the* Hamiltonian vector field *of h.*

Unfolding, we have a bilinear map {−,−} :*C*∞(*X*)×*C*∞(*X*) →*C*∞(*X*) that satisfies

$$\{\mathbf{g},f\} = -\{f,\mathbf{g}\};\tag{3.27}$$

$$\{f, \{\mathbf{g}, h\}\} + \{h, \{f, \mathbf{g}\}\} + \{\mathbf{g}, \{h, f\}\} = \mathbf{0};\tag{3.28}$$

$$\{f, gh\} = \{f, g\}h + \mathbf{g}\{f, h\}.\tag{3.29}$$

Bilinearity and the abstract properties (3.27) - (3.29) imply:

Proposition 3.7. *Each Poisson bracket on X defines a Lie algebra homomorphism*

$$\mathcal{C}^{\curvearrowleft}(X) \to \text{Der}(\mathcal{C}^{\curvearrowright}(X));\tag{3.30}$$

$$h \mapsto \delta\_h,\tag{3.31}$$

*or, equivalently, a Lie algebra homomorphism*

$$\mathcal{C}^{\infty}(X) \to \text{Vec}(X);\tag{3.32}$$

$$h \mapsto \mathfrak{F}\_h. \tag{3.33}$$

The time-honored example is *X* = R2*n*, with coordinates *x* = (*p*,*q*) and bracket

$$\{f, g\} = \sum\_{j=1}^{n} \left( \frac{\partial f}{\partial p\_j} \frac{\partial g}{\partial q^j} - \frac{\partial f}{\partial q^j} \frac{\partial g}{\partial p\_j} \right). \tag{3.34}$$

In that case, the Hamiltonian vector field of *h* is obviously given by

$$\mathfrak{H}\_h = \sum\_{j=1}^n \left( \frac{\partial h}{\partial p\_j} \frac{\partial}{\partial q^j} - \frac{\partial h}{\partial q^j} \frac{\partial}{\partial p\_j} \right). \tag{3.35}$$

The flow of ξ*h* gives the motion of a system with Hamiltonian *h*. Writing

$$
\mathfrak{q}\_l(p, q) = (p(t), q(t)),
$$

we see from (3.23) that this flow is given by *Hamilton's equations*

#### 3.2 Poisson brackets and Hamiltonian vector fields 89

$$\frac{dp\_j(t)}{dt} = -\frac{\partial h(p(t), q(t))}{\partial q^j};\tag{3.36}$$

$$\frac{dq^j(t)}{dt} = \frac{\partial h(p(t), q(t))}{\partial p\_j}.\tag{3.37}$$

Hamiltonians of the special form

$$h(p,q) = \frac{p^2}{2m} + V(q),\tag{3.38}$$

where *p*2 = ∑*j p*2 *j* , give *Newton's equation* "*F* = *ma*", where *Fj* = −∂*V*/∂*qj* , viz.

$$F\_j(q(t)) = m \frac{d^2 q^j(t)}{dt^2}.\tag{3.39}$$

Proposition 3.8. *For any vector field* ξ *on a manifold X, we say that a function f* ∈ *C*∞(*X*) *is* conserved *if f is constant along the flow of* ξ *. If X is a Poisson manifold and* ξ = ξ*h is Hamiltonian, then f is conserved iff* {*h*, *f* } = 0*.*

The proof is trivial. A Poisson bracket on *X* may also be defined in terms of a *Poisson tensor*. In coordinates, this is just an anti-symmetric matrix *Bi j*(*x*) that satisfies

$$\sum\_{l} \left( B^{l\bar{l}} \frac{\partial B^{jk}}{\partial \mathbf{x}\_{l}} + B^{l\bar{j}} \frac{\partial B^{ki}}{\partial \mathbf{x}\_{l}} + B^{lk} \frac{\partial B^{ij}}{\partial \mathbf{x}\_{l}} \right) = \mathbf{0},\tag{3.40}$$

for each (*i*, *j*, *k*). In terms of *B*, the Poisson bracket is then defined abstractly by

$$\{f, \mathbf{g}\} = B(df, d\mathbf{g}),\tag{3.41}$$

using standard notation of differential geometry, or, in coordinates, by

$$\{f, g\}(\mathbf{x}) = \sum\_{i, j} B^{ij}(\mathbf{x}) \frac{\partial f(\mathbf{x})}{\partial \mathbf{x}^{i}} \frac{\partial g(\mathbf{x})}{\partial \mathbf{x}^{j}}.\tag{3.42}$$

Conversely, a Poisson bracket must come from a Poisson tensor: for any derivation δ on *C*∞(*X*), the function δ(*g*) depends linearly on *dg*, so if δ*f*(*g*) = { *f*,*g*}, then δ*f*(*g*) = −δ*g*(*f*), so that { *f*,*g*} depends linearly on both *d f* and *dg*. This enforces (3.42), upon which (3.41) implies (3.40). A nice example is *X* = R3, with

$$\begin{split} \{f,g\}(\mathbf{x}) &= \mathbf{x} \left( \frac{\partial f}{\partial \mathbf{y}} \frac{\partial g}{\partial z} - \frac{\partial f}{\partial z} \frac{\partial g}{\partial \mathbf{y}} \right) + \mathbf{y} \left( \frac{\partial f}{\partial z} \frac{\partial g}{\partial \mathbf{x}} - \frac{\partial f}{\partial \mathbf{x}} \frac{\partial g}{\partial z} \right) + z \left( \frac{\partial f}{\partial \mathbf{x}} \frac{\partial g}{\partial \mathbf{y}} - \frac{\partial f}{\partial \mathbf{y}} \frac{\partial g}{\partial \mathbf{x}} \right); \\ \mathcal{B}^{ij}(\mathbf{x}) &= \sum\_{k} \mathbf{c}\_{kij} \mathbf{x}^{k}. \end{split} \tag{3.43}$$

Finally, we say that a Poisson manifold is*symplectic* if the corresponding Poisson tensor *B*(*x*) is given by an *invertible* matrix, for each *x* ∈ *X*. This requires *X* to be *even-dimensional*. For example, R2*n* with Poisson bracket (3.34) is symplectic.

#### 3.3 Symmetries of Poisson manifolds

Two equivalent notions of symmetries of classical physics suggest themselves: one is based on the idea of a Poisson *manifold* (*X*,*B*), the other comes from the equivalent notion of a Poisson *algebra* (*C*∞(*X*),{, }).

Definition 3.9. *1. A symmetry of a Poisson manifold* (*X*,*B*) *is a diffeomorphism* ϕ : *X* → *X (that is, an invertible smooth map with smooth inverse) satisfying*

$$
\varphi\_\* B = B.\tag{3.44}
$$

*2. A symmetry of a Poisson algebra* (*C*∞(*X*),{, }) *is an invertible linear map* α : *C*∞(*X*) → *C*∞(*X*) *that satisfies (for each f*,*g* ∈ *C*∞(*X*)*):*

$$
\mathfrak{a}(fg) = \mathfrak{a}(f)\mathfrak{a}(\mathfrak{g});\tag{3.45}
$$

$$\alpha(\{f,g\}) = \{\alpha(f), \alpha(g)\}.\tag{3.46}$$

Let us define the push-forward ϕ∗ in (3.44). We do this in terms of the *pullback* ϕ∗ of a smooth (i.e., infinitely often differentiable) map ϕ : *X* → *X*, defined as

$$\varphi^\* : \mathcal{C}^\circ(X) \to \mathcal{C}^\circ(X);\tag{3.47}$$

$$
\mathfrak{g}^\* f = f \circ \mathfrak{g}.\tag{3.48}
$$

If ϕ is a diffeomorphism, the *push-forward* ϕ∗ of ϕ, which acts on derivations, is

$$\mathfrak{q}\_\* : \text{Der}(\mathcal{C}^\infty(X)) \to \text{Der}(\mathcal{C}^\infty(X));\_{.} \tag{3.49}$$

$$(\mathfrak{q}\_\*\delta)(f) = \delta(\mathfrak{q}^\*f) \circ \mathfrak{q}^{-1};\tag{3.50}$$

this may be checked to define a derivation, as follows:

$$\begin{aligned} (\boldsymbol{\varrho}\_\*\boldsymbol{\delta})(f\cdot\boldsymbol{g}) &= (\boldsymbol{\varrho}^{-1})^\*\boldsymbol{\delta}(\boldsymbol{\varrho}^\*(f\cdot\boldsymbol{g})) \\ &= (\boldsymbol{\varrho}^{-1})^\*\boldsymbol{\delta}(\boldsymbol{\varrho}^\*(f)\boldsymbol{\varrho}^\*(\boldsymbol{g})) \\ &= (\boldsymbol{\varrho}^{-1})^\*(\boldsymbol{\delta}(\boldsymbol{\varrho}^\*(f))\boldsymbol{\varrho}^\*(\boldsymbol{g}) + \boldsymbol{\varrho}^\*(f)\boldsymbol{\delta}(\boldsymbol{\varrho}^\*(\boldsymbol{g}))) \\ &= (\boldsymbol{\varrho}\_\*\boldsymbol{\delta})(f)\cdot\boldsymbol{g} + f\cdot(\boldsymbol{\varrho}\_\*\boldsymbol{\delta})(g). \end{aligned}$$

If, given coordinates *x* = (*x*1,..., *xk*) on *X*, we now (without loss of generality) take our derivation δ to be a vector field ξ = ∑*j* ξ *j* ∂/∂ *x j* , and write ϕ(*x*) = (ϕ1(*x*),...,ϕ*l* (*x*)), for the image ϕ∗(ξ ) we obtain

$$\begin{aligned} (\Phi\_\*\xi)(f)(\mathbf{x}) &= (\xi(\mathfrak{q}^\*f))(\mathfrak{q}^{-1}(\mathbf{x})) \\ &= \sum\_j \xi^j (\mathfrak{q}^{-1}(\mathbf{x})) \left( \frac{\partial}{\partial \mathbf{x}^j} f \circ \mathfrak{q} \right) (\mathfrak{q}^{-1}(\mathbf{x})) \\ &= \sum\_{j,k} \xi^k (\mathfrak{q}^{-1}(\mathbf{x})) \frac{\partial f(\mathbf{x})}{\partial \mathbf{x}^j} \frac{\partial \mathfrak{q}^j}{\partial \mathbf{x}^k} (\mathfrak{q}^{-1}(\mathbf{x})), \end{aligned}$$

#### 3.3 Symmetries of Poisson manifolds 91

so that

ϕ∗ξ *j* (*x*) = ∑ *k* ∂ ϕ *j* ∂ *xk* (ϕ−1(*x*))ξ *k* (ϕ−1(*x*)), (3.51)

or, equivalently,

ϕ∗ξ *j* (ϕ(*x*)) = ∑ *k* ∂ ϕ *j* ∂ *xk* (*x*)ξ *k* (*x*), (3.52)

which only depends on ξ (*x*), so that for each *x* ∈ *X*, ϕ∗ may be localized to a linear map ϕ∗(*x*) : *TxX* → *T*ϕ(*x*)*X*. This may be done even if ϕ is not invertible. Physicists often write this as ϕ(*x*) ≡ *y* = *y*(*x*1,..., *xk*), ξ = *v*, ϕ∗ξ = *v* , so that we have a "covariant" transformation rule (*v* )*i* (*y*) = ∑*k j*=1 ∂ *yi* (*x*) ∂ *x j vj* (*x*).

Taking tensor products, one obtains similar rules for higher-order tensors. For example, if *N* = *X*, the transformation rule for the Poisson tensor *B* reads

ϕ∗*Bi j*(ϕ(*x*)) = *k* ∑ *m*,*n*=1 ∂ ϕ*i* (*x*) ∂ *xm* ∂ ϕ *j* (*x*) ∂ *xn Bmn*(*x*), (3.53)

so that, in coordinates, the invariance requirement (3.44) reads

$$\sum\_{m,n=1}^{k} \frac{\partial \phi^i(\mathbf{x})}{\partial \mathbf{x}^m} \frac{\partial \phi^j(\mathbf{x})}{\partial \mathbf{x}^n} B^{mn}(\mathbf{x}) = B^{ij}(\boldsymbol{\varphi}(\mathbf{x})).\tag{3.54}$$

Theorem 3.10. *The two parts of Definition 3.9 are equivalent, in that:*

*1. Given a diffeomorphism* ϕ : *X* → *X satisfying* (3.44)*, the map*

$$
\mathfrak{a} = \mathfrak{p}^\*,
\tag{3.55}
$$

Here an *anti-isomorphism* of groups is just an isomorphism that inverts the order of multiplication. This complication may be removed by writing ϕ−1 instead of ϕ in (3.55), but that change would make the next proposition a bit less natural.

*Proof.* The first claim is true by construction. The hard part is the second claim, which follows from a more general result about manifolds (note that in our terminology, manifolds are by definition assumed to be Hausdorff):

Proposition 3.11. *Let X and Y be a smooth manifolds. Then* (3.55) *establishes a bijective correspondence between linear maps* α :*C*∞(*X*) →*C*∞(*Y*) *satisfying* (3.45) *and smooth maps* ϕ : *Y* → *X.*

The proof is quite similar to a central part of the proof of Gelfand duality for commutative C\*-algebras, in which (3.55) establishes a bijective correspondence between C\*-homomorphisms α : *C*(*X*) → *C*(*Y*) and continuous maps ϕ : *Y* → *X*, where *X* and *Y* are compact Hausdorff spaces; see §C.3 and especially Proposition C.22.

For any commutative real algebra *A*, let Σ(*A*) be the space of non-zero algebra homomorphisms ω : *A* → R (these are just the non-zero multiplicative linear maps), equipped with the weakest topology that makes each function ˆ*a* : Σ(*A*) → R continuous, where ˆ*a*(ω) = ω(*a*). Furthermore, if *B* is another commutative real algebra, then any homomorphism α : *A* → *B* induces a continuous map α∗ : Σ(*B*) → Σ(*A*) in the obvious way, that is, by α∗ω = ω ◦ α. In the special case *A* = *C*∞(*X*) (and similarly if *A* = *C*(*X*)), one has a canonical map ev*X* : *X* → Σ(*C*(*X*)), given by ev*X x* (*f*) = *f*(*x*). The whole point (in which the entire difficulty of the proof lies) is that this map is a bijection (see Proposition C.21), which simultaneously equips *X* with a smooth structure that makes ev*X* a diffeomorphism (by definition of the smooth structure on Σ(*C*(*X*)). In view of all this, given a multiplicative linear map α : *C*∞(*X*) → *C*∞(*Y*), we obtain a continuous map ϕ : *Y* → *X* by

$$\mathfrak{q} = (\mathfrak{ev}^Y)^{-1} \circ \mathfrak{a}^\* \circ \mathfrak{ev}^X. \tag{3.56}$$

Eq. (3.55) then holds by construction. Smoothness of ϕ, then, is a consequence of the fact that α(*f*) = *f* ◦ϕ must be a smooth function on *Y* for any *f* ∈ *C*∞(*X*).

Applying this to the setting of Theorem 3.10 easily yields all claims. -

In what follows, we look at smooth actions of Lie groups on (Poisson) manifolds *X*, in other words, at homomorphisms ϕ : *G* → Diff(*X*) or ϕ : *G* → Diff(*X*,*B*), where *G* is a Lie group, Diff(*X*) is the group of all diffeomorphisms of a manifold, and Diff(*X*,*B*) is the group of all diffeomorphisms of a Poisson manifold preserving the Poisson structure. Foregoing the underlying differential geometry, we take a pragmatic attitude and only study *linear Lie groups*, defined as closed subgroups *G* of *GLn*(R) or *GLn*(C), with group multiplication given by matrix multiplication and hence group inverse being matrix inverse. Here one may think of *SU*(2) ⊂ *GL*2(C) or *SO*(3) ⊂ *GL*3(R), but also abelian Lie groups like the additive groups R*n* fall under this scope, since one may identify *a* ∈ R*n* with the 2*n*×2*n*-matrix

$$a \equiv \begin{pmatrix} 1 \ a \\ 0 \ 1 \end{pmatrix},\tag{3.57}$$

in which case matrix multiplication indeed reproduces addition. Similarly, the 2*n*+ 1-dimensional *Heisenberg group Hn* is the group of real (*n*+2)×(*n*+2)-matrices

$$(a,b,c) = \begin{pmatrix} 1 \ a^T \ c + \frac{1}{2} a^T b \\ 0 \ 1\_n & b \\ 0 \ 0 & 1 \end{pmatrix},\tag{3.58}$$

where *a*,*b* ∈ R*n*, *c* ∈ R, and *aT b* = *a*,*b*; this gives the multiplication rule

$$(a,b,c)\cdot(a',b',c')=(a+a',b+b',c+c'-\frac{1}{2}(\langle a,b'\rangle - \langle a',b\rangle)).\tag{3.59}$$

If *G* is a linear Lie group, its *Lie algebra* g may be defined as the vector space

$$\mathfrak{g} = \{ A \in M\_n(\mathbb{K}) \mid e^{tA} \in G \forall t \in \mathbb{R} \},\tag{3.60}$$

where K = R or C, as determined by the embedding *G* ⊂ *GLn*(R)) or *G* ⊂ *GLn*(C). Either way, g is seen as a *real* vector space, equipped with the *Lie bracket*

$$[A,B] = AB - BA.\tag{3.61}$$

This is trivially a bilinear antisymmetric map g×g → g satisfying the Jacobi identity

$$[A, [B, C]] + [C, [A, B]] + [B, [C, A]] = 0,\tag{3.62}$$

which in turn expresses the fact that for fixed *A* ∈ *g* the map δ*A* : g → g defined by

$$
\delta\_A(B) = [A, B] \tag{3.63}
$$

is a derivation of g with respect to its Lie bracket, i.e.,

$$\delta\_{\mathcal{A}}([B,C]) = [\delta\_{\mathcal{A}}(B),C] + [B,\delta\_{\mathcal{A}}(C)].\tag{3.64}$$

The *exponential map* exp : g → *G* is then just given by its usual power series, which for matrices is norm-convergent. Conversely, one may pass from *G* to g through

$$A = \frac{d}{dt}(e^{tA})\_{|t=0}.\tag{3.65}$$

If *G* = R*n*, we also have g = R*n*, and eq. (3.57) implies that exp is the identity map.

