Timeline for How to figure out the action of a circuit form its effect on logical operators?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 26 at 12:42 | comment | added | DaftWullie | 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 12:38 | comment | added | am567 | Yes, and then when you find the product of both terms divided by $2$: $\frac{1}{4}(I - Y \otimes Y)(I + Z \otimes X)$ and then find the $+1$ eigenvector of this product, I get $|00\rangle + |01\rangle + |10\rangle + |11\rangle$ as opposed to $\frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$? (Apologies for continued confusion) | |
| Feb 26 at 9:27 | history | edited | DaftWullie | CC BY-SA 4.0 |
added 13 characters in body
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| Feb 26 at 9:27 | comment | added | DaftWullie | It's the same factor of two (there's one for each term in the product) | |
| Feb 23 at 15:48 | comment | added | am567 | I thought we had already divided by $2$ for this reason when we found the projector of $\Pi_{i}(I + \bar{Z}_{i})/2 $ ? Or was there a different reason for dividing by $2$ at that stage? | |
| Feb 22 at 16:35 | comment | added | DaftWullie | 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 16:25 | comment | added | am567 | Actually, when I find the eigenvector corresponding to the $+1$ eigenvalue of the projector $\frac{1}{4}(I - Y \otimes Y)(I + Z \otimes X)$, I get $|00\rangle + |01\rangle + |10\rangle + |11\rangle$. Where does the $\frac{1}{2}$ multiplier come from? | |
| Feb 22 at 14:47 | comment | added | am567 | 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 14:44 | comment | added | am567 | Sorry, I think I understand. So if $\bar{Z_{i}}^{2} |\psi\rangle = \bar{Z_{i}}(\bar{Z_{i}} |\psi\rangle) = \bar{Z_{i}}(\lambda |\psi\rangle) = \lambda^{2} |\psi \rangle = I |\psi \rangle$, then $\lambda^{2} = 1 \rightarrow \lambda=\pm 1$ | |
| Feb 22 at 14:44 | comment | added | DaftWullie | 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 14:37 | comment | added | am567 | 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 14:26 | vote | accept | am567 | ||
| Feb 22 at 13:10 | history | answered | DaftWullie | CC BY-SA 4.0 |