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

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?

• I'm not exactly sure what you're asking... are you asking how you are supposed to find the state, given that you know it's a $+1$ eigenstate of both $\bar Z_1$ and $\bar Z_2$? Feb 22 at 12:37
• Yes, I don't understand how we obtain the mapping $|00\rangle \rightarrow \frac{1}{2}(|00\rangle + |01\rangle - |10\rangle +|11 \rangle)$ from the fact that it starts as an eigenvector of $Z_{1}$ and $Z_{2}$ and ends up as the $+1$ eigenvectors of $\bar{Z_{1}}$ and $\bar{Z_{2}}$. However, I do understand how we map $|01\rangle, |10\rangle$ by applying $\bar{X_{1}}, \bar{X_{2}}$ to the image of the first mapping. Feb 22 at 12:46

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*}
• The initial conditions, that we have a $+1$ eigenstate of a set of mutually commuting $\bar{Z}$ which satisfy $\bar{Z}^{2}=I$...where are these criteria coming from? Is it a theorem that tells us the the eigenvalues are $\pm1$? Specifically, why do we require that they must be mutually commuting? Feb 22 at 14:37
• We start by describing our initial state as the $+1$ eigenstate of mutually commuting projectors that square to identity. For example, if you start with the all zeros state, you have to operators $Z_i$. You can then prove that all those properties are preserved under unitary evolution. For example, if $Z^2=I$, then $(UZU^\dagger)^2=UZU^\dagger UZU^\dagger=UZ^2U^\dagger=I$. Feb 22 at 14:44
• Ah I see, so because we follow the unitary evolution of the logical operators, the properties of $Z_{1}, Z_{2}$ before the circuit are preserved and held by $\bar{Z_{1}}, \bar{Z_{2}}$ at the end of the circuit Feb 22 at 14:47
• The eigenvalues of $Z\otimes X$ are $\pm 1$, so the eigenvalues of $I+Z\otimes X$ are 0,2. We divide by 2 to make the eigenvalues 0,1. Feb 22 at 16:35
• Any eigenvector $|\psi\rangle$ is only defined up to normalisation: if $M|\psi\rangle=\lambda|\psi\rangle$ then $M(\alpha|\psi\rangle)=\lambda(\alpha|\psi\rangle)$. We just pick a convention for the normalisation such that the state is normalised (has length 1). Feb 26 at 12:42