For example, since *SO*(3) is the subgroup of *GL*3(R) consisting of matrices *R* that satisfy *RTR* = 13, its Lie algebra so(3) consists of all matrices *a* that satisfy *aT* = −*a*. As a vector space have so(3) ∼= R3, which follows by choosing a basis

$$J\_1 = \begin{pmatrix} 0 \ 0 \ 0 \\ 0 \ 0 \ -1 \\ 0 \ 1 \ 0 \end{pmatrix}, J\_2 = \begin{pmatrix} 0 & 0 \ 1 \\ 0 & 0 \ 0 \\ -1 & 0 \ 0 \end{pmatrix}, J\_3 = \begin{pmatrix} 0 \ -1 \ 0 \\ 1 \ 0 \ 0 \\ 0 \ 0 \ 0 \end{pmatrix}. \tag{3.66}$$

of the 3×3 real antisymmeric matrices. The commutators of these elements are

$$\left[J\_1, J\_2\right] = J\_3; \ \left[J\_3, J\_1\right] = J\_2; \ \left[J\_2, J\_3\right] = J\_1. \tag{3.67}$$

For the Lie algebra of the Heisenberg group we obtain h*n* = R2*n*+1, with basis

$$P\_i = \begin{pmatrix} 0 \ 0 & 0 \\ 0 \ 0 & -\mathbf{e}\_i \\ 0 \ 0 & 0 \end{pmatrix}, Q\_j = \begin{pmatrix} 0 \ \mathbf{e}\_j^T \ 0 \\ 0 \ \mathbf{0} \ 0 \\ 0 \ \mathbf{0} \ 0 \end{pmatrix}, Z = \begin{pmatrix} 0 \ 0 \ 1 \\ 0 \ 0 \ 0 \\ 0 \ 0 \ 0 \end{pmatrix},\tag{3.68}$$

where (e1,..., e*n*) is the usual basis of R*n*, satisfying commutation relations

$$[P\_l, Q\_j] = \delta\_{lj} Z; \ [P\_l, P\_j] = [\mathcal{Q}\_l, \mathcal{Q}\_j] = [P\_l, Z] = [\mathcal{Q}\_j, Z] = 0. \tag{3.69}$$

#### 3.4 The momentum map

Leaving out the Poisson structure for the moment, let *X* be a manifold, let *G* be a Lie group, and let ϕ : *G* → Diff(*X*) be a homomorphism; as already mentioned, this corresponds to a smooth action ϕ˜ : *G*×*X* → *X*, which we simply write as

$$
\gamma \cdot x \equiv \mathfrak{q}\_{\mathcal{I}}(x) \equiv \mathfrak{q}(\gamma, x) \,.
$$

In terms of the pullback ϕ∗ γ (*f*) = *f* ◦ϕγ , we then automatically have

$$
\mathfrak{op}\_{\mathcal{Y}}^\*(fg) = \mathfrak{op}\_{\mathcal{Y}}^\*(f)\mathfrak{op}\_{\mathcal{Y}}^\*(g). \tag{3.70}
$$

For each *A* ∈ g we then define a map δ*A* : *C*∞(*X*) → *C*∞(*X*) by

$$
\delta\_\mathcal{A} f(\mathbf{x}) = \frac{d}{dt} f(e^{-t\mathbf{A}} \cdot \mathbf{x})\_{|t=0}. \tag{3.71}
$$

This map is obviously linear. Moreover, it can be shown that δ is well behaved:

Proposition 3.12. *The map* δ : g → Der(*C*∞(*X*))*, A* → δ*A is a homomorphism of Lie algebra, i.e., each* δ*A is a derivation,* δ *is linear in A, and, for each A*,*B* ∈ g*,*

$$[\delta\_A, \delta\_B] = \delta\_{[A,B]}.\tag{3.72}$$

The proof relies on *Hadamard's Lemma*, which we only need for complete vector fields, or, equivalently, for derivations with complete flow (i.e., defined for all *t*).

Lemma 3.13. *If* δ *is a derivation of C*∞(*X*) *with complete flow* ϕ*, and f* ∈ *C*∞(*X*)*, then there is a function g*(*t*, *x*) ≡ *gt*(*x*) *such that for all x and t,*

$$\mathfrak{g}\_0(\mathfrak{x}) = \delta f(\mathfrak{x});\tag{3.73}$$

$$f(\mathfrak{g}\_l(\mathfrak{x})) = f(\mathfrak{x}) + t\mathfrak{g}\_l(\mathfrak{x}).\tag{3.74}$$

Indeed, if the flow is complete one may take

$$\mathbf{g}\_l(\mathbf{x}) = \int\_0^1 ds \dot{F}(\mathbf{st}, \mathbf{x}),\tag{3.75}$$

where *F*(*t*, *x*) = *f*(ϕ*t*(*x*)) and (in Newton's notation) *F*˙ is the time derivative of *F*.

*Proof.* To prove that δ*A* is linear in *A*, let ϕ be the flow of δ*A*, i.e., ϕ*t*(*x*) = *e*−*tAx*. For *B* ∈ g, Hadamard's Lemma with δ δ*A* and *x e*−*tBx* then gives us

$$f(e^{-tA}e^{-tB}\mathbf{x}) = f(\mathfrak{gl}(e^{-tB}\mathbf{x})) = f(e^{-tB}\mathbf{x}) + t\mathfrak{g}\_I(e^{-tB}\mathbf{x});$$

$$\Rightarrow \frac{d}{dt}f(e^{-tA}e^{-tB}\mathbf{x})|\_{t=0} = \mathfrak{G}\_{\mathfrak{B}}f(\mathbf{x}) + \mathfrak{g}\_0(\mathbf{x}) = \mathfrak{G}\_{\mathfrak{B}}f(\mathbf{x}) + \mathfrak{G}\_{\mathbf{A}}f(\mathbf{x}).\tag{3.76}$$

On the other hand, since *A* and *B* are matrices, we may use the CBH-formula

3.4 The momentum map 95

$$e^{-tA}e^{-tB} = e^{-t(A+B) + \frac{1}{2}t^2[A,B] + O(t^3)},\tag{3.77}$$

which gives *e*−*tAe*−*tB* = *e*−*t*(*A*+*B*) (1+*O*(*t* 2)), and hence

$$\frac{d}{dt}f(e^{-tA}e^{-tB}\mathbf{x})\_{|t=0} = \frac{d}{dt}f(e^{-t(A+B)}\mathbf{x})\_{|t=0} = \delta\_{A+B}f(\mathbf{x}).\tag{3.78}$$

Comparing (3.76) with (3.78) gives δ*A*+*B* = δ*A* +δ*B*. The property δ*sA* = *s*δ*A* is trivial. We now prove (3.72). Within the (matrix) Lie algebra g we have

$$\langle A, B \rangle = -\frac{d}{dt}(e^{-tA}Be^{tA})\_{|t=0} = -\lim\_{t \to 0} \frac{e^{-tA}Be^{tA} - B}{t}. \tag{3.79}$$

Furthermore, for any *g* ∈ *G* one has *egBg*−1 = *geBg*−1, so linearity of δ gives

$$\begin{split} \delta\_{[A,B]}f(\mathbf{x}) &= -\lim\_{t\to 0} \frac{1}{t} \left( \mathcal{S}\_{e^{-tA}Be^{tA}}f(\mathbf{x}) - \mathcal{S}\_{\mathbf{B}}f(\mathbf{x}) \right) \\ &= \lim\_{t\to 0} \frac{1}{t} \left( \frac{d}{ds} f(e^{-tA}e^{sB}e^{tA}\mathbf{x}) - \frac{d}{ds} f(e^{sB}\mathbf{x}) \right) \\ &= \lim\_{s,t\to 0} \frac{1}{st} \left( f(e^{-tA}e^{sB}e^{tA}\mathbf{x}) - f(e^{-tA}e^{tA}e^{sB}\mathbf{x}) \right) \\ &= \lim\_{s,t\to 0} \frac{1}{st} \left( f \circ \otimes\_{l} (e^{sB}e^{tA}\mathbf{x}) - f \circ \otimes\_{l} (e^{tA}e^{sB}\mathbf{x}) \right) \\ &= \lim\_{s,t\to 0} \left( \frac{1}{st} \left( f(e^{sB}e^{tA}\mathbf{x}) - f(e^{tA}e^{sB}\mathbf{x}) \right) + \frac{1}{s} \left( g\_{l}(e^{sB}e^{tA}\mathbf{x}) - g\_{l}(e^{tA}e^{sB}\mathbf{x}) \right) \right) \\ &= [\delta\_{l}, \delta\_{\mathbf{B}}]f(\mathbf{x}), \end{split}$$

since in the limit *t* → 0 the third term in the penultimate line cancels the fourth. -

Now suppose that, in addition, *X* is a Poisson manifold, and that each ϕγ acts on *X* as a Poisson symmetry, in that

ϕ∗ γ *B* = *B*, (3.80)

cf. (3.44), or, equivalently, cf. (3.46),

$$\mathfrak{op}\_{\mathcal{T}}^\*(\{f,\mathfrak{g}\}) = \{\mathfrak{op}\_{\mathcal{T}}^\*(f), \mathfrak{op}\_{\mathcal{T}}^\*(\mathfrak{g})\}.\tag{3.81}$$

This implies, for each *A* ∈ g, and each *f*,*g* ∈ *C*∞(*X*),

$$\delta\_{\mathcal{A}}(\{f,\mathcal{g}\}) = \{\delta\_{\mathcal{A}}(f), \mathcal{g}\} + \{f, \delta\_{\mathcal{A}}(\mathcal{g})\}.\tag{3.82}$$

Compare this with the following property δ*A* already has since it is a derivation:

$$
\delta\_\mathcal{A}(f\mathbf{g}) = \delta\_\mathcal{A}(f)\mathbf{g} + f\delta\_\mathcal{A}(\mathbf{g}).\tag{3.83}
$$

We may call a derivation δ : *C*∞(*X*) → *C*∞(*X*) satisfying the like of (3.82), i.e.,

$$\delta(\{f,\mathbf{g}\}) = \{\delta(f),\mathbf{g}\} + \{f,\delta(\mathbf{g})\},\tag{3.84}$$

a *Poisson derivation*. We are already familiar with a large class of Poisson derivations: for each *h* ∈ *C*∞(*X*), the corresponding map δ*h* defined by (3.26) is a Poisson derivation (this follows from the Jacobi identity). Let us call a Poisson derivation of the kind δ*h inner*. This raises the question if our derivations δ*A* are inner.

Definition 3.14. *A* momentum map *for a Lie group G acting on a Poisson manifold X is a map*

$$J: X \to \mathfrak{g}^\* \tag{3.85}$$

*such that for each A* ∈ g*,*

$$
\delta\_{\mathbf{A}} = \delta\_{I\_{\mathbf{A}}},\tag{3.86}
$$

*where the function JA* ∈ *C*∞(*X*) *is defined by by*

$$J\_A(\mathbf{x}) = \langle J(\mathbf{x}), A \rangle \equiv J(\mathbf{x})(A). \tag{3.87}$$

*In other words, for each A* ∈ g *and f* ∈ *C*∞(*X*) *we must have*

$$
\delta\_\mathcal{A}(f) = \{J\_\mathcal{A}, f\}. \tag{3.88}
$$

*A Lie group action admitting a momentum map is called* Hamiltonian*.*

Equivalently, a momentum map is a linear map

$$J^\*: \mathfrak{g} \to \mathcal{C}^\alpha(X) \tag{3.89}$$

such that δ*A* = δ*J*∗(*A*); the connection between the two definitions is given by

$$J\_A = J^\*(A). \tag{3.90}$$

The pullback notation *J*∗ would suggest that it is a map *C*∞(g∗) → *C*∞(*X*), which is not quite the case, but it is a near miss: we embed g → *C*∞(g∗) by *A* → *A*ˆ, where *A*ˆ(θ) = θ(*A*), so *J*∗ : g → *C*∞(*X*) is the restriction of the pullback *J*∗ to g. Another near miss would be to read *J*∗ as the adjoint to *J*, which maps g∗∗ ∼= g to the 'dual' *X*∗, but since *X* may not be a vector space, this dual cannot be defined as in linear algebra, so instead of all linear maps from *X* to R we might as well say that it consists of all smooth functions on *X*. Either way, the symbol *J*∗ seems justified.

Proposition 3.15. *Let G be a connected Lie group that acts on a Poisson manifold X. If this action is Hamiltonian (i.e., if it has a momentum map), then G acts on* (*X*,*B*) *by Poisson symmetries (in the sense that* (3.81) *holds).*

*Proof.* An easy computation shows that (3.82) holds. We omit the proof of the fact that for *connected* Lie groups this "infinitesimal" property is equivalent to (3.81); this relies on the fact that *G* is generated by the image of the exponential map. -

The converse is not true: if *G* acts by Poisson symmetries, the action is not necessarily Hamiltonian. For example, take *X* = R2, with the unusual Poisson bracket

$$\{f, g\}(p, q) = p \left( \frac{\partial f}{\partial p} \frac{\partial g}{\partial q} - \frac{\partial f}{\partial q} \frac{\partial g}{\partial p} \right), \tag{3.91}$$

and let *G* = R act on R2 by *b* ·(*p*,*q*)=(*p*,*q* + *b*). This action satisfies (3.81), and has a single generator δ = −∂/∂*q*. But there clearly is no function *J* ∈*C*∞(R2) such that {*J*, *f* } = −∂ *f* /∂*q* (it should be *J*(*p*,*q*) = −log(*p*), which is singular at *p* = 0).

However, in most "everyday situations" momentum maps exist:

	- a. Let *G* = R6 act on *X* by

$$(\mathbf{a}, \mathbf{b}) \cdot (\mathbf{p}, \mathbf{q}) = (\mathbf{p} + \mathbf{a}, \mathbf{q} + \mathbf{b}). \tag{3.92}$$

This action is Hamiltonian, with momentum map

$$J(\mathbf{p}, \mathbf{q}) = (\mathbf{q}, -\mathbf{p}). \tag{3.93}$$

b. Let *G* = *SO*(3) act on the same space *X* by

$$R \cdot (\mathbf{p}, \mathbf{q}) = (R\mathbf{p}, R\mathbf{q}).\tag{3.94}$$

Also this action is Hamiltonian, with momentum map

$$J(\mathbf{p}, \mathbf{q}) = \mathbf{p} \times \mathbf{q}. \tag{3.95}$$

2. Let *G* = *SO*(3) act on *X* = R3, equipped with the Poisson bracket (3.43), through its defining representation. This action has a momentum map

$$J(\mathbf{x}) = \mathbf{x},\tag{3.96}$$

where we have identified g with R3 by choosing the basis (3.66) of g, and have identified g∗ with g (and hence with R3 also) by the usual inner product on R3.

3. The previous example is a special case of the *Lie–Poisson structure*. Let *G* be a Lie group with Lie algebra g. Choose a basis (*Ta*) of g, with associated *structure constants Cc ab* defined by the Lie bracket on g as

$$[T\_a, T\_b] = \sum\_c C\_{ab}^c T\_c.\tag{3.97}$$

We write θ in the dual vector space g∗ as θ = ∑*a* θ*a*ω*a*, where (ω*a*) is the dual basis to a chosen basis (*Ta*) of g, i.e., ω*a*(*Tb*) = δ*ab*. In terms of these coordinates, the *Lie–Poisson bracket* on *C*∞(g∗) is defined by

$$\{f, g\}(\theta) = \mathcal{C}\_{ab}^{c} \theta\_c \frac{\partial f(\theta)}{\partial \theta\_a} \frac{\partial g(\theta)}{\partial \theta\_b}. \tag{3.98}$$

Equivalently, the Poisson bracket (3.98) may be defined by the condition

98 3 Classical physics on a general phase space

$$\{\hat{A},\hat{B}\} = \bar{[A,\hat{B}]},\tag{3.99}$$

where *A*,*B* ∈ g and *A*ˆ ∈ *C*∞(g∗) is the evaluation map *A*ˆ(θ) = θ(*A*). Now *G* canonically acts on g∗ through the *coadjoint representation*, defined by

$$(\boldsymbol{\mathfrak{x}} \cdot \boldsymbol{\mathfrak{e}})(\boldsymbol{A}) = \boldsymbol{\mathfrak{e}}(\boldsymbol{\mathfrak{x}}^{-1}\boldsymbol{A}\boldsymbol{\mathfrak{x}}).\tag{3.100}$$

This action is Hamiltonian with respect to the Lie–Poisson bracket (3.98), the associated momentum map simply being the identity map g∗ → g∗, as in (3.96). In other words, we have

$$J\_A = \hat{A},\tag{3.101}$$

whose correctness may be verified from the computation

$$\begin{aligned} \delta\_A \mathcal{B}(\boldsymbol{\theta}) &= \frac{d}{dt} \mathcal{B}(e^{-tA} \cdot \boldsymbol{\theta})\_{|t=0} = \frac{d}{dt} \boldsymbol{\Theta}(e^{tA} B e^{-tA})\_{|t=0} \\ &= \boldsymbol{\Theta}([A, \boldsymbol{\mathcal{B}}]) = \widehat{[A, \boldsymbol{\mathcal{B}}]}(\boldsymbol{\theta}) = \{\hat{A}, \hat{\boldsymbol{\mathcal{B}}}\}(\boldsymbol{\theta}) \\ &= \{J\_A, \hat{\boldsymbol{\mathcal{B}}}\}(\boldsymbol{\theta}). \end{aligned}$$

4. Let *X* = *T*∗*Q* for some manifold *Q*. e.g. *Q* = R*n* and hence *X* = R2*n*. We take

$$G = \text{Diff}(\mathcal{Q}),\tag{3.102}$$

i.e., the diffeomorphism group of *Q*. This is an infinite-dimensional Lie group (if described in the right way). The defining action of ϕ ∈ *G* on *Q* induces an action called ϕ∗ on *T*∗*Q*, given (in coordinates) by

$$\boldsymbol{\Phi}^\*(\boldsymbol{p}, \boldsymbol{q}) = (\boldsymbol{p}', \boldsymbol{q}');\tag{3.103}$$

$$(q^i)' = \mathfrak{q}^i(q);\tag{3.104}$$

$$p'\_i = \sum\_{j=1}^n \frac{\partial (\boldsymbol{\varrho}^{-1})^j(q)}{\partial q^i} p\_j. \tag{3.105}$$

This may be taken as a definition, but in the language of differential geometry this comes down to the neater prescription that if θ = ∑*j pjdqj* ∈ *T*∗ *q Q*, then ϕ∗θ ∈ *T*∗ ϕ(*q*) *Q* is the one-form that maps a vector *X* ∈ *T*ϕ(*q*)*Q* to θ(ϕ−1 ∗ (*X*)), i.e.,

$$(\mathfrak{g}^\*\theta)(X) = \theta(\mathfrak{g}\_\*^{-1}(X)),\tag{3.106}$$

where ϕ−1 ∗ (*X*) = ∑*j* ϕ−1 ∗ (*X*)*j* ∂/∂*qj* is given componentwise by, cf. (3.52),

ϕ−1 ∗ *X j* = ∑ *j* ∂ (ϕ−1)*j* (*q*) ∂*qk Xk* . (3.107)

If *Q* = R3 and ϕ = *R* ∈ *SO*(3), then, using *R*−1 = *RT* , we find that (3.104) - (3.105) simply become *R*∗(p,q)=(*R*p,*R*q), as in (3.94).

