Griffiths R.B. Consistent Quantum Theory
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144 Consistent histories
Ch. 9 are consistent, which justifies our having omitted any discussion of consis-
tency when introducing the Born rule.
The same argument works if we consider a more general situation in which the
initial state is a projector "
0
onto a subspace which could have a dimension greater
than 1, and instead of an orthonormal basis we consider a general decomposition
of the identity in projectors
I =
k
P
k
(10.31)
at time t
1
. The family of two-time histories
Y
k
= "
0
( P
k
, Z = (I − "
0
) ( I (10.32)
is again consistent, since for k = l
K(Y
k
), K(Y
l
)∝Tr
"
0
T(t
0
, t
1
)P
k
P
l
T(t
1
, t
0
)"
0
= 0 (10.33)
because P
k
P
l
= 0, while K(Y
k
), Z=0 follows, as in (10.30), from cycling
operators inside the trace. (This argument is a special case of the general result in
Sec. 11.3 that any family based on just two times is automatically consistent.) The
probability of Y
k
is given by
Pr(Y
k
) = Tr
P
k
T(t
1
, t
0
)"
0
T(t
0
, t
1
)
/ Tr
(
"
0
)
, (10.34)
which we shall refer to as the generalized Born rule. The factor of 1/ Tr("
0
) is
needed to normalize the probability when "
0
projects onto a space of dimension
greater than 1.
Another situation in which the consistency conditions are automatically satis-
fied is that of a unitary family as defined in Sec. 8.7. For a given initial state such a
family contains one unitary history, (8.41), obtained by unitary time development
of this initial state, and various nonunitary histories, such as (8.42). It is straight-
forward to show that the chain operator for any nonunitary history in such a family
is zero, and that the chain operators for unitary histories with different initial states
(belonging to the same decomposition of the identity) are orthogonal to one an-
other. Thus the consistency conditions are satisfied. If the initial condition assigns
probability 1 to a particular initial state, the corresponding unitary history occurs
with probability 1, and zero probability is assigned to every other history in the
family.
To find an example of an inconsistent family, one must look at histories defined
at three or more times. Here is a fairly simple example for a spin-half particle. The
10.3 Examples of consistent and inconsistent families 145
five history projectors
Y
0
= [z
−
] ( I ( I,
Y
1
= [z
+
] ( [x
+
] ( [z
+
],
Y
2
= [z
+
] ( [x
+
] ( [z
−
],
Y
3
= [z
+
] ( [x
−
] ( [z
+
],
Y
4
= [z
+
] ( [x
−
] ( [z
−
]
(10.35)
defined at the three times t
0
< t
1
< t
2
form a decomposition of the history identity,
and thus a sample space of histories. However, for trivial dynamics, T = I, the
family is inconsistent. To show this it suffices to compute the chain operators using
(10.10). In particular,
K(Y
1
) =|z
+
|x
+
|
2
·|z
+
z
+
|,
K(Y
3
) =|z
+
|x
−
|
2
·|z
+
z
+
|
(10.36)
are not orthogonal, since |z
+
|x
+
|
2
and |z
+
|x
−
|
2
are both equal to 1/2; indeed,
K(Y
1
), K(Y
3
)=1/4. (10.37)
Similarly, K(Y
2
) and K(Y
4
) are not orthogonal, whereas K(Y
1
) is orthogonal
to K(Y
2
), and K(Y
3
) to K(Y
4
). In addition, K(Y
0
) is orthogonal to the chain
operators of the other histories. Since consistency requires that all pairs of chain
operators for distinct histories in the sample space be orthogonal, this is not a
consistent family.
On the other hand, the same five histories in (10.35) can form a consistent family
if one uses a suitable dynamics. Suppose that there is a magnetic field along the y
axis, and the time intervals t
1
− t
0
and t
2
− t
1
are chosen in such a way that
T(t
1
, t
0
) = T(t
2
, t
1
) = R, (10.38)
where R is the unitary operator such that
R|z
+
=|x
+
, R|z
−
=|x
−
,
R|x
+
=|z
−
, R|x
−
=−|z
+
,
(10.39)
where the second line is a consequence of the first when one uses the definitions
in (4.14). With this dynamics, Y
2
is a unitary history whose chain operator is
orthogonal to that of Y
0
, because of the orthogonal initial states, while the chain
operators for Y
1
, Y
3
, and Y
4
vanish. Thus the consistency conditions are satisfied.
