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

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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?

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    $\begingroup$ 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 $\bar Z_2$? $\endgroup$
    – DaftWullie
    Feb 22 at 12:37
  • $\begingroup$ 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. $\endgroup$
    – am567
    Feb 22 at 12:46

1 Answer 1

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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*}

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  • $\begingroup$ 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? $\endgroup$
    – am567
    Feb 22 at 14:37
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    $\begingroup$ 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 $(UZU^\dagger)^2=UZU^\dagger UZU^\dagger=UZ^2U^\dagger=I$. $\endgroup$
    – DaftWullie
    Feb 22 at 14:44
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    $\begingroup$ 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 $\endgroup$
    – am567
    Feb 22 at 14:47
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    $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Feb 22 at 16:35
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    $\begingroup$ 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). $\endgroup$
    – DaftWullie
    Feb 26 at 12:42

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