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% Roger Penrose
% Angular momentum: an approach to combinatorial space-time
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% in:
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% Quantum theory and beyond
% Essays and discussions arising from a colloquium
% edited by Ted Bastin
% Cambridge University Press 1971
% ISBN 0 521 07956 X
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% pages: 151-180
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\begin{document}
\begin{center}
{\large
ANGULAR MOMENTUM: AN APPROACH TO
COMBINATORIAL SPACE-TIME}
\footnotetext{This paper originally appeared in
\textsl{Quantum Theory and Beyond},
edited by Ted Bastin,
Cambridge University Press 1971,
pp.\ 151--180.}
\end{center}
\begin{center}
ROGER PENROSE
\end{center}
I want to describe an idea which is related to other things that were
suggested in the colloquium, though my approach will be quite
different. The basic theme of these suggestions have been to try to get
rid of the continuum and build up physical theory from discreteness.
The most obvious place in which the continuum comes into physics
is the structure of space-time. But, apparently independently of
this, there is also another place in which the continuum is built into
present physical theory. This is in quantum theory, where there is the
superposition law: if you have two states, you're supposed to be able to
form any linear combination of these two states. These are complex
linear combinations, so again you have a continuum coming in---namely
the two-dimensional complex continuum---in a fundamental way.
My basic idea is to try and build up both space-time and quantum
mechanics simultaneously---from \emph{combinatorial} principles---but not
(at least in the first instance) to try and change physical theory. In
the first place it is a \emph{reformulation}, though ultimately, perhaps, there
will be some changes. Different things will suggest themselves in
a reformulated theory, than in the original formulation. One scarcely
wants to take \emph{every} concept in existing theory and try to make it
combinatorial: there are too many things which look continuous in
existing theory. And to try to eliminate the continuum by \emph{approximating}
it by some discrete structure would be to change the theory.
The idea, instead, is to concentrate only on things which, in fact, \emph{are}
discrete in existing theory and try and use them as \emph{primary concepts}---then
to build up other things using these discrete primary concepts as
the basic building blocks. Continuous concepts could emerge in a
limit, when we take more and more complicated systems.
The most obvious physical concept that one has to start with,
where quantum mechanics says something is discrete, and which is
connected with the structure of space-time in a very intimate way, is
in \emph{angular momentum}. The idea here, then, is to start with the concept
of angular momentum---here one has a \emph{discrete} spectrum---and use
the rules for combining angular momenta together and see if in some
sense one can construct the concept of \emph{space} from this.
One of the basic ideas here springs from something which always
used to worry me. Suppose you have an electron or some other spin
$\frac{1}{2}\hbar$ particle. You ask it about its spin: is it spinning up or down? But
how does it know which way is `up' and which way is `down'? And
you can equally well ask the question whether it spins right or left.
But whatever question you ask it about directions, the electron has
only just \emph{two} directions to choose from. Whether the alternatives are
`up' and `down', or `right' and `left' depends on how things are
connected with the macroscopic world.
Also you could consider a particle which has \emph{zero} angular momentum.
Quantum mechanics tells us that such a particle has to be
spherically symmetrical. Therefore there isn't really any choice of
direction that the particle can make (in its own rest-frame). In effect
there \emph{is} only one `direction'. So that a thing with zero angular
momentum has just one `direction' to choose from and with spin one-half
it would have \emph{two} `directions' to choose from. Similarly, with spin
one, there would always be just three `directions' to choose from, etc.
Generally , there would be $2s + 1$ `directions' available to a spin $s$ object.
Of course I don't mean to imply that these are just directions in
space in the ordinary sense. I just mean that these are the choices
available to the object as regards its state of spin. That is, however
we may choose to interpret the different possibilities when viewed on
a macroscopic scale, the object itself is `aware' only that these are the
different possibilities that are open to it. Thus, if the object is in an
$s$-state, there is but one possibility open to it. If it is in a $p$-state there
are three possibilities, etc., etc. I don't mean that these possibilities
are things that from a macroscopic point of view we would necessarily
think of as directions in all cases. The $s$-state is an example of
a case where we would not!
So we oughtn't at the outset to have the concept of macroscopic
space-direction built into the theory. Instead, we ought to work with
just these discrete alternatives open to particles or to simple systems.
Since we don't want to think of these alternatives as referring to pre-existing
directions of a background space---that would be to beg the
question---we must deal only with \emph{total} angular momentum ($j$-value)
rather than spin in a direction ($m$-value).
Thus, the primary concept here has to be the concept of \emph{total
angular momentum} not the concept of angular momentum in, say,
the $z$-direction, because: which is the $z$-direction?
Imagine, then, a universe built up of things like that shown in fig.~1.
These lines may be thought of as the world-lines of particles. We can
view \emph{time} as going in one direction, say, from the bottom of the
diagram to the top. But it turns out, really, that it's irrelevant which
way time is going. So I don't want to worry too much about this.
I'm going to put a number on each line. This number, the \emph{spin-number}
%% Fig. 1
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-1.eps}\\
Fig.~1
\end{center}
will have to be an integer. It will represent twice the angular
momentum, in units of $\hbar$. All the information I'm allowed to know
about this picture will be just this diagram (fig.~2): the network of
connections and spin numbers 3, 2, 3, $\ldots$ like that. I should say
% The original indeed says "3, 2, 3", for some reason. - JB
%% Fig. 2
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-2.eps}\\
Fig.~2
\end{center}
that the picture I want to give here is just a model. Although
it does describe a type of \emph{idealized} situation exactly according to
quantum theory, I certainly don't want to suggest that the universe
`is' this picture or anything like that. But it is not unlikely that
\emph{some} essential features of the model that I am describing could still
have relevance in a more complete theory applicable to more
realistic situations.
I have referred to these line segments as representing, in some way,
the world-lines of particles. But I don't want to imply that these
lines stand just for \emph{elementary} particles (say). Each line could represent
some compound system which separates itself from other such
systems for long enough that (in some sense) it can be regarded as
isolated and stationary, with a well-defined total angular momentum
$n\times\frac{1}{2}\hbar$. Let us call such a system or particle an $n$-unit. (We allow
$n=0,1,2,\ldots$) For the precise model I am describing, we must also
imagine that the particles or systems are not moving relative to one
another. They just transfer angular momentum around, regrouping
themselves into different subsystems, perhaps annihilating one
another, perhaps producing new units. In the diagram (fig.~2), the
$3$-unit at the bottom on the left splits into a $2$-unit and another $3$-unit.
