Week 6:
1) Quantum cosmology, talk given at Texas/Pascos 1992 at Berkeley by
Alexander Vilenkin, preprint available in TeX form as grqc/9302016
This is, as Vilenkin notes, an elementary review of quantum cosmology.
It won't be news to anyone who has kept up on that subject (except
perhaps for a few speculations at the end), but for those who haven't
been following this stuff, like myself, it might be a good way to get started.
Let's get warmed up....
Quantizing gravity is mighty hard. For one thing, there's the "problem
of time"  the lack of a distinguished time parameter in *classical*
general relativity means that the usual recipe for quantizing a
dynamical system  "represent time evolution by the unitary operators
exp(iHt) on the Hilbert space of states, where t is the time and H, the
Hamiltonian, is a selfadjoint operator"  breaks down! As Wheeler so
picturesquely put it, in general relativity we have "manyfingered
time"; there are lots of ways of pushing a spacelike surface forwards in
time.
But if we simplify the heck out of the problem, we might make a little
progress. (This is a standard method in physics, and whether or not
it's really justified, it's often the only thing one can do!)
For one thing, note that in the big bang cosmology there is a
distinguished "rest frame" (or more precisely, field of timelike
vectors) given by the galaxies, if we discount their small random
motions. In reality these are maybe not so small, and maybe not so
random  such things as the "Virgo flow" show this  but we're talking
strictly theory here, okay?  so don't bother us with facts! So, if we
imagine that things go the way the simplest big bang models predict, the
galaxies just sit there like dots on a balloon that is being inflated,
defining a notion of "rest" at each point in spacetime. This gives
a corresponding notion of time, since one can measure time using
clocks that are at rest relative to the galaxies. Then, since we are
pretending the universe is completely homogeneous and isotropic  and
let's say it's a closed universe in the shape of a 3sphere, to be
specific  the metric is given by
dt^2  r(t)^2[(d psi)^2 + (sin psi)^2{(d theta)^2 + (sin theta)^2 (d phi)^2}]
What does all this mean? Here r(t) is the radius of the universe as a
function of time, the following stuff is just the usual metric on the
unit 3sphere with hyperspherical coordinates psi, theta, phi
generalizing the standard coordinates on the 2sphere we all learn in
college:
(d psi)^2 + (sin psi)^2{(d theta)^2 + (sin theta)^2 (d phi)^2}
and the fact that the metric on spacetime is dt^2 minus a bunch of stuff
reflects the fact that spacetime geometry is "Lorentzian," just as
in flat Minkowski space the metric is
dt^2  dx^2  dy^2  dz^2.
The name of the game in this simple sort of big bang cosmology is thus
finding the function r(t)! To do this, of course, we need to see what
Einstein's equations reduce to in this special case, and since
Einstein's equations tell us how spacetime curves in response to the
stressenergy tensor, this will depend on what sort of matter we have
around. We are assuming that it's homogeneous and isotropic, whatever
it is, so it turns out that all we need to know is its density rho and
pressure P (which are functions of time). We get the equations
r''/r = (4pi/3)(rho + 3P) (r')^2 = (8pi/3) rho r^2  1
Here primes denote differentiation with respect to t, and I'm using
units in which the gravitational constant and speed of light are equal to 1.
Let's simplify this even more. Let's assume our matter is "dust," which
is the technical term for zero pressure. We get two equations:
r''/r = (4pi/3)rho (r')^2 = (8pi/3) rho r^2  1. (1)
Now let's take the second one, differentiate with respect to t,
2r'' r' = (8pi/3)(rho' r^2 + 2 rho r r')
plug in what the first equation said about r'',
(8pi/3) rho r r' = (8pi/3)(rho' r^2 + 2 rho r r')
clear out the crud, and lo:
3 rho r' =  rho' r
or, more enlighteningly,
d(rho r^3)/dt = 0.
This is just "conservation of dust"  the dust density times the volume
of the universe is staying constant. This, by the way, is a special
case of the fact that Einstein's equations *automatically imply*
local conservation of energy (i.e., that the stressenergy tensor is
divergencefree).
Okay, so let's say rho r^3 = D, with D being the total amount of dust.
Then we can eliminate rho from equations (1) and get:
r'' = 4piD/3r^2 (r')^2  (8pi/3) D/r =  1 (2)
What does this mean? Well, the first one looks like it's saying
there's a force trying to make the universe collapse, and that the
strength of this force is proportional to 1/r^2. Sound vaguely
familiar? It's actually misleadingly simple  if we had put in
something besides dust it wouldn't work quite this way  but as long as we
don't take it too seriously, we can just think of this as gravity trying
to get the universe to collapse. And the second one looks like it's
saying that the "kinetic" energy proportional to (r')^2, plus the
"potential" energy proportional to 1/r, is constant! In other words,
we have a nice analogy between the big bang cosmology and a very
oldfashioned system, a classical particle in one dimension attracted to
the origin by a 1/r^2 force!
