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Say I have a code, for example the $ [[5,1,3]] $ code, and I want to (fault tolerantly) prepare the logical $ |+ \rangle $ state. Is that any harder than preparing the logical $ | 0 \rangle $ state? Or what if I want to (fault tolerantly) measure in the logical $ |+ \rangle $ basis (in other words measure in the basis consisting of the $ 32 $ simultaneous eigenvectors of $ XXXXX, XZZXI, IXZZX, XIXZZ, ZXIXZ $). Is that any harder than measuring in the logical $ | 0 \rangle $ basis (in other words measuring in the basis consisting of the $ 32 $ simultaneous eigenvectors of $ ZZZZZ, XZZXI, IXZZX, XIXZZ, ZXIXZ $)?

In general my question is that when it comes to actually (fault tolerantly) implementing a quantum circuit is it any harder to work with (i.e. do state preparation and measurement) the $ |+ \rangle $ basis as opposed to the the $ |0 \rangle $ basis? For a code with transversal Hadamard it should be easy to fault tolerantly switch between the $ |+ \rangle $ basis and the $ |0 \rangle $ basis and so they should be equally easy to work with. But what about a code like the $ [[5,1,3]] $ that doesn't have transversal Hadamard?

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It entirely depends on the code you are using. For CSS codes, Z-basis and X-basis states are typically equally easy. They both have transversal initialization, the CNOT treats them symmetrically, etc.

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  • $\begingroup$ How about the $ [[5,1,3]] $ code I gave in the example? Does that still have transversal initialization of some states? Could you maybe explain more about why transversal initialization of Z-basis and X-basis states always works for a CSS code? $\endgroup$ Commented Jul 4 at 22:16
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$[[5, 1, 3]]$ is tri-symmetric for $X,Y,Z$ basis so I believe it won't showcase any bias for different basis choices.

For CSS codes (without transversal $H$), the answer seems quite apparent when your $d_X\neq d_Z$. A quick example is the repetition code where we only have Z stabilizers. Logical $|0\rangle$ is easier to prepare as it's just the tensor product of physical $|0\rangle$.

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  • $\begingroup$ What about a CSS code like the $ [[15,1,3]] $ triorthogonal code? That does not have X,Y,Z symmetry from the Facet gate or X,Z symmetry from the Hadamard gate. Then how does working with $ | + \rangle $ basis compare to working with $ | 0 \rangle $ basis? $\endgroup$ Commented Jul 5 at 14:33

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