That the family (10.35) is consistent for one choice of dynamics and inconsistent
for another serves to emphasize the important fact, noted earlier, that consistency
depends upon the dynamical law of time evolution. This is not surprising given
that the probabilities assigned to histories depend upon the dynamical law.
146 Consistent histories
A number of additional examples of consistent and inconsistent histories will
be discussed in Chs. 12 and 13. However, checking consistency by the process
of finding chain operators for every history in a sample space is rather tedious and
inefficient. Some general principles and various tricks explained in the next chapter
make this process a lot easier. However, the reader may prefer to move on to the
examples, and only refer back to Ch. 11 as needed.
10.4 Refinement and compatibility
The refinement of a sample space of histories was discussed in Sec. 8.6. In essence,
the idea is the same as for any other quantum sample space: some or perhaps all of
the projectors in a decomposition of the identity are replaced by two or more finer
projectors whose sum is the coarser projector. It is important to note that even if the
coarser family one starts with is consistent, the finer family need not be consistent.
Suppose that Z ={Z
β
} is a consistent sample space of histories, Y ={Y
α
} is a
refinement of Z, and that
Z
1
= Y
1
+ Y
2
. (10.40)
Then, by linearity,
K(Z
1
) = K(Y
1
) + K(Y
2
). (10.41)
When a vector is written as a sum of two other vectors, the latter need not be
perpendicular to each other, and, by analogy, there is no reason to suppose that the
terms on the right side of (10.3) are mutually orthogonal, K(Y
1
), K(Y
2
)=0, as
is necessary if Y is to be a consistent family. Another way in which Y may fail to
be consistent is the following. Since Y is a refinement of Z, any projector in the
sample space Z, for example Z
3
, belongs to the Boolean algebra generated by Y.
Because they represent mutually exclusive events, Z
3
Z
1
= 0, and because Y
1
and
Y
2
in (10.40) are projectors, this means that
Z
3
Y
1
= 0 = Z
3
Y
2
. (10.42)
In addition, the consistency of Z implies that
K(Z
3
), K(Z
1
)=K(Z
3
), K(Y
1
)+K(Z
3
), K(Y
2
)=0. (10.43)
However, this does not mean that either K(Z
3
), K(Y
1
) or K(Z
3
), K(Y
2
) is
zero, whereas (10.21) implies, given (10.42), that both must vanish in order for Y
to be consistent.
An example which illustrates these principles is the family Y whose sample
space is (10.35), regarded as a refinement of the coarser family Z whose sample
space consists of the three projectors Y
0
, Y
1
+Y
2
, and Y
3
+Y
4
. As the histories in
10.4 Refinement and compatibility 147
Z depend (effectively) on only two times, t
0
and t
1
, the consistency of this family is
a consequence of the general argument for the first example in Sec. 10.3. However,
the family Y is inconsistent for T(t
, t) = I.
In light of these considerations, we shall say that two or more consistent families
are compatible if and only if they have a common refinement which is itself a
consistent family. In order for two consistent families Y and Z to be compatible,
two conditions, taken together, are necessary and sufficient. First, the projectors
for the two sample spaces, or decompositions of the history identity, {Y
α
} and
{Z
β
} must commute with each other:
Y
α
Z
β
= Z
β
Y
α
for all α, β. (10.44)
Second, the chain operators associated with distinct projectors of the form Y
α
Z
β
must be mutually orthogonal:
K(Y
α
Z
β
), K(Y
ˆα
Z
ˆ
β
)=0ifα = ˆα or β =
ˆ
β. (10.45)
Note that (10.45) is automatically satisfied when Y
α
Z
β
= 0, so one only needs
to check this condition for nonzero products. Similar considerations apply in an
obvious way to three or more families. Consistent families that are not compatible
are said to be (mutually) incompatible.
11
Checking consistency
11.1 Introduction
The conditions which define a consistent family of histories were stated in Ch. 10.
