
Mathematics of Topological Quantum Computing - seycombi
https://arxiv.org/abs/1705.06206
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zmgehlke
So they touch on 4-dimensional theories only briefly, but one thing I'm very
curious about is if higher dimensional theories provide a richer computational
structure, in any of several senses.

Every 1-knot has a sort of 2-knot analog (actually, a whole family of them)
made by "revolving" the knot, but there are 2-knots not generated in this
manner. (The same holds true for higher dimensional knots.)

As we go higher, there are knot moves not even possible in lower dimensions
(such as twists).

Just a thought.

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Entalpi
Someone explain this to as you would a five year old.

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kmill
My approximate understanding: Some problems reduce to calculating the trace of
a matrix. The matrix might be really complicated, but sometimes you can
approximate it by having particles braid around each other and measuring how
likely the particles end up in a particular state. In theory it can be put
into practice using materials called "topological insulators."

The Jones polynomial is a thing which can tell knots apart. The quantum
computer can calculate the Jones polynomial by having particles trace out the
knot. The Jones polynomial usually takes exponential time to calculate.

The "topological" in "topological quantum computing" refers to the braiding
and the fact that they're computing topological invariants of a space.

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p1esk
I think he meant a 5 year old _without_ a PhD in mathematics.

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thret
"Elementary particles are elementary excitations of the vacuum, which explains
why they are identical." (pg 4 ref 3) Is this now accepted fact, everything is
a wave and there are no particles?

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nnfy
This isn't really a question, all particles exhibit both wavelike and particle
like behaviors. What matters is the scale (mass and energy) which determines
which behavior will dominate.

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gaze
Has someone found a particle for which some exchange yields something like a
magic state? I've been told that known nonabelian anyons can only be exchanged
up to Cliffords

