From Classical Mechanics to Quantum Field Theory

Mathematical Foundations of Quantum Mechanics 157

Proof. (a)is obvious from Proposition 2.3.23, as restricting a stateρonL(H)to
LR(H) we still obtain a state as one can immediately verify. Let us prove (b).
Evidently, everyρ|L(Hk)is a positive measure with 0≤ρ(Pk)≤1. We can apply
Gleason’s theorem findingTk∈B(Hk) withTk≥0andTrTk=ρ(Pk) such that
ρ(Q)=tr(TkQ)ifQ∈L(Hk). Notice also that||Tk||≤ρ(Pk) because

||Tk||=sup

λ∈σp(Tk)

|λ|=sup

λ∈σp(Tk)

λ≤

∑

λ∈σp(Tk)

dλλ=TrTk=ρ(Pk).

IfQ∈LR(H),Q=

∑

kQk,whereQk:=PkQ∈L(Hk),QkQh=0ifk=hand
thus, byσ-additivity,

ρ(Q)=

∑

k

ρ(Qk)=

∑

k

tr(TkQk)

sinceHk⊥Hh, this identity can be rewritten as

ρ(Q)=tr(TQ)

providedT:=⊕kTk∈B 1 (H). It is clear thatT∈B(H) because, ifx∈Hand
||x||=1then,asx=

∑

kxkwithxk∈Hk,||Tx||≤

∑

k||Tk||||xk||≤

∑

∑ k||Tk||^1 ≤
kρ(Pk) = 1. In particular||T|| ≤1. T ≥0 because eachTk ≥0. Hence
|T|=

√

T†T=

√

TT=Tvia functional calculus, and also|Tk|=Tk.Moreover,
using the spectral decomposition ofT, whose PVM commutes with eachPk,one
easily has|T|=⊕k|Tk|=⊕kTk. The condition

1=ρ(I)=

∑

k

ρ(Pk)=

∑

k

tr(TkPk)=

∑

k

tr(|Tk|Pk)

is equivalent to say thattr|T|= 1 using a Hilbertian basis ofHmade of the
union of bases in eachHk. We have obtained, as wanted, thatT∈B 1 (H),T≥0,
tr T=1andρ(Q)=tr(TQ) for allQ∈LR(H).
(c)For, the proof straightforwardly follows the formLRk(Hk)=L(B(Hk))
becauseRk =B(Hk) and, evidently,ρT 1 =ρT 2 if and only ifρT 1 |L(B(Hk))=
ρT 2 |L(B(Hk))for allk∈K. Regarding (d) it is clear that ifρencompasses more
than one componentρ|L(Hk)= 0 cannot be extremal because it is, by construction,
a convex combination of other states which vanishes in some of the given coherent
subspace. Therefore only states such that only one restrictionρ|L(Hk 0 )does not
vanish may be extremal. Now (a) of Proposition 2.3.28 implies that, among these
states, the extremal ones are precisely those of the form said in (d) of the thesis.

Remark 2.3.45.
(a)Takeψ=

∑

k∈Kckψkwhere theψk∈Hkare unit vectors and also suppose
that||ψ||^2 =

∑

k|ck|^2 =1. This vector induces a stateρψ onRby means of