It's easy enough to solve this equation, and easier still to figure it out
qualitatively. The key thing is that since the total "energy" in
the second equation of (2) is negative, there won't be enough "energy"
for r to go to infinity, that is, there'll be a big bang and then a big
crunch. Here's r as a function of t, roughly:
 . .
 . .
r  . . Figure 1
 . .
 . .

t
What goes up, must come down! This curve, which I haven't drawn too
well, is just a cycloid, which is the curve traced out by a point on the
rim of rolling wheel. So, succumbing to romanticism momentarily we
could call this picture ONE TURN OF THE GREAT WHEEL OF TIME.... But
there is *no* reason to expect further turns, because the
differential equation simply becomes singular when r = 0. We may either
say it doesn't make sense to speak of "before the big bang" or "after
the big crunch"  or we can look for improved laws that avoid these
singularities. (I should repeat that we are dealing with unrealistic models
here, since for example there is no evidence that there is enough matter
around to "close the universe" and make this solution qualitatively
valid  it may well be that there's a big bang but no big crunch. In
this case, there's only one singularity to worry about, not two.)
People have certainly not been too ashamed to study the *quantum*
theory of this system (and soupedup variants) in an effort to get a
little insight into quantum gravity. We would expect that quantum
effects wouldn't matter much until the radius of the universe is very
small, but when it *is* very small they would matter a lot, and maybe 
one might hope  they would save the day, preventing the nasty
singularities. I'm not saying they DO  this is hotly debated  but
certainly some people hope they do. Of course, serious quantum gravity
should take into account the fact that geometry of spacetime has all
sorts of wiggles in it it isn't just a symmetrical sphere. This may make
a vast difference in how things work out. (For example, the big crunch
would be a lot more exciting if there were lots of black holes around by
then.) The technical term for the space of all metrics on space is
"superspace" (sigh), and the toy models one gets by ignoring all but
finitely many degrees of freedom are called "minisuperspace" models.
Let's look at a simple minisuperspace model. The simplest thing
to try is to take the classical equations of motion (2) and try to
quantize them just like one would a particle in a potential. This is a
delicate business, by the way, because one can't just take some
classical equations of motion and quantize them in any routine way.
There are lots of methods of quantization, but all of them require a
certain amount of casebycase finesse.
The idea of "canonical quantization" of a classical system with one
degree of freedom  like our big bang model above, where the one degree
of freedom is r  is to turn the "position" (that's r) into a
multiplication operator and the "momentum" (often that's something like
r', but watch out!) into a differentation operator, say i hbar d/dr, so that
we get the "canonical commutation relations"
[i hbar d/dr, r] = i hbar.
We then take the formula for the energy, or Hamiltonian, in terms of
position and momentum, and plug in these operators, so that the
Hamiltonian becomes an operator. (Here various "operatorordering"
problems can arise, because the position and momentum commuted in the
original classical system but not anymore!) To explain what I mean,
why don't I just do it!
So: I said that the formula
(r')^2  (8pi/3) D/r =  1 (3)
looks a lot like a formula of the form "kinetic energy plus potential
energy is constant". Of course, we could multiply the whole equation
by anything and get a valid equation, so it's not obvious that the
``right'' Hamiltonian is
(r')^2  (8pi/3) D/r
or (adding 1 doesn't hurt)
(r')^2  (8pi/3) D/r + 1
In fact, note that multiplying the Hamiltonian by some function of r
just amounts to reparametrizing time, which is perfectly fine in general
relativity. In fact, Vilenkin and other before him have decided it's
better to multiply the Hamiltonian above by r^2. Why? Well, it has to
do with figuring out what the right notion of "momentum" is
corresponding to the "position" r. Let's do that. We use the old
formula
p = dL/dq'
relating momentum to the Lagrangian, where for us the position, usually
called q, is really r.
The Lagrangian of general relativity is the "Ricci scalar" R  a measure of
curvature of the metric  and in the present problem it turns out to be
R = 6 (r''/r + (r')^2/r^2)
But we are reducing the full field theory problem down to a problem with
one degree of freedom, so our Lagrangian should be the above integrated
over the 3sphere, which has volume 16 pi r^3/3, giving us
32pi (r'' r^2 + (r')^2 r)
However, the a'' is a nuisance, and we only use the integral of the
Lagrangian with respect to time (that's the action, which classically is
extremized to get the equations of motion), so let's do an integration
by parts, or in other words add a total divergence, to get the Lagrangian
L = 32pi (r')^2 r.