This second $3$-unit combines with a $1$-unit (produced in the break up of
a $2$-unit into two $1$-units) to make a new $2$-unit, etc., etc. It is only
the \emph{topological} relationship between the different segments, together
with the spin-number values, which is to have significance. The time-ordering
of events will actually play no role here (except conceptually).
We could, for example, read the diagram as though time
increased from the left to the right, rather than from the bottom to
the top, say.
Angular momentum conservation will be involved when I finally
give the rules for these diagrams. These rules, though combinatorial,
are actually derived from the standard quantum mechanics for
angular momentum. Thus, in particular, the conservation of total
angular momentum must be built into the rules.
Now, I want to indicate answers to two questions. First, what \emph{are}
these combinatorial rules and how are we to interpret them?
Secondly, how does this enable us to build up a concept of space out
of total angular momentum? In order not to get bogged down at this
stage with too much detail, I shall defer, until later on, the complete
definition of the combinatorial rules that will be used. All I shall say
at this stage is that every diagram, such as fig.~2 (called a \emph{spin-network})
will be assigned a non-negative integer which I call its \emph{norm}.
In some vague way, we are to envisage that the norm of a diagram
gives us a measure of the frequency of occurrence of that particular
spin-network in the history of the universe. This is not actually quite
right---I shall be more precise later---but it will serve to orient our
thinking. We shall be able to use these norms to calculate the probabilities
of various spin values occurring in certain simple `experiments'.
These probabilities will turn out always to be rational
numbers, arising from the fact that the norm is always an integer.
Given any spin-network, its norm can be calculated from it in a
purely combinatorial way. I shall give the rule later.
But first let me say something about the answer to the second
question. How can I say anything about \emph{directions} in space, when
I only have the non-directional concept of total angular momentum?
How do I get `$m$-values' out of $j$-values, in other words?
Clearly we can't do quite this. In order to know what the `$m$-value'
of an $n$-unit is, we would require knowledge of which direction in
space is the `$z$-direction'. But the `$z$-direction' has no physical
meaning. Instead, we may ask for the `orientation' of one of our
$n$-units \emph{in relation to some larger structure} belonging to the system
under consideration. We need some larger structure which in fact
does give us something that we may regard as a well-defined `direction
in space' and which could serve in place of the `$z$-direction'. As
we have seen, a structure of spin zero, being spherically symmetrical,
is no good for this; spin $\frac{1}{2}\hbar$ is not much better; spin $\hbar$ only a little
better; and so on. Clearly we need a system involving a fairly large
total angular momentum number if we are to obtain a reasonably
well-defined `direction' against which to test the `spin direction' of
the smaller units. We may imagine that for a large total angular
momentum number $N$, we have the potentiality, at least, to define
a well-defined direction as the \emph{spin axis} of the system. Thus, if we
\emph{define} a `direction' in space as something associated with an $N$-unit
with a large $N$ value (I call this a \emph{large unit}), then we can ask how to
define \emph{angles} between these `directions'. And if we can decide on a
good way of measuring angles, we can then ask the question whether
the angles we get are consistent with an interpretation in terms of
directions in a Euclidean three-dimensional space, or perhaps in some
other kind of space.
How, then, are we to define an angle between two large units?
Well, we do this by performing an `experiment'. Suppose we detach
a $1$-unit (e.g. an electron, or any other spin $\frac{1}{2}\hbar$-particle) from a large
$N$-unit in such a way as to leave it as an ($N-1$)-unit. We can then
re-attach the $1$-unit to some other large unit, say an $M$-unit. What
does the $M$-unit do? Well (according to the rules we are allowed here)
it can either become an ($M-1$)-unit or an ($M+1$)-unit. There will be
a certain probability of one outcome and a certain probability of the
other. Knowing these probability values, we shall have information
as to the angle between the $N$-unit and the $M$-unit. Thus, if our two
units are to be `parallel', we would expect zero probability for the
$M-1$ value and certainty for the $M + 1$ value. If the two units are to
be `anti-parallel' we would expect exactly the reverse probabilities.
If they are `perpendicular', then we would expect equal probability
values of $\frac{1}{2}$, for each of the two outcomes. Generally, for an angle $\theta$
between the directions of the two large units we would expect a
%% Fig. 3
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-3.eps}\\
Fig.~3
\end{center}
probability $\frac{1}{2}-\frac{1}{2}\,\cos\theta$ for the $M$-unit to be reduced to an ($M-1$)-unit
and a probability $\frac{1}{2}+\frac{1}{2}\,\cos\theta$ for it to be increased to an ($M+1$)-unit.
Let me draw a diagram to represent this experiment (fig.~3). Here
$\kappa$ represents some known spin-network. By means of a precise
(combinatorial) calculational procedure---which I shall describe
shortly---we can calculate, from knowledge of the spin-network $\kappa$, the
probability of each of the two possible final outcomes. Hence, we
have a way of getting hold of the concept of Euclidean angle, starting
from a purely combinatorial scheme.
As I remarked earlier, these probabilities will always turn out to be
\emph{rational} numbers. You might think, then, that I could only obtain
angles with rational cosines in this way. But this would be a somewhat
misleading way of viewing the situation. With a finite spin-network
with finite spin-numbers, the angle can never be quite well-enough
defined. I \emph{can} work out numerical values for these `cosines
of angles' for a finite spin-network, but these `angles' would normally
not quite agree with the actual angle of Euclidean space until I go to
the limit.
The view that I am expressing here is that rational probabilities
are to be regarded as something which can be more primitive than
ordinary real number probabilities. I don't need to call upon the full
continuum of probability values in order to proceed with the theory.
A rational probability $p=m/n$ might be thought of as arising because
the universe has to make a choice between $m$ alternative possibilities
of one kind and $n$ alternative possibilities of another---all of which are
to be equally probable. Only in the limit, when numbers go to infinity
do we expect to get the full continuum of probability values.
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-4.eps}\\
Fig.~4
\end{center}
As a matter of fact, it was this question of rational values for
primitive probabilities arising in nature, which really started me off
on this entire line of thought concerning spin-networks, etc. The idea
was to find some situation in nature which one might reasonably
regard as giving rise to a `pure probability', I am not really sure
whether it is fair to assume that `pure probabilities' exist in nature,
but by these I mean probabilities (necessarily quantum mechanical)
whose values are determined by nature alone and not in principle
influenced by our ignorance of initial conditions, etc. I suppose
I might have thought of branching ratios in particle decays as a
possible example. Instead, I was led to consider a situation of the
following type.
Two spin zero particles each decay into pairs of spin $\frac{1}{2}\hbar$ particles.