The sample space must consist of a collection of mutually orthogonal projectors
that add up to the history identity, and the chain operators for different members of
the sample space must be mutually orthogonal, (10.20). Checking these conditions
is in principle straightforward. In practice it can be rather tedious. Thus if there are
n histories in the sample space, checking orthogonality involves computing n chain
operators and then taking n(n − 1)/2 operator inner products to check that they are
mutually orthogonal. There are a number of simple observations, some definitions,
and several “tricks” which can simplify the task of constructing a sample space of a
consistent family, or checking that a given sample space is consistent. These form
the subject matter of the present chapter. It is probably not worthwhile trying to
read through this chapter as a unit. The reader will find it easier to learn the tricks
by working through examples in Ch. 12 and later chapters, and referring back to
this chapter as needed.
The discussion is limited to families in which all the histories in the sample space
are of the product form, that is, represented by a projector on the history space
which is a tensor product of quantum properties at different times, as in (8.7). As
in the remainder of this book, the “strong” consistency conditions (10.20) are used
rather than the weaker (10.25).
11.2 Support of a consistent family
A sample space of histories and the corresponding Boolean algebra it generates
will be called complete if the sum of the projectors for the different histories in
the sample space is the identity operator
˘
I on the history Hilbert space, (8.23). As
noted at the end of Sec. 10.1, it is possible for the chain operator K(Y) to be zero
even if the history projector Y is not zero. The weight W(Y) =K(Y), K(Y) of
148
11.3 Initial and final projectors 149
such a history is obviously 0, so the history is dynamically impossible. Conversely,
if W(Y) = 0, then K(Y) = 0; see the discussion in connection with (10.13). The
support of a consistent family of histories is defined to be the set of all the histories
in the sample space whose weight is strictly positive, that is, whose chain operators
do not vanish. In other words, the support is what remains in the sample space if
the histories of zero weight are removed. In general the support of a family is not
complete, as that term was defined above, but one can say that it is dynamically
complete.
When checking consistency, only histories lying in the support need be consid-
ered, because a chain operator which is zero is (trivially) orthogonal to all other
chain operators. Using this fact can simplify the task of checking consistency in
certain cases, such as the families considered in Ch. 12. Zero-weight histories are
nonetheless of some importance, for they help to determine which histories, in-
cluding histories of finite weight, are included in the Boolean event algebra. See
the comments in Sec. 11.5.
11.3 Initial and final projectors
Checking consistency is often simplified by paying attention to the initial and final
projectors of the histories in the sample space. Thus suppose that two histories
Y = F
0
( F
1
(···F
f
,
Y
= F
0
( F
1
( ···F
f
(11.1)
are defined for the same set of times t
0
< t
1
< ···t
f
. If either F
0
F
0
= 0or
F
f
F
f
= 0, then one can easily show, by writing out the corresponding trace and
cyclingoperators around the trace, that K(Y), K(Y
)=0. Consequently, one can
sometimes tell by inspection that two chain operators will be orthogonal, without
actually computing what they are.
If the sample space consists of histories with just two times t
0
< t
1
, then the
family is automatically consistent. The reason is that the product of the history
projectors for two different histories in the sample space is 0 (as the sample space
consists of mutually exclusive possibilities). But in order that
(F
0
( F
1
) · (F
0
( F
1
) = F
0
F
0
( F
1
F
1
(11.2)
be 0, it is necessary that either F
0
F
0
or F
1
F
1
vanish. As we have just seen, either
possibility implies that the chain operators for the two histories are orthogonal. As
this holds for any pair of histories in the sample space, the consistency conditions
are satisfied.
For families of histories involving three or more times, looking at the initial
and final projectors does not settle the problem of consistency, but it does make
150 Checking consistency
checking consistency somewhat simpler. Suppose, forexample, we are considering
a family of histories based upon a fixed initial state "
0
(see Sec. 11.5), with two
possible projectors at the final time based upon the decomposition
t
f
: I = P +
˜
P. (11.3)
Then the sample space will consist of various histories, some of whose projectors
will have P at the final time, and some
˜
P. The chain operator of a projector with
a final P will be orthogonal to one with a final
˜
P. Thus we only need to check
whether the chain operators for the histories ending in P are mutually orthogonal
among themselves, and, similarly, the mutual orthogonality of the chain operators
for histories ending in
˜
P. If the decomposition of the identity at the final time t
f
involves more than two projectors, one need only check the orthogonality of chain
operators for histories which end in the same projector, as it is automatic when the
final projectors are different.
