"Very realistic," said the Wiz. "If you keep it up, it'll become impossible to tell you're a fictional character. But cut out the 'How was that?' business -- it's a dead giveaway. Okay. Last time we proved a version of Schur's Lemma, which said that a spin network with one spin-j edge coming in and one spin-k edge going out:
|
j|
__|__
| |
| |
| S |
| |
|_____|
|
k|
|
is zero if j is different from k,
and a constant multiple of the identity operator if j = k.
We even determined the constant:
| ______ | | __/__ \ | j| | | | j| | | | | | | | S | |j | __|__ | | | | | | |_____| | | | | \______/ | | S | = ______________ | | | _____ | |_____| / \ | | | |j | | | | | | \_____/ | j| | | |Now let's see if we can do something like this for spin networks with more edges coming in and out."
"What about fewer edges coming in and out?" asked Toby.
"Very good!" said the Wizard. "That's exactly the sort of thing you'd say. Let's start with a spin network with no edges in or out:
_____
| |
| |
| S |
| |
|_____|
In other words, a closed spin network!
Here we already know the answer:
any such spin network evaluates to a number.
Next, a spin network with just one edge coming in:
|
j|
__|__
| |
| |
| S |
| |
|_____|
Here can use our tricks from last time to see that
unless j = 0, this must equal zero.
Just think: after we expand all the symmetrizers,
we get a sum over pictures with 2j spin-1/2 edges coming in on top.
They can't just stop dead; they have to go out somewhere.
All they can do is double back and go out the way they came in!
But we saw last time that this gives zero. For example:
2|
|
-------
/ / \ \
/ / \ \ = 0
\ \ / /
\ \_/ /
\___/
If j = 0, we can actually rip off the spin-0 edge coming in
without changing the meaning of the spin network,
reducing it to the previous case.
Of course, we can always change inputs to outputs and vice versa using caps and cups, so the same remarks apply to spin networks with just one edge going out:
_____
| |
| |
| S | = 0 unless j = 0
| |
|_____|
|
j|
|
All that really matters in this game is the total number of edges
coming in or going out.
So the next interesting case is spin networks with
3 edges coming in or going out,
like this for example:
\ /
j1\ /j2
_\_/_
| |
| |
| S |
| |
|_____|
|
j3|
|
What does this equal? What could it equal?"
Said Jay, "I guess it could just be
\ /
j1\ /j2
\ /
|
j3|
|
or any multiple of this."
Miguel added, "Well, at least if (j1,j2,j3) is an admissible triple!"
"Right!" said the Wiz. "If they add up to an integer and satisfy the triangle condition, we call them an "admissible triple", and we saw in Week 8 that we get an operator like this
\ /
j1\ /j2
\ /
|
j3|
|
So let's see if we can show that
\ / \ /
j1\ /j2 j1\ /j2
_\_/_ \ /
| | \ /
| | |
| S | = c |
| | |
|_____| |
| |
j3| j3|
| |
for some constant c when (j1,j2,j3) is admissible, and
\ /
j1\ /j2
_\_/_
| |
| |
| S | = 0
| |
|_____|
|
j3|
|
otherwise.
How can we do this?"
There was a pause, and then Toby's eyes brightened strangely and he said: "Let's use the technology we developed last week! After we expand the symmetrizers, we get a big sum of terms. Each term is a picture made of spin-1/2 strands, tangled up like a big bowl of spaghetti. We've got a bunch of strands coming in from the top left, a bunch coming in from the top right, and a bunch going out the bottom. We may also have closed loops..."
Some joker shouted "Oh-oh! Spaghettios!"
"... but we can eliminate these using the formula
______
/ \
/ \
| | = -2
\ /
\______/
Then, disentangling what's left, each term becomes proportional
to something like this:
j1 | j2 |
| |
--------- -----------
\..\ \..\ /../ /.../
\..\ \..\ /../ /.../
\..\ \..\___/../ /.../
\..\ \......./ /.../
\..\ \__A__/ /.../
\..\ |...|
|..|B C|...|
------------------
|
j3 |
where A strands go from top left to top right,
B go from top left to the bottom,
and C go from top right to the bottom.
But we saw in Week 8 that the numbers A,B,C are unique,
and they exist only if (j1,j2,j3) is an admissible triple.
In this case we have, by definition:
j1| |j2 j3\ /j2
| | \ /
--------- ----------- \ /
\..\ \..\ /../ /.../ \ /
\..\ \..\ /../ /.../ \ /
\..\ \..\___/../ /.../ \ /
\..\ \......./ /.../ \ /
\..\ \__A__/ /.../ \ /
\..\ |...| \ /
|..|B C|...| = |
------------------ |
| |
j3| j3|
So we're done."
"Right!" said the Wiz. "Couldn't have said it better myself! And of course if (j1,j2,j3) is not admissible, the same argument shows
\ /
j1\ /j2
_\_/_
| |
| |
| S | = 0
| |
|_____|
|
j3|
|
since there isn't anything nonzero for it to be!