Two of the spin $\frac{1}{2}\hbar$ particles then come together, one from each pair,
and combine to form a new particle (fig.~4). What is the spin of this
new particle? Well, it must be either zero or $\hbar$, with respective probabilities
$\frac{1}{4}$ and $\frac{3}{4}$ (assuming no orbital components contribute, etc.).
Although you can see that there are objections even here to regarding
this as giving a `pure probability', at least the example served as a
starting point. (This example was to some extent stimulated by
Bohm's version of the Einstein-Rosen-Podolsky thought experiment,
which it somewhat resembles.) The idea, then, is that any
`pure probability' (if such exists) ought to be something arising ultimately
out of a choice between equally probable alternatives. All
`pure probabilities' ought therefore, to be rational numbers.
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-5.eps}\\
Fig.~5
\end{center}
But let me leave all this aside since it doesn't affect the rest of the
discussion. Actually, I haven't quite finished my `angle measuring
experiment', so let me return to this.
Let us consider the following particular situation. Suppose we have
a number of disconnected systems, each producing a large $N$-unit.
There are to be absolutely no connections between them (fig.~5).
%% Fig 6
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-6.eps}\\
Fig.~6
\end{center}
Let me try to measure the `angle' between two of them by doing one
of the `experiments' I described earlier. I detach a $1$-unit from one
of the $N$-units in such a way as to leave it as an ($N - 1$)-unit. Then
I reattach the $1$-unit to one of the other $N$-units (fig.~6). According
to the rules (cf.\ later) it will follow that the probability of the second
$N$-unit to become an ($N\pm1$)-unit is $\frac{1}{2}\,(N + 1 \pm 1)/(N+1)$. These two
probabilities become equal in the limit $N \rightarrow \infty$. Thus, if we are to
assign an `angle' between these units, then, for $N$ large, this would
have to be a \emph{right-angle}. This is just using the probability blindly.
I would similarly have to say, of any other pair of the $N$-units, that
they are at right-angles. It would seem that I could put any number
of $N$-units at right-angles to each other. In this instance I have
drawn five. Does this mean that we get a five-dimensional space?---or
an $\infty$-dimensional space?
Clearly I have not done things quite right. There are no connections
between any of the $N$-units here, so one would like to think of the
probabilities that arise out of one of these experiments as being not
\emph{just} due to the \emph{angle} between the $N$-units (if they have
an angle in some
sense), but also due to the `ignorance' implicit in the set-up. That is,
we think of the probabilities as arising in two different ways. In the
first instance, probabilities can arise in this type of experiment, if we
have a \emph{definite angle} between two spinning bodies (as we have seen).
These are the genuine quantum mechanical probabilities. But, in the
second instance, we may just be \emph{uncertain} as to what the angle is
between the two bodies. This lack of knowledge, concerning the
history (or origins) of the two bodies, will give us a contribution to the
probability value---an \emph{ignorance} factor---which will serve to obscure
the meaning of the probability in terms of angles. In the present
instance, we are allowed absolutely no information concerning the
interconnections between the different $N$-units, so the probability is
%%Fig. 7
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-7.eps}\\
Fig.~7
\end{center}
not really due to `angle' at all. In this extreme case, the probability
is \emph{entirely} `ignorance factor'. In general, the two effects will be mixed
up, so we shall need a means of separating them.
Let me change the picture a bit. I'll put in some `known' connecting
network (now denoted by $\kappa$) and have two large units coming
out, as in fig.~7. I do one of these experiments, but then \emph{repeat} the
experiment. Suppose the $N$-unit is reduced to an ($N-1$)-unit and
then to an ($N-2$)-unit. The $M$-unit becomes an ($M\pm1$)-unit and
then an ($M \pm 2$)-unit, or an $M$-unit again (fig.~8). The question is:
%% Fig. 8
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-8.eps}\\
Fig.~8
\end{center}
is the probability of the second experiment influenced by the result
of the first experiment? If this is essentially an `ignorance' situation,
where one doesn't initially know such about how the spin axes are
pointing, then the \emph{result} of the first experiment provides us with some
\emph{information} as to the relative directions of the spin axes. Therefore,
the probabilities in the second experiment will be altered by the
knowledge of the result of the first experiment.\footnote{
It should be borne in mind that all these probability values are simply calculated
here, from knowledge of the spin-networks involved. The `experiments' are really
theoretical constructions. However, it would be possible (in principle---with the
usual reservations) to measure these probabilities experimentally, by simply
repeating the experiment many times, each time reconstructing the spin-network
afresh.}
If the probabilities calculated for the second experiment are \emph{not}
substantially altered by the knowledge of the result of the first experiment,
then I say that the angle between the two large units is essentially
well-defined. If they \emph{are} substantially altered by the results of the
first experiment, then there is a large `ignorance' factor involved, and
the probabilities arise not just from `angle'.
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-9.eps}\\
Fig.~9
\end{center}
Suppose, now, we have the system shown in fig.~9, which has a
number of large units emerging, and suppose that it happens to be
the case that the angle between any two of them is well-defined in
the sense I just described. (All the numbers $A,B,\ldots$ are large compared
with unity; I can do a few odd experiments which do not much
change these numbers.) Then there is a theorem which can be proved
to the effect that these angles are all consistent with angles between
directions in Euclidean three-dimensional space.
Now, should I be in any way surprised by this result? Admittedly
I should have been surprised if the method gave me any different
space; but on the other hand, it is not completely clear to me that the
result is something I could genuinely have inferred beforehand. Let
me mention a number of curious features of the theory in this context.
In the first place, suppose I set the situation up with wave functions
and everything, and work according to ordinary quantum mechanical
rules. I have these particles (or systems) with large angular momentum,
and I finally find out that I get these angles consistent with
directions in Euclidean three-dimensional space. I never, at any
stage, specified that these large angular momentum systems should,
in fact, correspond to bodies which do have well-defined directions
(as rotation axes). There are states with large total angular momentum
(e.g.\ $m = 0$ states) which point all over the place, not necessarily in
any one direction.
I can start from some \emph{given} Euclidean $3$-space and use an ordinary
Cartesian description in terms of $x,y,z$. I can use particles (or
systems) with large total angular momentum, but which do \emph{not}
happen to give well-defined directions in the original space. Then
I work out the `angles' between them and find that these angle do
not correspond to anything I can see as angles in the original description,
but they are nevertheless consistent with the angle between
directions in \emph{some} abstract Euclidean three-dimensional space.
I therefore take the view that the Euclidean three-dimensional space
that I get out of all this, using probabilities, etc. is the \emph{real space}, and
that the original space, with its $x,y,z$'s that I wrote down, is an
irrelevant convenience, like co-ordinates in general relativity, where
one writes down any co-ordinates which don't necessarily mean anything.
