How to figure out the action of a circuit form its effect on logical operators?

Question

I am looking at the following circuit from "The Heisenberg Representation of Quantum Computers" by Daniel Gottesman.

I understand how the circuit acts on the logical operators. I understand how we got: $$\bar{X_{1}}= Z \otimes I$$ $$\bar{X_{2}} = - I \otimes Y$$ $$\bar{Z_{1}} = -Y \otimes Y$$ $$\bar{Z_{2}} = Z \otimes X.$$

I also understand that the circuit maps $$|00\rangle \rightarrow \frac{1}{2} (|00\rangle + |01 \rangle + |10 \rangle + |11 \rangle).$$ However, I only understood the above action of the circuit by converting the gates to their matrix form and seeing how they operate on the state vectors. I don't understand how to figure out this mapping from the action of the circui on the logical operators?

The paper states that: "the initial state $|00\rangle$ starts as an eigenvector of $Z \otimes I$ and $I \otimes Z$, with both eigenvalues $+1$. Therefore, after the network, it will still be the $+1$ eigenvectors of both $\bar{Z_{1}} = -Y \otimes Y$ and $\bar{Z_{2}} = Z \otimes X$. This we deduce that this network maps $$|00\rangle \rightarrow (|00\rangle + |01\rangle -|10 \rangle + |11 \rangle).$$ In addition, $|01\rangle = (I \otimes X)|00\rangle$, so $|01\rangle$ will map to $\bar{X_{2}}$ applied to the image of $|00\rangle$. Since $\bar{X_{2}} \rightarrow -I \otimes Y$, $$|01\rangle \rightarrow \frac{i}{2} (-|01\rangle + |00\rangle + |11\rangle + |10\rangle)$$

I don't understand how we got these mappings from the action of the circuit on the logical operators?

Answer 1

Most of the time, you don't actually need to calculate the state, but, sure, it's helpful when you're learning the topic to relate back to things you already know about.

The key ingredients that you have is that you state is a $+1$ eigenstate of a set of mutually commuting $\bar Z$ which all satisfy $\bar Z^2=I$. This tells you that the eigenvalues of $\bar Z$ are $\pm 1$. We can therefore construct a projector onto the +1 eigenstate by simplify evaluating $(I+\bar Z)/2$. The projector onto our state is then just $$ |\psi\rangle\langle\psi|=\prod_i\left((I+\bar Z_i)/2\right). $$ So, one option that you have is just to take $$ \frac14(I-YY)(I+ZX) $$ and find its eigenvector with +1 eigenvalue.

Alternatively, just guess a state $|\phi\rangle$. If you calculate $$ (|\psi\rangle\langle\psi|)|\phi\rangle=\langle\psi|\phi\rangle\ |\psi\rangle, $$ then so long as you got lucky enough that $\langle\psi|\phi\rangle\neq 0$, the output you get is proportional to the state you're trying to identify. For example, I could take $|\phi\rangle=|00\rangle$: \begin{align*} (I-YY)(I+ZX)|00\rangle&=(I-YY)(|00\rangle+|01\rangle) \\ &=|00\rangle+|01\rangle-(-|11\rangle+|10\rangle) \\ &=|00\rangle+|01\rangle-|10\rangle+|11\rangle. \end{align*}