8
$\begingroup$

The Toric code Hamiltonian is:

$\sum_{x,y}\left( \prod_{i\in p(x,y)} Z_{ixy} + \prod_{i\in v(x,y)} X_{ixy} \right),$

where the $v$ and $p$ are defined according to this picture (courtesy of James Wooton's contribution to Wikipedia):

Toric Code Lattice courtesy of James Wooton from Wikipedia

At the moment we have an infinite 2D lattice:

$x\rightarrow \pm \infty$
$y\rightarrow \pm \infty$.

But if we set periodic boundary conditions such that (and feel free to edit the question if I am incorrect about this):

$p(x+10,y)=p(x,y)$
$v(x,y+10)=v(x,y)$,

We get the follownig torus (image courtesy of James Wooton's contribution to Wikipedia) :

Toric Code Torus courtesy of James Wooton from Wikipedia

Now in my periodic boundary conditions, I chose to add $+10$ but could have added some other number instead. How does this "size of the torus" affect the function of the toric code?

$\endgroup$
5
$\begingroup$

The Toric code is an error correcting code. The distance of the code (I.e. the number of local operations required to convert one logical state into an orthogonal one) is equal to $N$, where the Toric code is defined on an $N\times N$ grid.

One of the places that the performance of the Toric code really wins out is that although it is only distance $N$, the vast majority of sets of $N$ single qubit errors can be corrected, and it is only once you get $O(N^2)$ errors that you get killed. That means that as $N\rightarrow\infty$, those $O(N)$ terms vanish, and you get a finite per-qubit error rate as a threshold for error correction. For finite $N$, the error correcting threshold will be lower.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.