Furthermore, if ϕ(q) = q + b, then the partial derivatives in (3.105) form the identity matrix, so that ϕ∗(p,q)=(p,q+b). To show that the action of Diff(*Q*) on *T*∗*Q* is Hamiltonian and compute its momentum map, we need to know that the Lie algebra of Diff(*Q*) is the space Vec(*X*) of all vector fields on *Q*, with its canonical Lie bracket (3.61)! We will not prove this, but the exponential map exp : g → *G* is given through the flow ϕ of the vector field ξ on *Q* by (cf. (3.20))

$$e^{l\frac{\pi}{\xi}} = \mathfrak{q}\_l.\tag{3.108}$$

Theorem 3.16. *The action of* Diff(*Q*) *on T*∗*Q has momentum map*

$$J\_X(p,q) = -\sum\_j p\_j X^j(q),\tag{3.109}$$

*and hence is Hamiltonian. Moreover, this momentum map satisfies*

$$\{J\_{\xi}, J\_{\eta}\}\_{\xi} = -J\_{[\xi, \eta]}.\tag{3.110}$$

*Proof.* First note that ϕ−1 *t* = ϕ−*t*, so from (3.71), (3.108), and (3.104) - (3.105),

$$\begin{split} \delta\_{\xi} f(p,q) &= \frac{d}{dt} f(\Phi^{\*}\_{-t}(p,q))\_{|t=0} \\ &= \sum\_{i,j} \frac{\partial f}{\partial p\_{i}}(p,q) \frac{d}{dt} \left(\frac{\partial \Phi^{j}\_{t}(q)}{\partial q^{i}}\right)\_{|t=0} p\_{j} + \sum\_{i} \frac{\partial f}{\partial q^{i}}(p,q) \frac{d}{dt} \Phi^{i}\_{-t}(q)\_{|t=0} \\ &= \sum\_{i,j} p\_{j} \frac{\partial X^{j}(q)}{\partial q^{i}} \frac{\partial f}{\partial p\_{i}}(p,q) - \sum\_{j} X^{j}(q) \frac{\partial f}{\partial q^{j}}(p,q). \end{split}$$

From this and (3.109), using the canonical Poisson bracket (3.34) we find

$$\{J\_{\xi},f\} = \delta\_{\xi}f.$$

Finally, verifying (3.110) is a simple exercise. -

Thus the momentum map is a generalization of (minus) the momentum, whence its name; the quantity in (3.95) is (minus) the angular momentum. These annoying minus signs could be removed by putting a minus sign in (3.86), but that would have other negative (*sic*) consequences. For example, with our sign choice one often has

$$\{J\_A, J\_B\} = J\_{[A,B]},\tag{3.111}$$

in which case the accompanying map (3.89) is a homomorphism of Lie algebras, or, equivalently, *J* is a morphism with respect to the given Poisson bracket on *X* and the Lie–Poisson bracket on g∗. Such a momentum map is called *infinitesimally equivariant*, for if *G* is connected, (3.111) is equivalent to the equivariance property

$$J(\mathbf{g} \cdot \mathbf{x}) = \mathbf{g} \cdot J(\mathbf{x}). \tag{3.112}$$

.

Here the *G*-action on g∗ on the right-hand side is the coadjoint representation.

All of this is true for our examples (3.95), (3.96), (3.101), and (3.109); in the latter case we note that the Lie bracket in the Lie algebra of Diff(*Q*) is *minus* the commutator of vector fields. However, (3.111) does not always hold (in which case *a fortiori* also (3.112) fails). For example, it fails for (3.93): if we take the usual basis (e,f) ≡ (*e*1, *e*2, *e*3, *f*1, *f*2, *f*3) of g = R6 and relabel *e j* ≡ *Qj* and *fi* ≡ −*Pi*, then

$$J\_{P\_l}(\mathbf{p}, \mathbf{q}) = p\_l;\tag{3.113}$$

$$J\_{\mathcal{Q}\_f}(\mathbf{p}, \mathbf{q}) = q\_f,\tag{3.114}$$

cf. (3.93), and hence, although [*Pi*,*Pj*]=[*Qi*,*Qj*]=[*Pi*,*Qj*] = 0, we obtain

$$\{J\_{\mathcal{P}\_l}, J\_{\mathcal{P}\_j}\} = \{J\_{\mathcal{Q}\_l}, J\_{\mathcal{Q}\_j}\} = 0;\tag{3.115}$$

$$\{J\_{\mathbb{P}}, J\_{\mathbb{Q}\_{\rangle}}\} = \delta\_{ij}\mathbf{1}\_{\mathbb{R}^6}.\tag{3.116}$$

Fortunately, in cases like that one can often find a central extension *G*ϕ of *G* (see §5.10 below for notation) that acts on *X* through its quotient group *G* and does have an infinitesimally equivariant momentum map. In the case at hand, the Heisenberg group *H*3 does the job, whose central elements (0,0, *c*) then act trivially on R6. In terms of the generators (3.68) we take *JPi* and *JQj* as in (3.113) - (3.114), and add *JZ* = 1R6 ; according to (3.69) and (3.115) - (3.116) we then have (3.111), as desired.

Finally, the above formalism leads to a clean formulation of *Noether's Theorem*, providing the well-known link between symmetries and conserved quantities:

Theorem 3.17. *Let X be a Poisson action equipped with a Hamiltonian action of some Lie group G (so that there is a momentum map J* : *X* → g∗*). Suppose h* ∈*C*∞(*X*) *is G-invariant, in that h*(γ · *x*) = *h*(*x*) *for each* γ ∈ *G and x* ∈ *X. Then for each A* ∈ g*, the function JA is constant along the flow of the vector field Xh. In other words,*

$$J\_A(\mathfrak{q}\_l(\mathfrak{x})) = J\_A(\mathfrak{x})\tag{3.117}$$

*for any x* ∈ *X and any t* ∈ R *for which the flow* ϕ*t*(*x*) *of Xh is defined.*

*Proof.* Using all assumptions as well as the definition of a flow, we compute:

$$\frac{d}{dt}J\_A(\mathfrak{q}\_l(\mathbf{x})) = X\_h(J\_A)(\mathfrak{q}\_l(\mathbf{x})) = \delta\_h(J\_A)(\mathfrak{q}\_l(\mathbf{x})) $$

$$= \{h, J\_A\}(\mathfrak{q}\_l(\mathbf{x})) = -\{J\_A, h\}(\mathfrak{q}\_l(\mathbf{x})) $$

$$= -\delta\_A(h)(\mathfrak{q}\_l(\mathbf{x})) = \frac{d}{ds}h(e^{sA}\mathfrak{q}\_l(\mathbf{x}))\_{|s=0} $$

$$= \frac{d}{ds}h(\mathfrak{q}\_l(\mathbf{x}))\_{|s=0} = 0. \tag{7} $$

For example, a Hamiltonian (3.38) has conserved (angular) momentum if the potential *V* is translation (rotation) invariant, reflecting (3.93) and (3.95), respectively.

#### Notes

The traditional symplectic approach to classical mechanics, culminating in the momentum map, is exhaustively covered in Guillemin & Sternberg (1984) and Abraham & Marsden (1985). A founding paper for Poisson geometry is Weinstein (1983). The modern Poisson approach to mechanics may be found in Marsden & Ratiu (1994), from which most of the material in this chapter originates.

Our proof of Proposition 3.11 is based on Navarro Gonzalez & Sancho de Salas ´ (2003), §2.1. Burtscher (2009) is a nice survey of many similar results.

## Chapter 4 Quantum physics on a general Hilbert space

In this chapter we generalize the results of Chapter 2 to infinite-dimensional Hilbert spaces. So let *H* be a Hilbert space and let *B*(*H*) be the set of all *bounded* operators on *H*. Here a notable point is that linear operators on *finite-dimensional* Hilbert spaces are automatically bounded, whereas in general they are not. Thus we impose boundedness as an extra requirement, beyond linearity. This is very convenient, because as in the finite-dimensional case, *B*(*H*) is a C\*-algebra, cf. §C.1. At the same time, assuming boundedness involves no loss of generality whatsoever, since we can alway replace closed unbounded operators by bounded ones through the *bounded transform*, as explained in §B.21. Nonetheless, even the relatively easy setting of bounded operators leads to some technical complications we have to deal with. First, Definition 2.1 must be adjusted as follows:

#### Definition 4.1. *Let H be a Hilbert space.*

As shown in Corollary B.88, if *H* is finite-dimensional this notion of a spectrum reduces to the set of eigenvalues of *a*. Even *H* is infinite-dimensional, the spectrum of a self-adjoint operator *a* is real (i.e., σ(*a*) ⊂ R); this is also true if *a* is unbounded (see Theorem B.93). For any *H*, unit vectors ψ still define special density matrices *e*ψ, as in (2.7); we will later see that these are pure states on *B*(*H*), although the set of pure states is no longer exhausted by such density matrices. Finally, quantum events *in H* still bijectively correspond with projections *on H*; see Proposition B.76. The Born rule as well as the correspondence between density matrices and states require a separate discussion, to which we now turn.

#### 4.1 The Born rule from Bohrification (II)

In this section we extend the characterization of the Born rule in §2.5, which was restricted to finite phase spaces *X* and finite-dimensional Hilbert spaces *H*, to the general case. Recall that a *probability space* is a measure space (*X*,Σ,μ) for which μ(*X*) = 1, and that, for compact *X*, a state on *C*(*X*) is a positive map ϕ : *C*(*X*) → C that is positive and satisfies ϕ(1*X* ) = 1. Theorem B.15 and Corollary (B.17) yield:

Theorem 4.2. *Let X be a compact Hausdorff space. There is a bijective correspondence between probability measures* μ *on X and states* ω *on C*(*X*)*, given by*

$$\mathfrak{w}(f) = \int\_X d\mu \, f, \, f \in \mathcal{C}(X). \tag{4.1}$$

More precisely, the correspondence in question is between complete regular probability spaces (*X*,Σ,μ) and states on *C*(*X*), and this is understood in what follows.

Second, we recall that if *H* is a Hilbert space and *a* ∈ *B*(*H*), then *C*∗(*a*) is the C\*-algebra generated by *a* and 1*H* (i.e., the norm-closure of the algebra of all polynomials in *a*). Theorems B.84, B.94, and B.93 give the following spectral theorem:

Theorem 4.3. *If a*∗ = *a* ∈ *B*(*H*)*, then C*∗(*a*) *is commutative,* σ(*a*) ⊂ R *is compact, and there is an isomorphism of (commutative) C\*-algebras*

$$\mathcal{C}(\sigma(a)) \cong \mathcal{C}^\*(a),\tag{4.2}$$

*written f* → *f*(*a*)*, which is unique if it is subject to the following conditions:*

*Furthermore, this* continuous functional calculus *satisfies the rules*

$$(tf+g)(a) = tf(a) + g(a);\tag{4.3}$$

$$(fg)(a) = f(a)g(a);\tag{4.4}$$

$$f(a)^{\*} = f^{\*}(a). \tag{4.5}$$

Combining Theorems 4.2 and 4.3 gives a result of great importance:

Corollary 4.4. *Let H be a Hilbert space, let a*∗ = *a* ∈ *B*(*H*)*, and let* ψ ∈ *H be a unit vector. There exists a unique probability measure* μψ *on the spectrum* σ(*a*) *such that*

$$
\langle \Psi, f(a)\Psi \rangle = \int\_{\sigma(a)} d\mu\_{\Psi} f, \ f \in C(\sigma(a)). \tag{4.6}
$$

*In terms of the spectral projections e*Δ = 1Δ (*a*) *(defined for Borel sets* Δ ⊆ σ(*a*)*) constructed in* (B.305) *-* (B.307) *and Theorem B.102, the Born measure is given by*

$$\mu\_{\Psi}(\Delta) = \left\| e\_{\Delta} \Psi \right\|^{2}. \tag{4.7}$$

*More generally, a density operator* ρ ∈ D(*H*) *induces a unique probability measure* μρ *on* σ(*a*) *for which*

$$\operatorname{Tr}\left(\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{?}}}}}}}}}}}}}}\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{?}}}}}}}}}}}}}}}}\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{?}}}}}}}}}}}}}}}}}\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{\mathfrak{?}}}}}}}}}}}}}}}}}}}}$$

$$
\mu\_{\mathcal{P}}(\Delta) = \text{Tr}\,(\rho e\_{\Delta}).\tag{4.9}
$$

*This measure on* σ(*a*) *is called the* Born measure *(defined by a and* ψ *or* ρ*).*

*Proof.* The point is that the map *f* → ψ, *f*(*a*)ψ defines a state on *C*(σ(*a*)):

To prove (4.7), use Lemma B.97 to approximate 1Δ by functions *fn* ∈ *C*(σ(*a*)) as stated. By Theorem B.13.2 (i.e., the Lebesgue Monotone Convergence Theorem), we have σ(*a*) *d*μψ *fn* → σ(*a*) *d*μψ 1Δ = μψ(Δ), whereas by (B.315) with *an* = *fn*(*a*), one has ψ, *fn*(*a*)ψ→ψ, *e*Δψ = *e*Δψ2. Hence (4.7) follows from (4.6).

The proof for density operators is analogous. -

Defining the mean value *a*ψ of *a* with respect to the Born measure μψ by

$$
\langle a \rangle\_{\Psi} = \int\_{\sigma(a)} d\mu\_{\Psi}(\mathbf{x}) \,\mathbf{x}, \tag{4.10}
$$

and similarly for ρ, using Theorem 4.3.2 we easily obtain

$$
\langle a \rangle\_{\Psi} = \langle \Psi, a\Psi \rangle;\tag{4.11}
$$

$$
\langle a \rangle\_{\mathcal{P}} = \text{Tr}(\not p a). \tag{4.12}
$$

As an important special case, suppose that σ(*a*) = σ*p*(*a*) (i.e., each λ ∈ σ(*a*) is an eigenvalue); this always happens if *H* is finite-dimensional. Eq. (A.57) then gives

$$\langle \Psi, f(a)\Psi \rangle = \sum\_{\lambda \in \sigma(a)} f(\lambda) \cdot \left||e\_{\lambda}\Psi||^{2},\right|$$

where *e*λ is the projection onto the eigenspace *H*λ = {ψ ∈ *H* | *a*ψ = λψ}. Thus

$$\mu\_{\Psi}(\mathcal{X}) = \left\| e\_{\lambda} \Psi \right\|^{2},\tag{4.13}$$

and using the notation *P*ψ(*a* = λ) for μψ(λ), eq. (4.11) just becomes

$$\langle a \rangle\_{\Psi} = \sum\_{\lambda \in \sigma(a)} \lambda \cdot P\_{\Psi}(a = \lambda). \tag{4.14}$$

It is customary to extend the Born measure on σ(*a*) ⊂ R to a (probability) measure μ ψ on all of R by simply stipulating that

$$
\mu\_{\Psi}'(\Delta) = \mu\_{\Psi}(\Delta \cap \sigma(a));
\tag{4.15}
$$

we will often assume this and omit the prime. This obviously implies that μψ(Δ) = 0 for any Borel set Δ ⊂ R disjoint from σ(*a*); in particular, if σ(*a*) is discrete, then μψ is concentrated on the eigenvalues λ of *a*, in that

$$\mu\_{\Psi}(\Delta) = \sum\_{\lambda \in \Delta \cap \sigma(a)} \mu\_{\Psi}(\lambda). \tag{4.16}$$

To state an interesting property of the Born measure we need Hausdorff's solution to the relevant special case of the famous *Hamburger Moment Problem*:

Theorem 4.5. *If K* ⊂ R *is compact, then any finite measure* μ *on K is determined by its* moments

$$\alpha\_n = \int\_K d\mu(\mathbf{x}) \mathbf{x}^n. \tag{4.17}$$

Using *f*(*x*) = *xn* in (4.6), we therefore obtain:

Corollary 4.6. *The Born measure* μψ *is determined by its moments*

$$
\alpha\_n = \langle \Psi, a^n \Psi \rangle. \tag{4.18}
$$

More precisely, we need to be sure that numbers (α*n*) of the kind (4.18) are the moments of some (probability) measure. This follows from the spectral theorem by running the above argument backwards, but one may also use the general solution of the Hamburger Moment Problem, which we here state without proof:

Theorem 4.7. *A sequence of real numbers* (α*n*) *forms the moments of some measure* μ *on* R *iff for all N* ∈ N *and* (β1,...,β*N*) ∈ C*N one has* ∑*N n*,*m*=0 β*n*β*m*α*n*+*m* ≥ 0*. Furthermore, if there are constants C and D such that* |α*n*| ≤ *CDnn*!*, then* μ *is uniquely determined by its moments* (α*n*)*.*

These conditions are easily checked from (4.18).

