Griffiths R.B. Consistent Quantum Theory

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154 Checking consistency

we can write (11.17) as



α∈S

K(Y

) =

, (11.22)

and since

K(Y) is, like K(Y), a linear function of its argument Y, the argument

leading to

K(Y



) = 0, obviously equivalent to K(Y



) = 0, is somewhat more

transparent.

11.6 Initial pure state. Chain kets

If the initial projector of Sec. 11.5 projects onto a pure state,

= [ψ

] =|ψ

ψ

|, (11.23)

where we will assume that |ψ

 is normalized, there is an alternative route for cal-

culating weights and checking consistency which involves using chain kets rather

than chain operators. Since it is usually easier to manipulate kets than it is to carry

out the corresponding tasks on operators, using chain kets has advantages in terms

of both speed and simplicity. Suppose that Y

in (11.11) has the form given in

(8.30),

= [ψ

] ( P

( P

(···P

, (11.24)

with projectors at t

, t

, etc. drawn from decompositions of the identity of the type

(8.25). Then it is easy to see that the corresponding chain operator is of the form

K(Y

) =|αψ

|, (11.25)

where the chain ket |α is given by the expression

|α=P

T(t

, t

f −1

) ···P

T(t

, t

T(t

, t

)|ψ

. (11.26)

That is, start with |ψ

, integrate Schr

odinger’s equation from t

to t

, and apply

the projector P

to the result in order to obtain

|φ

=P

T(t

, t

)|ψ

. (11.27)

Next use |φ

 as the starting state, integrate Schr

odinger’s equation from t

to t

and apply P

. Continuing in this way will eventually yield |α, where the symbol

α stands for (α

,α

,...α

The inner product of two chain operators of the form (11.22) is the same as the

inner product of the corresponding chain kets:

K(Y

), K(Y

)=Tr



†

)K(Y

)



= Tr



|ψ

α|βψ



=α|β. (11.28)

11.7 Unitary extensions 155

Consequently, the consistency condition becomes

α|β=0forα = β, (11.29)

while the weight of a history is

W(Y

) =α|α. (11.30)

In the special case in which one of the projectors at time t

projects onto a pure

state |α

, the chain ket will be a complex constant, which could be 0, times |α

.

If two or more histories in the sample space have the same ﬁnal projector onto a

pure state |α

, then consistency requires that at most one of these chain kets can

be nonzero.

The analog of the argument in Sec. 11.5 following (11.17) leads to the following

conclusion. Suppose one has a collection S of nonzero chain kets of the form

(11.26) with the property that



α∈S

|α=T(t

, t

)|ψ

. (11.31)

That is, they add up to the state produced by the unitary time evolution of |ψ



from t

to t

. Suppose also that for the collection S the consistency conditions

(11.29) are satisﬁed. Then one knows that the collection of histories {Y

: α ∈ S}

is the support of a consistent family: there is at least one way (and usually there

are many different ways) to add histories of zero weight to the support S in order

to have a sample space satisfying (11.13), with "

= [ψ

]. Nonetheless, for the

reasons discussed towards the end of Sec. 11.5, it is sometimes a good idea to go

ahead and construct the zero-weight histories explicitly, in order to have a Boolean

algebra of history projectors with certain speciﬁc properties, rather than relying on

a general existence proof.

11.7 Unitary extensions

For the following discussion it is convenient to use the Heisenberg representa-

tion introduced in Sec. 11.4, even though the concept of unitary extensions works

equally well for the ordinary (Schr

odinger) representation. Unitary histories were

introduced in Sec. 8.7 and deﬁned by (8.38). An equivalent deﬁnition is that the

corresponding Heisenberg operators be identical,

=···

, (11.32)

where we have used t

as the initial time rather than t

as in Sec. 8.7. It is obvi-

ous from (11.9) that the Heisenberg chain operator

K for a unitary history is the

projector

156 Checking consistency

Next suppose that in place of (11.32) we have

=···

m−1

=

m+1

=···

, (11.33)

where m is some integer in the interval 1 ≤ m ≤ f . We shall call this a “one-jump

history”, because the Heisenberg projectors are not all equal; there is a change, or