The central idea is that \emph{the system defines the geometry}. If you
like, you can use the conventional description to fit the thing into the
`ordinary space-time' to begin with, but then the geometry you get
out is not necessarily the one you put into it. So I don't know whether
I should be surprised or not by the fact that I actually get the right
geometry in the end.
There is a second aspect of this work that I think I regarded as
slightly surprising at first. This is the fact that although no complex
numbers are ever introduced into the scheme, we can still build up
the full \emph{three}-dimensional array of directions, rather than, say, a
two-dimensional subset. To represent all possible directions as states
of spin of a spin $\frac{1}{2}\hbar$ particle, we need to take \emph{complex} linear combinations
(in the conventional formalism). Here we only use rational
numbers---and complex numbers cannot be approximated by rational
numbers alone! Again, the answer seems to be that the space I end up
with is not really the `same' space as the ($x, y, z$)-space that I could
start with---even though both are Euclidean $3$-spaces.
One might ask whether corresponding rules might be invented
which lead to other dimensional schemes. I don't in fact see \emph{a priori}
why one shouldn't be able to invent rules, similar to the ones I use,
for spaces of other dimensionality. But I'm not quite sure how one
would do this. Also, it's not obvious that the whole scheme for getting
the space out in the end would still work. The rules I use are derived
from the irreducible representations of SO(3). These have some rather
unique features.
Now, from what I've said so far, you might wonder whether you
would just scatter the numbers on the network at random. Actually,
you can if you like, but unless you are a bit careful the resulting spin-network
will have zero norm. And if the norm is zero, then the situation
represented by the spin-network a not realizable (i.e.\ zero
probability) according to the rules of quantum mechanics.
There are, in fact, two simple necessary requirements which must
be satisfied at each vertex of a spin-network, for its norm to be non-zero.
Notice first that all the spin-networks that I have explicitly
drawn have the property that precisely \emph{three} edges (i.e.\ units) come
together at each vertex. (This isn't one of the `requirements' I am
referring to. It's just that I don't know how one would handle more
general types of vertex within the scheme.) Suppose we have a vertex
at which an $a$-unit, a $b$-unit and a $c$-unit come together (fig.~10).
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-10.eps}\\
Fig.~10
\end{center}
Then for a spin-network containing the vertex to have a non-zero
norm, it is necessary that the triangle inequality hold:
\[
a+b+c \ge 2 \, \textrm{max}(a, b, c) \quad ;
\]
and furthermore that there be conservation of fermion number
(mod 2):
\[
a+b+c \quad \textrm{\emph{is even}.}
\]
These are, of course, properties that one would want to hold in real
physical processes, with the interpretations that I have given to the
spin-networks.
But even if these requirements hold at every vertex, the spin-network
may still have zero norm. For example, each of the two types
of spin-network shown in fig.~11 has zero norm, where $n \ne 0$ in the
first case and $n \ne m$ the second. In each case, the shaded portion
represents some spin-network with no other free ends. In fact, the
%fig 11, fig 12
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-11.eps}\\
Fig.~11
\end{center}
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-12.eps}\\
Fig.~12
\end{center}
first is effectively a special case of the second, with $m = 0$. This is
because any $0$-unit can be omitted from a spin-network (if we also
suitably delete the relevant vertices) without changing the norm. We
may interpret the vanishing of the norm whenever $n \ne m$ in the
second case as an expression of \emph{conservation of total angular momentum}.
In addition to these cases, there are many particular spin-networks
which turn out to have zero norm. One example is shown in fig.~12.
But so far I have only been giving particular cases. Let us now pass
to the general rule.
I shall give the definition of the norm in terms of a closely related
concept, namely, what I shall call the \emph{value} of a closed oriented spin-network.
I call a spin-network \emph{closed} if it has no free ends (e.g.\
analogous to a disconnected vacuum process). A spin-network which
is not closed will not be assigned a value. The definition of \emph{orientation}
for a spin-network is a little difficult to give concisely. Any spin-network
can be assigned two alternative orientations. Fixing the
orientation of a closed spin-network will serve to define the \emph{sign} of
its value (which can be positive or negative). Roughly speaking the
orientation assigns a cyclic order to the three units attached to each
vertex---but if we reverse the cyclic order at any \emph{even} number of
vertices this is to leave the orientation unchanged. The orientation
will change, on the other hand, if the cyclic order is reversed at an
\emph{odd} number of vertices.
I shall adopt the convention, when drawing spin-networks, that
the orientation is to be fixed by the way that the spin-network is
depicted on the plane. At each vertex we specify `\emph{counter-clockwise}'
as the cyclic order for the three units attached to the vertex. This
defines the spin-network's orientation. The diagrams in fig.~13
illustrate an example of a closed spin-network with its two possible
orientations.
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-13.eps}\\
Fig.~13
\end{center}
It will also be convenient to use the representation of a spin-network
as a drawing on a plane, in order to keep track of signs
properly when defining the value. This may have the effect of making
the definition \emph{seem} less `combinatorial' than it really is. Of course, the
definition could be reformulated without the use of such a drawing
if desired. Consider, then, a closed spin-network $\alpha$ depicted as a
%%fig 14
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-14.eps}\\
Fig.~14
\end{center}
drawing on a plane. Now, imagine each $n$-unit to be replaced in the
drawing by $n$ parallel strands. At each vertex, the strand ends must
be connected together in pairs, but no two strands associated with the
same $n$-unit are to be connected together. Let us call such a connection
scheme a \emph{vertex connection}. One such vertex connection is illustrated
in fig.~14, while fig.~15 shows a non-allowable connection,
%% Fig. 15
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-15.eps}\\
Fig.~15
\end{center}
since two strands of the $7$-unit are connected to one another. The
\emph{sign} of a vertex connection is defined most simply as $(-1)^x$ where
$x$ is the number of intersection points between different strands at
the vertex, as drawn on the plane. (These intersection points must be
counted correctly if more than two strands cross at a point, or if two
strands touch: and ignored if a strand crosses itself. It is simplest on
the other hand, just to avoid such features by drawing the strands
in general position and not allowing any strand connection to cross
itself.) The sign of a vertex connection, in fact, does not depend on
the details of how it is drawn, but only on the pairing off of the
strands. The allowable vertex connection depicted above has $-1$ as
its sign, since there are thirteen crossing points.