Differentiating with respect to r' we get the momentum "conjugate to r",
p = 64pi r'r.
Now I notice that Vilenkin uses as the momentum simply r'r, somehow
sweeping the monstrous 64pi under the rug. I have the feeling that
this amounts to pushing this factor into the definition of hbar in the
canonical commutation relations. Since I was going to set hbar to 1 in
a minute anyway, this is okay (honest). So let's keep life simple and
use
p = r'r.
Okay! Now here's the point, we want to exploit the analogy with good
old quantum mechanics, which typically has Hamiltonians containing
something like p^2. So let's take our preliminary Hamiltonian
(r')^2  (8pi/3) D/r + 1
and multiply it by r^2, getting
H = p^2  (8pi D/3)r + r^2.
Hey, what's this? A harmonic oscillator! (Slightly shifted by the
term proportional to r.) So the universe is just a harmonic
oscillator... I guess that's why they stressed that so much in all my
classes!
Actually, despite the fact that we are working with a very simple model
of quantum cosmology, it's not quite *that* simple. First of all,
recall our original classical equation, (3). This constrained the
energy to have a certain value. I.e., we are dealing not with a
Hamiltonian in the ordinary sense, but a "Hamiltonian constraint" 
typical of systems with time reparametrization invariance. So our
quantized equation says that the "wavefunction of the universe," psi(r),
must satisfy
H psi = 0.
Also, unlike the ordinary harmonic oscillator we have the requirement
that r > 0. In other word, we're working with a problem that's like a
harmonic oscillator and a "wall" that keeps r > 0. Think of a particle
in a potential like this:

 .

V(r)  Figure 2
 .
 .
..
. r
Here V(r) =  (8pi D/3)r + r^2. The minimum of V is at r = 4 pi D/3
and the zeroes are at r = 0 and 8 pi D/3. Classically, a particle
with zero energy starting at r = 0 will roll to the right and make it
out to r = 4 pi D/3 before rolling back to r = 0. This is basically the
picture we had in Figure 1, except that we've reparametrized time so we
have simple harmonic motion instead of cycloid.
Quantum mechanically, however one must pick boundary conditions at r = 0
to make the problem welldefined!
This is where the fur begins to fly!! Hawking and Vilenkin have very
different ideas about what the right boundary conditions are. And note
that this is not a mere technical issue, since they determine the
wavefunction of the universe in this approach! I will not discuss this
since Vilenkin does so quite clearly, and if you understand what I have
written above you'll be in a decent position to understand him. I will
just note that Vilenkin, rather than working with a universe full of
"dust," considers a universe in which the dominant contribution to the
stressenergy tensor is the cosmological constant, that is, the negative
energy density of a "false vacuum", which believers in inflation (such
as Vilenkin) think powered the exponential growth of the universe at an
early stage. So his equations are slightly different from those above (and
are only meant to apply to the early history of the universe).
[Let me just interject a question to the experts if I may  since I've
written this long article primarily to educate myself. It would seem to
me that the equation H psi = 0 above would only have a normalizable
solution if the boundary conditions were finetuned! I.e., maybe the
equation H psi = 0 itself determines the boundary conditions! This
would be very nice; has anyone thought of this? It seems reasonable
because, with typical boundary conditions, the operator H above will
have pure point spectrum (only eigenvalues) and it would be rather
special for one of them to be 0, allowing a normalizable solution of
H psi = 0. Also, corrections and education of any sort are welcomed.
I would love to discuss this with some experts.]
Anyway, suppose we find some boundary conditions and calculate psi, the
"wavefunction of the universe." (I like repeating that phrase because
it sounds so momentous, despite the fact that we are working with a
laughably oversimplified toy model.) What then? What are the
implications for the man in the street?
Let me get quite vague at this point. Think of the radius of the
universe as analogous to a particle moving in the potential of Figure 2.
In the current state of affairs classical mechanics is an excellent
approximation, so it seems to trace out a classical trajectory. Of
course it is really obeying the laws of quantum mechanics, so the
trajectory is really a "wave packet"  technically, we use the WKB
approximation to see how the wave packet can seem like a classical
trajectory. But near the big bang or big crunch, quantum mechanics
matters a lot: there the potential is rapidly varying (in our simple
model it just becomes a "wall") and the wave packet may smear out
noticeably. (Think of how when you shoot an electron at a nucleus it
bounces off in an unpredictable direction  it's wavefunction just tells
you the *probability* that it'll go this way or that!) So some quantum
cosmologists have suggested that if there is a big crunch, the universe
will pop back out in a highly unpredictable, random kind of way!
