Claire's paper

Using my basis instead of Claire's. Her "tau-tau-tau" vertex is the fourth root of phi times mine.

q = e^i phi/5

Irrational ancilla

This is on page 7. I ran my program on claireirrational.txt
compare:

| n |
% %
% %
% %
| y |
| y

and

(5q + 4q^3 + q^5 + 7q^7) || +
(7q^2 + q^4 + 4q^6 + 5q^8) H

Yes, they're equal!

Those coefficients obviously have the same norm. 

 What's their ratio? I get:

(5 + 4q^2 + q^4 + 7q^6)
over
(7 + q^2 + 4q^4 + 5q^6)
=
(12 + 10q^2 + 3q^4 + 6q^6)/11

This is obviously not a root of unity. I think the only roots of unity in Q[e^i pi/5] are the tenth roots of unity. Otherwise they'd generate a cyclotomic field for a root of unity of order a multiple of 10, which would have to have a larger Euler totient function, hence larger degree extension of Q.

Bell procedure

I'd call this an "unreliable Pauli sigma_x gate". It's only for use on ancillas.


It works, according to clairebell.txt

Irrational phase gate

This is the section "from ancilla to gate".

Start with (input tensor ancilla).

Cap the middle two strands. Then cap the middle two strands again. Done.

First recovery

Suppose the first cap actually fused the two strands into one. The paper said something about measure the last four strands and try again.

Or...

Bring the right two strands together. If they cap then, after one F move, we have

1/phi times input tensor <1-qubit | ancilla >

If they fuse then, after two F moves, we have

1/phi times split-split-input tensor <tau-qubit | ancilla>.

Then do a couple of fuses. In either case, we're back to the input.

Second recovery

Suppose the first cap worked but the second fused.

Introduce a cap between the third and fourth strand. Take a measurement from the middle of that cap to the far right. If you measure empty then after a couple of F-moves you have

1/phi times H-input tensor <tau-qubit | ancilla>

That "H-input" is the input with a letter H on its right two strands. We'll deal with that later.

Now suppose you measured tau.

Ugh - look like circular logic...

Fibonacci anyons

I'm trying to understand Claire's quantum gates for Fibonacci anyons. There's one kind of particle (not counting the vacuum), and trivalent vertices. She told me the F-matrix, and I get the key relation

Claire: H = sqrt(phi) cup-cap - 1/sqrt(phi) ||

I fudge my vertex = phi^(-1/4) times Claire's. I get

Then I get the Jones-Wenzl at q=exp(i pi/5). A crossing is

sigma = q^3 || + q^-3 cup-cap

An fuse-split (like a letter I) is equal to the Jones-Wenzl idempotent p_2.

Kuperberg-esque relations:

bubble = phi
digon = 1
triangle = -1/phi
square = 1/phi^2 (cup-cap + ||)
pentagon = -1/phi (break-one-side + break-the-two-adjacent-sides-instead)

It's a graph planar algebra. The colors are 0 and 1, with weights phi and 1 respectively. For a vertex with colors 001, the 1 gets 180 degrees. There are fudge constants

000: -1/phi
001: 1

Bowling balls

I'm quixotically trying to prove the volume conjecture using my bowling ball representation of braid groups.

N is a number.
q is e^(2 pi i / N)
[k] is 1 + ... + q^(k-1).
[k]! is [1]*...*[k]
[k choose r] is [k]! / [r]![k-r]!

We have a positive braid, made of lanes of a bowling alley. Each lane can hold up to N-1 balls. The chance c balls fall when a pass over b is f^a_b(c) =

[a choose c] * [N-1-b choose c] * [c]! * (1-q)^c * q^(a-c)(N-1-b-c).

I want to let N go to infinity, and have "liquid bowling balls".

So, what's the argument of f^a_b(c)?

• [k] = sqrt(q)^k-1 |[k]|
• [k]! = sqrt(q)^kchoose2 |[k]!|
• [k choose r] = sqrt(q)^r(k-r) |[k choose r]|
• 1-q = sqrt(-q) |1-q| = sqrt(q)^(1 - N/2) |1-q|

Altogether I get the fourth root of q to the power of

c(2b + c + 3) - a(4b + 2c + 4) + cN    (mod 4N)

Now let's figure out the change of argument when you move one strand from the right to the left side of a whole region of the braid diagram.

For a crossing on the right, a and c go down

2a + 2b + 1 - N 

On the left, b goes up and c goes down

-2a - 2b - 3 - N

On the top, a and c go up, b goes down

2a - 2b - 2c - 1 + N

On the bottom, c goes up

2pi(-c/N - 1/4)

-2a + 2b + 2c + 3 + N

--------------------------------------------------------------------
c(a-c)theta + c(N-b-c)theta + c-choose-2 theta + (theta-pi/2)c + 2(a-c)(N-b-c)theta
= pi/2N times -4ab - 2ac + 4aN + 2bc + c^2 - 3cN + c

Now let's figure out the change of argument when you move one strand from the right to the left side of a whole region of the braid diagram.

For a crossing on the right of the region, a and c goes down and b stays constant. So let's take a kind of negative derivative:

pi times -1/2 + a/N + b/N - 1/2N

For a crossing on the left side of the region, b goes up and c goes down.

pi times 3/2 - a/N - b/N - 1/2N

For the crossing on the bottom, c goes up.

pi times -3/2 - a/N + b/N + c/N + 1/2N

For the crossing on the top, a goes up, b goes down, c goes up.

pi times 1/2 + a/N - b/N - c/N + 1/2N

We want the derivative on each face to be zero. Ignore the 1/N terms.


For sigma1 sigma2 sigma1, with a,b,c at the bottom, d,e,f at top, x,y,z in the middle reading from bottom to top, I get

x = -3/2 + b + e + f
y = 3/2 + d - b - c
z = -3/2 + b + c + e

For sigma2 sigma1 sigma2

x = -1/2 + b + d + e
y = 1/2 + f - a - b
z = -1/2 + a + b + e

The fudges seem to be wrong. Call them L,R,T,B.

sigma1 sigma2 sigma1:
x = b+e+f + (R+T+B)
y = a-e-f - (R+T+B)
z = e+b+c + (R+T+B)

sigma2 sigma1 sigma2:
x = b+d+e - (T+L+B)
y = f-a-b + (T+L+B)
z = a+b+e - (T+L+B)

It all seems to work if T+B = LR = 0.

There must be a mistake.

Let [N]=0, and a'=a-c, b'=b+c. I get
f^a_b(c) = [a choose a'] [b' - 1]! (1-q)^(b'-b) q^(-a'b') / [b-1]!
Does that help?

welcome

Since I have a google account through the department, I may as well use it to blog. Think of this as a private journal that I'm leaving lying around for no particular reason.