“jump” between t

m−1

and t

. In a one-jump history there are precisely two types of

Heisenberg projectors, with all the projectors of one type occurring at times which

are earlier than the ﬁrst occurrence of a projector of the other type. The chain

operator for a history with one jump is

K =

. (If, as is usually the case,

and

do not commute,

K is not a projector.) Similarly, a history with two jumps

is characterized by

=···

m−1

=

=···



−1

=



=···

, (11.34)

with m and m



two integers in the range 1 ≤ m < m



≤ f , and its chain operator

K =

. (It could be the case that

.) Histories with three or more

jumps are deﬁned in a similar way.

A unitary extension of a unitary history (11.32) is obtained by adding some ad-

ditional times, which may be earlier than t

or between t

and t

or later than t

; the

only restriction is that the new times do not appear in the original list t

, t

,...t

At each new time the projector for the event is chosen so that the corresponding

Heisenberg projector is identical to those in the original history, (11.32). Hence, a

unitary extension of a unitary history is itself a unitary history, and its Heisenberg

chain operator is

, the same as for the original history.

A unitary extension of a history with one jump is obtained by including addi-

tional times, and requiring that the corresponding Heisenberg projectors are such

that the new history has one jump. This means that if a new time t



precedes t

m−1

in (11.33), the corresponding Heisenberg projector



will be

, whereas if it fol-

lows t



will be

. If additional times are introduced between t

m−1

and t

, then

the Heisenberg projectors corresponding to these times must all be

, or all

,or

if some are

and some are

, then all the times associated with the former must

precede the earliest time associated with the latter. The Heisenberg chain operator

of the extension is the same as for the original history,

Unitary extensions of histories with two or more jumps follow the same pattern.

One or more additional times are introduced, and the corresponding Heisenberg

projectors must be such that the number of jumps in the new history is the same

as in the original history. As a consequence, the Heisenberg chain operator is left

unchanged. By using a limiting process it is possible to produce a unitary extension

of a history in which events are deﬁned on a continuous time interval. However, it

is not clear that there is any advantage to doing so.

11.8 Intrinsically inconsistent histories 157

The fact that the Heisenberg chain operator is not altered in forming a unitary

extension means that the weight W of an extended history is the same as that of

the original history. Likewise, if the chain operators for a collection of histories

are mutually orthogonal, the same is true for the chain operators of the unitary

extensions. These results can be used to extend a consistent family of histories to

include additional times without having to recheck the consistency conditions or

recalculate the weights.

There is a slight complication in that while the histories obtained by unitary

extension of the histories in the original sample space form the support of the new

sample space, one needs additional zero-weight histories so that the projectors will

add up to the history identity (or the projector for an initial state). The argument

which follows shows that such zero-weight histories will always exist. Imagine

that some history is being extended in steps, adding one more time at each step.

Suppose that t



has just been added to the set of times, with



the corresponding

Heisenberg projector. We now deﬁne a zero-weight history which has the same

set of times as the newly extended history, and the same projectors at these times,

except that at t



the projector



is replaced with its complement



= I −



. (11.35)

What is



for the history containing



? Since the unitary extension had the

same number of jumps as the original history,



must occur next to an



in the

product which deﬁnes



, and this means that



= 0, since





= 0. Thus

we have produced a zero-weight history whose history projector when added to

that of the newly extended history yields the projector for the history before the

extension, since





= I. Consequently, by carrying out unitary extensions

in successive steps, at each step we generate zero-weight histories of the form

needed to produce a ﬁnal sample space in which all the history projectors add up

to the desired answer. While the procedure just described can always be applied

to generate a sample space, there will usually be other ways to add zero-weight

histories, and since the choice of zero-weight histories can determine what events

occur in the ﬁnal Boolean algebra, as noted towards the end of Sec. 11.5, one may

prefer to use some alternative to the procedure just described.

