Why Is This the Ground State of a Toric Code?

I am currently trying to study the ground state of the Toric Code. I am currently reading this paper. The Hamiltonian is given by the following, where $$A_s$$'s are the star operators made out a tensor product of Pauli X matrices and $$B_p$$'s are the plaquette operators made out of a tensor product of the Pauli Z matrices:

$$H = -\sum_s A_s-\sum_pB_p$$

I can clearly see that all these operators $$A,B$$'s commute with one another and therefore each star (and plaquette) operator commutes with the Hamiltonian. Hence the star (plaquette) operators and the Hamiltonian are simultaneously diagonalizable.

The paper mentions that one of the ground states is given by the following:

$$\propto \prod_s (1 + A_s)|0\rangle$$

where $$|0\rangle$$ is a tensor product of k single-qubit states $$|0\rangle$$, where the dimensions of the square lattice is $$k \times k$$.

Question 1: Why is this a ground state?

I applied the Hamiltonian to this state. For instance, consider hitting the ground state with the operator $$A_{s'}$$, where $$s'$$ is a particular site. Since all the star operators commute, I see that

$$A_{s'}\prod_s (1 + A_s) |0\rangle = \bigg[ \prod_s (1 + A_s)\bigg]A_{s'}|0\rangle$$.

I see that $$A_{s'}|0\rangle$$ creates a state where the qubits on the links adjacent to site $$s'$$ are flipped from $$0$$ to $$1$$. Thus when I apply $$\sum_s A_s$$ to the claimed ground state, the resulting state is a sum of states of the form $$\prod_s(1 + A_s)| \text{mixture of 0s and 1s}\rangle$$

I also see that applying $$\sum_p B_p$$ to the claimed ground state gives me a sum of states, where each state/term is the the claimed ground state.

When I combine these together, I do not see the why the result is the ground state. Can anyone show me (eg. a proof) why the above ground state formula is the ground state of the toric code?

Question 2: How can I generate the other ground states of the Toric Code with the above formula for the ground state?

The crucial point it seems you are missing is to recognize that $$A_s^2=1$$ (for that matter, $$B_p^2=1$$ also) and therefore $$(1+A_s)/2$$ is a projector onto the $$+1$$ eigenspace of $$A_s$$. We would then have $$A_s(1+A_s) = (1+A_s)$$.

Thus the purported state should be thought of as "projecting the $$|000\cdots\rangle$$ into the simultaneous eigenstates of $$A_s=+1$$ for all $$s$$". (A subtlety is that we should check that this is not a vanishing state).

Assuming it is not the zero vector, this state $$|\psi\rangle$$ obeys $$A_s |\psi\rangle = +1 |\psi\rangle$$ by construction.

Now, since $$B_p$$s commute with all $$A_s$$s, we can move the action of a $$B_p$$ past all the projectors and onto $$|000\cdots \rangle$$. But clearly $$B_p|000\cdots\rangle = +1 |000\cdots\rangle$$. So $$|\psi\rangle$$ is also the +1 eigenstate of all $$B_p$$. Thus it is the ground state of the Hamiltonian as all terms are minimized.

To find other ground states, we should be careful and count the number of independent stars and plaquettes. On a torus, the product of all $$B_p = 1$$, and also the product of $$A_s = 1$$. This means that guaranteeing all local $$B_p = +1$$ and all local $$A_s=+1$$ does not uniquely specify a state and we need two more pieces of information (more precisely, we need to find 2 more operators which cannot be made from products of $$B_p$$s and $$A_s$$s which nevertheless commute with them and with each other). This means the ground state degeneracy on a torus is 4.

You can convince yourself that an operator $$G_1$$ made from a product of $$Z$$s which wraps around the torus vertically satisfies the above desired property. Similarly a string $$G_2$$ of $$Z$$s which wraps around the torus horizontally does too. Note $$G_1^2 = 1$$ and $$G_2^2=1$$ so they each have eigenvalues $$\pm1$$.

Therefore to get all four ground states of the topic code, we may further project the state we had before, $$|\psi$$, into definite quantum numbers of $$G_1,G_2$$, via the projection $$(1 + g_1 G_1)/2 \times (1+g_2 G_2)/2$$, with $$g_1,g_2 \in \{-1,+1\}$$.