Understanding the (2,3) threshold quantum secret sharing scheme in Cleve et al. 1999

Question

In the paper "How to share quantum secret" by R.Cleve et al (arxiv).

There is an example of secret sharing using qutrit in a (2,3) threshold scheme. $$\alpha\vert0\rangle+\beta\vert1\rangle+\gamma\vert2\rangle\mapsto$$ $$\alpha(\vert000\rangle+\vert111\rangle+\vert222\rangle)+\beta(\vert012\rangle+\vert120\rangle+\vert201\rangle)+\gamma(\vert021\rangle+\vert102\rangle+\vert210\rangle)$$

Where the secret encoding maps the initial qutrit as shown above, and any combination of the two can be used to determine the secret qutrit back.

Given the first two shares (for instance), add the value of the first share to the second (modulo three), and then add the value of the second share to the first, to obtain the state below.

$$(\alpha\vert0\rangle+\beta\vert1\rangle+\gamma\vert2\rangle)\vert00\rangle+\vert12\rangle+\vert21\rangle)$$ I am unable to understand what exactly is happening here, any help would be really appreciated.

Answer 1

add the value of the first share to the second (modulo three), and then add the value of the second share to the first

means you first perform, on the 2 shares, the 2-qutrit unitary which maps

$$|00\rangle\to|00\rangle\\ |01\rangle\to|01\rangle\\ \ldots\\ |22\rangle\to|21\rangle$$ (i.e. adding the first share to the second modulo 3, so $2+2 = 1 \mod 3$, hence the mapping of 22 to 21)

Then you perform the unitary which adds the second to the first $\mod 3$: $$|00\rangle\to|00\rangle\\ |01\rangle\to|11\rangle\\ \ldots\\ |22\rangle\to|12\rangle$$

Apply both and, e.g., you map the term $\beta|120\rangle\to\beta|100\rangle\to\beta|100\rangle$. The encoded state turns into the final state you show in your question, with the secret reproduced on the first qutrit.