11.8 Intrinsically inconsistent histories

A single history is said to be intrinsically inconsistent, or simply inconsistent,if

there is no consistent family which contains it as one of the elements of the Boolean

algebra. The smallest Boolean algebra of histories which contains a history pro-

158 Checking consistency

jector Y consists of 0, Y,

Y =

I − Y, and the history identity

I. Since Y

Y = 0,

K(Y), K(

Y)=0, (11.36)

see (10.21), is a necessary and sufﬁcient condition that Y be intrinsically inconsis-

tent.

If one restricts attention to histories which are product projectors, (8.6), no his-

tory involving just two times can be intrinsically inconsistent, so the simplest pos-

sibility is a three-time history of the form

Y = A ( B ( C. (11.37)

Given Y,deﬁne the three histories



= A (

B ( C,



= A ( I (



A ( I ( I,

(11.38)

where, as usual,

P stands for I − P. Then it is evident that

Y + Y



+ Y



+ Y



= I ( I ( I =

I, (11.39)

so that

Y = Y



+ Y



+ Y



, (11.40)

and thus

Y) = K(Y



) + K(Y



) + K(Y



). (11.41)

By considering initial and ﬁnal projectors, Sec. 11.3, it is at once evident that

K(Y



) and K(Y



) are orthogonal to K(Y). Consequently,

K(Y), K(

Y)=K(Y), K(Y



), (11.42)

so that Y is an inconsistent history if the right side of this equation is nonzero.

As an example, consider the histories in (10.35), and let Y = Y

. Then Y



= Y

and (10.37), which was used to show that (10.35) is an inconsistent family, also

shows that Y

is intrinsically inconsistent. The same is true of Y

, Y

,andY

. The

same basic strategy can be applied in certain cases which are at ﬁrst sight more

complicated; e.g., the histories in (13.19).

Examples of consistent families

12.1 Toy beam splitter

Beam splitters are employed in optics, in devices such as the Michelson and Mach–

Zehnder interferometers, to split an incoming beam of light into two separate

beams propagating perpendicularly to each other. The analogous situation in a

neutron interferometer is achieved using a single crystal of silicon as a beam split-

ter. The toy beam splitter in Fig. 12.1 can be thought of as a model of either an

optical or a neutron beam splitter. It has two entrance channels (or ports) a and b,

and two exit channels c and d. The sites are labeled by a pair mz, where m is an

integer, and z is one of the four letters a, b, c,ord, indicating the channel in which

the site is located.

The unitary time development operator is T = S

, where the action of the oper-

ator S

is given by

|mz=|(m + 1)z, (12.1)

with the exceptions:

|0a=



+|1c+|1d



√

|0b=



−|1c+|1d



√

(12.2)

The physical signiﬁcance of the states |0a, |1c, etc., is not altered if they are

multiplied by arbitrary phase factors, see Sec. 2.2, and this means that (12.2) is not

the only possible way of representing the action of the beam splitter. One could

equally well replace the states on the right side with



i|1c+|1d



√



|1c+i|1d



√

2, (12.3)

or make other choices for the phases. There are two other exceptions to (12.1) that

are needed to supply the model with periodic boundary conditions which connect

the c channel back into the a channel and the d channel back into the b channel (or

159

160 Examples of consistent families

−1a 0a

−1b

1d 2d 3d

Fig. 12.1. Toy beam splitter.

c into b and d into a if one prefers). It is not necessary to write down a formula,

since we shall only be interested in short time intervals during which the particle

will not pass across the periodic boundaries and come back to the beam splitter.

That S

is unitary follows from the fact that it maps an orthonormal basis of the

Hilbert space, namely the collection of all kets of the form |mz, onto another

orthonormal basis of the same space; see Sec. 7.2.

