Decoherence and Consistent Histories

7/28/2019 Decoherence and Consistent Histories

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3. Decoherence and Consistent Histories

asic Idea: Classical probability theory is based on classical(Boolean) logic. The probabilities

fined by the Born Rule in quantum mechanics are based on quantum(non-Boolean) logic. In

rticular, they do not satisfy the classical or-additionaxiom of classical probability theory.

Topics:I. Quantum Probabilities and InterferenceII. DecoherenceIII.Consistent Histories

Quantum Probabilities and Interference

assical Probability Theory and the ClassicalOr-AdditionRule

asic Idea: Begin with a set of simple events (the sample space). Then form a collection F o

mpound events by taking all possible (classical) logical combinations of the simple events. The

fine a probability function Prcthat maps elements F to the real interval [0, 1]. Finally slap on

nch of axioms Prcmust satisfy.

o: A classical probability theory is given by a triple: (, F, Prc).

he probability function p is required to satisfy the following axioms:

ASIDE: F is a

Boolean algebraformed by closingunder the operationsof set-theoreticcomplement andcountable union.

(C1) Prc() = 0

(C2) Prc(A) = 1 Prc(A)

(C3) Prc(AA') = Prc(A) + Prc(A') Prc(AA')

3) is the Classical Or-Addition Rule. It says, "The probability ofeitherAorA'occurring is

ual to the sum of the probability ofA occuring and the probability ofA'occurring, minus the

obability of both A and A'occurring." (IfA and A'are mutually exclusive, then this latter

obability is zero.)

ample: Let = {1, 2, 3, 4, 5, 6} represent the simple events corresponding to all possible results of

ingle roll of a die. Now form F by taking all possible classical logical combinations of these events:

"1", "2", ..., etc., "1 or2", "1 or3", ...., etc., "not1", "not2", etc... .

ese correspond to the sets:

{1}, {2}, ..., etc., {1} {2}, {1} {3}, ..., etc., {1}, {2}, etc...

{1, 2} {2, 3, 4, 5, 6}

ASIDE: Again, recall that "or"and "not" form a complete set ofclassical logic operators (youcan form all other releventoperators, like "and", "if, then",etc, just from appropriatecombinations of "or" and "not").


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uantum Probability Theory and "Interference"

asic Idea: Replace the classical sample space set with a Hilbert space H. Now form a

llection L of subspaces ofH (the "compound events") by taking all possible quantum logical

mbinations of the rays in H (the simple events). Define the probability function Prqon them b

e Born Rule.

o: A quantum probability theory is given by a triple: (H, L, Prq),

where Prq is defined by

ASIDE: Recall from thelecture on quantum logic thathe collection of subspacesLnot a Boolean algebra.Technically, it forms what'scalled a lattice, and is obtainby closingH (thought of as acollection of rays) under theoperations of orthocomplemand linear span).

Prq(|ai, |) = |ai||2 , where |ai, | are elements ofH

ecall: This is the probability that the system possesses the value aiof a property represented b

an operator with eigenvector |ai, when the system is in the state represented by |.

Main Result: Quantum probabilities, so-defined, do not in general satisfy C3! They do satisfy

e following (where V, Ware subspaces ofH

and0

is the "zero" subspace):

(Q1) Prq(0) = 0

(Q2) Prq(V) = 1 Prc(V)

(Q3) Prq(VW) = Prq(V) + Prq(W), when VW

ecall: Linear span does notcorrespond to classical "or".

nally, define the probability function to be Prc({i}) = 1/6, for i = 1...6.

en, for instance, the probability of getting either 1 or 3 on a single roll is given by:

Prc({1} {3}) = Prc({1}) + Prc({3}) Prc({1} {3}) Axiom C3

= 1/6 + 1/6 0 = 1/3

e probability of getting either a value in the range {1, 2, 3} or a value in the range {3, 4, 5} on a single roll is:

