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ALGEBRAIC CONSISTENT QUANTUM THEORY MARK SINNETT 1. Introduction Let A be a bounded, self-adjoint operator on a Hilbert space H, representing a physical quantity. Let Σ be the spectrum of A. Let E : B (Σ) → P (H) be the spectral measure (B Borel measurable subsets, P projections). The map E (X) is a property, X ∈ B (Σ). Each normal state is defined by a density operator ρ0 on H, which is positive, trace-class, kρ0 k1 = 1. Let ωρ0 : B (H) → C be the function defined by ωρ0 (B) = T r (ρ0 B) . The trace-class ideal I 1 (H) = {trace class operators}. In general, the trace-class norm kAk1 of a trace-class operator A is equal to T r (|A|). The probability wrt ρ0 of E (X) is Pρ10 [E (X)] = ωρ0 (E (X)) , the probability of that the measurement of the observable A will be in X. Let A1 , ..., An be observables with corresponding spectra Σ1 , ..., Σn and spectral measures E1 , ..., En and a sequence of properties E1 (X1 ) , ..., En (Xn ) corresponding to times t1 < ... < tn . The finite sequence (E1 (X1 ) , ..., En (Xn )) is called a history. Think of this is an apparatus that makes measurements. Another proposal is that we should think of a history as a product h = E1 (X1 ) ... En (Xn ) , where has the algebraic properties of the tensor product. We simplify Fj = Ej (Xj ). In the physics literature, the nth order probability of the history h is defined to be Pρn0 (h) = T r (Fn Fn−1 ...F1 ρ0 F1 ...Fn−1 ) . For example, if h0 = F 1 F 2 , then Pρ20 (h0 ) = T r (F2 F1 ρ0 F1 ) , 1 2 MARK SINNETT so then F1 ρ0 F1 kF1 ρ0 F1 k F1 ρ0 F1 = T r (F1 ρ0 F1 ) F1 ρ0 F1 = T r (ρ0 F1 ) F1 ρ0 F1 = Pρ10 (F1 ) ρ0 (X1 ) = is a density operator. Then Pρ20 (h0 ) = T r (F2 ρ0 (X1 )) Pρ10 (F1 ) = Pρ10 (X1 ) (F2 ) Pρ10 (F1 ) , the first factor of which is like a conditional probability. Let h0 = F1 F2 Think of X1 × X2 as a product, with X1 = X11 ∪ X12 . Then E1 (X1 ) = E1 X11 ∨ E1 X12 = F11 ∨ F12 , so that h0 = F11 F2 ∨ F12 F2 = h1 ∨ h2 Is it true that Pρ20 (h0 ) = Pρ20 (h1 ∨ h2 ) = Pρ20 (h1 ) + Pρ20 (h2 )? Well, not generally, because Pρ20 (h0 ) = T r F2 F11 ∨ F12 ρ0 F11 ∨ F12 = T r F2 F11 ρ0 F11 + T r F2 F12 ρ0 F12 = Pρ20 (h1 ) + Pρ20 (h2 ) +T r F2 F11 ρ0 F12 + T r F2 F12 ρ0 F11 . The