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In a lot of quantum computing formalism, it is relatively easy to create $\vert 0\rangle$, $\vert 1\rangle$, $\vert +\rangle$ and $\vert -\rangle$. However, it is hard to create $\vert i\rangle$. Why is this so?

For example, here it is claimed that transversal initialization works for eigenstates of $X$ and $Z$ gates but one has to do complex stuff for the $Y$ gate eigenstates. Fundamentally, why is it harder?

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In CSS codes, the reason this happens is because the Z observable is protected by the Z stabilizers and the X observable is protected by the X stabilizers. But the Y observable is protected by the combination of the X and Z stabilizers.

When you transversally initialize in X, you get all the X stabilizers right away, so you can protect the X observable.

When you transversally initialize in Z, you get all the Z stabilizers right away, so you can protect the Z observable.

When you transversally initialize in Y, you typically get none of the stabilizers right away. So you can't protect anything when initializing in the Y basis.

In some CSS codes you can form Y stabilizers as small products of X and Z stabilizers. In these codes you may learn enough to start protecting something. For example, in the color code the X and Z stabilizers are right on top of each other so you can take their product to form Y stabilizers. When you transversally initialize into Y, you get these small-product stabilizers. And you can keep computing them from products of stabilizer measurements in future rounds. That's enough to protect the Y observable. So the color code ends up with transversal Y initialization. But it relies on this special can-get-protective-Y-stabilizers-from-products property, which is not shared by all CSS codes.

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  • $\begingroup$ Just quick question. We are still able to prepare Y basis states with H and S gates and these are Clifford ones. So, what is meant by it is hard to prepare? $\endgroup$ Commented May 18 at 6:45
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    $\begingroup$ @MartinVesely Hard as in it is more expensive; takes more time or space. Codes where Y prep is non-trivial also won't have trivial H and S gates. $\endgroup$ Commented May 18 at 11:14

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