If *a* is unbounded, but still assumed to be self-adjoint (in the sense appropriate for unbounded operators, cf. Definition B.70), the spectrum σ(*a*) remains real (see Theorem B.93) but it is no longer compact. Nonetheless, the Born measure on σ(*a*) may be constructed in almost exactly the same way as in the bounded case, this time invoking Corollary B.21 and Theorem B.158 instead of Theorems 4.2 and B.94, respectively. Corollary 4.4 then holds almost *verbatim* for the unbounded case:

Corollary 4.8. *Let H be a Hilbert space, let a*∗ = *a, and let* ψ ∈ *H be a unit vector. There exists a unique probability measure* μψ *on the spectrum* σ(*a*) *such that*

$$
\langle \Psi, f(a)\Psi \rangle = \int\_{\sigma(a)} d\mu\_{\Psi} f, \ f \in \mathcal{C}\_0(\sigma(a)). \tag{4.19}
$$

*Also, eqs.* (4.7) *and* (4.9) *hold, as does* (4.8)*, with f* ∈ *C*0(σ(*a*))*.*

There is no need to worry about domains, since even if *a* is unbounded, *f*(*a*) is bounded for *f* ∈ *Cb*(σ(*a*)), and hence also for *f* ∈ *C*0(σ(*a*)).

The physical relevance of the Born measure is given by the *Born rule*:

*If an observable a is measured in a state* ρ*, then the probability P*ρ (*a* ∈ Δ) *that the outcome lies in* Δ ⊂ R *is given by the Born measure* μρ *defined by a and* ρ*, i.e.,*

$$P\_{\mathcal{P}}(a \in \Delta) = \mu\_{\mathcal{P}}(\Delta). \tag{4.20}$$

As in the finite-dimensional case, the Born measure may be generalized to families (*a*1,...,*an*) of commuting self-adjoint operators. Assuming these are bounded, the C\*-algebra *C*∗(*a*1,...,*an*) is defined in the obvious way, i.e., as the smallest C\* algebra containing each *ai*, or, equivalently, as the norm-closure of the algebra of all finite polynomials in the (*a*1,...,*an*). This C\*-algebra is commutative, as a simple approximation argument shows: polynomials in the *ai* obviously commute, and this property extends to the closure by continuity of multiplication. However, even in the bounded case, the correct notion of a joint spectrum is not obvious. In order to motivate the following definition, it helps to recall Definition 1.4, Theorem C.24, and especially the last sentence before the proof of the latter, making the point that the spectrum σ(*a*) of a single (bounded) self-adjoint operator coincides with the image of the Gelfand spectrum Σ(*C*∗(*a*)) in C under the map ω → ω(*a*).

Definition 4.9. *1. The* joint spectrum σ(*a*) = σ(*a*1,...,*an*) ⊂ R*n of a finite family a* = (*a*1,...,*an*) *of commuting bounded self-adjoint operators is the image of the Gelfand spectrum* Σ(*C*∗(*a*1,...,*an*)) = Σ(*C*∗(*a*)) *under the map*

$$\Sigma(\mathbb{C}^\*(a\_1, \dots, a\_n)) \to \mathbb{R}^n, \text{ } \mathfrak{o} \mapsto (\mathfrak{o}(a\_1), \dots, \mathfrak{o}(a\_n)).\tag{4.21}$$

*Since* ω(*ai*) *only utilizes the restriction of* ω *to C*∗(*ai*) ⊂ *C*∗(*a*)*, we have* ω(*ai*) ∈ σ(*ai*) ⊂ R*, so that* Σ(*C*∗(*a*)) ⊆ σ(*a*1)×···×σ(*an*) *is a compact subset of* R*n.*

To justify this definition, we note that:

$$\lim\_{k \to \infty} \left\|(a\_i - \lambda\_i)\Psi\_k\right\| = 0,\tag{4.22}$$

for each *i* = 1,...,*n*. The proof is similar.

One way to see the second claim is to use Proposition C.14 joined with the observation that, as in the case of *A* = *B*(*H*) for finite-dimensional *H*, any pure state on a finite-dimensional C\*-algebra *A* ⊂ *B*(*H*) is a vector state (2.42), too. To see this, we first specialize Theorem C.133 to the finite-dimensional case (where the proof becomes elementary), so that each state on *C*∗(*a*) takes the form (2.33). Subsequently, we use the spectral decomposition (2.6), and use the definition of purity: suppose ω(*b*) = Tr(ρ*b*) = ∑*i pi*υ*i*,*b*υ*i* ≡ ∑*i pi*ωυ*i* (*b*) is pure, where *b* ∈ *C*∗(*a*). Then ωυ*i* = ω for each *i*, so that ω is a vector state, say ω(*b*) = ψ,*b*ψ where ψ is one of the υ*i*. Once we know this, suppose λ = (λ1,...,λ*n*) ∈ σ(*a*), with λ*i* = ω(*ai*). Multiplicativity of ω implies that for any finite polynomial in *n* real variables we have ψ, *p*(*a*)ψ = *p*(λ), which easily gives *ai*ψ = λ*i*ψ for each *i*; for example, take *p*(*x*)=(*xi* −λ*i*)2, so that the previous equation gives (*ai* −λ*i*)ψ2 = 0.

Conversely, if λ is a joint eigenvalue of *a*, then by definition there exists a joint eigenvector ψ whose vector state ω(*b*) = ψ,*b*ψ on *C*∗(*a*) is multiplicative.

Using this (perhaps contrived) notion of a joint spectrum, Theorem 2.19 now holds by construction also if dim(*H*) = ∞, where the pertinent isomorphism *f* → *f*(*a*) is given as in the single operator case, that is, by starting with polynomials and using a continuity argument to pass to arbitrary continuous functions.

Theorem 2.18 and Corollary 4.4 then generalize to:

Theorem 4.10. *Let H be a Hilbert space, let a* = (*a*1,...,*an*) *be a finite family of commuting bounded self-adjoint operators, and let* ψ ∈ *H be a unit vector. There exists a unique probability measure* μψ *on the joint spectrum* σ(*a*) *such that*

$$\langle \Psi, f(\underline{a})\Psi \rangle = \int\_{\sigma(\underline{a})} d\mu\_{\Psi} f, \; f \in \mathcal{C}(\sigma(\underline{a})), \tag{4.23}$$

*or, equivalently, for special Borel sets* Δ = Δ1 ×···×Δ*n* ⊆ σ(*a*)*, where* Δ*i* ⊂ σ(*ai*)*,*

$$\mu\_{\Psi}(\underline{\mathbf{A}}) = \left\| e\_{\Delta\_{1}} \cdots e\_{\Delta\_{n}} \Psi \right\|^{2},\tag{4.24}$$

*where the e*Δ*i* = 1Δ*i* (*ai*) *are the pertinent spectral projections (which commute).*

Similarly for density operators instead of pure states.

If (some of) the operators *ai* are unbounded, we use the trick of §B.21 and pass to their bounded transforms *bi*, see Theorem B.152. We say that the *bi commute* iff the corresponding bounded operators *bi* do; this is equivalent to commutativity of all spectral projections of the *ai*. We then define, in self-explanatory notation,

$$\sigma(\underline{a}) = \{ \underline{\lambda} (1 - \underline{\lambda}^2)^{-1/2} \mid \underline{\lambda} \in \sigma(\underline{b}) \cap (-1, 1)^n \}. \tag{4.25}$$

This leads to Born measures on σ(*a*) defined either as in (4.23), with *f* ∈ *C*(σ(*a*)) replaced by *f* ∈ *C*0(σ(*a*)), cf. (4.19), or as in (4.24).

For example, if *H* = *L*2(R*n*) and *ai*ψ(*x*) = *xi*ψ(*x*), defined on the domain

$$D(a\_l) = \{ \Psi \in L^2(\mathbb{R}^n) \mid \int\_{\mathbb{R}^n} d^n \mathbf{x} \mathbf{x}\_l^2 |\Psi(\mathbf{x})|^2 < \infty \},\tag{4.26}$$

as in (B.242), then *bi*ψ(*x*) = *xi*(1+*x*2 *i* )−1/2ψ(*x*), so that σ(*b*)=[−1,1] *n* and hence σ(*a*) = R*n*. For a measurable region Δ ⊂ R*n* we then have Pauli's famous formula

$$
\mu\_{\Psi}(\underline{\Delta}) = \int\_{\underline{\Delta}} d^n x \, |\Psi(x)|^2 \tag{4.27}
$$

for finding the particle in the region Δ, given that the system is in a pure state ψ.

#### 4.2 Density operators and normal states

Definition 2.4 of a state still makes good sense in the infinite-dimensional case, as it simply specializes the general definition of a state on a C\*-algebra *A* to the case *A* = *B*(*H*). Thus we continue to say that a state on *B*(*H*) is a complex-linear map ω : *B*(*H*) → C satisfying ω(*b*∗*b*) ≥ 0 for each *b* ∈ *B*(*H*) and ω(1*H*) = 1. Despite this lack of novelty in the definition of a state (i.e., compared to finite-dimensional Hilbert spaces), Theorem 2.7 no longer holds if *H* is infinite-dimensional: although it (almost trivially) remains true that density operators ρ on *H* define states on *B*(*H*) through the fundamental correspondence ω(*a*) = Tr(ρ*a*), *a* ∈ *B*(*H*), cf. (2.33), there are (many) states that are *not* given in that way (see below). Fortunately, states that *do* arise through (2.33) can be characterized in a simple way.

Definition 4.11. *A state* ω : *B*(*H*) → C *is called* normal *if for each orthogonal family* (*ei*) *of projections (i.e., e*∗ *i* = *ei and eiej* = δ*i jei) one has*

$$\text{co}\left(\sum\_{i} e\_{i}\right) = \sum\_{i} \text{co}(e\_{i}).\tag{4.28}$$

*Here* ∑*i ei is defined as the projection on the smallest closed subspace K of H that contains each eiH (that is,* ∑*i ei* = ∨*iei, i.e., the supremum in the poset* P(*H*) *of all projections on H with respect to the partial order e* ≤ *f iff eH* ⊆ *f H). Furthermore, the sum over i on the right-hand side is defined by* (B.11)*, i.e., as the supremum (in* R*) of the set of all sums* ∑*i*∈*F* ω(*ei*) *over finite subsets F* ⊂ *I of the index set I in which i takes values. It is finite because* ∑*i*∈*F ei* ≤ 1*H and hence, since* ω *is positive,*

$$\sum\_{i \in F} \mathcal{o}(e\_i) \le \mathcal{o}(1\_H) = 1.$$

For example, let (υ*i*) be a basis of *H* with associated one-dimensional projections

$$e\_l = |\mathfrak{v}\_l\rangle\langle\mathfrak{v}\_l|.\tag{4.29}$$

If ω is assumed to be a state, then the additivity condition (4.28) implies

$$\sum\_{i} \mathcal{O}(e\_i) = 1,\tag{4.30}$$

or, equivalently, using Definition B.6 etc. as well as the notation *eF* ≡ ∑*i*∈*F ei*,

$$\lim\_{F} \mathfrak{o}(e\_F) = 1.\tag{4.31}$$

If *H* is separable, any orthogonal family (*ei*) of projections is necessarily countable, and (4.28) is analogous to the countable additivity condition defining a measure.

Theorem 4.12. *A state* ω *on B*(*H*) *takes the form* ω(*a*) = Tr(ρ*a*) *for some (unique) density operator* ρ ∈ D(*H*) *iff it is normal.*

*Proof.* First, eq. (2.33) implies (4.28). To see this, take the trace with respect to some basis (υ*j*) of *H* that is *adapted* to the family (*ei*) in the sense that for each *j*, either *ei*υ*j* = υ*j* (i.e., υ*j* ∈ *eiH*) for one value of *i*, or *ei*υ*j* = υ*j* for all *i*. Then

$$\rho a \left(\sum\_{i} e\_i\right) = \text{Tr}\left(\rho \sum\_{i} e\_i\right) = \sum\_{j} \langle \upsilon\_j, \rho \sum\_{i} e\_i \upsilon\_j \rangle = \sum\_{j}' \langle \upsilon\_j, \rho \upsilon\_j \rangle,$$

where the sum ∑ *j* is over those *j* for which υ*j* ∈ *K* ≡ ∨*ieiH*. On the other hand, since the basis is adapted, we have υ*j* ∈ *K* iff there is an *i* for which *ei*υ*j* = υ*j* (since otherwise *ei*υ*j* = 0 and hence υ*j* ⊥ *eiH* for each *i*, so that υ*j* ∈ *K*⊥), so

$$\sum\_{i} \mathfrak{o}(e\_i) = \sum\_{i} \text{Tr} \left( \mathfrak{o} e\_i \right) = \sum\_{i} \sum\_{j} \langle \mathfrak{v}\_j, \mathfrak{o} e\_i \mathfrak{v}\_j \rangle = \sup\_{F \subset I} \sum\_{j \in J\_F} \langle \mathfrak{v}\_j, \mathfrak{o} \mathfrak{v}\_j \rangle = \sum\_{j}^{\prime} \langle \mathfrak{v}\_j, \mathfrak{o} \mathfrak{v}\_j \rangle,$$

where *JF* consists of those *j* for which υ*j* ∈ ∑*i*∈*F eiH*. This gives (4.28).

Conversely, assume ω is normal. For the *ei* in (4.28) we now take the projections (4.29) determined by some basis (υ*i*). For each *a* ∈ *B*(*H*) we then have

$$\mathfrak{so}(a) = \lim\_{F} \mathfrak{o}(e\_F a). \tag{4.32}$$

Indeed, using Cauchy–Schwarz for the positive semi-definite form (*a*,*b*) = ω(*a*∗*b*), as in (C.197), and using ∑*i ei* = 1*H* and hence ω(*a*) = ω(∑*i eia*) we have

$$|\mathfrak{a}(a) - \mathfrak{a}(e\_F a)|^2 = |\mathfrak{a}(e\_{F^c} a)|^2 \le \mathfrak{a}(a^\* a)\mathfrak{a}(e\_{F^c}) \le \|a\|^2 \mathfrak{a}(e\_{F^c}),\tag{4.33}$$

since *eFc* ≡ ∑*i*∈/*F ei* is a projection. Since ω(*eF*) +ω(*eFc* ) = ω(1*H*) = 1, eq. (4.31) gives lim*F* ω(*eFc* ) = 0, so that (4.33) gives (4.32). For each finite *F* ⊂ *I*, the operator *eFa* has finite rank and hence is compact. According to Theorem B.146, the restriction of ω : *B*(*H*) → C to the C\*-algebra *B*0(*H*) of compact operators on *H* is induced by a trace-class operator ρ, which (from the requirement that ω be a state) must be a density operator. Hence ω(*eFa*) = Tr(ρ*eFa*), and we finally have

$$\mathfrak{o}(a) = \lim\_{F} \mathfrak{o}(e\_F a) = \lim\_{F} \mathrm{Tr}(\mathfrak{\rho} e\_F a) = \mathrm{Tr}(\mathfrak{\rho} a). \tag{4.34}$$

To derive the final equality, we rewrite Tr(ρ*eFa*) = Tr(*eFa*ρ), cf. (A.78) and Proposition B.144, note that *a*ρ ∈ *B*1(*H*), as shown in Corollary B.147, and observe that for any *b* ∈ *B*1(*H*) we have lim*F* Tr(*eFb*) = Tr(*b*). To see this, simply compute the trace in the basis (υ*i*) defining the projections *ei* through (4.29), so that Tr(*eFb*) = ∑*i*∈*F*υ*i*,*b*υ*i*, and note that by Definition B.6,

$$\lim\_{F} \sum\_{i \in F} \langle \mathfrak{v}\_i, b\mathfrak{v}\_i \rangle = \sum\_i \langle \mathfrak{v}\_i, b\mathfrak{v}\_i \rangle = \text{Tr} \,(b).$$

Finally, suppose ω(*a*) = Tr(ρ1*a*) = Tr(ρ2*a*) for each *a* ∈ *B*(*H*) and hence for each *a* ∈ *B*0(*H*). It follows from (B.476) that Tr(ρ*a*) = 0 for all *a* ∈ *B*0(*H*) iff ρ = 0. Hence ρ1 = ρ2, i.e., a normal state ω uniquely determines a density operator ρ. -

If ω is normal, we may therefore use the spectral resolution (2.6) of the corresponding density operator ρ, i.e., ρ = ∑*i pi*|υ*i*υ*i*|, where (υ*i*) is some basis of *H* consisting of eigenvectors of ρ (which exists because ρ is compact and self-adjoint), and the corrsponding eigenvalues satisfy *pi* ≥ 0 and ∑*i pi* = 1; see the explanation after Definition B.148. Computing the trace in the same basis gives

$$\operatorname{Tr}\left(\rho a\right) = \sum\_{i} p\_{i} \langle \mathfrak{v}\_{i}, a\mathfrak{v}\_{i} \rangle. \tag{4.35}$$

We may characterize normality in a number of other ways. First note that because of the duality *B*1(*H*)∗ ∼= *B*(*H*) of Theorem B.146, cf. (B.477), we may equip *B*(*H*) with the *w*∗-topology in its role as the dual of the trace-class operators *B*1(*H*), see §B.9; this means that *a*λ → *a* iff Tr(ρ*a*λ ) → Tr(ρ*a*) for each ρ ∈ *B*1(*H*), or, equivalently, for each ρ ∈ D(*H*), since each trace-class operator is a linear combination of at most four density operators, as follows from Lemma C.53 with (C.8) - (C.9). The *w*∗-topology on *B*(*H*), seen as the dual of *B*1(*H*), is called the σ*-weak topology*. By Proposition B.46, the σ-weakly continuous linear functionals ϕ on *B*(*H*) are just those given by ϕ(*a*) = Tr(ρ*b*) for some trace-class operator *b* ∈ *B*1(*H*).

Secondly, *B*(*H*) is *monotone complete*, in the sense that each net (*a*λ ) of positive operators that is bounded (i.e., 0 ≤ *a*λ ≤ *c* · 1*H* for some *c* > 0 and all λ ∈ Λ) and increasing (in that *a*λ ≤ *a*λ whenever λ ≤ λ ) has a supremum *a* with respect to the standard ordering ≤ on *B*(*H*)+, which supremum coincides with the strong limit of the net (i.e., limλ *a*λψ = *a*ψ for each ψ ∈ *H*); the proof is the same as for Proposition B.98, and also here we write *a*λ \$ *a* to describe this entire situation.