Suppose that at t = 0 the particle starts off in the state

|ψ

=|0a, (12.4)

that is, it is in the a channel and about to enter the beam splitter. Unitary time

development up to a time t > 0 results in

|ψ

=S

|ψ

=



|tc+|td



√

2 =|t ¯a, (12.5)

where

|m¯a :=



|mc+|md



√

2, |m

b :=



−|mc+|md



√

2 (12.6)

are the states resulting from unitary time evolution when the particle starts off in

|0a or |0b, respectively.

Let us consider histories involving just two times, with an initial state |ψ

=

|0a at t = 0, and a basis at some time t > 0 consisting of the states {|mz}, z = a,

b, c,ord, corresponding to a decomposition of the identity

I =



m,z

[mz]. (12.7)

12.1 Toy beam splitter 161

By treating |ψ

 as a pre-probability, see Sec. 9.4, one ﬁnds that

Pr([mc]

) = (1/2)δ

= Pr([md]

), (12.8)

while all other probabilities vanish; that is, at time t the particle will be either in the

c output channel at the site tc, or in the d channel at td. Here [mc] is a projector

onto the ray which contains |mc, and the subscript indicates the time at which the

event occurs.

If, on the other hand, one employs a unitary history, Sec. 8.7, in which at time

t the particle is in the state |t ¯a, one cannot say that it is in either the c or the d

channel. The situation is analogous to the case of a spin-half particle with an initial

state |z

 and trivial dynamics, discussed in Sec. 9.3. In a unitary history with

=+1/2 at a later time it is not meaningful to ascribe a value to S

, whereas by

using a sample space in which S

at the later time makes sense, one concludes that

=+1/2orS

=−1/2, each with probability 1/2.

The toy beam splitter is a bit more complicated than a spin-half particle, because

when we say that “the particle is in the c channel”, we are not committed to saying

that it is at a particular site in the c channel. Instead, being in the c channel or

being in the d channel is represented by means of projectors

C =



|mcmc|=



[mc], D =



[md]. (12.9)

Neither of these projectors commutes with a projector [m¯a] corresponding to the

state |m¯a deﬁned in (12.6), so if we use a unitary history, we cannot say that the

particle is in channel c or channel d. Note that whenever it is sensible to speak of

a particle being in channel c or channel d, it cannot possibly be in both channels,

since

CD= 0; (12.10)

that is, these properties are mutually exclusive. A quantum particle can lack a

deﬁnite location, as in the state |m ¯a, but, as already pointed out in Sec. 4.5, it

cannot be in two places at the same time.

The fact that the particle is at the site tc with probability 1/2 and at the site td

with probability 1/2 at a time t > 0, (12.8), might suggest that with probability

1/2 the particle is moving out the c channel through a succession of sites 1c,2c,

3c, and so forth, and with probability 1/2 out the d channel through 1d,2d, etc.

But this is not something one can infer by considering histories deﬁned at only two

times, for it would be equally consistent to suppose that the particle hops from 2c

to 3d during the time step from t = 2tot = 3, and from 2d to 3c if it happens

to be in the d channel at t = 2. In order to rule out unphysical possibilities of this

sort we need to consider histories involving more than just two times.

162 Examples of consistent families

Consider a family of histories based upon the initial state [0a] and at each time

t > 0 the decomposition of the identity (12.7), so that the particle has a deﬁnite

location. The histories are then of the form, for a set of times t = 0, 1, 2,... f ,

Y = [0a] ( [mz] ( [m



] (···[m



], (12.11)

with a chain operator of the form K(Y) =|φ0a|, Sec. 11.6, where the chain ket

|φ=|m



···m



|mzmz|S

|0a. (12.12)

From (12.2) it is obvious that the term mz|S

|0a is 0 unless m = 1 and z = c

or d, and given m = 1, it follows from (12.1) that m



|mz vanishes unless



= 2andz



= z. By continuing this argument one sees that |φ, and therefore

K(Y), will vanish for all but two histories, which in the case f = 4 are

= [0a] ( [1c] ( [2c] ( [3c] ( [4c],

= [0a] ( [1d] ( [2d] ( [3d] ([4d].