When the vertex connections have been completed at every vertex
of a closed spin-network, then we shall have a number of \emph{closed loops},
with no open-ended strands remaining. Consider, now, every possible
way of allowably completing the vertex connection for the spin-network
$\alpha$. We form the expression
\[
\textrm{value of }\alpha = \frac{\sum\pm(-2)^c}{\prod n!}
\]
where the summation extends over all possible completed allowable
connection schemes, where the `$\pm$' stands for the product of the
signs of all the vertex connections, where $c$ is the number of closed
loops resulting from the vertex connections and where the product in
the denominator ranges over all the units of the spin-network,
$n$ being the spin-number of the unit. The value of any closed spin-network
always turns out to be an integer.
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-16.eps}\\
Fig.~16
\end{center}
\begin{eqnarray}
\rightarrow\,\textrm{value}
&=& \frac{1}{2!\,1!\,1!}\,\{+(-2)-(-2)^2-(-2)^2+(-2)\} \nonumber \\
&=& -6. \nonumber
\end{eqnarray}
Let us consider a simple example, given in fig.~16. Note that the
`accidental' intersection, arising from the crossing of the two $1$-units
in the first drawing of the spin-network, does not contribute to the
sign of the terms in the sum. Only the intersections at the vertex
connections count.
The definition of the value of a closed oriented spin-network that
I have just given is perhaps the simplest to state, but it is by no
means the most useful to use in actual calculations. When the spin-networks
become even slightly more complicated than the simple one
evaluated above, the detailed calculations can become very unwieldy.
A more useful procedure is to employ certain reduction formulae
which can be used to express complicated networks in terms of
simpler ones.\footnote{
Diagrams closely related to spin-networks were introduced by Ord-Smith and
Edmonds for the graphical treatment of quantum mechanical angular momentum.
(See reference (l) for a detailed discussion.)}
For this purpose, it will be necessary to introduce
a slight variation of the spin-network theme; I shall consider the
related concept of a \emph{strand-network}.
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-17.eps}\\
Fig.~17
\end{center}
A strand-network is a series of connections relating objects (which
I shall still refer to as $n$-units) an example of which is depicted in fig.~17.
The units are `tied together' at various places, as indicated by the
thick bar. Any spin-network can be translated into strand-network
terms, by replacing each vertex according to the scheme shown in
fig.~18. I thus introduce three more (`virtual') units at each vertex.
A strand-network is \emph{closed} if it has no free ends. Any closed (oriented)
strand-network will have a \emph{value} which is an integer (positive,
negative or zero). This value will be chosen to agree with that defined
for a spin-network, in the case of closed strand-networks obtained by
means of the above replacement. Generally, to obtain the value of
s closed strand-network $\beta$ we employ the same formula as before:
\[
\textrm{value of }\beta = \frac{\sum\pm(-2)^c}{\prod n!}
\]
% Fig. 18
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-18.eps}\\
Fig.~18
\end{center}
where now the `$\pm$' sign refers to the product of all the signs of all the
permutations involved in each strand connection at which the strands
come together. For example, one possible connection scheme for
% Fig. 19
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-19.eps}\\
Fig.~19
\end{center}
`strand vertex' of fig.~19, would be that shown in fig.~20. This connection
scheme would contribute a minus sign, since an odd permutation
is involved. (There are nine crossing points---this is essentially
% Fig. 20
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-20.eps}\\
Fig.~20
\end{center}
the `Aitken diagram' method of determining the sign of a permutation.)
Notice that for a connection scheme to be possible at all, we
require that the total of the spin-numbers entering at one side must
equal the total of the spin-numbers leaving at the other. This \emph{one}
requirement now takes the place of the `triangle inequality' and
`fermion conservation' that we had earlier.
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-21.eps}\\
Fig.~21
\end{center}
\begin{eqnarray}
% ugly LaTeX-hack for AMS's \therefore
\begin{array}{c} {}_\cdot \\ {}^{\cdot\;\,\cdot}\end{array}\,
\textrm{value}
&=& \frac{1}{1!\,1!\,1!\,1!}\,\{+(-2)^2-(-2)-(-2)+(-2)\} \nonumber \\
&=& 2. \nonumber
\end{eqnarray}
Let us evaluate the simple closed strand-network of fig.~21 as an
example. Again there is an `accidental' intersection depicted (where
two $1$-units cross) which does not contribute to the sign of the terms
in the sum.
Let me list a number of relations and reduction formulae which are
useful in evaluating strand-networks (fig.~22). (I am not going to
prove anything here, but most of the relations are not hard to verify.)
These relations may be substituted into any closed strand-network
and a valid relation between values is obtained. Finally, let me make
the remark that the value is \emph{multiplicative}, that is to say, the value
of the union of two \emph{disjoint} strand-networks or spin-networks is equal
to the product of their individual values.
I now come to the definition of the \emph{norm} of a spin-network.
A strand-network will likewise have a norm. This is simply obtained
by drawing \emph{two} copies of the spin-network (or strand-network),
joining together the corresponding free end units to make a closed
network, and then taking the modulus of the value of this resulting
closed network. As an example, the norm of the spin-network
consisting of a single vertex is found in fig.~23. An even simpler
example, depicted in fig.~24, is the norm of a single isolated $n$-unit.
Finally, let me describe how the norm may be used in the calculation
of \emph{probabilities} for spin-numbers, in the type of `experiment' that
we have been considering. (Again I shall give no proofs.) Suppose we
start with a spin-network $\alpha$, with an $a$-unit and a $b$-unit among its
% Fig 22
% Fig 23
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-22.eps}\\
Fig.~22
\end{center}
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-23.eps}\\
Fig.~23
\end{center}
free ends (fig.~25). Suppose the $a$-unit and the $b$-unit come together
to form an $x$-unit, the resulting spin-network being denoted by $\beta$ (see
fig.~26). We wish to know (given $\alpha$) what are the various probabilities
for the different possible values of the spin-number $x$. Let $\gamma$ denote
the spin-network representing the coming together of the $a$-unit and
%Fig. 24
%Fig. 25
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-24.eps}\\
Fig.~24
\end{center}
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-25.eps}\\
Fig.~25
\end{center}
the $b$-unit to form the $x$-unit. Let $\xi$ denote the spin-network
consisting
of the $x$-unit alone. These are illustrated in fig.~27. Then the
required probability for the resulting spin-number to be $x$ is
\[
\textrm{probability} = \frac{\textrm{norm}\beta\,\textrm{norm}\xi}
{\textrm{norm}\alpha\,\textrm{norm}\gamma}\;.
