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...