Yet another way of reducing the work involved in checking consistency can also
be illustrated using (11.3). Suppose that at t
f −1
there is a decomposition of the
identity of the form
t
f −1
: I =
m
Q
m
, (11.4)
and suppose that we have already checked that the chain operators for the different
histories ending in P are all mutually orthogonal. In that case we can be sure that
the chain operators for two histories with projectors
Y = "
0
(···Q
m
(
˜
P,
Y
= "
0
(···Q
m
(
˜
P
(11.5)
ending in
˜
P will also be orthogonal to each other, provided m
= m. The reason
is that by cycling operators around the trace in a suitable fashion one obtains an
expression for the inner product of the chain operators in the form
K(Y), K(Y
)=Tr
···Q
m
T(t
f −1
, t
f
)
˜
PT(t
f
, t
f −1
)Q
m
= Tr
···Q
m
Q
m
− Tr
···Q
m
T(t
f −1
, t
f
)PT(t
f
, t
f −1
)Q
m
, (11.6)
where ··· refers to the same product of operators in each case. The second line of
the equation is obtained from the first by replacing
˜
P by (I −P), using the linearity
of the trace, and noting that T(t
f −1
, t
f
)T(t
f
, t
f −1
) is the identity operator, see
(7.40). The trace of the product which contains Q
m
Q
m
vanishes, because m
= m
means that Q
m
Q
m
= 0. The final trace vanishes because it is the inner product of
the chain operators for the histories obtained from Y and Y
in (11.5) by replacing
˜
P at the final position with P; by assumption, the orthogonality of these has already
11.4 Heisenberg representation 151
been checked. Thus the right side of (11.6) vanishes, so the chain operators for Y
and Y
are orthogonal.
If the decomposition of the identity at t
f
is into n > 2 projectors, the trick just
discussed can still be used; however, it is necessary to check the mutual orthog-
onality of the chain operators for histories corresponding to each of n − 1 final
projectors before one can obtain a certain number of results for those ending in the
nth projector “for free”. If, rather than a fixed initial state "
0
, one is interested in
a decomposition of the identity at t
0
involving several projectors, there is an anal-
ogous trick in which the projectors at t
1
play the role of the Q
m
in the preceding
discussion.
11.4 Heisenberg representation
It is sometimes convenient to use the Heisenberg representation for the projectors
and the chain operators, in place of the ordinary or Schr
¨
odinger representation
which we have been using up to now. Suppose F
j
is a projector representing an
event thought of as happening at time t
j
.Wedefine the corresponding Heisenberg
projector
ˆ
F
j
using the formula
ˆ
F
j
= T(t
r
, t
j
)F
j
T(t
j
, t
r
), (11.7)
where the reference time t
r
is arbitrary, but must be kept fixed while analyzing a
given family of histories. In particular, t
r
cannot depend upon j. One can, for
example, use t
r
= t
0
, but there are other possibilities as well. Given a history
Y = F
0
( F
1
(···F
f
(11.8)
of events at the times t
0
< t
1
< ···t
f
, the Heisenberg chain operator is defined
by:
ˆ
K(Y) =
ˆ
F
f
ˆ
F
f −1
···
ˆ
F
0
= T(t
r
, t
f
)K(Y)T(t
0
, t
r
), (11.9)
where the second equality is easily verified using the definition of K(Y) in (10.5)
along with (11.7). Note that
ˆ
K(Y), like K(Y), is a linear function of its argument.