Prc({1, 2, 3} {3, 4, 5}) = Prc({1, 2, 3}) + Prc({3, 4, 5}) Prc({1, 2, 3} {3, 4, 5}) Axiom C3

= [Prc({1}) + Prc({2}) + Prc({3})] + [Prc({3}) + Prc({4}) + Prc({5})] Prc({3}

= [1/6 + 1/6 + 1/6] + [1/6 + 1/6 + 1/6] 1/6 = 5/6


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|A(x)|2 + |B(x)|

2

A

B

A

B

|A(x) + B(x)|2

ith both slits open, experiments indicate that the probability that eis located at xis given by

A(x) + B(x)|2. This is notequal to the sum |A(x)|

2 + |B(x)|2, which, according to (C3),

presents the probability that the electron eitherwent through slit Aorslit B(assuming these

e mutually exclusive events).

A- or B-distribution (what doesn't happen) Interference distribution (what does happen)

xample: 2-slit probabilities and interference

|A(x)|2

A

B

A

B

|B(x)|2

With Slit A open,Prq(electron is located at x in stateA(x)) = |A(x)|

2With Slit Bopen,Prq(e is at x in stateB(x)) = |B(x)|

2

A-distribution B-distribution

ASIDE: The Born Rule tells us that the state corresponding to theprobability distribution|A(x) + B(x)|

2 is the superpositionA(x) +B(x).This is in the subspace V W which is the linear span of the subspace Vcontaining the state

A(x) and the subspace W containing the state

B(x).


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wherejP|aj= 1

Prq(value of A is ai in state|) = |ai||2

= |aiai|

=

j|P

|aj

|aia

i

|

=j|ajaj|aiai|

=jaj|aiai||aj

=Tr(|aiai||)

= Tr(P|aiP|)

he Born Rule can now be rewritten in terms of projection operators:

Prq(value of A is ai in state|) = Tr(P|aiP|)O:

he trace has the following properties:

Tr(A) = Tr(A) where is any number

Tr(A + B) = Tr(A) + Tr(B)

Tr(AB) = Tr(BA)

Def. Suppose Qis a linear operator on an N-dim vector space H with orthonormal basis

|b1, |b2, ... |bN. Then the traceTr(Q) ofQis given by:

Tr(Q) b

iQ b

ii=1

N

ow Consider: The "trace" of an operator.

ote: The trace is just the sum of the diagonal elements of any matrix representation of Q. It turns out that all

such representations have this sum in common! So the trace is independent of the basis it's calculated in.

et's see how this works in the more general setting of projection operators.

rst Recall:

) The projection operator P|ai = |aiai| corresponds to the 1-dim subspace defined by |ai (i.e.

the ray in which |ai is pointing).

) iP|ai= 1.


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ow: Consider the composite state m-eof a Hardnessmeasuring device and a blackelectron:

| = {|"hard"m|harde+ |"soft"m|softe} call it simply {|"h"|h + |"s"|s}

s statistical operator P| = || is given by:

o: Prq(value of A is ai in state|)= Tr(P|aiP|)

= Tr(P|aiP|"hard"P|hard) + Tr(P|aiP|"soft"P|soft)

+ Tr(P|ai|"hard""soft"||hardsoft|) + Tr(P|ai|"soft""hard"||softhard|)

= Prq(value of A is ai in state|"hard"|hard) + Prq(value of A is ai in state|"soft"|soft)

+ interference terms

= {|"h"|h + |"s"|s}{"h"|h| + "s"|s|}

= {|"h"|h"h"|h| + |"s"|s"s"|s|+ |"h"|h"s"|s| + |"s"|s"h"|h|}

= {|"h""h"||hh| + |"s""s"||ss|+ |"h""s"||hs| + |"s""h"||sh|}

= {P|"hard"P|hard + P|"soft"P|soft + |"h""s"||hs| + |"s""h"||sh|}

making sure m-ketsbelong to m-bras and

e-kets belong to e-bras!

statistical operator

for|"hard"m|harde

statistical operator

for|"soft"m|softe

interference terms!

erminology: The projection operator corresponding to a state is called the statistical

operator(alternatively, density matrix) for the state.