Okay, great. Now what?"
"Spin networks with a total of 4 edges coming in or going out?" asked John.
"Yes. But first, a little point of business: given that
\ / \ /
j1\ /j2 j1\ /j2
_\_/_ \ /
| | \ /
| | |
| S | = c |
| | |
|_____| |
| |
j3| j3|
| |
we should really get a nice formula for the constant c.
How can we do that?"
"Take the trace of both sides," said Miguel, "and solve for c."
"That's basically right," said the Wiz. "Of course, we usually define the trace for an operator from a vector space to itself, but here we can close up the diagrams and get numbers as follows:
___ ___ ___ ___
/ \ / \ / \ / \
| j1\ /j2 | | j1\ /j2 |
| _\_/_ | | \ / |
| | | | | \ / |
| | | | | | |
| | S | | = c | | |
| | | | | | |
| |_____| | | j3| |
| | | | | |
| j3| | | | |
| | | | | |
\_____/ \_____ / \______/ \______/
so solving for c and plugging it back in, we get
\ / ___ ___ \ /
j1\ /j2 / _\_/_ \ j1\ /j2
_\_/_ | | | | \ /
| | | | S | | \ /
| | | |_____| | |
| S | j1| j3| |j2 |
| | \____|____/ |
|_____| = ________________ |
| _____ |
j3| / | \ j3|
| / j3| \ |
| \ | / |
| j1\__|__/j2 |
Cool, eh? Physicists call this the Wigner-Eckhart Theorem."
The class cheered. It was fun to watch the Wizard prove results by literally grabbing ahold of diagrams made from strands of pure light, then bending and manipulating them according to rules they had worked out in previous classes. Jay particularly enjoyed it whenever the Wiz teased apart a spin-j strand into its constituent spin-1/2 strands, which were woven together like the filaments of a cable, but completely indistinguishable, thanks to the symmetrizer.
The Wizard began to ham it up. "And now for my next trick...."
"Spin networks with 4 strands coming in or out!"
"Right! Here, let me make one:
a\ b| /c
\ | /
\ | /
_\_|_/_
| |
| |
| S |
| |
|_______|
|
k|
|
I'm getting sick of subscripts, so I'll call the spins a, b, c, and k.
Don't ask me why I chose those letters!
Now, what does this equal? What could it equal?"
The class thought a minute.
"Well," said Miguel, "it could equal something like this:
a\ b/ c/
\ / /
\/ /
\ /
j\ /
\ /
|
|
k|
Or any multiple of this!"
"Right. But what's j?"
"Just some spin... such that (a,b,j) and (j,c,k) are admissible."
"Right! Of course, there may be a lot of choices. So for the first time, we must resort to linear combinations. In other words, we can try to show that
a\ b| /c a\ b/ /c
\ | / \ / /
\ | / \ / /
_\_|_/_ \/ /
| | \ /
| | j\ /
| S | = sum c(j) \ /
| | j |
|_______| |
| |
k| k|
| |
for some coefficients c(j).
In fact, this is true for a unique choice of such coefficients!
In other words, these spin networks
a\ b/ c/
\ / /
\/ /
\ /
j\ /
\ /
|
|
k|
form a basis of spin networks from a (x) b (x) c to k.
I won't prove this here, but it's not hard.
By symmetry, these also form a basis:
a\ b\ c/
\ \ /
\ \/
\ /
\ /i
\ /
|
|
k|
The matrix for changing from one basis to the other
is called the "6j symbols", since it depends on 6 spins:
a\ b/ c/ a\ b\ c/
\ / / \ \ /
\/ / \ \/
\ / \ /
j\ / = sum {a b i} \ /i
\ / i {c k j} \ /
| |
| |
k| k|
Can we work out a formula for these 6j symbols?
Yes! Our spin network technology makes it easy.
Here's how it goes. First we close up both sides of the equation like this:
_________ _________
/ \ / \
k| | k| |
| | | |
/ \ | / \ |
/ \m | / \m |
/ \ | / \ |
/ / \ | / /\ |
/ / \ | / / \ |
/ / \ | = sum {a b i} / / \ |
a\ b/ c/ | i {c k j} a\ b\ c/ |
\ / / | \ \ / |
\/ / | \ \/ |
\ / | \ / |
j\ / | \ /i |
\ / | \ / |
| | | |
k| | k| |
\_________/ \_________/
Then we use Schur's Lemma on the right-hand side: it says that
| ______ |
m| / \ i|
| /\ | |
| b/ \c |i |
/ \ \ / | |
b/ \c \/ | |
/ \ \______/ |
\ / = _______________ |
\ / _____ |
\ / / \ |
| / \i |
| \ / |
i| \_____/ |
| |
if m = i, and
|
m|
|
/ \
b/ \c
/ \
\ / = 0
\ /
\ /
|
i|
|
otherwise.