Corollary 4.13. *The following conditions on a state* ω ∈ *S*(*B*(*H*)) *are equivalent:*

*Proof.* We have seen 1 ↔ 3 ↔ 4, and 2 → 1 is obvious, so establishing 3 → 2 would complete the proof. To this effect, we first note that because the sum (4.35) is convergent, for ε > 0 we may find a finite subset *F* ⊂ *I* for which ∑*i*∈/*F pi* < ε/2*a* (assuming *a* = 0). Since 0 ≤ *a*λ ≤ *a* also implies *a*λ ≤ *a*· 1*H* (since *a* ≤ *a*· 1*H*), we therefore have |∑*i*∈/*F pi*υ*i*,(*a*λ −*a*)υ*i*| < 2ε/3, uniformly in λ. Moreover, since *F* is finite and *a*λ → *a* strongly, we can find λ0 such that for all λ ≥ λ0 we have

$$|\sum\_{i \in F} p\_i \langle \mathfrak{v}\_i, (a\_\lambda - a)\mathfrak{v}\_i \rangle| < \mathfrak{e}/3. \tag{4.36}$$

Consequently, for such λ,

$$|\mathrm{Tr}\left(\mathfrak{p}(a\_{\lambda}-a)\right)| \leq |\sum\_{i \in F} p\_i \langle \mathfrak{v}\_i, (a\_{\lambda}-a)\mathfrak{v}\_i \rangle| + |\sum\_{i \notin F} p\_i \langle \mathfrak{v}\_i, (a\_{\lambda}-a)\mathfrak{v}\_i \rangle| \\ < \frac{2}{3}\mathfrak{c} + \frac{1}{3}\mathfrak{c} = \mathfrak{c}.$$

This shows that limλ |Tr(ρ(*a*λ −*a*))| = 0, so that assumption 3 implies no. 2. -

We denote the *normal state space* of *B*(*H*), i.e., the set of all normal states on *B*(*H*) by *Sn*(*B*(*H*)). It is easy to see from Definition B.148 that *Sn*(*B*(*H*)) is a convex (but not necessarily compact!) subset of the total state space *S*(*B*(*H*)).

Corollary 4.14. *The relation* ω(*a*) = Tr(ρ*a*) *induces an isomorphism*

$$\mathcal{S}\_n(B(H)) \cong \mathcal{Q}(H) \tag{4.37}$$

*of convex sets (i.e.,* ω ↔ ρ*). Furthermore, for the corresponding pure states we have*

$$P\_n(B(H)) \cong \mathcal{P}\_1(H),\tag{4.38}$$

*i.e., any pure state* ω *on B*0(*H*)*, as well as any normal pure state on B*(*H*)*, is given by* ω = ωψ *for some unit vector* ψ ∈ *H, where* ω(*a*) = ψ,*a*ψ*, cf.* (2.42)*.*

The proof of (4.38) is practically the same as in the finite-dimensional case. From Theorem B.146 we obtain another characterization of *Sn*(*B*(*H*)) and hence of D(*H*):

Corollary 4.15. *If B*0(*H*) *is the C\*-algebra of compact operators on H, we have*

$$S(B\_0(H)) = S\_n(B(H));\tag{4.39}$$

$$P(B\_0(H)) = P\_n(B(H)),\tag{4.40}$$

*in the sense that any (pure) state* ω *on B*0(*H*) *has a unique* normal *extension to a (pure) state* ω *on B*(*H*)*, given by the same density operator* ρ *that yields* ω*.*

It can be shown that any state ω ∈ *S*(*B*(*H*)) has a convex decomposition

$$
\alpha \bullet \alpha = t \bullet \mathfrak{a}\_{\mathfrak{t}} + (1 - t) \mathfrak{a}\_{\mathfrak{s}}, \tag{4.41}
$$

where *t* ∈ [0,1], ω*n* is a normal state, and ω*s* is called a *singular state*. In particular, since for *t* ∈ (0,1) the state ω is mixed, *a pure state is either normal or singular*.

Singular states are not as aberrant as the terminology may suggest: such states are routinely used in the physics literature and are typically denoted by |λ, where λ lies in the continuous spectrum of some self-adjoint operator (that has to be maximal for this notation to even begin to make sense, see §4.3 below). Examples of such "improper eigenstates" are |*x* and |*p*, which many physicists regard as idealizations. However, mathematically such states are at least defined, namely as singular pure states on *B*(*H*). The key to the existence of such states lies in Proposition C.15 and its proof, which should be reviewed now; we only need the case *a*∗ = *a*.

Proposition 4.16. *Let a* = *a*∗ ∈ *B*(*H*) *have non-empty continuous spectrum, so that there is some* λ ∈ σ(*a*) *that is not an eigenvalue of a. Then* ωλ (*f*(*a*)) = *f*(λ) *defines a pure state on A* = *C*∗(*a*)*, whose extension to B*(*H*) *by any pure state is singular.*

*Proof.* Normal pure states on *B*(*H*) take the form ωψ(*b*) = ψ,*b*ψ, where ψ ∈ *H* is a unit vector and *b* ∈ *B*(*H*). We know from Proposition C.14 that ωλ is multiplicative on *C*∗(*a*). However, if some multiplicative state ω on *C*∗(*a*) has the form ω = ωψ, then ψ must be eigenvector of *a*; cf. the proof of Proposition 2.3. -

#### 4.3 The Kadison–Singer Conjecture

To obtain deeper insight into singular pure states, and as a matter of independent interest, we return to the Kadison–Singer problem, cf. §2.6. Recall that this problem asks if some abelian unital C\*-algebra *A* ⊂ *B*(*H*) has the *Kadison–Singer property*, stating that a pure state ω*A* on *A* has a *unique* pure extension ω to *B*(*H*). Here the issue is uniqueness rather than existence, since at least one such extension exists: since *A* is necessarily unital (with 1*A* = 1*H*) and ω*A* is a state on *A*, so that in particular ω*A*(1*A*) = ω*A* = 1, Corollary B.41 gives the existence of a bounded extension ω satisfying ω(1*H*) = ω = 1, which by Proposition C.5 is a state on *B*(*H*). Proposition 2.22 then gives the existence of a *pure* extension ω. As in the finite-dimensional case, the Kadison–Singer property forces *A* to be maximal (in the poset C (*B*(*H*)) of all abelian unital C\*-subalgebras of *B*(*H*), ordered by inclusion):

Proposition 4.17. *If some abelian unital C\*-subalgebra A of B*(*H*) *has the Kadison– Singer property, then A is necessarily maximal.*

*Proof.* We use the Gelfand isomorphism *A* ∼= *C*(*P*(*A*)), where *P*(*A*) is the pure state space of *A*, cf. Theorem C.8 and Proposition C.14. If *A* has the Kadison–Singer property and *A* ⊆ *B* ⊂ *B*(*H*), where *B* is an abelian unital C\*-subalgebra *A* of *B*(*H*), then ω*A* has a unique pure extension ω on *B*(*H*), which restricts to some state ω*B* on *B*. The same reasoning as in the proof of Proposition 2.22 shows that ω*B* is a pure state on *B*, so that we obtain a unique map

$$P(A) \longleftrightarrow P(B);\tag{4.42}$$

ω*A* → ω*B*. (4.43)

The inverse of this map is simply the pullback of the inclusion *A* → *B*, i.e., ω*B* ∈ *P*(*B*) defines ω*A* ∈ *P*(*A*) by restriction, so that we have a bijection *P*(*A*) ∼= *P*(*B*), ω*A* ↔ ω*B*. Since for any pair of C\*-algebras *A* ⊆ *B* the pullback *S*(*B*) → *S*(*A*) is continuous (in the pertinent *w*∗-topology), the map ω*B* → ω*A* is continuous. As in Lemma C.20, this implies that it is in fact a homeomorphism, so that *A* ∼= *B* through the inclusion *A* → *B*. This gives *A* = *B*, and hence *A* is maximal. -

Maximality of *A* implies *A* = *A*, so that *A* is a von Neumann algebra, sharing the unit of *B*(*H*). To see the relevance of singular states for the Kadison–Singer problem, we first settle the normal case. We know what it means for a state on *B*(*H*) to be normal (cf. Definition 4.11 and Corollary 4.13); for arbitrary von Neumann algebras *A* ⊂ *B*(*H*) the situation is exactly the same: we *define* normality by (4.28) and *characterize* it by the equivalent properties in Corollary 4.13, where the σ-weak topology on *A* may be defined either as the one inherited from *B*(*H*), or, more intrinsically, and the *w*∗-topology from the duality *A* = *A*∗ ∗, where the Banach space *A*∗ is the so-called predual of *A*, e.g., -∞ ∗ ∼= -1 and *L*∞(0,1)∗ = *L*1(0,1), cf. §B.9.

Theorem 4.18. *Let H be a separable Hilbert space and let* ω*A be a* normal *pure state on a maximal commutative unital C\*-algebra A in B*(*H*)*. Then* ω*A has a unique extension to a state* ω *on B*(*H*)*, which is necessarily pure and normal.*

*Proof.* As noted after (4.41), a pure state on *B*(*H*) is either normal or singular. The possibility that ω*A* is normal whereas ω is singular is excluded by Corollary 4.13.3, so ω must be normal and hence given by a density operator. The proof of uniqueness is then the same as in the finite-dimensional case, cf. Theorem 2.21. -

We now recall the classification of maximal maximal abelian ∗-algebras (and hence of maximal abelian von Neumann algebras) *A* in *B*(*H*) up to unitary equivalence (cf. Theorem B.118). This classification is the relevant one for the Kadison– Singer problem, since, as is easily seen, *A* ⊂ *B*(*H*) has the Kadison–Singer property iff *uAu*−1 ⊂ *B*(*uH*) has it. The uniqueness of the finite-dimensional case will be lost:

Theorem 4.19. *If H is separable and infinite-dimensional, and A* ⊂ *B*(*H*) *is a maximal abelian* ∗*-algebra, then A is unitarily equivalent to exactly one of the following:*

*1. L*∞(0,1) ⊂ *B*(*L*2(0,1))*; 2.* -∞ ⊂ *B*(-2)*; 3. L*∞(0,1)⊕-∞(κ) ⊂ *B*(*L*2(0,1)⊕-2(κ))*,*

*where* -∞ ≡ -∞(N)*,* -2 ≡ -2(N)*, and* κ *is either* {1,...,*n*}*, in which case* -2(κ) = C*n and* -∞(κ) = *Dn*(C)*, or* κ = N*, in which case* -2(κ) = -2 *and* -∞(κ) = -∞*.*

This classification sheds some more light on Theorem 4.18. Since *L*∞(0,1) has no pure normal states and *Dn*(C) has been dealt with in Theorem 2.21, the interesting case is -∞. Using Corollary 4.13.3 (or the analysis below), it is easy to check that the normal pure states on -∞ are given by ω*A*(*f*) = *f*(*x*) for some *x* ∈ N; these are vector state of the kind ω*A*(*f*) = ψ,*mf*ψ with ψ = δ*x*, or, in other words, they are given by ω*A*(*f*) = Tr(ρ*mf*) with ρ = |δ*x*δ*x*|. We now invoke a fairly deep result:

Proposition 4.20. *A pure state* ω *on B*(*H*) *is singular iff one (and hence all) of the following equivalent conditions is satisfied:*

One direction is easy: a normal pure state certainly does not satisfy the condition in question. For example, given (2.42) one may take *a* = |ψψ|, which as a onedimensional projection lies in *B*0(*H*), so that ωψ(*a*) = 1. We omit the other direction of the proof. We conclude from this proposition that a pure singular state on *B*(-2) cannot restrict to a normal pure state on -∞, which reconfirms Theorem 4.18.

We now study the Kadison–Singer property for each of the three cases in Theorem 4.19 (where the third will be an easy corollary of the first and the second). Since the proofs of the first two cases are formidable, we just sketch the argument.

Theorem 4.21. • *There exist (necessarily singular) pure states on L*∞(0,1) *that do not have a unique extension to B*(*L*2(0,1))*, and similarly for L*∞(0,1)⊕-∞(κ)*.*

• *Any pure state on* -∞ *has a unique extension to B*(-2)*.* The statement about -∞ is the *Kadison–Singer Conjecture*, which dates from 1959 but was only proved in 2013. The first claim (which was already known to Kadison and Singer themselves) is equally remarkable, however, as is the contrast between the two parts of Theorem 4.21. In particular, Dirac's notation |λ may be ambiguous.

The key to the proof of the first claim lies in the choice of a total countable family of normal states on *L*∞(0,1), from which all pure states may be constructed by a limiting operation. Here we call a (countable) family (ω*n*)*n*∈N of states on some C\*-algebra *A total* if, for any self-adjoint *a* ∈ *A*, the conditions ω*n*(*a*) ≥ 0 for each *n* imply *a* ≥ 0 (the converse is trivial). For example, the well-known *Haar basis* (*hn*) of *L*2(0,1) provides such a family. The functions forming this basis are defined via some bijection β between the set of pairs (*k*,*l*) and N, e.g., β(*k*,*l*) = *k* +2*l* , by

$$h\_n = \chi\_{\beta^{-1}(n)}, \ (n \in \mathbb{N} = \{1, 2, \ldots\});\tag{4.44}$$

$$\mathfrak{X}\_{k,l}(\mathbf{x}) = \mathfrak{Z}^{k/2} \mathfrak{g}(\mathfrak{Z}^k \mathbf{x} - l), \ (k \in \mathbb{N} \cup \{0\}, \mathbf{0} \le l < \mathfrak{Z}^k);\tag{4.45}$$

$$\mathbf{g(x)} = \mathbf{1}\_{[0,1/2)} - \mathbf{1}\_{[1/2,1]}.\tag{4.46}$$

Basic analysis then shows that the Haar functions *hn* form a basis of *L*2(0,1) and that the associated vector states ω*n* on *L*∞(0,1) form a total set, where obviously

$$\langle \alpha\_n(f) = \langle h\_n, m\_f h\_n \rangle = \int\_0^1 h\_n^2 f. \tag{4.47}$$

The relevance of total sets to the conjecture is explained by the following lemma.

Lemma 4.22. *If T* ⊂ *S*(*A*) *is a total set of states on a unital C\*-algebra A, then*

$$S(A) = \text{co}(T)^{-};\tag{4.48}$$

$$P(A) \subseteq T^-,\tag{4.49}$$

*where* co(*T*)− *is the w*∗*-closure of the convex hull of T in A*∗ *or in S*(*A*)*.*

*Proof.* The inclusion co(*T*)− ⊆ *S*(*A*) is obvious, since *T* ⊆ *S*(*A*) and *S*(*A*) is a compact (and hence a closed) convex set. To prove the converse inclusion, suppose *a* = *a*∗ ∈ *A* and *s* ∈ R are such that ω(*a*) ≥ *s* for each ω ∈ *T*. Then ω(*a*−*s*· 1*A*) ≥ 0 and hence ω(*a*) ≥ *s* for each ω ∈ *S*(*A*). Using Theorem B.43 (of Hahn–Banach type), this property would lead to a contradiction if *S*(*A*) were not contained in co(*T*)−.

The second claim, which is the one we will use, follows from the first through a corollary of the Krein–Milman Theorem B.50, stating that if *T* ⊂ *K* is any subset of a compact convex set *K* such that *K* = co(*T*)−, then ∂*eK* ⊆ *T* −. This corollary may be proved (by contradiction) from Theorem B.43 in a similar way. -

Our next aim is to get rid of the closure in (4.49). The Haar basis yields a map

$$h: \mathbb{N} \to S(L^{\infty}(0, 1)); \tag{4.50}$$

$$m \mapsto \mathfrak{o}\_n,\tag{4.51}$$

with image *T*, i.e., the set of Haar states. Since *S*(*A*) is a compact Hausdorff space (in its *w*∗-topology), the universal property (B.135) of the Cech–Stone compactification ˇ βN of N implies that *h* extends (uniquely) to a continuous map

$$
\beta h: \mathcal{B} \mathbb{N} \to S(A),
$$

whose image is compact and hence closed (since βN is compact). Since *T* = *h*(N) ⊂ *S*(*A*) we have *T* ⊆ β*h*(βN) and hence *T* − ⊆ β*h*(βN), so that, from (4.49),

$$P(L^{\infty}(0,1)) \subseteq \mathcal{B}h(\mathcal{B}\mathbb{N}).\tag{4.52}$$

Hence each pure state ω*c* ≡ ω*L*∞(0,1) on *L*∞(0,1) takes the form ω*c* = ω(*U*) *c* , where

$$\mathfrak{gl}\_c^{(U)}(f) = \lim\_{U} \mathfrak{o}\_{\mathfrak{n}}(f) = \bigcap\_{A \in U} \{ \mathfrak{o}\_{\mathfrak{n}}(f) \mid n \in A \}^-, \ f \in L^\infty(0, 1), \tag{4.53}$$

and *U* ∈ βN is some ultrafilter on N, cf. (B.136). The point of this analysis, then, is that ω*U* can immediately be extended to *B*(*L*2(0,1)) by the same formula, i.e.,

$$\mathfrak{o}^{(U)}(a) = \lim\_{U} \mathfrak{o}\_{\mathfrak{n}}(a) = \bigcap\_{A \in U} \{ \mathfrak{o}\_{\mathfrak{n}}(a) \mid n \in A \}^{-}, \ a \in B(L^2(0, 1)), \tag{4.54}$$

where ω*n*(*a*) = *hn*,*ahn*. If *L*∞(0,1) had the Kadison–Singer property, this were the unique extension of ω*U* , and we will show that this leads to a contradiction.