consistency condition for second order histories is that T r F2 F11 ρ0 F12 + T r F2 F12 ρ0 F11 = 0 for every measurable decomposition of X1 as X11 ∪ X12 . We will consider A1 , ..., An observables, A1 , ..., An maximal abelian von Neumann algebras. Each of these has Gel’fand spectrum S1 , ..., Sn and is endowed with a unique spectral integral Pi : B (Si ) → Ai and spectral measure Ei : B (Si ) → P (Ai ) . CONSISTENT QUANTUM THEORY 3 Then An1 Sn1 Pn1 En1 : : : : = A1 ⊗...⊗An = S1 × ... × Sn = P1 ⊗...⊗Pn = E1 ⊗...⊗En Then E1n (Yn1 ) = E1 (Y1 ) ... En (Yn ) = F1 ... Fn = Fn1 Then Pρn0 [Fn1 ] = T r (Fn Fn−1 ...F1 ρ0 F1 ...Fn−1 ) = ωρ0 (F1 ) T r (Fn Fn−1 ...F2 ρ0 (Y1 ) F2 ...Fn−1 ) = ωρ0 (F1 ) ωρ0 (Y1 ) (F2 ) ...ωρ0 (Yn−1 ) (Fn ) 1 n n n n−1 n = ωρ0 Y1 [E1 (Y1 )] = ωρ0 Yn−1 ◦ En1 [Yn1 ] 1 n = µnρ0 Yn−1 (Y1 ) . 1 Notation and Summary of Results for Lecture 2 (1) Elementary history: E1n (Y1n ) = F1n (2) kth initial density operator: ρk = ρ0 (Y1k ) := Ek (Yk )ρ0 (Y1k−1 )Ek (Yk ) T r{Ek (Yk )ρ0 (Y1k−1 )} (3) History state with respect to ρ0 generated by Y1n−1 : ωρn0 (Y1n−1 ) = ωρk0 (Y1k−1 ) ⊗ ωρk ⊗ ωρk (Yk+1 ) ⊗ · · · ⊗ ωρk (Y n−1 ) k+1 = n−1 ωρk0 (Y1k−1 ) ⊗ ωρn−k (Yk+1 ) k (4) History measure with respect to ρ0 generated by Y1n−1 : µnρ0 [Y1n−1 ] = µkρ0 [Y1k−1 ] ⊗ µρk ⊗ µρk (Yk+1 ) ⊗ · · · ⊗ µρk (Y n−1 ) k+1 k−1 n−1 k n−k = µρ0 Y1 ⊗ µρk Yk+1 Theorem 3.1. The nth order probability of E1n (Y1n ) with respect to ρ0 is given by pnρ0 [F1n ] = ωρn0 (Y1n−1 )[F1n ] = µnρ0 [Y1n−1 ](Y1n ). (5) Measurable decomposition: Yk = Yk1 t Yk2 for any k, 1 6 k < n. m m (6) For i = 1, 2 : Y1m([ik) := Y1k−1 ×Yki ×Yk+1 , F1m(ik) = E1m(Y1m(ik)) = F1k−1 ×Fki ×Fk+1 , for k < m 6 n. ρik := Ek (Yki )ρ0 (Y1k−1 )Ek (Yki ) Fki ρ0 (Y1k−1 )Fki = . T r{Ek (Yki )ρ0 (Y1k−1 )} T r{Fki ρ0 (Y1k−1 )} (7) 4 MARK SINNETT Definition. An elementary history E1n (Y1n ) = F1n is consistent with respect to a density operator ρ0 if the probabiltiy is additive. That is, for each binary measurable decomposition Yk = Yk1 t Yk2 for any k, 1 6 k 6 n, pnρ0 [F1n ] = ωρn0 (Y1n−1 )[F1n ] = ωρn0 Y1n−1 1k (F1n 1k) + ωρn0 Y1n−1 2k (F1n 2k) = pnρ0 (F1n 1k) + pnρ0 (F1n 2k) . ηki := ωρ0 (Y k−1 ) (Fki ) 1 ωρ0 (Y k−1 ) (Fk ) = 1 µρ0 (Y k−1 ) (Yki ) 1 µρ0 (Y k−1 ) (Yk ) ; 