Now let Y
be a history similar to Y, except that each F
j
in (11.8) is replaced by
an event F
j
(which may or may not be the same as F
j
). Then it is easy to show that
K(Y
), K(Y)=
ˆ
K(Y
),
ˆ
K(Y)=Tr
ˆ
F
0
ˆ
F
1
···
ˆ
F
f
ˆ
F
f
ˆ
F
f −1
···
ˆ
F
0
. (11.10)
(Note that the inner product of the Heisenberg chain operators does not depend
upon the choice of the reference time t
r
.) Thus one obtains quite simple expres-
sions for weights of histories and inner products of chain operators by using the
Heisenberg representation. While this is not necessarily an advantage when doing
152 Checking consistency
an explicit calculation — time dependence has disappeared from (11.9), but one
still has to use it to calculate the
ˆ
F
j
in terms of the F
j
, (11.7) — it does make some
of the formulas simpler, and therefore more transparent. One disadvantage of using
Heisenberg projectors is that, unlike ordinary (Schr
¨
odinger) projectors, they do not
have a direct physical interpretation: what they signify in physical terms depends
both on the form of the operator and on the dynamics of the quantum system.
11.5 Fixed initial state
A family of histories for the times t
0
< t
1
< ···t
f
based on an initial state "
0
was
introduced in Sec. 8.5, see (8.30). Let us write the elements of the sample space in
the form
Y
α
= "
0
( X
α
, (11.11)
where for each α, X
α
is a projector on the space
¯
H of histories at times t
1
< t
2
<
···< t
f
, with identity
¯
I,and
α
X
α
=
¯
I, (11.12)
so that
α
Y
α
= "
0
(
¯
I. (11.13)
The index α may have many components, as in the case of the product of sample
spaces considered in Sec. 8.5. Since the Y
α
do not add up to
˘
I, we complete the
sample space by adding another history
Z = (I − "
0
) (
¯
I, (11.14)
as in (8.31).
The chain operator K(Z) is automatically orthogonal to the chain operators of all
of the histories of the form (11.11) because the initial projectors are orthogonal, see
Sec. 11.3. Consequently, thenecessary and sufficientcondition that the consistency
conditions are satisfied for this sample space is that
K("
0
( X
α
), K("
0
( X
β
)=0 for α = β. (11.15)
As one normally assigns "
0
probability 1 and
˜
"
0
probability 0, the history Z can
be ignored, and we shall henceforth assume that our sample space consists of the
histories of the form (11.11).
One consequence of (11.12) and the fact that the chain operator K(Y) is a linear
11.5 Fixed initial state 153
function of Y, (10.7), is that
α
K(Y
α
) = K("
0
(
¯
I) = T(t
f
, t
0
)"
0
. (11.16)
Of course, this is still true if we omit all the zero terms from the sum on the left
side, that is to say, if we sum only over histories in the support S of the sample
space (as defined in Sec. 11.2):
α∈S
K(Y
α
) = K("
0
(
¯
I) = T(t
f
, t
0
)"
0
. (11.17)
One can sometimes make use of the result (11.17) in the following way. Suppose
that we have found a certain collection S of histories of the form (11.11), repre-
sented by mutually orthogonal history projectors (that is, X
α
X
β
= 0ifα = β)
with nonzero weights. Suppose that, in addition, (11.17) is satisfied, but the X
α
for
α ∈ S do not add up to
¯
I, (11.12). Can we be sure of finding a set of zero-weight
histories of the form (11.11) so that we can complete our sample space in the sense
that (11.13) is satisfied? Generally there are several ways of completing a sample
space with histories of zero weight. One way is to define
X
=
¯
I −
α∈S
X
α
, Y
= "
0
( X
. (11.18)
Then, since
Y
+
α∈S
Y
α
= "
0
(
¯
I, (11.19)
it follows from the linearity of K, see (11.17), that
K(Y
) = 0. (11.20)
Consequently, Y
as defined in (11.18) is a zero-weight history of the correct type,
showing that there is at least one solution to our problem.
However, Y
might not be the sort of solution we are looking for. The point is
that while zero-weight histories never occur, and thus in some sense they can be
ignored, nonetheless they help to determine what constitutes the Boolean algebra
of histories, since this depends upon the sample space. Sometimes one wants to
discuss a particular item in the Boolean algebra which occurs with finite probabil-
ity, but whose very presence in the algebra depends upon the existence of certain
zero-weight histories in the sample space. In such a case one might need to use a
collection of zero-weight history projectors adding up to Y
rather than Y
by itself.
The argument which begins at (11.16) looks a bit simpler if one uses the Heisen-
berg representation for the projectors and the chain operators. In particular, since
ˆ
K("
0
(
¯
I) =
ˆ
"
0
, (11.21)