Apart from the use of ultrafilters, the technically most challenging part of the argument disproving the Kadison–Singer property for *L*∞(0,1) is as follows. If *A* = *C*([0,1]), for any *f* ∈ *A* and any pure state ω ∈ *P*(*A*) there is some *x* ∈ [0,1] such that ω(*f*) = *f*(*x*); see Propositions C.14 and C.19. For *A* = *L*∞(0,1) the situation is not that simple due to measure zero complications. Nonetheless, it is easy to show that for each *positive f* ∈ *L*∞(0,1) and ω*c* ∈ *P*(*L*∞(0,1)) and each ε > 0 one has

$$\mu(\{x \in (0,1) \mid f(x) \in [a\mathfrak{a}\_c(f) - \mathfrak{e}, a\mathfrak{a}\_c(f) + \mathfrak{e}]\}) > 0. \tag{4.55}$$

where μ is Lebesque measure on (0,1). Taking the projection

$$e = \mathbf{1}\_{\{\mathbf{x} \in (0,1) \mid f(\mathbf{x}) \in [a\mathbf{a}\_{\mathbb{C}}(f) - \mathfrak{e}/2, \mathbf{a}\mathbf{a}\_{\mathbb{C}}(f) + \mathfrak{e}/2\} \text{ }\mathbf{x}}$$

it follows that for each positive *f* ∈ *L*∞(0,1), ω ∈ *P*(*L*∞(0,1)) and ε > 0 there exists a projection *e* ∈ P(*L*∞(0,1)) with ω(*e*) = 1 and *e f* −*e*ω*c*(*f*) < ε. Hard analysis then generalizes this property from *L*∞(0,1) to *B*(*L*2(0,1)), as follows:

Lemma 4.23. *If* ω*c* ∈ *P*(*L*∞(0,1)) *has a unique extension* ω *to B*(*L*2(0,1)) *(which is necessarily pure if it is unique), then for each a* ∈ *B*(*L*2(0,1)) *and* ε > 0 *there exists a projection e* ∈ P(*L*∞(0,1)) *with* ω*c*(*e*) = 1 *and*

$$\|ea - e\mathfrak{o}(a)\| < \mathfrak{e}.\tag{4.56}$$

To derive a contradiction between (4.54) and (4.56), we use a bijection *b* : N → N that cyclically permutes the ordered subsets (2*k* +1,...,2*k*+1), *k* = 0,1,..., that is, (1,2), (3,4), (5,6,7,8), (9,...,16), etc. This bijection induces a unitary operator

$$
u: L^2(0,1) \to L^2(0,1);
\tag{4.57}$$

$$
u h\_n = h\_{b(n)},\tag{4.58}$$

which is easily shown to have the following properties:

$$a\mathfrak{o}\_n(\mu) = 0, \ n \in \mathbb{N};\tag{4.59}$$

$$\|\|eue\|\| = 1, \ e \in \mathcal{P}(L^{\infty}(0,1)), e \neq 0. \tag{4.60}$$

To show that *L*∞(0,1) fails to have the Kadison–Singer property, suppose it does, so that any ω*c* ∈ *P*(*L*∞(0,1)) has a unique extension ω ∈ *P*(*B*(*L*2(0,1))). As already noted, we may then assume that ω*c* = ω(*U*) *c* , as in (4.53), whilst ω = ω(*U*) , as in (4.54). Taking *a* = *u* then gives ω(*u*) = 0, see (4.59), so that *eu* < ε by (4.56). But this contradicts (4.60), finishing the sketch of the proof of the first claim in Theorem 4.21. The remark about *L*∞(0,1)⊕-∞(κ) follows from the one about *L*∞(0,1).

We now pass to the (even) more difficult case of -∞ ⊂ *B*(-2). Although this will not be used in the proof, it gives some insight to know which states on -∞ we are actually talking about, i.e., the singular pure states, and compare this with (4.53).

Theorem 4.24.

This follows from case no. 5 in §B.9, notably eqs. (B.153) - (B.154). In other words, the pure states ω*d* on -∞ are given by ultrafilters *U* on N through

$$\mathfrak{op}\_d^{(U)}(f) = \mathcal{B}f(U) = \lim\_{U} f(n);\tag{4.62}$$

the analogy with (4.53) is even clearer if we write *f*(*n*) = δ*n*,*mf* δ*n* ≡ ω*n*(*f*). If *U* = *Un* is a principal ultrafilter, *n* ∈ N, we thus recover the normal pure states

$$\mathfrak{op}\_d^{(U\_n)}(f) = f(n). \tag{4.63}$$

As in (4.54), we find at least one natural extension ω(*U*) of ω(*U*) *d* to *B*(-2), namely

$$\mathfrak{o}\mathfrak{o}^{(U)}(a) = \lim\_{U} \mathfrak{o}\_{\mathfrak{n}}(a). \tag{4.64}$$

We now show that that -∞ has the Kadison–Singer property, making ω(*U*) the *only* extension of ω(*U*) *d* . The proof relies on an extremely difficult lemma from linear algebra (formerly known as a *paving conjecture*). We first define a linear map *D* : *Mn*(C) → *Dn*(C) by *D*(*a*)*ii* = *aii*, *i* = 1,...,*n*, and *D*(*a*)*i j* = 0 whenever *i* = *j*.

Lemma 4.25. *For any* ε > 0 *there exist l* ∈ N *such that for all n* ∈ N *and a* ∈ *Mn*(C) *with D*(*a*) = 0*, there are l projections* (*e*1,..., *el*) *in Dn*(C) *such that*

$$\sum\_{k=1}^{l} e\_k = 1\_n;\tag{4.65}$$

$$\|e\_i a e\_l\| \le \varepsilon \|a\|, \ i = 1, \ldots, l. \tag{4.66}$$

Since this estimate is uniform in *n*, the lemma extends to -2, where *D* : *B*(-2) → -∞ is defined analogously, i.e., *D*(*a*) is diagonal in the canonical basis (δ*n*) of -2 with

$$D(a)\delta\_n = a\mathfrak{o}\_n(a)\delta\_n, \ n \in \mathbb{N}.\tag{4.67}$$

Lemma 4.26. *For any* ε > 0 *there exist l* ∈ N *such that for all a* ∈ *B*(-2) *with D*(*a*) = 0*, there are l projections* (*e*1,..., *el*) *in* -∞ *such that*

$$\sum\_{k=1}^{l} e\_k = 1\_H;\tag{4.68}$$

$$\|\|e\_i a e\_i\|\| \le \varepsilon \|a\|, \ i = 1, \ldots, l. \tag{4.69}$$

Now suppose that ω*d* ∈ *P*(-∞), that ω ∈ *S*(*B*(-2)) extends ω*d*, and that *a* ∈ *B*(-2) has *D*(*a*) = 0. Let *ei* be one of the projections in Lemma 4.26. Using Cauchy–Schwarz for the sesquilinear form (*a*,*b*) = ω(*a*∗*b*), we obtain (using *e*2 *i* = *e*∗ *i* = *ei*)

$$|\mathfrak{o}(e\_i a e\_j)|^2 \le \mathfrak{o}(e\_i)\mathfrak{o}(e\_j a^\* a e\_l);\tag{4.70}$$

$$\left|\mathfrak{o}(e\_i a e\_j)\right|^2 \le \mathfrak{o}(a^\* e\_i a)\mathfrak{o}(e\_j). \tag{4.71}$$

Since ω(*ei*) = ω*d*(*ei*) and ω*d* is a pure state (and hence is multiplicative), we have ω(*ei*) ∈ {0,1}, since *ei* is a projection. Moreover, in view of (4.68) and the normalization ω(1*H*) = 1, there must be exactly one value of *i* = 1,...,*l*, say *i* = *i*0, such that ω(*ei*0 ) = 1, and ω(*ei*) = 0 for all *i* = *i*0. Eqs. (4.70) - (4.71) therefore imply that ω(*eiae j*) = 0 iff *i* = *j* = *i*0. Using (4.68) once more, we see that ω(*a*) = ∑*i*, *j* ω(*eiaej*) = ω(*ei*0 *aei*0 ), so that |ω(*a*)|≤ω*ei*0 *aei*0 ≤ 1 · ε*a* by (4.66). Letting ε → 0, we proved:

Lemma 4.27. *If* ω ∈ *S*(*B*(-2)) *extends* ω*d* ∈ *P*(-∞)*, and D*(*a*) = 0*, then* ω(*a*) = 0*.* Since *D*2 = *D*, we have *D*(*a*−*D*(*a*)) = 0, so that for any *a* ∈ *B*(-2), we have

$$\mathfrak{o}(a) = \mathfrak{o}(D(a)) = \mathfrak{o}\_d(D(a)),\tag{4.72}$$

provided that ω extends ω*d*, as before. This shows that ω is determined by ω*d* and hence is unique, completing the proof (sketch) of Theorem 4.21.

#### 4.4 Gleason's Theorem in arbitrary dimension

To a large extent the thrust and difficulty of the proof of Gleason's Theorem 2.28 already lies in its finite-dimensional version, but some care is needed in the general case, and also Corollary 2.29 needs to be refined. A major point here is that Definition 2.23 has no unambiguous generalization to arbitrary Hilbert spaces.

Definition 4.28. *Let H be an arbitrary Hilbert space with unit sphere H*1*.*

*1. A* probability distribution *on* P(*H*) *is a map p* : *H*1 → [0,1] *that satisfies*

$$\sum\_{i \in I} p(\mathfrak{v}\_i) = 1,\text{ for any basis } (\mathfrak{v}\_i) \text{ of } H,\tag{4.73}$$

*where, as in* §*B.12, the sum (over a possibly uncountable index set) is meant as in Definition B.6. In particular, if H is separable and the basis is labeled and ordered by I* = N*, then it is an ordinary convergent sum of the kind* ∑∞ *i*=1 ···*. 2. A map P* : P(*H*) → [0,1] *that satisfies P*(1*H*) = 1 *is called a:*

#### *a.* finitely additive probability measure *if*

$$P\left(\sum\_{j\in J} e\_j\right) = \sum\_{j\in J} P(e\_j) \tag{4.74}$$

*for any* finite *collection* (*ej*)*j*∈*J of mutually orthogonal projections on H (i.e., ejH* ⊥ *ekH, or equivalently, e jek* = 0*, whenever j* = *k); this is equivalent to the condition P*(*e*+ *f*) = *P*(*e*) +*P*(*f*) *whenever e f* = 0*, cf. Definition 2.23.2.*

Thus a probability measure is by definition σ-additive in the usual sense of measure theory; the other two cases are unusual from that perspective. However, if *H* is separable, then *J* can be at most countable, so that complete additivity is the same as σ-additivity and hence any probability measure is completely additive. Surprisingly, assuming the *Continuum Hypothesis* (CH) of set theory, it can be shown that this is even the case for arbitrary Hilbert spaces. The fundamental distinction, then, is between *finitely* additive probability measures and probability measures (which by definition are *countably* additive). As we shall see, this reflects the distinction between *arbitrary* and *normal* states on *B*(*H*), respectively, cf. §4.2. In what follows, in dealing with non-separable Hilbert spaces we assume CH, in which case probability distributions on *H* are equivalent to probability measures on P(*H*).

The proof is the same as in finite dimension (taking into account that infinite sums over projections are defined strongly). Even without CH, Gleason's Theorem still holds for non-separable Hilbert spaces if we assume *P* to be completely additive, and probability distributions are equivalent to completely additive probability measures on P(*H*). For separable Hilbert spaces, CH is irrelevant and unnecessary altogether.

We then have the following generalization (and bifurcation) of Theorem 2.28.

Theorem 4.29. *Let H be a Hilbert space of dimension* > 2*.*

*1. Each probability measure P on* P(*H*) *is induced by a unique normal state on B*(*H*) *via* (2.122)*, i.e.,*

$$P(e) = \text{Tr}\left(\rho e\right),\tag{4.75}$$

*where* ρ *is a density operator on H uniquely determined by P. Equivalently, each probability distribution p on* P(*H*) *is given by* (2.123)*, or*

$$p(\mathfrak{v}) = \langle \mathfrak{v}, \mathfrak{p}\mathfrak{v} \rangle. \tag{4.76}$$

*Conversely, each density operator* ρ *on H defines a probability measure P on* P(*H*) *via* (4.75)*, as well as as a probability distribution p on* P(*H*) *via* (4.76)*.*

*2. Each finitely additive probability measure P on* P(*H*) *is induced by a unique state* ω *on B*(*H*) *via*

$$P(e) = \mathcal{O}(e),\tag{4.77}$$

*and similarly each probability distribution p on* P(*H*) *is given by*

$$p(\mathfrak{v}) = \mathfrak{o}(e\_{\mathfrak{v}}).\tag{4.78}$$

*Conversely, each state* ω *on H defines a probability measure P on* P(*H*) *via* (4.77)*, as well as as probability distribution p on* P(*H*) *via* (4.78)*.*

*Proof.* The proof of part 1 is practically the same as in finite dimension, except for the fact that in the proof of Lemma 2.33 the reference to Proposition A.23 should be replaced by Proposition B.79, upon which one obtains a bounded positive operator ρ for which (2.123) holds. The normalization condition (2.110) then yields Tr(ρ) = 1 if the trace is taken over any basis of *H*, and since ρ is positive this implies ρ ∈ *B*1(*H*), see §B.20 (complete additivity of *P* is just necessary to relate it to *p*).

Unfortunately, the proof of part 2 exceeds the scope of this book (see Notes). -

In infinite dimension, Corollary 2.29 becomes more complicated, too; for one thing, Definition 2.26 of a quasi-state bifurcates into two possibilities. The one given still makes perfect sense and is natural from the point of view of Bohrification; to avoid confusion we call a map ω : *B*(*H*) → C satisfying the conditions in Definition 2.26 a *strong quasi-state*. In the context of Gleason's Theorem, a slightly different notion is appropriate: a *weak quasi-state* on *B*(*H*) satisfies Definition 2.26, except that linearity is only required on commutative C\*-algebras in *B*(*H*) of the form *C*∗(*a*), where *a* = *a*∗ ∈ *B*(*H*) (these are *singly generated*). Since commutative unital C\*-subalgebras of *B*(*H*) are not necessarily singly generated, and a specific counterexample exists, weak quasi-states are not necessarily strong quasi-states.

Proposition 4.30. *The map* ω → ω|P(*H*) *gives a bijective correspondence between weak quasi-states* ω *on B*(*H*) *and finitely additive probability measures on* P(*H*)*.*

*Proof.* For some finite family (*e*1,..., *en*) of mutually orthogonal projections on *H*, add *e*0 = 1*H* − ∑*j ej* if necessary and let *a* = ∑*n j*=0 λ*jej*, with all λ*j* ∈ R different. Then σ(*a*) = {λ0,...,λ*n*}, so that *C*∗(*a*) ∼= *C*(σ(*a*) ∼= C*n*+1 (cf. Theorem B.94) coincides with the linear span of the projections *e j*. If ω is a weak quasi-state, then it is linear on *C*∗(*a*) and hence also on the *ej*, so that ω|P(*H*) is finitely additive.

Conversely, let μ be a finitely additive probability measure on P(*H*). If *a* = *a*∗ ∈ *B*(*H*) is given, using the notation (B.328) we symbolically define ω on *a* by

$$
\mathfrak{so}(a) = \int\_{\sigma(a)} d\mu(e\_{\lambda}) \,\lambda. \tag{4.79}
$$

More precisely, for any ε > 0 we use Corollary B.104 to define ωε (*a*) = ∑*n i*=1 λ*i*μ(*eAi* ) and let ω(*a*) = limε→0ωε (*a*); it follows from Lemma B.103 (or the theory underlying the Riemann–Stieltjes integral (4.79)) that this limit exists. Now let *b*, *c* ∈*C*∗(*a*), so that *b* = *f*(*a*) and *c* = *g*(*a*) for certain *f*,*g* ∈*C*(σ(*a*)), and *b*+*c* = (*f* +*g*)(*a*), cf. Theorem B.94. By (B.325) we therefore have ωε (*b* + *c*) = ∑*n i*=1(*f* + *g*)(λ*i*)μ(*eAi* ), which, since (*f* + *g*)(λ*i*) = *f*(λ*i*) + *g*(λ*i*), again by (B.325) equals ωε (*b*) + ωε (*c*). Since this holds for every ε > 0, letting ε → 0 we obtain ω(*b*+*c*) = ω(*b*) +ω(*c*), making ω linear on *C*∗(*a*). It is clear that the quasi-state ω thus obtained, on restriction to P(*H*) reproduces μ, making the map ω → ω|P(*H*) surjective. Finally, injectivity of this map follows from Corollary B.104. -.

Corollary 4.31. *If* dim(*H*) > 2*, then each weak quasi-state on B*(*H*) *(and* a fortiori *each strong quasi-state) is linear and hence is actually a state.*

This is immediate from Theorem 4.29.2. and Proposition 4.30.

Another corollary of Gleason's Theorem is the *Kochen–Specker Theorem*, which we will explain in detail in Chapter 6, where it will also be proved in a different way.

Theorem 4.32. *If* dim(*H*) > 2*, there are no weak quasi-states* ω : *B*(*H*) → C *whose restriction to each C\*-subalgebra C*∗(*a*) ⊂ *B*(*H*) *is pure (where a* = *a*∗ ∈ *B*(*H*)*).*

Equivalently, there are no nonzero maps ω : *B*(*H*)sa → R that are:

Cf. Definitions 6.1 and 6.3. To see that these conditions are equivalent to those stated in Theorem 4.32 (despite the impression that linearity on all commuting self-adjoint operators seems stronger than linearity on each *C*∗(*a*)), extend ω to ω : *B*(*H*) → C by complex linearity, as in Definition 2.26.1, and note that dispersion-freeness implies positivity and hence continuity on each subalgebra*C*∗(*a*) (cf. Theorem C.52 and Lemma C.4). We then see that the two conditions just stated imply that ω is multiplicative on *C*∗(*a*), and hence pure, see Proposition C.14, which conversely implies that pure states on *C*∗(*a*) are dispersion-free. We now prove Theorem 4.32.

*Proof.* If *e* is a projection, then *e*2 = *e*, so that ω(*e*2) = ω(*e*). Since ω is dispersionfree (as just explained), we also have ω(*e*2) = ω(*e*)2, whence ω(*e*)2 = ω(*e*) and hence ω(*e*) ∈ {0,1}. Furthermore, since ω is a state by Corollary 4.31, we may apply the GNS-construction, see Theorem C.88 (whose notation we use). In particular, for any projection *e*, using the fact that πω(*e*) = πω(*e*)∗πω(*e*), by (C.196) we have

$$\mathfrak{so}(e) = \langle \mathfrak{Q}\_{\mathfrak{so}}, \pi\_{\mathfrak{op}}(e)\mathfrak{Q}\_{\mathfrak{op}}\rangle = \left||\pi\_{\mathfrak{op}}(e)\mathfrak{Q}\_{\mathfrak{op}}\right||^2. \tag{4.80}$$

If ω(*e*) = 0, then πω(*e*)Ωω = 0 from the second equality. If ω(*e*) = 1, then πω(*e*)Ωω = Ωω from the first inequality and Cauchy–Schwarz (in which we have equality, so that πω(*e*)Ωω = *z*Ωω for some *z* ∈ T, upon which (4.80) forces *z* = 1).