\]
Using the explicit expressions for norm $\gamma$ and norm $\xi$ that were just
given as examples (using a slightly different notation), we can rewrite
this as
\[
\textrm{probability}
= \frac{\textrm{norm}\beta\,
(x+1)\,\{\frac{1}{2}(a+b+x)+1\}!}
{\textrm{norm}\alpha\,\{\frac{1}{2}(a+b-x)\}!\,
\{\frac{1}{2}(b+x-a)\}!\,
\{\frac{1}{2}(x+a-b)\}!}
\]
From the combinatorial nature of the definition of norm, it is clear
that these probabilities must all be rational numbers. And with the
interpretation of `angle' that I have given, the three-dimensional
Euclidean nature of the `directions in space' that are obtained, is
a \emph{consequence} of these combinatorial probabilities.
\begin{center}
\includegraphics[width=20pc,keepaspectratio]{fig-26.eps}\\
Fig.~26
\end{center}
\begin{center}
\includegraphics[width=30pc,keepaspectratio]{fig-27.eps}\\
Fig.~27
\end{center}
I should emphasize again that the space that I get out in the end
is the one \emph{defined} by the system itself and is not really the same space
as the one that might have been introduced at the start if a conventional
formalism had been used. Thus, although undoubtedly the
reason that we end up with directions in a Euclidean three-dimensional
space is intimately \emph{related} to the fact that we start with representations
of the rotation group SO(3), the precise logical connections are not
at all clear to me. When I come to consider the generalization of all
this to a relativistic scheme in a moment, this question will again
present itself. I shall also need to consider the spatial locations of
objects, not just their orientations. My model works with objects and
the interrelations between objects. An object is thus `located', either
directionally or positionally in terms of its relations with other
objects. One doesn't really need a space to begin with. The notion of
space comes out as a \emph{convenience} at the end.
Essentially, I have so far been using a non-relativistic scheme.
The angular momentum is not relativistic angular momentum. From
a four-dimensional viewpoint, the directions I get are those orthogonal
to a given timelike direction, i.e.\ directions in three-dimensional
space. All the particles are going along in this same timelike direction.
Perhaps they can knock each other a little bit, but they are not really
moving very much. They just transfer angular momentum backwards
and forwards. All the particles are, strictly speaking in the
same place, not moving relative to one another. Consequently, one
does not have any problem of mixing between orbital and spin
angular momentum. Once I allow orbital contributions to come in,
then I must drastically change the scheme, since now not only is the
question of `direction' and `angle' involved, but so also is `position'
and `distance'. Thus, if one thinks of \emph{real} particles moving relative
to each other, then there is the problem not only of doing things
relativistically, but also of bringing in actual displacements between
particles. Consider two particles in relative motion. Suppose they
come together and combine to form a system with a well-defined
total spin. Then to obtain the spin of the combined system, we
cannot just add up the individual spins because we have to bring in
the orbital component. There is a mixture of the actual displacements
in space with the angular momentum concept. So I spent a
long time thinking how one should combine rotations and displacements
together into an appropriate relativistic scheme. Eventually
I was led to consider a certain algebra for space-time which treats
linear displacements on the same footing as it treats rotations. Thus,
linear momentum is treated on a similar footing to angular
momentum.
Now you might raise the objection that linear momentum has a
continuous spectrum, while it is only for angular momentum that one
has discreteness. This is a problem of some significance to us. My
answer to it is roughly the following: each particle has its own
discrete spectrum for its angular momentum. When two particles are
considered together as a unit, then again there is a discrete spectrum
for the combined system. The way these `spins' add up implicitly
brings in the relative motion between the two particles. So the
momentum is brought in through the back door, in a sense, for one
could be always talking in terms of `bound systems'.
I consider momentum states as being linear combinations of
angular momentum states. There is indeed a problem to see how this
continuous momentum should be built up from something discrete,
but, in principle, there is nothing against it. In effect, the idea is that
the momentum should be brought in indirectly. I would propose that,
in a sense, there should \emph{not} be a well-defined distinction between
momentum and angular momentum---except in the limit. Individual
particles and simple systems would not really `know' what momentum
is. Like the idea of `direction' that I considered earlier, it would be
only in the limit of large systems that the concept of momentum
really attains a well-defined meaning. Smaller systems might retain
a combined concept of momentum and angular momentum, but these
things would only sort themselves out properly in the limit.
The algebra I have used to treat linear displacements and rotations
together, or linear and angular momentum together, I call the
algebra of \emph{twistors}. I have used the term `twistor' to denote a `spinor'
for the six-dimensional ($+\,+\,-\,-\,-\,-$) pseudo-orthogonal group
O(2,4). The twistor group is the ($+\,+\,-\,-$) pseudo-unitary group
SU(2,2), which is locally isomorphic with O(2,4). In turn, O(2,4) is
locally isomorphic with the fifteen-parameter (local) conformal group
of space-time. Under a conformal transformation of space-time, the
twistors will transform (linearly) according to a representation of the
group SU(2,2).
The basic twistor is a four-complex-dimensional object. We can
thus describe it by means of four complex components $Z^\alpha$:
\[
(Z^\alpha) = (Z^0,\,Z^1,\,Z^2,\,Z^3)\;.
\]
The complex conjugate of the twistor $Z^\alpha$ is an object $\bar{Z}_\alpha$ whose
components
\[
(\bar{Z}_\alpha) = (\bar{Z}_0,\,\bar{Z}_1,\,\bar{Z}_2,\,\bar{Z}_3)
\]
are given (according to a convenient co-ordinate system) by
\[
\bar{Z}_0 = \overline{(Z^2)}, \quad
\bar{Z}_1 = \overline{(Z^3)}, \quad
\bar{Z}_2 = \overline{(Z^0)}, \quad
\bar{Z}_3 = \overline{(Z^1)}.
\]
This implies that the Hermitian form $Z^\alpha\,\bar{Z}_\alpha$ (summation convention
assumed) has signature ($+\,+\,-\,-$), which is required in order to
give SU(2,2). (I have already described these objects$^{(2)}$ and their
geometrical significance in Minkowski space-time, and a later paper$^{(3)}$
goes into some further developments, including some of the quite
surprising aspects of the theory which arise when one starts to
describe physical fields in twistor terms.)
When $Z^\alpha\,\bar{Z}_\alpha=0$ I call $Z^\alpha$ a \emph{null} twistor. A null twistor $Z^\alpha$ has
a very direct geometrical interpretation in space-time terms. In fact,
$Z^\alpha$ defines a \emph{null straight line}, which we can think of as the world line
of a zero rest-mass particle. (An important aspect of twistor theory
is that \emph{zero rest-mass} is to be regarded as more fundamental than
a finite rest-mass. Finite mass particles are viewed as composite
systems, the mass arising from interactions.) The twistor $\lambda Z^\alpha$, for any
non-zero complex number $\lambda$, defines the same null line as does $Z^\alpha$.