Aside on terminology: Suppose the exact state of a system is known only to fall within

some set {|k} to which weights wkcan be assigned such that kwk= 1. Then the statistical

operatorWfor the system is defined by W=kwkP|k(this need not be a projection

operator). A pure stateis one for which Wreduces to a single term. This is the case in

which the exact state of the system is known, say |, so the statistical operator is just the

projection operator P|. A mixed stateis one for which Whas more than one term.

gain: P|ai is the projection operator corresponding to the state |ai (more precisely, the 1-dim

subspace defined by |ai). P| is the projection operator corresponding to the state |

(more precisely, the 1-dim subspace defined by |).


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. Decoherence and Interference

laim: Interference effects are destroyed when a system interacts with its environment. In

rticular, when an observer ends up in an entangled state with a measuring device, environment

teractions effectively "decohere" the state into one in which the observer effectively records a

finite measurement outcome.

asic Argument:

is experimentally impossible to distinguish between:

here |hard, |soft are states of the environment Ein which it is correlated with a hard electro

d a soft electron, respectively.

ecall: To distinguish between (1) and (2), we would need a very complex multi-particle

property that (1) possesses and that neither state in (2) possesses. Given that E

realistically has a huge number of degrees of freedom, it is experimentally impossible to

measure such a property. So (1) and (2) are indistinguishable for all practical purposes!

No collapse:no definite outcome.

) The state:

|"hard"m|harde|hard+ |"soft"m|softe|soft

) Either of the states:

|"hard"m|harde|hard or |"soft"m|softe|soft The reulst of collapse:a definite outcome.

What this is supposed to mean:

henever the post-measurement state of a composite system is of the form of (1), it does, for all

actical purposes, describe a situation in which a definite measurement outcome occurred. The

vironment, for all practical purposes, collapses the entangled superposition.

What this means:

he probability associated with the superposed state is not in general the sum of the probabilities

sociated with its terms. There are "interference terms" that are not in general zero. Quantum

obabilities do not obey the classical or/addition rule (C3).


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et's see how this works using statistical operators.

he statistical operator P| = || for the state in (1) is:

= {|"h"|h|Eh + |"s"|s|Es}{"h"|h|Eh| + "s"|s|Es|}

= {|"h"|h|Eh"h"|h|Eh| + |"s"|s|Es"s"|s|Es|

+ |"h"|h|Eh"s"|s|Es| + |"s"|s|Es"h"|h|Eh|}

= {|"h""h"||hh||EhEh| + |"s""s"||ss||EsEs|

+ |"h""s"||hs||EhEs| + |"s""h"||sh||EsEh|}

= P|"hard"P|hardP|Eh + P|"soft"P|softP|Es + |"h""s"||hs||EhEs|

+ |"s""h"||sh||EsEhstatistical operator for|"hard"m|harde|hardE

statistical operator for|"soft"m|softe|softE

interference terms!

Now: Take the "partial trace" ofP| with respect to the Environment basis |Eh, |Es:

TrE(P|) = Eh|P||Eh + Es|P||Es

= P|"hard"P|hard + P|"soft"P|soft

esult: "Tracing over the environment" kills the interference terms: They include the operators

hEs| and |EsEh|. When you slap |Eh or |Esto both sides of these operators, you get zero!

hen you do the same to the operators P|"hard"P|hardP|Eh and P|"soft"P|softP|Es, you just kill

e Environment operators at the end. What we're left with is just the statistical operator for the

ates |"hard"m|hardeor |"soft"m|softe.

asic Claim: "Tracing over the environment" is the mathematical expression of having the

nvironment decohere the entangled superposed state in (1).