So all the terms in the sum vanish except for m = i, and we get
_________ _________
/ \ / \
k| | k| |
| | | |
/ \ | / \ |
/ \i | / \i |
/ \ | / \ |
/ / \ | / /\ |
/ / \ | / / \ |
/ / \ | = {a b i} / / \ |
a\ b/ c/ | {c k j} a\ b\ c/ |
\ / / | \ \ / |
\/ / | \ \/ |
\ / | \ / |
j\ / | \ /i |
\ / | \ / |
| | | |
k| | k| |
\_________/ \_________/
______ ______
/ \ / \
/\ | /\ |
b/ \c |i a/ \i |k
\ / | \ / |
\/ | \/ |
\______/ \______/
= {a b i} ______________________________
{c k j} _____
/ \
/ \i
\ /
\_____/
Solving this for the 6j symbols, we get:
_________
/ \
/\i |
/ \ | _____
/ /\ | / \
a/ b/ \c |k / \i
\ / / | \ /
\/ / | \_____/
\ / |
j\/ |
\_________/
{a b i} = ___________________________________________
{c k j}
______ ______
/ \ / \
/\ | /\ |
b/ \c |i a/ \i |k
\ / | \ / |
\/ | \/ |
\______/ \______/
Voila!"
The class erupted in applause, and the Wizard took a low bow and tipped his hat.
"Now, this little guy in the numerator:
_____
/ \
/ \i
\ /
\_____/
is just the superdimension of the spin-i representation --
plus or minus 2i+1, depending on whether i is an integer or half-integer.
The guys in the denominator:
______ ______
/ \ / \
/\ | /\ |
b/ \c |i a/ \i |k
\ / | \ / |
\/ | \/ |
\______/ \______/
are called "3j symbols" or "theta nets", because we can
redraw them to look like the Greek letter theta:
_____ _____
/ b \ / a \
/ \ / \
/_________\ /_________\
\ c / \ i /
\ / \ /
\_____/ \_____/
i k
These are the simplest closed spin networks, important but not so exciting.
The real star of the show is this guy:
_________
/ \
/\i |
/ \ |
/ /\ |
a/ b/ \c |k
\ / / |
\/ / |
\ / |
j\/ |
\_________/
If you stare at it, perhaps you can see how it summarizes the
process of going from
a\ b/ /c
\/ /
j\ /
\ /
|
k|
to
a\ b\ /c
\ \/
\ /i
\ /
|
k|
Later we'll see this process is related to the "associator" --
the isomorphism
(a (x) b) (x) c -> a (x) (b (x) c)between two ways of parenthesizing a triple tensor product. But for now, let's ponder its geometric meaning:
_________
/ \
/\i |
/ \ |
/ /\ |
a/ b/ \c |k
\ / / |
\/ / |
\ / |
j\/ |
\_________/
What shape is this?"
The class looked at it suspiciously, never having seen a shape quite like this.
"Hmm. What if I bend the edge labelled "k" back behind the rest...
/|\
/ | \i
/ k| \
/ | / \
/ |/ \
a/ / \c
\ b/| /
\ / | /
\/ | /
\ | /
j\ | /
\|/
Recognize it now?"
Toby frowned. "Don't you have to worry about minus signs when you twist those trivalent vertices like that?"
"True," said the Wiz, "But don't let a measly minus sign get in the way now! Does anyone recognize this shape?"
The class continued to stare.
"Hey! It's a tetrahedron!" cried Miguel.
"Right!" said the Wiz. "Here, I'll bend it around some more, to make it obvious:
/|\
/ | \
/ | \i
/ k| \
/ | \
a/___________\
\ b /
\ | /
\ | /c
j\ | /
\ | /
\|/
See?"
"Wow!" said Jay. "Now that's what I call a tetrahedron."
"Indeed. So: apart from some puny fudge factors, the 6j symbols come from evaluating a tetrahedral spin network -- sometimes called a "tet net". Even better, there's a deep relation between the 6j symbols, the associator, and the tetrahedron!
This is why 3d quantum gravity works so nicely. In any quantum field theory there's an important number called the "partition function", which depends on the choice of spacetime. With suitable finagling, all physics can be extracted from this. In 3d quantum gravity, to compute the partition function of a 3d spacetime we first chop it into tetrahedra and label all the edges with spins. This gives a bunch of tetrahedral spin networks! Then we evaluate each of these tetrahedra using our rules and multiply all the numbers we get. Finally, we sum over all labellings. Modulo some details I don't want to talk about yet, this gives the partition function. The really cool part is that this number is independent of how we chopped spacetime into tetrahedra! Miraculously, the 6j symbols satisfy just the right identities to make this work. I want to explain this, and explain why this "miracle" is no accident. But it will take a while: you'll need to learn some category theory to really understand the relation between 6j symbols, associators, and tetrahedra.
Okay, time for a break!"