But we can distinguish $Z^\alpha$ from $\lambda Z^\alpha$ by assigning to the particle
a $4$-momentum (in its direction of motion) and, in addition, a sort of
`polarization' direction (both constant along the world line of the
particle). When $Z^\alpha$ is replaced by $r Z^\alpha$ ($r$ real) the momentum is
multiplied by $r^2$. When $Z^\alpha$ is replaced by $e^{i\theta}Z^\alpha$ ($\theta$ real) the `polarization'
plane is rotated through an angle $2\theta$. If $Y^\alpha$ is another null
twistor, the condition for the null lines represented by $Y^\alpha$ and $Z^\alpha$ to
\emph{meet} is
\[
Y^\alpha\,\bar{Z}_\alpha=0,
\]
i.e.\ this is the condition for the two particles to `collide'. (To be
strictly accurate we have to include the possibility that they may
`meet at infinity'. In addition, some of the null twistors describe
`null lines at infinity' rather than actual null lines.)
The \emph{non-null} twistors are divided into two classes according as
$Z^\alpha\,\bar{Z}_\alpha$, is positive or negative. If $Z^\alpha\,\bar{Z}_\alpha>0$, I call $Z^\alpha$ right-handed; if
$Z^\alpha\,\bar{Z}_\alpha<0$, left-handed. In Minkowski space-time, one can give an
interpretation of a non-null twistor in terms of a twisting system of
null lines. The helicity of the twist is defined by the sign of $Z^\alpha\,\bar{Z}_\alpha$. In
more physical terms, the twistor $Z^\alpha$ (up the phase) describes the
momentum and angular momentum structure of a zero rest-mass
particle.\footnote{Using a convenient co-ordinate system, we can relate the momentum $P_\alpha$ and the
angular momentum $M^{ab}(=-M^{ba})$ to the twistor variables $Z^\alpha$, $\bar{Z}_\alpha$ a by:
\arraycolsep1.5pt
\begin{eqnarray}
P_0+P_1 = 2^\frac{1}{2}\,Z^0\,\bar{Z}_2,\quad
P_0-P_1 &=& 2^\frac{1}{2}\,Z^0\,\bar{Z}_3,\quad
P_2+i\,P_3 = 2^\frac{1}{2}\,Z^1\,\bar{Z}_2 \nonumber\\
%
P_2-i\,P_3 = 2^\frac{1}{2}\,Z^0\,\bar{Z}_3,\quad
M^{23}+i\,M_{01} &=& Z^1\,\bar{Z}_1-Z^0\,\bar{Z}_0,\quad
M^{23}-i\,M_{01} = Z^3\,\bar{Z}_3-Z^2\,\bar{Z}_2 \nonumber\\
M^{13}+M^{03}+i\,M_{02}+i\,M_{12} = 2\,Z^2\,\bar{Z}_2,
&&
M^{13}-M^{03}+i\,M_{02}-i\,M_{12} = 2\,Z^3\,\bar{Z}_2 \nonumber\\
M^{13}+M^{03}-i\,M_{02}-i\,M_{12} = 2\,Z^1\,\bar{Z}_0,
&&
M^{13}-M^{03}-i\,M_{02}+i\,M_{12} = 2\,Z^0\,\bar{Z}_1 \nonumber
\end{eqnarray}
}
We can make the interpretation that $Z^\alpha\,\bar{Z}_\alpha$, is (twice) the
\emph{intrinsic spin} of the particle, measured in suitable units, with a \emph{sign}
defining the helicity. If $Z^\alpha\,\bar{Z}_\alpha\ne0$, then it is not possible actually to
localize the particle as a null straight line. Only if $Z^\alpha\,\bar{Z}_\alpha=0$ do we
get a uniquely defined null line which we can think of as the world
line of the particle; otherwise the particle to some extent spreads
itself throughout space.
The twistor co-ordinates $Z^0$, $Z^1$, $Z^2$, $Z^3$, together with their complex
conjugates $\bar{Z}_0$, $\bar{Z}_1$, $\bar{Z}_2$, $\bar{Z}_3$, can be used in place of the usual $x$, $y$, $z$, $t$
and their canonical conjugates $p_x$, $p_y$, $p_z$, $E$. In fact, anything that
can be written in normal Minkowski space terms can be rewritten in
terms of twistors. However, in principle, the twistor expressions for
even quite simple physical processes may turn out to be very complicated.
But in fact it emerges that the basic elementary processes that
one requires, can actually be expressed very simply if one goes about
it in the right way. Analytic (holomorphic) functions in the $Z^\alpha$
variables play a key role. So does contour integration.
We can regard the $Z^\alpha$ as quantum operators under suitable circumstances.
Then $\bar{Z}_\alpha$, can be regarded as the canonical conjugate of $Z^\alpha$.
(I shall go into the reasons for this a little more later.) We have commutation
rules
\[
Z^\alpha\,\bar{Z}_\beta - \bar{Z}_\beta\,Z^\alpha = {\delta^\alpha}_\beta\,\hbar
\]
Then, since $Z^\alpha$ and $\bar{Z}_\alpha$ do not commute, we must re-interpret the
expression for the \emph{spin-helicity} $\frac{1}{2}n$ as the \emph{symmetrized} quantity,
\[
\frac{1}{4}(Z^\alpha\,\bar{Z}_\alpha - \bar{Z}_\alpha\,Z^\alpha)
= \frac{1}{2}n\hbar
\]
Only zero rest-mass states can be eigenstates of this operator. The
eigenvalues of $n$ are $\ldots -2,-1,0,1,2, \dots$ The operators for the ten
components of momentum and angular momentum (together with
those for the five extra components arising from the conformal
invariance of zero rest-mass fields) are
\[
Z^\alpha\,\bar{Z}_\beta - \frac{1}{4}{\delta^\alpha}_\beta\,Z^\gamma\,\bar{Z}_\gamma
\]
in twistor notation. The usual commutation rules for momentum and
angular momentum are then a consequence of the twistor commutation
rules.
One idea would now be to use this fact and simply let the twistors
take the place of the \emph{two-component spinors} that lay `behind the
scenes' in my previous non-relativistic approach, and then to attempt
to build a concept of a four-dimensional space-time from whatever
graphical algebra arises from the twistor rules. I have not attempted
to do quite this, as yet, since I am not sure that it is exactly the right
thing to do. There are certain other aspects of twistor theory which
should really be taken into account first.
Let me mention one particular point. It is a rather remarkable one.