Aside: This is called a "partial trace" because we are not using a "full" basis to calculate the

diagonal elements of P|. The "full" basis should be a "3-particle" basis like |"h"|h|Eh, |"s"|s|Es,

|"h"|h|Es, |"h"|s|Es, etc (eight such basis vectors). In this case, we're only using a "1-particle"

basis|Eh, |Es. This particular "partial trace" results, not in a number, but a 2-particle operator!


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oes decoherence solve the measurement problem? NO!

hen we "trace out the environment", we're left with the statistical operator

P|"hard"P|hard + P|"soft"P|soft

his is the statistical operator for a "mixed state". This is how the state of a system is represente

hen its exact form is known only to lie within a set of possible states. In this case, the state of

e system is eitherof the pair {|"hard"m|harde, |"soft"m|softe}, each with equal weight .

ut: The result of a measurement (as given by the Projection Postulate and by our experience)

a definiteoutcome. In this case, the result is either|"hard"m|hardeor|"hard"m|harde. It

definitelyone of these two alternatives. It's not a weighted sum of them both!

Aside: This is the essential criticism of the decoherence approach to the measurement problem.

A less severe criticism is the observation that the environment states |Eh and |Es will in realistic

situations never be exactlyorthogonal to each other. The way in which the environment

correlates itself to a pointer pointing to "hard" may "overlap" the way in which it correlates itself

to a pointer pointing to "soft". Thus, in such realistic cases, when we "trace over the

environment", we will notkill allthe interference terms. However, one can show that the "off-

diagonal" interference terms that survive in realistic cases have very short life-spans and go to

zero very quickly. Of course this doesn't address the basic fact that, even when allinterference

terms are killed by decoherence, the state that results does not describe a definite outcome.

I. Consistent Histories

Example: P1(t1) might be P|hard which represents the property "The value of

Hardnessis hard". Or it might be P|a, which represents the property

"The value of the property represented by A is a".

Def 1. A historyhis a time-sequence of facts, represented by time-indexed

projection operators:

h= {P1(t1), P2(t2), ..., Pn(tn)}

side: Projection operators evolve viathe Schrdinger dynamics: P(t) = eiHt/P(0)eiHt/.


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ide: A more detailed motivation for Def. 2:

nsider the simple history h= {P|a(t1), P|b(t2)}. Suppose our system is in an initial state |.

ow: Calculate the conditional probabilitythat it has the value b (of some operator B) at t2, given it had the valuea(of some other operator A) at time t1. To do this, note that, if it possessed aat t1, then at t2, it will no

longer be in the state |! Measuring | for aat t1 changes | to a new state, call it |'. |' is the

projection of | onto the subspace spanned by |a. It's (normalized) statistical operator is given by

P=

PaPP

a

Tr(PaPP

a)

o: Prq(value of B is b in |', given value of A is a in |)

= Prq(value of B is b in |')Prq(value of A is a in |)

= Tr(P|bP|')Tr(P|aP|)

= TrP

bPaPPa

Tr(PaPPa)

Tr P

aP( )

= TrP

bPaPPa

Tr(PaP )

Tr PaP( )

= Tr PbPaPPa( )

Tr PaP( )

Tr PaP( )

= Tr Pb(t

2)P

a(t

1)P

Pa(t

1)P

b(t

2)( )

sinceTr(P|aP|P|a) = Tr(P|aP|aP|) = Tr(P|aP|)

sinceTr(A) = Tr(A), for any number

sinceTr(P|bP|aP|P|a) = Tr(P|bP|bP|aP|P|a)

= Tr(P|bP|aP|P|aP|b)

ow extend this to histories of arbitrary length, and we get Def. 2.

Def 2. The probabilityassociated with a history his given by:

Prq(h) = Tr(Pn(tn)...P2(t2)P1(t1)P|P1(t1)P2(t2)...Pn(tn))

where P| is the statistical operator associated with an initial state |.