By the spectral theorem (e.g. in the form Corollary B.104) or the theory of von Neumann algebras, the linear span of P(*H*) is norm-dense in *B*(*H*). Since Ωω is cyclic for πω(*B*(*H*)) by the GNS-construction, it must be that *H*ω = C · Ωω, and hence πω(*a*) = ω(*a*)· 1*H*ω for any *a* ∈ *B*(*H*). Since πω(*ab*) = πω(*a*)πω(*b*) by the GNS-construction, this gives ω(*ab*) = ω(*a*)ω(*b*) for all *a*,*b* ∈ *B*(*H*). However, such multiplicative statesω on *B*(*H*) cannot exist if dim(*H*) > 1. This is clear if ω is normal, cf. Proposition 2.10, so that the following argument (which also covers the normal case) is especially meant for the case where ω is singular.

1. If dim(*H*) = *n* < ∞, there are *n* one-dimensional projections (*e*1,..., *en*) such that ∑*j ej* = 1*H*. (indeed, we may assume that *B*(*H*) = *Mn*(C) and take diagonal matrices *e*1 = diag(1,0,...,0), etc.). Now for any pair (*ei*, *e j*) there is some *v* ∈ *B*(*H*) (which by definition is a partial isometry) such that *ei* = *vv*∗, *ej* = *v*∗*v* (in the above case *ei* and *ej* are thus related if *vi j* = 1 and *vi j* = 0 otherwise). Hence

$$\mathfrak{o}(e\_l) = \mathfrak{o}(\nu \boldsymbol{\nu}^\*) = \mathfrak{o}(\nu)\mathfrak{o}(\boldsymbol{\nu}^\*) = \mathfrak{o}(\boldsymbol{\nu}^\* \boldsymbol{\nu}) = \mathfrak{o}(e\_j), \tag{4.81}$$

since ω is multiplicative. But ω is also additive, which implies

$$\sum\_{j=1}^{n} \mathfrak{o}(e\_i) = \mathfrak{o}\left(\sum\_{j=1}^{n} e\_j\right) = \mathfrak{o}(1\_H) = 1. \tag{4.82}$$

Since also ω(*ei*) ∈ {0,1}, eqs. (4.81) - (4.82) are clearly contradictory.

2. If dim(*H*) = ∞, separable or not, a similar contradiction arises from the *halving lemma*, which states that there is a projection *e* and an operator *v* such that *e* = *vv*∗, 1*H* −*e* = *v*∗*v*. For example, in the separable case assume *H* = -2 and take *e* the projection onto the closed linear span -2 *e* of the basis vectors (δ*x*) with *x* ∈ N even, so that 1*H* − *e* projects onto the closed linear span -2 *o* of the basis vectors (δ*x*) with *x* ∈ N odd. Then -2 = -2 *e* ⊕ -2 *o*; take *v* = 0 on -2 *e* and *v* : -2 *o* → -2 *e* any unitary operator. In general, a similar method works, for if *I* is a set indexing some basis of *H* one may find a subset *E* ⊂ *I* that has the same cardinality as its complement *I*\*E*, upon which -2(*E*) ∼= -2(*I*\*E*), cf. Theorem B.63.

Multiplicativity of ω then leads to similar contradiction between the properties ω(*e*) = ω(1*H* −*e*), as in (4.81), and ω(*e*)+ω(1*H* −*e*) = ω(1*H*) = 1, as in (4.82): if ω(*e*) = 0 one finds 0 = 1, whereas ω(*e*) = 1 implies 2 = 1. -

#### Notes

## §4.1. The Born rule from Bohrification (II)

The Born measure (and its construction along the lines of this section) is well known in functional analysis, cf. Pedersen (1989), §4.5. For the Hamburger Moment Problem see, for example, Reed, M. & Simon, B. (1975), *Methods of Modern Mathematical Physics. Vol II. Fourier Analysis, Self-adjointness* (New York: Academic Press), Theorem X.4, p. 145 and Example 4, p. 205. In fact, the proof uses spectral theory! Corollary 4.6 was suggested by the treatment of the Born rule in Hall (2013). Definition 4.9 of the joint spectrum goes back (at least) to Arens (1961) and Hormander (1966), ¨ §3.1.13.

## §4.2. Density operators and normal states

These are really results about von Neumann algebras and come from the pertinent literature; our proofs derive from Li (1992), §1.8 and Takesaki (2002), Ch. III.

## §4.3. The Kadison–Singer Conjecture

As already mentioned in the notes to §2.6, the Kadison–Singer Conjecture was first discussed in Kadison & Singer (1959) and was finally proved by Marcus, Spielman, & Srivastava (2014ab), following important intermediate contributions by e.g. Anderson (1979) and Weaver (2004). For an introduction including a complete proof see Stevens (2016), and for applications of the conjecture and its proof to other areas of mathematics see Casazza et al (2005) as well as Casazza & Tremain (2016). Proposition 4.20 is due to Glimm (1960).

## §4.4. Gleason's Theorem in arbitrary dimension

The extension of Gleason's Theorem to non-separable Hilbert space assuming complete additivity of *P* is due to Maeda (1980). Maeda (1990) generalizes this result to von Neumann algebras without summands of type *I*2. The proof that assuming CH countable additivity implies complete additivity (and hence Gleason's Theorem) was given by Eilers & Horst (1975). Proposition 4.30 is due to Aarens (1970), whose Theorem 1 is wrong: see Aarens (1991). The proof of Theorem 4.32 is due to Doring (2004), using results of Hamhalter (1993). ¨

## Chapter 5 Symmetry in quantum mechanics

Roughly speaking, a *symmetry* of some mathematical object is an invertible transformation that leaves all relevant structure as it is. Thus a symmetry of a set is just a bijection (as sets have no further structure, whence invertibility is the only demand on a symmetry), a symmetry of a topological space is a homeomorphism, a symmetry of a Banach space is a linear isometric isomorphism, and, crucially important for this chapter, a symmetry of a Hilbert space *H* is a *unitary operator*, i.e., a linear map *u* : *H* → *H* satisfying one and hence all of the following equivalent conditions:

The discussion of symmetries in quantum physics is based on the above idea, but the mathematically obvious choices need not be the physically relevant ones. Even in elementary quantum mechanics, where *A* = *B*(*H*), i.e., the C\*-algebra of all bounded operators on some Hilbert space *H*, the concept of a symmetry is already diverse. The main structures whose symmetries we shall study in this chapter are:

Each of these structures comes with its own notion of a symmetry, but the main point of this chapter will be to show these notions are equivalent, corresponding in all cases to either unitary or—surprisingly—*anti-unitary* operators, both merely defined up to a phase. The latter subtlety will open the world of *projective* unitary group representation to quantum mechanics (without which the existence of spin- 1 2 particles such as electrons, and therewith also of ourselves, would be impossible).

#### 5.1 Six basic mathematical structures of quantum mechanics

We first recall the objects just described in a bit more detail. We have:

$$\mathcal{O}^{\theta} \_1(H) = \{ e \in B(H) \mid e^2 = e^\* = e, \text{Tr}\,(e) = \text{dim}(eH) = 1\};\tag{5.1}$$

$$\mathcal{O}(H) = \{ \mathfrak{p} \in \mathcal{B}(H) \mid \mathfrak{p} \ge 0, \text{Tr}(\mathfrak{p}) = 1 \};\tag{5.2}$$

$$B(H)\_{\rm sa} = \{ a \in B(H) \mid a^\* = a \};\tag{5.3}$$

$$\mathcal{S}^{\mathbb{C}}(H) = \{ a \in \mathcal{B}(H) \mid 0 \le a \le 1\_H \};\tag{5.4}$$

$$\mathcal{O}^{\emptyset}(H) = \{ e \in \mathcal{B}(H) \mid e^2 = e^\* = e \};\tag{5.5}$$

$$\mathcal{NC}(B(H)) = \{ \mathbf{C} \subset B(H) \mid \mathbf{C} \text{ commutative } \mathbf{C}^\*\text{-algebra}, 1\_H \in \mathbf{C} \}. \tag{5.6}$$

The point is that each of these sets has some additional structure that defines what it means to be a symmetry of it, as we now spell out in detail.

Definition 5.1. *Let H be a Hilbert space (not necessarily finite-dimensional).*

*1. A*Wigner symmetry *(of H) is a bijection*

$$\mathcal{W} \colon \mathcal{P}\_1(H) \to \mathcal{P}\_1(H) \tag{5.7}$$

*that satisfies*

$$\operatorname{Tr}(\mathsf{W}(e)\mathsf{W}(f)) = \operatorname{Tr}(ef),\ e,f \in \mathcal{P}\_1(H). \tag{5.8}$$

*2. A* Kadison symmetry *is an* affine *bijection*

$$
\mathbb{K}: \mathcal{Q}(H) \to \mathcal{Q}(H),
\tag{5.9}
$$

*i.e. a bijection* K *that preserves convex sums: for t* ∈ (0,1) *and* ρ1,ρ2 ∈ D(*H*)*,*

$$
\mathsf{K}(t\mathfrak{p}\_1 + (1-t)\mathfrak{p}\_2) = t\mathsf{K}\mathfrak{p}\_1 + (1-t)\mathsf{K}\mathfrak{p}\_2.\tag{5.10}
$$

#### *3. a. A* Jordan symmetry *is an invertible* Jordan map

$$\downarrow : B(H)\_{\text{sa}} \to B(H)\_{\text{sa}},\tag{5.11}$$

*i.e., an* R*-linear bijection that satisfies the equivalent conditions*

$$\mathbb{J}(a \circ b) = \mathbb{J}(a) \circ \mathbb{J}(b);\tag{5.12}$$

$$\mathbb{J}(a^2) = \mathbb{J}(a)^2. \tag{5.13}$$

*Here*

$$a \circ b = \frac{1}{2}(ab + ba) \tag{5.14}$$

*is the* Jordan product *on B*(*H*)sa*, which turns the (real) vector space B*(*H*)sa *into a* Jordan algebra*, cf.* §*C.25.*

*b. A* weak Jordan symmetry *is an invertible* weak Jordan map*, i.e., a bijection* (5.11) *of which the restriction* J|*C*sa *is a Jordan map for each C* ∈ C (*B*(*H*))*.*

#### *4. A* Ludwig symmetry *is an affine order isomorphism*

$$
\mathsf{L}: \mathscr{E}(H) \to \mathscr{E}(H). \tag{5.15}
$$

*5. A* von Neumann symmetry *is an order isomorphism*

$$\mathbb{N}: \mathcal{P}(H) \to \mathcal{P}(H) \tag{5.16}$$

*preserving orthocomplementation, i.e.* N(1−*e*) = 1−N(*e*) *for each e* ∈ P(*H*)*. 6. A* Bohr symmetry *is an order isomorphism*

$$\mathcal{B}: \mathcal{C}(B(H)) \to \mathcal{C}(B(H)).\tag{5.17}$$

In nos. 3 and 5–6, an *order isomorphism* O of the given poset is a bijection that preserves the partial order ≤ (i.e., if *x* ≤ *y*, then O(*x*) ≤ O(*y*)) and whose inverse O−1 does so, too; cf. §D.1. The names in question have been chosen for historical reasons and (except perhaps for the first and third) are not standard.

Let us note that any Jordan map has a unique extension to a C-linear map

$$\mathbb{J}\_{\mathbb{C}}: \mathcal{B}(H) \to \mathcal{B}(H);\tag{5.18}$$

$$\mathsf{J}\_{\mathbb{C}}(a^\*) = \mathsf{J}\_{\mathbb{C}}(a)^\*,\tag{5.19}$$

which satisfies (5.12) for all *a*,*b*, as well as

$$\mathbb{J}\_{\mathbb{C}}(a+ib) = \mathbb{J}(a) + i\mathbb{J}(b),\tag{5.20}$$

with notation as in Proposition 2.6. Conversely, such a Jordan map (5.18) defines a real Jordan map (5.11) by J = J|*B*(*H*)sa . Similarly, a weak Jordan symmetry is equivalent to a map (5.18) that satisfies (5.19), preserves squares as in (5.13), and is linear on each subspace *C* of *B*(*H*), with *C* ∈ C (*B*(*H*)). In other words (in the spirit of Bohrification), JC is a homomorphism of C\*-algebras on each commutative unital C\*-subalgebra *C* ⊂ *B*(*H*). Therefore, either way J and JC are essentially the same thing, and if no confusion may arise we call it J. Note that a weak Jordan map J *a priori* satisfies (5.12) only for *commuting* self-adjoint *a* and *b*. It follows that weak (and hence ordinary) Jordan symmetries are unital: since

$$\mathbb{J}(b) = \mathbb{J}(1\_H \circ b) = \mathbb{J}(1\_H) \circ \mathbb{J}(b) \tag{5.21}$$

for any *b*, we may pick *b* = J−1(1*H*) to find, reading (5.21) from right to left,

$$\mathbb{J}(1\_H) = \mathbb{J}(1\_H) \diamond 1\_H = 1\_H. \tag{5.22}$$

The special role of unitary operators *u* now emerges: each such operator defines the relevant symmetry in the obvious way, namely, in order of appearance:

$$\mathcal{W}(e) = ueu^\*;\tag{5.23}$$

$$\mathsf{K}(\mathfrak{p}) = \mathsf{u}\mathfrak{p}u^\*;\tag{5.24}$$

$$
\mathsf{L}(a) = \mathsf{u}au^\*; \tag{5.25}
$$

$$\mathbf{J}(a) = \mu a u^\*;\tag{5.26}$$

$$\mathsf{N}(e) = \mathsf{u}e\mathsf{u}^\*;\tag{5.27}$$

$$\mathbb{B}(C) = \mathfrak{u}C\mathfrak{u}^\*,\tag{5.28}$$

where *a* ∗ = *a* in (5.26). If not, this formula remains valid also for the map *J*C. Furthermore, in (5.28) the notation *uCu*∗ is shorthand for the set {*uau*∗ | *a* ∈ *C*}, which is easily seen to be a member of C (*B*(*H*)). Here, as well as in the other three cases, it is easy to verify that the right-hand side belongs to the required set, that is,

$$
u
u^\* \in \mathcal{P}\_1(H), \ u \mathfrak{p} \mu^\* \in \mathcal{O}(H), \ u \mathfrak{p} \mu^\* \in \mathcal{E}(H),\tag{5.29}$$

$$
\mu u u^\* \in B(H)\_{\text{sa}}, \ u \rho u^\* \in \mathcal{O}'(H), \ u Cu^\* \in \mathcal{O}'(B(H)), \tag{5.30}
$$

respectively, provided, of course, that

$$a \in \mathcal{P}\_1(H), \ p \in \mathcal{O}(H), \ a \in \mathcal{E}(H) \ a \in \mathcal{B}(H)\_{\mathsf{sa}}, \ e \in \mathcal{P}(H), \ \mathcal{C} \in \mathcal{C}(\mathcal{B}(H)).$$

Indeed, if, in (5.23), *e* = *e*ψ = |ψihψ| for some unit vector ψ ∈ *H*, then

$$
\mu e\_{\Psi} \mu^\* = e\_{\mu \Psi}.\tag{5.31}
$$

If ρ ≥ 0 in that hψ,ρψi ≥ 0 for each ψ ∈ *H*, then clearly also *u*ρ*u* ∗ ≥ 0, and if Tr(ρ) = 1, then also Tr(*u*ρ*u* ∗ ) = 1. If *a* ∗ = *a*, then

$$(\mu au^\*)^\* = u^{\*\*}a^\*u^\* = \mu au^\*.\tag{5.32}$$

However, one may also choose *u* in these formulae to be *anti-unitary*, as follows:

Definition 5.2. *1. A real-linear operator u* : *H* → *H is* anti-linear *if*

$$
\mu(z\Psi) = \overline{z}\Psi \text{ (}z \in \mathbb{C}\text{)}.\tag{5.33}
$$

*2. An anti-linear operator u* : *H* → *H is* anti-unitary *if it is invertible, and*

$$
\langle \mu \mathfrak{p}, \mu \mathfrak{v} \rangle = \overline{\langle \mathfrak{p}, \mathfrak{v} \rangle}^{\cdot} (\mathfrak{p}, \mathfrak{v} \in H). \tag{5.34}
$$

*The adjoint u*∗ *of a (bounded) anti-linear operator u is defined by the property*

$$<\langle \mu^\*\boldsymbol{\varphi}, \boldsymbol{\Psi} \rangle = \overline{\langle \boldsymbol{\Phi}, \boldsymbol{\mu}\boldsymbol{\Psi} \rangle} \ (\boldsymbol{\Phi}, \boldsymbol{\Psi} \in H), \tag{5.35}$$

in which case *u* ∗ is anti-linear, too. Hence we may equally well say that an anti-linear operator is anti-unitary if *uu*∗ = *u* ∗*u* = 1*H*. The simplest example is the map

5.1 Six basic mathematical structures of quantum mechanics 129

$$J: \mathbb{C}^n \to \mathbb{C}^n;$$

$$Jz = \overline{z},\tag{5.36}$$

i.e., if *z* = (*z*1,...,*zn*) ∈ C*n*, then (*Jz*)*i* = *zi*. Similarly, one may define

$$J: \ell^2 \to \ell^2;$$

$$J\Psi = \overline{\Psi},\tag{5.37}$$

and likewise on *L*2, where complex conjugation is defined pointwise, that is,

$$(J\Psi)(\mathbf{x}) = \overline{\Psi(\mathbf{x})}.\tag{5.38}$$

For any Hilbert space one may pick a basis (υ*i*) and define *J* relative to this basis by

$$J\left(\sum\_{i} c\_{i}\mathbf{u}\_{i}\right) = \sum\_{i} \overline{c}\_{i}\mathbf{u}\_{i}.\tag{5.39}$$

For future use, we state two obvious facts.

Proposition 5.3. *1. The product of two anti-unitary operators is unitary.*

*2. Any anti-unitary operator u* : *H* → *H takes the form u* = *Jv, where v is unitary and J is an anti-unitary operator on H of the kind constructed above.*

It is an easy verification that (5.23) - (5.28) still define symmetries if *u* is antiunitary. Note that in terms of the complexification JC, eq. (5.26) should read

$$\mathbb{J}\_{\mathbb{C}}(a) = \mu a^\* \mu^\*. \tag{5.40}$$

The goal of the following sections is to show that these are the only possibilities:

Theorem 5.4. *Let H be a Hilbert space, with* dim(*H*) > 1*.*

*where in all cases the operator u is either unitary or anti-unitary, and is uniquely determined by the symmetry in question up to a phase (that is, u and u implement the same symmetry by conjugation iff u* = *zu, where z* ∈ T*).*

As we shall see, the reason why the case *H* = C2 is exceptional with regard to weak Jordan symmetries, von Neumann symmetries, and Bohr symmetries is that in those cases the proof relies on Gleason's Theorem, which fails for *H* = C2.