If the twistor approach is going to have any fundamental significance
in physical theory, then it ought, in principle at least, to be possible
to carry the formalism over and apply it to a \emph{curved} space-time,
rather then just a flat space-time. These objects, as originally defined,
are very much tied up with the Minkowski flat space-time concept.
How can we carry them over into a curved space-time? Actually,
a twistor for which $Z^\alpha\,\bar{Z}_\alpha=0$ carries over very well. Its interpretation
is now simply as a \emph{null geodesic} (i.e.\ world line of a freely moving
massless particle) with a momentum (pointing along the world line)
and a `polarization' direction (both covariantly constant along the
world line). On the other hand, it does not seem to be possible to
interpret a \emph{non-null} twistor, in a general curved space-time, in
precise classical space-time terms. Nevertheless it turns out to be
convenient to \emph{postulate} the existence of these non-null twistors---as
objects with no classical realization in curved space-time terms. (In
a sense, twistors are more appropriate to the treatment of \emph{quantized}
gravitation\footnote{
Since this lecture was delivered, there have been some developments in twistor
theory of relevance to this discussion. It seems to be possible to express quantized
gravitational theory in twistor form. By means of $3k$-dimensional contour integrals
in spaces of many twistor variables, one can apparently calculate scattering amplitudes
for processes involving gravitons, photons and other particles. Diagrams arise
which can be used to replace the spin-networks of the formalism described here. It
is not impossible that the calculations can be reduced to a set of comparatively
simple combinatorial rules, but it is unclear, as yet, whether this is so. The
work is
still very much at a preliminary stage of development and many queries remain
unanswered.}
than of classical general relativity.)
Let us concentrate attention, for the moment, on the \emph{null} twistors
only so that we can consider purely geometrical questions. We are
interested in properties of null geodesics which refer to each geodesic
as a \emph{whole} and not to the neighbourhood of some point on a geodesic.
Consider, for example, the condition of orthogonality between
twistors. We have seen that in flat space-time, the condition of
orthogonality $Y^\alpha\bar{Z}_\alpha=0$ between two twistors $Y^\alpha$, $Z^\alpha$ corresponds to
the \emph{meeting} of the corresponding null lines. In curved space-time this
is not really satisfactory, because although I can tell whether two
null geodesics are going to meet if I look in the neighbourhood of the
intersection point, if I look somewhere else, I can't see whether or
not they will meet, because the curvature may have bent them
away from each other. So, in fact, the orthogonality property is not
something which is preserved, as an invariant concept, when one
turns to curved space-time. On the other hand, certain things \emph{are}
preserved; and, somewhat surprisingly, they correspond to assigning
a \emph{symplectic} structure to the twistor space.
This symplectic structure is expressed (in appropriate co-ordinates)
as the invariance of the 2-form
\[
dZ^\alpha\,\wedge\,d\bar{Z}_\alpha
\]
(using Cartan notation). Strictly speaking, this requires the postulated
non-null twistors, in addition to the null ones. The null twistors
only form a seven-real-dimensional manifold, whereas a symplectic
manifold must be even-dimensional. The null twistor manifold must
be embedded in the eight-real-dimensional manifold consisting of \emph{all}
twistors. The structure of the null twistor manifold is that induced
by the embedding in this eight-dimensional symplectic manifold. In
addition to the symplectic structure (and closely related to it), the
expressions
\[
Z^\alpha\bar{Z}_\alpha, \quad
Z^\alpha d\bar{Z}_\alpha, \quad
Z^\alpha \, \frac{\partial}{\partial Z^\alpha}
\]
are also invariant. All these invariant quantities can be interpreted,
to some extent, in terms of the geometrical properties of null geodesics.
But it will not be worthwhile for me to go into all this here.
The invariance of the symplectic structure of the twistor space for
curved space-time can be re-expressed as the invariance of the
\emph{Poisson brackets}
\[
[\psi,\chi] = i\,\frac{\partial \psi}{\partial Z^\alpha}\,
\frac{\partial \chi}{\partial \bar{Z}_\alpha}
-i\,\frac{\partial \chi}{\partial Z^\alpha}\,
\frac{\partial \psi}{\partial \bar{Z}_\alpha} \quad .
\]
This strongly suggests that in the passage to quantum theory, $\bar{Z}_\alpha$
should be regarded as the conjugate variable to $Z^\alpha$. Thus, we are led
to the commutation rules I mentioned earlier, relating quantum
operators $Z^\alpha$ and $\bar{Z}_\alpha$. These commutation rules in turn give us the
commutation relations for momentum and angular momentum, as
I indicated before. This suggests that there may possibly be some
deep connection between these commutation rules (or perhaps some
slight modifications of them) and the curvature of space-time.
The picture that one gets is that in some sense the curvature of
space-time is to do with canonical transformations between the
twistor variables $Z^\alpha$, $\bar{Z}_\alpha$. Suppose we start in some region of space-time
where things are essentially flat. Then we can interpret $Z^\alpha$ and
$\bar{Z}_\alpha$ in a straightforward way in terms of geometry and angular
momentum, etc. Suppose we then pass through a region of curvature
to another region where things are again essentially flat. We then
find that our interpretations have undergone a `shift' corresponding
to a canonical transformation between $Z^\alpha$ and $\bar{Z}_\alpha$. In effect the
`twistor position' (i.e.\ $Z^\alpha$) and `twistor momentum' (i.e.\ $\bar{Z}_\alpha$) have got
mixed up. Somehow it is this mixing up of the `twistor position' and
`twistor momentum' which corresponds to what we see as space-time
curvature.
Going back to my original combinatorial approach for building
space up from angular momentum, we can ask now whether such
a combinatorial scheme could be applied to the twistors. Might it be
that, instead of ending up with a flat space, we could end up with
a curved space-time? Even if I start with the commutation rules
appropriate just to the Poincar\'{e} group, or perhaps the conformal
group, it is obvious that I must end up with essentially the same space
that I `start' with? One has to define the things with which one builds
up geometry (e.g.\ points, angle, etc.), in \emph{terms} of the physical objects
under consideration. It is not at all clear to me that the geometry that
is built up in one region will not be `shifted' with respect to the
geometry built up in some other region. Is it then not possible that
a space-time possessing curvature might be the result? That is really
the final point I wanted to make.
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REFERENCES
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(l) Yutsis, A.~P., Levinson, I.~B. and Vanagas, V. V. \emph{Mathematical
apparatus of the theory of angular momentum} (Jerusalem: I.P.S.T.,
1962).\\
(2) Penrose, R. \emph{J. Math. Phys.} (1967) {\bf 8}, 345.\\
(3) Penrose, R. \emph{Int. Jl Theor. Phys.} (1968) {\bf 1}, 61.
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