(12.13)

The fact that the ﬁnal projectors [4c] and [4d] in (12.13) are orthogonal to each

other means that the chain operators K(Y

) and K(Y

) are orthogonal, in accor-

dance with a general principle noted in Sec. 11.3. Since the chain operators of all

the other histories are zero, it follows that Y

and Y

form the support, as deﬁned in

Sec. 11.2, of a consistent family. It is straightforward to show, either by means of

chain kets as discussed in Sec. 11.6 or by a direct use of W(Y) =K(Y), K(Y),

that

W(Y

) = 1/2 = W(Y

), (12.14)

and hence, assuming an initial state of [0a] with probability 1, the two histories

and Y

each have probability 1/2, while all other histories in this family have

probability 0.

The fact that the only histories with ﬁnite probability are Y

and Y

means that

if the particle arrives at the site 1c at t = 1, it continues to move out along the c

channel, and does not hop to the d channel, and if the particle is at 1d at time t = 1,

it moves out along the d channel. Thus by using multiple-time histories one can

eliminate the possibility that the particle hops back and forth between channels c

and d, something which cannot be excluded by considering only two-time histories,

as noted earlier. A formal argument conﬁrming what is rather obviousfrom looking

at (12.13) can be constructed by calculating the probability

Pr(D

| [1c]

) = Pr(D

∧ [1c]

)/ Pr([1c]

) (12.15)

that the particle will be in the d channel at some time t > 0, given that it was at the

12.1 Toy beam splitter 163

site [1c]att = 1. Here D

is a projector on the history space for the particle to be

in channel d at time t. For example, for t = 2,

= I ( I ( D ( I ( I, (12.16)

and thus

∧ [1c]

= I ( [1c] ( D ( I ( I. (12.17)

This projector gives 0 when applied to either Y

or Y

, the only two histories with

positive probability, and therefore the numerator on the right side of (12.15) is 0.

Thus if the particle is at 1c at t = 1, it will not be in the d channel at t = 2. The

same argument works equally well for other values of t, and analogous results are

obtained if the particle is initially in the d channel. Thus one has

Pr(D

| [1c]

) = 0 = Pr(C

| [1d]

Pr(C

| [1c]

) = 1 = Pr(D

| [1d]

)

(12.18)

for any t ≥ 1, where C

is deﬁned in the same manner as D

, with C in place of D.

(Since we are considering a family which is based on the initial state [0a], the

preceding discussion runs into the technical difﬁculty that C

and D

do not belong

to the corresponding Boolean algebra of histories when the latter is constructed in

the manner indicated in Sec. 8.5. One can get around this problem by replacing

and D

with the operators C

∧[a0]

and D

∧[a0]

, and remembering that the

probabilities in (12.15) and (12.18) always contain the initial state [a0] at t = 0as

an (implicit) condition. Also see the remarks in Sec. 14.4.)

Another family of consistent histories can be constructed in the following way.

At the times t = 1, 2 use, in place of (12.7), a three-projector decomposition of the

identity

I = [t ¯a] + [t

b] + J

, (12.19)

where the states |t ¯a, |t

b are deﬁned in (12.6), and

= I − [t ¯a] − [t

b] = I − [tc] −[td] (12.20)

is a projector for the particle to be someplace other than the two sites tc or td.At

later times t ≥ 3 use the decomposition (12.7). It is easy to show that in the case

f = 4 the two histories

= [0a] ( [1¯a] ([2¯a] ( [3c] ( [4c],

= [0a] ( [1¯a] ([2¯a] ( [3d] ( [4d],

(12.21)

each with weight 1/2, form the support of the sample space of a consistent family;

all other histories have zero weight.

The histories

and

in (12.21) have the physical signiﬁcance that at t =