Download ALGEBRAIC CONSISTENT QUANTUM THEORY 1. Introduction Let

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 ) ,
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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 ) .
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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)
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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)
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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
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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