Motivation: All the terms inside the trace commute with each other (one of the properties of th

ace), so you can rearrange them into Tr(Pn(tn)Pn(tn)...P2(t2)P2(t2)P1(t1)P1(t1)P|). And sinceojection operators are idempotent, this is equal to Tr(Pn(tn)...P2(t2)P1(t1)P|), which can be

ought of as the Born Rule for the probability that the system, in the state |, has the "historic

operty" represented by the operator Pn(tn)...P2(t2)P1(t1).


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What this means: The projection operators in any such set {Pii(ti)} represent all the possible

lues of the property associated with Pi(ti).

ote: Histories can be embedded in families of histories:

P|t= 0

t= t1 P11(t1) P1

2(t1) .... .... P1N(t1)

P21(t2) P2

2(t2) .... .... P2N(t2)

Pn1(tn) Pn

2(tn) .... .... PnN(tn)

t= t2

t= tn

.

.

.

.

.

.

.

.

.

.

.

.

h

h'

h= {P|

, P1

1(t1), P

2

2(t2), ..., P

n

1(tn)}

h'= {P|, P12(t1), P2

N(t2), ..., Pn2(tn)}

hand h'are distinct histories within the family {{P|}, {P11(t1)}, {P2

2(t2)}, ..., {Pnn(tn)}}

Def 3. Afamily ofhistoriesis a time-sequence of sets of "exhaustive" facts. Each set of

"exhaustive" facts is represented by a set of projection operators whose sum is the

identity.

{{P11(t1)}, {P2

2(t2)}, ..., {Pnn(tn)}}, where each i= 1, ..., N

where set {Pii(ti)} consists ofNprojection operators, Pi

1(ti), Pi2(ti), ..., Pi

N(ti),

such that Pi1(ti) + Pi

2(ti) + ... + PiN(ti) = IN.

{P|}

{P11(t1)}

{P22(t2)}

{Pnn(tn)}

.

.

.

ote: We can assign probabilities to histories within a family by means ofDef. 2.

ut: Since this is based on the Born Rule, these will be quantumprobabilities that exhibit

"interference" effects.

re there histories within a given family that can be assigned classical probabilties?

N = dim ofHilbert space

Aside: Def. 3 defines what we'll call below a "fine-grainehistory in which all projection operators are 1-dim. Therecan be "coarse-grained" histories with projection operatorswith dim > 1. We still require their sum be the identity,but now there need not be exactlyl N terms in this sum.


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rst: Need a general expression for the disjunction, horh', of two histories h, h'. Suppose hA

d hBare histories that differ only in the property at t= ti:

hA = {P1(t1), ..., PiA(ti), ..., Pn(tn)}

hB= {P1(t1), ..., PiB(ti), ..., Pn(tn)}

ow let the history, hAorhB, be given by:

hAorhB= {P1(t1), ..., PiA(ti) + Pi

B(ti), ..., Pn(tn)}

hen, for an initial state |:

q(hAorhB) = Tr(Pn(tn)...[PiA(ti) + Pi

B(ti)]...P1(t1)P|P1(t1)...[PiA(ti) + Pi

B(ti)]...Pn(tn))

= Tr(Pn(tn)...PiA(ti)...P1(t1)P|P1(t1)...Pi

A(ti)...Pn(tn))

+ Tr(Pn(tn)...PiB(ti)...P1(t1)P|P1(t1)...Pi

B(ti)...Pn(tn))

+ Tr(Pn(t

n)...P

i

A(ti)...P

1(t

1)P

|P

1(t

1)...P

i

B(ti)...P

n(t

n))

+ Tr(Pn(tn)...PiB(ti)...P1(t1)P|P1(t1)...Pi

A(ti)...Pn(tn))

= Prq(hA) + Prq(hB) + interference terms

o: The probabilities assigned to hA and hBby Def. 2 will be classical (i.e., obey the classical Or

Addition Rule) just when the interference terms vanish.