0 6 ηki 6 1; and ηk1 + ηk2 = 1. (8) 1 Proposition 4.8(i). ρk = ηk1 ρ1k + ηk2 ρ2k . Corollary 6.4. µnρ0 Y1n−1 ik µnρ0 (Y1n−1 ) everywhere in B(S1n ). n−1 n−1 It follows that µn−k [Yk+1 ] µn−k ρk (Yk+1 ), with Radon-Nikodym derivative denoted by ρi n−1 δρn−k [Yk+1 ]. i k k Theorem [4.1, 4.9, 6.5]. For an elementary history E1n (Y1n ) and a density operator ρ0 the following four conditions are equivalent: (i) E1n (Y1n ) is consistent with respect to ρ0 ; (ii) (CH1) T r{G∗k Gk [Fk1 ρ0 (Y1k−1 )Fk2 + Fk2 ρ0 (Y1k−1 )Fk1 ]} = 0, where Gk = Fn · · · Fk+1 ; n−1 n (iii) (CH4s) ωηn−k 1 1 2 2 (Yk+1 )[Fk+1 ] k ρk +ηk ρk n o n−1 n−1 n = ηk1 ωρn−k (Yk+1 ) + ηk2 ωρn−k (Yk+1 ) [Fk+1 ]; 1 2 k k and, n−1 (iv) (CH5) the Radon-Nikodym derivatives δρn−k [Yk+1 ] satisfy the relations: i k n−1 δρn−k [Yk+1 ] 1 k + n−1 ] δρn−k [Yk+1 2 k =1 n almost everywhere on Yk+1 , and n−1 n−1 [Yk+1 ] + ϕYk2 ⊗ δρn−k [Yk+1 ]=0 ϕYk1 ⊗ δρn−k 2 1 k almost everywhere on k Ykn . Note that the condition for a second order history is that T r F2 F11 ρ0 F12 + F12 ρ0 F11 = 0, and (CH1) is a generalization of this. Proof: n−1 n−k+1 n n−2 k−2 Pρn0 [F1n ] = ωρk−1 Y F ω Y [Fk ] , 1 1 ρ k 0 k−1 CONSISTENT QUANTUM THEORY 5 and Ykn−2 [Fnk ] = T r Gk Fk ρ0 Y1k−1 Fk1 ∨ Fk2 G∗k ωρn−k+1 k−1 = T r Gk Fk1 ρ0 Y1k−1 Fk1 G∗k + T r Gk Fk2 ρ0 Y1k−1 Fk2 G∗k +cross terms So the cross terms have to be zero. This implies (CH1). For i = 1, 2, 1 n ωρ0 Y1k−1 (Fki ) n−k n−1 n−1 i i n−k+1 = ηk ωρ1 (Yk ω Y Fk+1 k−1 (Fk ) ω i k ) Fk k+1 ρk k ωρ0 Y1k−1 (Fk ) ρ0 (Y1 ) n i n−1 Yk+1 Fk k = ωρ0 Y1k−1 Fki ωρn−k i k Ykn−1 ik Fnk ik = ωρn−k+1 k−1 Then ηk1 ωρn−k+1 (Ykn−1 1 1 1 1 n 1 n−1 2 n−1 1 n−1 2 2 n−k+1 n−k+1 n−k+1 ) F + η ω (Y ) F = ω Y F + ω Y 1 k k k ρ k k ρk−1 k k k ρk−1 k k k k k k n−k+1 n 1 n−k+1 n 2 = Pρk−1 Fk k + Pρk−1 Fk k . Multiplying through by ωρk−1 Y1k−2 Fk−1 we obtain the sum of probabilities 1 0 n o n−1 n−1 Pρk0 [Fn1 ] = ηk1 ωρn−k (Yk+1 ) [Fn1 ] + ηk2 ωρn−k (Yk+1 ) [Fn1 ] . 1 2 k k Then (CH4s) follows. Corollary 6.9. If E1n (Y1n ) is consistent with respect to ρ0 then n−1 n−1 n−1 1 n−k ωηn−k (Yk+1 ) + ηk2 ωρn−k (Yk+1 ) 1 ρ1 +η 2 ρ2 (Yk+1 ) = ηk ωρ1 2 k k k k k k everywhere on Ank+1 . Department of Mathematics, Texas Christian University, Fort Worth, Texas 76129, USA