I should note that Vilenkin has a very different picture. Since this
stuff makes large numbers of assumptions with very little supporting
evidence, it is science that's just on the brink of being mythology.
Still, it's very interesting.
2) Finite, diffeomorphism invariant observables in quantum gravity,
by Lee Smolin, preprint available in LaTeX form as grqc/9302011.
The big problem in canonical quantization of gravity, once one
gets beyond "mini" and "midisuperspace" models, is to find enough
diffeomorphisminvariant observables. There is a certain amount of
argument about this stuff, and various approaches, but one common
viewpoint is that the "physical" observables, that is, the really
observable observables, in general relativity are those that are
invariant under all diffeomorphisms of spacetime. I.e., those that are
independent of any choice of coordinates. For example, saying "My
position is (242,2361,12,17)" is not diffeomorphisminvariant, but
saying "I'm having the time of my life" is. It's hard to find lots of
(tractable) diffeomorphism invariant observables  or even any! Try
figuring out how you would precisely describe the shape of a rock without
introducing any coordinates, and you'll begin to see the problem. (The
quantum mechanical aspects make it harder.)
Rovelli came up a while back with a very clever angle on this problem.
It's rather artificial but still a big start. Using a "field of
clocks" he was able to come up with interesting diffeomorphism
invariant observables. The idea is simply that if you had clocks all
around you could say "when the bells rang 2 a.m. I was having the time of
my life"  and this would be a diffeomorphisminvariant statement, since
rather than referring to an abstract coordinate system it expresses the
coincidence of two physical occurences, just like "the baseball broke
through the window". Then he pushed this idea to define "evolving
constants of motion"  a deliberate oxymoron  to deal with the famous
"problem of time" in general relativity: how to treat time evolution in
a coordinatefree manner on a spacetime that's not flat and, worse,
whose geometry is "uncertain" a la Heisenberg? This is treated, by the
way, in
Time in quantum gravity: An hypothesis, by Carlo Rovelli, Phys. Rev.
D43 (1991), 442456.
Also, an excellent and very thorough review of the problem of time and
various proposed solutions, including Rovelli's, is given in
Canonical Quantum Gravity and the Problem of Time,
Chris J. Isham, 125 pages of LaTeX output, preprint available as
grqc/9210011.
Anyway, in a paper I very briefly described in the file "week"
(see below for how to get it):
Time, measurement and information loss in quantum cosmology, by Lee
Smolin, preprint now available as grqc/9301016,
Smolin showed, how, using a clever trick sketched in the present paper
to get "observables" invariant under spatial but not temporal
observables, together with Rovelli's idea, one could define lots of
REAL observables, invariant under spacetime diffeomorphisms that is,
thus making a serious bite into this problem.
I warn the reader that there is a fair amount that is not too realistic
about these methods. First there's the "clock field"  this can actually be
taken as a free massless scalar field, but in so doing there is the
likelihood of serious technical problems. Some of these are discussed
in
Comment on "Time in quantum gravity  an hypothesis" by P. Hajicek,
Phys. Rev. D44 (1991), 13371338.
(But I haven't actually read this, just Isham's description.) Also, the
clever trick of the present paper is to couple gravity to an
antisymmetric tensor gauge field so that in addition to having loops
as part of ones "loop representation," one has surfaces  a "surface
representation". But this antisymmetric tensor gauge field is not the
sort of thing that actually seems to arise in physics (unless I'm
missing something). Still, it's a start. I think I'll finish by quoting
Smolin's abstract:
Two sets of spatially diffeomorphism invariant operators are
constructed in the loop representation formulation of quantum
gravity. This is done by coupling general relativity to an anti
symmetric tensor gauge field and using that field to pick out
sets of surfaces, with boundaries, in the spatial three manifold.
The two sets of observables then measure the areas of these surfaces
and the Wilson loops for the selfdual connection around their
boundaries. The operators that represent these observables are
finite and background independent when constructed through a
proper regularization procedure. Furthermore, the spectra of the
area operators are discrete so that the possible values that one
can obtain by a measurement of the area of a physical surface in
quantum gravity are valued in a discrete set that includes integral
multiples of half the Planck area. These results make possible the
construction of a correspondence between any three geometry whose
curvature is small in Planck units and a diffeomorphism invariant state
of the gravitational and matter fields. This correspondence relies on
the approximation of the classical geometry by a piecewise flat Regge
manifold, which is then put in correspondence with a diffeomorphism invariant
state of the gravitymatter system in which the matter fields
specify the faces of the triangulation and the gravitational field
is in an eigenstate of the operators that measure their areas.