Prq(hor h') = Prq(h) + Prq(h').

other words, are there histories within a given family that do not "interfere" with each other?

hese would be histories h, h'whose probabilities obey the classical Or-Addition Rule:

Tr(Pn(tn)...P1(t1)P|P1'(t1)...Pn'(tn)) = 0

ow: Consider the general case:

h= {P1(t1), ..., Pn(tn)}

h'= {P1'(t1), ..., Pn'(tn)}

horh'= {[P1(t1) + P1'(t1)], ..., [Pi(ti) + Pi'(ti)], ..., [Pn(tn) + Pn'(tn)]}

he probabilities assigned to hand h'by Def. 2 will be classical (i.e., obey the classical Or-

ddition Rule) just when the general interference term vanishes:


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Def. 4. Two histories h= {P1(t1), ..., Pn(tn)}, h'= {P1'(t1), ..., Pn'(tn)} are consistent

just when Tr(Pn(tn)...P1(t1)P|P1'(t1)...Pn'(tn)) = 0.

Def. 5. A consistent family of historiesis a family of histories such that any two

histories embeddable in it are consistent.

ow add the notion of decoherence...

an now say:

Fine-grained histories cannot in general be assigned classical probabilities.

Course-grained histories can be assigned approximateclassical probabilities, and these get bette

(more classical) as Tr(Pn(tn)...P1(t1)P|P1'(t1)...Pn'(tn)) 0.

As Tr(Pn(tn)...P1(t1)P|P1'(t1)...Pn'(tn)) 0, such coarse-grained histories "decohere".

What this means: A consistent family of histories is a collection of histories that defines a

assical sample space! You can assign classical probabilities to its members.

Def. 6.

(1) his afine-grained historyjust when all projection operators in hare 1-dim.

(2) h'is a coarse-graining ofhjust when some projection operators in h'are sums of

projection operators in h.

What this is supposed to mean:

oarse-graining a family of histories corresponds to "tracing out the environment". The

vironment interacts with the coarse-grained histories to damp out the interference effects,

ndering the family approximately consistent.

Def. 7. Two histories h= {P1(t1), ..., Pn(tn)}, h'= {P1'(t1), ..., Pn'(tn)} are

decoherentjust when Tr(Pn(tn)...P1(t1)P|P1'(t1)...Pn'(tn)) 0.

Def. 5. A decoherent family of historiesis a family of histories such that any two

histories embeddable in it are decoherent.


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Characteristics of the Consistent/Decoherent Histories (CH) Approach

Replaces statesof a physical system with historiesa physical system.

The properties (projection operators) that make up a history evolve onlyviathe

Schrdinger dynamics (no Projection Postulate).

Identifies a way to associate a probability with a history (Def. 2).

Identifies a condition that picks out those families of histories that are classical (or

approximately classical) (Defs. 4, 5).

roblems:

How are alternative histories within a decoherent family to be interpreted?

Is one such history our actual history and the others just possible? This is our experience, but

CH is silent as to how to specify the actual history from possible alternatives.

Are all histories within a decoherent family occurent? If so, then how are probabilitiesexplained? (This is the Problem of Probabilitiesthat Many Worlds faces.)

How are alternative decoherent families to be interpreted?

ny history hcan be embedded in many different mutually incompatible decoherent families (any

e of which defines an approximately classical probability space). Which do we choose in order

calculate the probability ofh? (This is the Preferred Basis Problemthat Many Worlds faces.)

General Problem with the Notion of Decoherenceracing over the environment" (or "coarse-graining" histories) does not pick out a unique

easurement/interaction outcome. It does not effect a "collapse" of superposed states (or

nterfering" histories). So it cannot be appealed to in order to reconcile superpositions (or

nterfering" histories) with our experience of unique outcomes.

oblems 1 & 2 Combined:

em to indicate that CH isn't fundamentally different from Many Worlds. All CH does is replac

orld-talk with history-talk, and adds a criterion for identifying histories that behave "classically"

Decoherence and Consistent Histories

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