To see this more explicitly, and also to prove the positive cases (i.e., nos. 1–4a) in a simple situation without invoking higher principles, before proving Theorem 5.4 in general it is instructive to first illustrate it in the two-dimensional case *H* = C2.

## 5.2 The case *H* = C2

We start with some background. Any complex 2×2 matrix *a* can be written as

$$a = a(\mathbf{x}\_0, \mathbf{x}\_1, \mathbf{x}\_2, \mathbf{x}\_3) = \frac{1}{2} \sum\_{\mu=0}^{3} \mathbf{x}\_{\mu} \sigma\_{\mu} \ (\mathbf{x}\_{\mu} \in \mathbb{C});\tag{5.41}$$

$$
\sigma\_0 = \begin{pmatrix} 1 \ 0 \\ 0 \ 1 \end{pmatrix}, \sigma\_1 = \begin{pmatrix} 0 \ 1 \\ 1 \ 0 \end{pmatrix}, \sigma\_2 = \begin{pmatrix} 0 \ -i \\ i \ 0 \end{pmatrix}, \sigma\_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \qquad (5.42)
$$

i.e., the *Pauli matrices*. Furthermore, if we equip the vector space *M*2(C) of complex 2 × 2 matrices with the canonical inner product (2.34), then the rescaled matrices σ μ = σμ / √ 2 form a basis (≡ orthonormal basis) of the ensuing Hilbert space.

Writing x = (*x*1, *x*2, *x*3), some interesting special cases are:

The first case follows because *SU*(2) consist of all matrices of the form

$$\left(\begin{array}{c} \alpha \quad \beta\\ -\overline{\beta} \ \overline{\alpha} \end{array}\right), \ \alpha, \beta \in \mathbb{C}, \ |\alpha|^2 + |\beta|^2 = 1. \tag{5.43}$$

The second case is obvious, and the third follows from Proposition 2.9.

Assume the third case, so that *a* = *e* with *e*2 = *e*∗ = *e* and Tr(*e*) = 1. If a linear map *u* : C2 → C2 is unitary, then simple computations show that *e* = *ueu*∗ is a onedimensional projection, too, given by *e* = 1 2 ∑3 μ=0 *x* μσμ with *x* 0 = 1, x ∈ R3, and x = 1. Writing x = *R*x for some map *R* : *S*2 → *S*2, we have

$$
\mu(\mathbf{x} \cdot \boldsymbol{\sigma})\mu^\* = (R\mathbf{x}) \cdot \boldsymbol{\sigma}, \tag{5.44}
$$

where x · σ = ∑3 *j*=1 *xj*σ*j*. This also shows that *R* extends to a linear isometry *R* : R3 → R3. Using the formula Tr(σ*i*σ*j*) = 2δ*i j*, the matrix-form of *R* follows as

$$\mathcal{R}\_{ij} = \frac{1}{2} \text{Tr} \left( \mu \sigma\_l \mu^\* \sigma\_j \right). \tag{5.45}$$

Define *U*(2) as the (connected) group of all unitary 2×2 matrices (whose connected subgroup *SU*(2) of elements with unit determinant has just been mentioned). Also, recall that *O*(3) is the group of all real orthogonal 3×3 matrices *M*, a condition that may be expressed in (at least) four equivalent ways (like unitarity):

$$\bullet \quad MM^T = M^M M = 1\_3;$$

This implies det(*M*) = ±1 (as can be seen by diagonalizing *M*; being a real linear isometry, its eigenvalues can only be ±1, and det(*M*) is their product). Thus *O*(3) breaks up into two parts *O*±(3) = {*R* ∈ *O*(3) | det(*R*) = ±1}, of which *O*+ ≡ *SO*(3) consists of rotations. Using an explicit parametrization of *SO*(3), e.g., through Euler angles, or, using surjectivity of the exponential map (from the Lie algebra of *SO*(3), which consist of anti-symmetric real matrices), it follows that *O*±(3) are precisely the two connected components of *O*(3), the identity of course lying in *O*+(3).

Proposition 5.5. *The map u* → *R defined by* (5.44) *is a homomorphism from U*(2) *onto SO*(3)*. In terms of SU*(2) ⊂ *U*(2)*, this map restricts to a two-fold covering*

$$\mathfrak{A}: SU(2) \to SO(3),\tag{5.46}$$

*with discrete kernel*

$$\ker(\tilde{\mathfrak{x}}) = \{1\_2, -1\_2\}.\tag{5.47}$$

*Proof.* As a finite-dimensional linear isometry, *R* is automatically invertible (this also follows from unitarity and hence invertibility of *u*), hence *R* ∈ *O*(3). It is obvious from (5.44) that *u* → *R* is a continuous homomorphism (of groups). Since *U*(2) is connected and *u* → *R* is continuous, *R* must lie in the connected component of *O*(3) containing the identity, whence *R* ∈ *SO*(3). To show surjectivity of π˜, take some unit vector u ∈ R3 and define *u* = cos( 1 2θ)+*i*sin( 1 2θ)u·σ. The corresponding rotation *R*θ (u) is the one around u by an angle θ, and such rotations generate *SO*(3).

Finally, it follows from (5.44) that *u* ∈ ker(π˜) iff *u* commutes with each σ*i* and hence, by (5.41), with all matrices. Therefore, *u* = *z*·12 for some *z* ∈ C, upon which the the condition det(*u*) = 1 (in that *u* ∈ *SU*(2)) enforces *z* = ±1. -

Note that the covering (5.46) is topologically nontrivial (i.e., *SU*(2) = *SO*(3)×Z2), since *SU*(2) ∼= *S*3 is simply connected, whereas *SO*(3) is doubly connected: a closed path *t* → *R*2π*t*(u), *t* ∈ [0,1] in *SO*(3) (starting and ending at 13) lifts to a path

$$t \mapsto \cos(\pi t) + i \sin(\pi t) \mathbf{u} \cdot \boldsymbol{\sigma}$$

in *SU*(2) that starts at the unit matrix 12 and ends at −12.

To incorporate *O*−(3), let *Ua*(2) be the set of all anti-unitary 2 × 2 matrices. These do not form a group, as the product of two anti-unitaries is unitary, but the union *U*(2)∪*Ua*(2) is a disconnected Lie group with identity component *U*(2).

Proposition 5.6. *The map u* → *R defined by* (5.44) *is a surjective homomorphism*

$$\mathfrak{A}': U(2) \cup U\_a(2) \to O(3), \tag{5.48}$$

*with kernel U*(1)*, seen as the diagonal matrices z*· 12*, z* ∈ T*. Moreover,* π˜ *maps U*(2) *onto SO*(3) *and maps Ua*(2) *onto O*−(3)*.*

*Proof.* The map *u* → *R* in (5.44) sends the anti-unitary operator *u* = *J* on C2 to *R* = diag(1,−1,1) ∈ *O*−(3). Since*Ua*(2) = *J* ·*U*(2) and similarly *O*−(3) = *R*·*SO*(3), the last claim follows. The computation of the kernel may now be restricted to *U*(2), and then follows as in the last step op the proof of the previous proposition. - We now return to Theorem 5.4 and go through its special cases one by one.

Part 1 of Theorem 5.4 is *Wigner's Theorem*, which in the case at hands reads:

Theorem 5.7. *Each bijection* W : P1(C2) → P1(C2) *that satisfies*

$$\operatorname{Tr}\left(\mathsf{W}(e)\mathsf{W}(f)\right) = \operatorname{Tr}\left(ef\right) \tag{5.49}$$

*for each e*, *f* ∈ P1(C2) *takes the form* W(*e*) = *ueu*∗*, where u is either unitary or anti-unitary, and is uniquely determined by* W *up to a phase.*

To prove, this we transfer the whole situation to the two-sphere, where it is easy:

Proposition 5.8. *The pure state space* P1(C2) *corresponds bijectively to the sphere*

$$S^2 = \{(\mathbf{x}, \mathbf{y}, z) \in \mathbb{R}^3 \mid \mathbf{x}^2 + \mathbf{y}^2 + z^2 = 1\},$$

*in that each one-dimensional projection e* ∈ P1(C2) *may be expressed uniquely as*

$$e(\mathbf{x}, \mathbf{y}, \mathbf{z}) = \frac{1}{2} \begin{pmatrix} 1+z \ \mathbf{x} - i\mathbf{y} \\ \mathbf{x} + i\mathbf{y} \ 1-z \end{pmatrix},\tag{5.50}$$

*where* (*x*, *y*,*z*) ∈ R3 *and x*2 +*y*2 +*z*2 = 1*. Under the ensuing bijection*

$$\mathcal{P}\_1(\mathbb{C}^2) \cong \mathbb{S}^2,\tag{5.51}$$

*Wigner symmetries* W *of* C2 *turn into orthogonal maps R* ∈ *O*(3)*, restricted to S*2*.*

*Proof.* The first claim restates Proposition 2.9. If ψ and ψ are unit vectors in C2 with corresponding one-dimensional projections *e*ψ(*x*, *y*,*z*) and *e*ψ(*x* , *y* ,*z* ) then, as one easily verifies, the corresponding transition probability takes the form

$$\operatorname{Tr}(e\_{\Psi}e\_{\Psi'}) = \frac{1}{2}(1 + \langle \mathbf{x}, \mathbf{x'} \rangle) = \cos^2(\frac{1}{2}\theta(\mathbf{x}, \mathbf{y})),\tag{5.52}$$

where θ(x,y) is the arc (i.e., geodesic) distance between x and y. Consequently, if W : P1(C2) → P1(C2) satisfies (5.8), then the corresponding map *R* : *S*2 → *S*2 (defined through the above identification P1(C2) ∼= *S*2) satisfies

$$
\langle \mathcal{R}(\mathbf{x}), \mathcal{R}(\mathbf{x'}) \rangle = \langle \mathbf{x}, \mathbf{x'} \rangle \text{ (}\mathbf{x}, \mathbf{x'} \in \mathbb{S}^2\text{)}.\tag{5.53}
$$

Lemma 5.9. *If some bijection R* : *S*2 → *S*2 *satisfies* (5.53)*, then R extends (uniquely) to an orthogonal linear map (for simplicity also called) R* : R3 → R3*.*

*Proof.* With (u1,u2,u3) the standard basis of R3, define a 3×3 matrix by

$$R\_{kl} = \langle \mathbf{u}\_k, R(\mathbf{u}\_l) \rangle. \tag{5.54}$$

It follows from (5.53) that *R*−1(u*j*)*k* = *Rjk*, which implies *R*−1(u*j*),x = ∑*k Rjkxk*, or, once again using (5.53), *R*(x)*j* = ∑*k Rjkxk*. Hence the map x → ∑*j*,*k Rjkxk*u*j*, i.e., the usual linear map defined by the matrix (5.54), extends the given bijection *R*. Orthogonality of this linear map is, of course, equivalent to (5.53). -

Wigner's Theorem then follows by combining Propositions 5.6 and 5.8: given the linear map *R* just constructed, read (5.44) from right to left, where *u* exists by surjectivity of the map (5.48), and the precise lack of uniqueness of *u* as claimed in Theorem 5.4 is just a restatement of the fact that (5.48) has *U*(1) as its kernel. -

*Kadison's Theorem* is part 2 of Theorem 5.4. Explicitly, for *H* = C2 we have:

Theorem 5.10. *Each affine bijection* K : D(C2) → D(C2) *is given as* K(ρ) = *u*ρ*u*∗*, where u is unitary or anti-unitary, and is uniquely determined by* K *up to a phase.*

*Proof.* We once again invoke Proposition 2.9, implying that any density matrix ρ on C2 takes the form

$$\mathcal{P} = \frac{1}{2} \left( 1\_2 + \sum\_{\mu=1}^{3} \mathbf{x}\_{\mu} \sigma\_{\mu} \right), \tag{5.55}$$

with x ≤ 1. Moreover, the ensuing bijection D(C2) ∼= *B*3, ρ → x, is clearly affine, in that a convex sums *t*ρ + (1−*t*)ρ of density matrices correspond to convex sums *t*x+ (1−*t*)x of the corresponding vectors in R3.

Lemma 5.11. *Any affine bijection* K *of the unit ball B*3 *in* R3 *is given by an orthogonal linear map R* ∈ *O*(3)*.*

*Proof.* First, K must map the boundary ∂*eB*3 = *S*2 to itself (necessarily bijectively): if x ∈ *S*2 and K(x) = *t*x + (1−*t*)x, then x = *tK*−1(x )+(1−*t*)*K*−1(x), whence

$$K^{-1}(\mathbf{x}') = K^{-1}(\mathbf{x}''),\tag{5.56}$$

since x is pure, whence x = x, so that also K(x) is pure.

Second, the basis of all further steps is the property

$$\mathbb{K}(\mathbf{0}) = \mathbf{0}.\tag{5.57}$$

This is because 0 is intrinsic to the convex structure of *B*3: it is the unique point with the property that for any x ∈ *S*2 there exists a unique x such that 1 2 x+ 1 2 x = 0, namely x = −x. Thus 0 must be preserved under affine bijections. For a formal proof (by contradiction), suppose K(0) = 0, and define y = K(0)/K(0) ∈ *S*2. Then K(0) has an extremal decomposition K(0) = *t*y + (1 −*t*)y , with y = −y and *t* = 1 2 (1+K(0)). Applying the affine map K−1 then gives

$$\|\|\mathsf{K}^{-1}(\mathbf{y}')\|\| = \|\mathsf{K}^{-1}(\mathbf{y})\| \cdot \frac{1 + \|\mathsf{K}(\mathbf{0})\|}{1 - \|\mathsf{K}(\mathbf{0})\|}.$$

Now y ∈ *S*2 and hence K−1(y) ∈ *S*2 by part one of this proof (applied to K−1), so that K−1(y) = 1. But this implies K−1(y ) > 1, which is impossible because y ∈ *S*2 and hence K−1(y ) = 1.

Third, for x ∈ *B*3 and *t* ∈ [0,1] the preceding point implies that

$$\mathsf{K}(t\mathbf{x}) = \mathsf{K}(t\mathbf{x} + (1 - t)\mathbf{0}) = t\mathsf{K}(\mathbf{x}) + (1 - t)\mathsf{K}(\mathbf{0}) = t\mathsf{K}(\mathbf{x}).\tag{5.58}$$

The same then holds for x ∈ *B*3 and all *t* ≥ 0 as long as *t*x ∈ *B*3: for take *t* > 1, so that *t* −1 ∈ (0,1), and use the previous step with x *t*x and *t t* −1 to compute

$$
\mathsf{K}(t\mathbf{x}) = t\mathfrak{t}^{-1}\mathsf{K}(t\mathbf{x}) = t\mathsf{K}(t^{-1}t\mathbf{x}) = t\mathsf{K}(\mathbf{x}).
$$

Also, (5.58) and affinity imply that for any x,y ∈ *B*3 for which x+y ∈ *B*3, we have

$$\mathcal{K}(\mathbf{x} + \mathbf{y}) = 2\mathsf{K}(\frac{1}{2}\mathbf{x} + \frac{1}{2}\mathbf{y}) = 2 \cdot (\frac{1}{2}\mathsf{K}(\mathbf{x}) + \frac{1}{2}\mathsf{K}(\mathbf{y})) = \mathsf{K}(\mathbf{x}) + \mathsf{K}(\mathbf{y}).\tag{5.59}$$

With our earlier result (5.57), this also gives

$$
\mathbb{K}(-\mathbf{x}) = -\mathbb{K}(\mathbf{x}).\tag{5.60}
$$

For some nonzero x ∈ R3, take *s* ≥ x and *t* ≥ x. Then by (5.58) we have

$$s\mathbb{K}(\mathbf{x}/s) = s\mathbb{K}\left(\frac{t}{s}\frac{\mathbf{x}}{t}\right) = t\mathbb{K}(\mathbf{x}/t).$$

We may therefore define a map *R* : R3 → R3 by

$$R(\mathbf{0}) = \mathbf{0};\tag{5.61}$$

$$R(\mathbf{x}) = s \cdot \mathsf{K}(\mathbf{x}/s) \ (\mathbf{x} \neq \mathbf{0}),\tag{5.62}$$

for any choice of *s* ≥ x. For x ∈ *B*3 we may take *s* = 1, so that *R* extends K.

To prove that *R* is linear, for x ∈ R3 and *t* ≥ 0 pick some *s* ≥ *t*x and compute

$$R(t\mathbf{x}) = s\mathbb{K}\left(\frac{t}{s}\mathbf{x}\right) = s\mathbb{K}\left(\|\mathbf{x}\|\frac{t}{s}\frac{\mathbf{x}}{\|\mathbf{x}\|}\right) = s \cdot \|\mathbf{x}\|\frac{t}{s}\mathbb{K}\left(\frac{\mathbf{x}}{\|\mathbf{x}\|}\right) = tR(\mathbf{x}).\tag{5.63}$$

For *t* < 0, we first show from (5.60) and (5.62) that

$$R(-\mathbf{x}) = -R(\mathbf{x}),\tag{5.64}$$

upon which (5.63) gives

$$R(t\mathbf{x}) = R(|t| \cdot (-\mathbf{x})) = |t|R(-\mathbf{x}) = -|t|R(\mathbf{x}) = -tR(\mathbf{x}).\tag{5.65}$$

Furthermore, for given x,y ∈ *B*3, pick *s* > 0 such that *s* ≥ x and *s* ≥ y, so that *s* = 2*s* ≥ x+y by the triangle inequality, and use (5.59) to compute

$$\begin{split} \mathcal{R}(\mathbf{x} + \mathbf{y}) &= s\mathbb{K}\left(\frac{\mathbf{x} + \mathbf{y}}{s}\right) = s\mathbb{K}\left(\frac{\mathbf{x}}{s} + \frac{\mathbf{y}}{s}\right) = s\mathbb{K}(\mathbf{x}/s) + s\mathbb{K}(\mathbf{y}/s) \\ &= \mathcal{R}(\mathbf{x}) + \mathcal{R}(\mathbf{y}). \end{split} \tag{5.66}$$