Previous editions of "This Week's Finds," and other expository posts
on mathematical physics, are available by anonymous ftp from
math.princeton.edu, thanks to Francis Fung. They are in the directory
/pub/fycfung/baezpapers. For info on grqc and hepth see the
sci.physics FAQ.
And, just for fun:
"In the Space and Time marriage we have the greatest Boy meets Girl
story of the age. To our greatgrandchildren this will be as poetical a
union as the ancient Greek marriage of Cupid and Psyche seems to us."
Lawrence Durrell, _Balthazar_

Note added in proofs: While walking through the rain tonight, I think I
partially figured out my problem. Initially there is a 4dimensional
phase space for this problem, coordinatized by r, r', rho, and rho'.
The timereparametrization invariance of the problem gives a Hamiltonian
constraint, (r')^2 = (8pi/3) rho r^2  1. ^C
(Interrupt  one more to kill letter)
^Cguitar177 >mail mods
Subject: Addendum to post
I tried to post something myself to sci.physics.research a while back and
it was apparently swallowed by a black hole.
I posted another thing tonight and then had an addendum. If one of you
got the original thing and deem it postworthy, please tack on the following
addendum:

Note added in proofs: While walking through the rain tonight, I think I
partially figured out my problem. Initially there is a 4dimensional
phase space for this problem, coordinatized by r, r', rho, and rho'.
To quantize this we should follow Dirac's procedure for quantizing
systems with gauge symmetries (hence constraints)... or in classical
terms, we do MarsdenWeinstein reduction by 1) finding solutions
to the constraint equation:
(r')^2 = (8pi/3) rho r^2  1
and 2) and modding out by the time evolution given by
r''/r = (4pi/3)rho.
Is this right?

And just in case, here's the original post:

In the forthcoming "This Week's Finds" I want to explain some basic
stuff on quantum cosmology as a warmup for Vilenkin's review paper on it
that just appeared on grqc. Also I'm talking about this a little in
the class I'm giving on knots and quantum gravity. And, lo and behold,
I discovered that I don't understand it as well as I'd like  so maybe I
can get a little help.
Let's say we've got a RobertsonWalker metric for an R x S^3 universe
full of dust. It satisfies
r''/r = (4pi/3)rho (r')^2 = (8pi/3) rho r^2  1.
where r(t) is the radius of the universe at time t and rho(t) is the
density of dust. Now we can obtain that rho r^3 is a constant and
solve for rho in terms of r if we want, say rho = Dr^{3}, so
r'' = 4piD/3r^2 (r')^2  (8pi/3) D/r =  1
and we see that the first equation is just a consequence of the 2nd.
(By the way, I'll explain this all a lot better in my real post;
here I'm mainly just asking for help from people who know this
infinitely better than I do.)
Now I want to quantize this model as if it were a dynamical system.
We've picked a particular parametrization of time already so I'd think
we have an honest Hamiltonian, not just a Hamiltonian constraint. No?
Maybe not. What *is* the Hamiltonian? The fact that it's conserved
should have a whole lot to do with
(r')^2  (8pi/3) D/r =  1
It's actually tempting  at least for the foolish  to hope that the
left hand side *is* the Hamiltonian (since then we have a familiar sort
of problem: a particle in a 1/r potential  but with r > 0 only).
But then what the heck is this equation doing saying the Hamiltonian has
to equal 1? It looks a lot more like a Hamiltonian constraint. I'm
confused. Enlightenment, anyone?
Anyway, the people I'm reading seem to know what they're doing and they
treat it as a Hamiltonian constraint:
H = 0 where H = (r')^2  (8pi/3) D/r + 1
But actually Vilenkin (who's dealing with a different model), multiplies
this by r^2; he says
H = 0 where H = (rr')^2  (8pi/3) Dr + r^2
This is of course equally good on the classical level, and it's nicer
in a way since when one works out the variable canonically conjugate to
r one gets p = rr'. (Actually I get a number times this; this is what
Vilenkin gets, but I guess it all comes out in the wash.) Thus
Vilenkin gets
H = p^2 + (8pi/3) Dr + r^2
(or he *would* get this if he was looking at this model) and quantizing
amounts to finding a solution of
H psi = 0
where psi is a function of r (for r > 0) and now we think of p as
id/dr.
Question: how do you justify working with the Hamiltonian
(rr')^2  (8pi/3) Dr + r^2
rather than something like
(r')^2  (8pi/3) D/r + 1 ?
jb