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  • In this paper, http://arxiv.org/abs/1301.4504 in equation 4.1 in what sense are the two states a "9-qubit state"? I did not understand this counting.

  • Can someone explain what are the different $X_i$ and $Z_i$ in 4.2? How is say $X_1$ Pauli matrix different from $X_8$ and so on?

  • And what is the non-trivial thing that happened in equation 4.2? (aren't the two kets on the LHS and the RHS the same?)

It would be great to hear of any other general insights/explanation people might have about this section 4.

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  • $\begingroup$ Seems to me that since $|000\rangle$ is a 3-cubit state, then the product of 3 of these should give a 9-cubit state. $\endgroup$
    – Kyle Kanos
    Commented Jul 21, 2014 at 15:53

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$X_i,Y_i,Z_i$ are three Pauli matrices acting on the $i$-th qubit where $i=1,2,3,4,5,6,7,8,9$ labels the qubit. In equation 4.1, the state is a superposition of tensor product of three states similar to $|000\rangle$. The latter is a state of three qubits, so if one takes the third power, it is a state of $3\times 3 = 9$ qubits.

$X_1$ differs from $X_8$ by the fact that it only nontrivially acts on the first qubit, the first 0 or 1 from the left, while it acts as the identity operator on the eighth qubit. For $X_8$, the action on the two qubits is reversed.

The equation 4.2 doesn't have an LHS or a RHS.

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  • $\begingroup$ @Lubos Motl In 4.2 I am not getting the notation. How is $\vert \bar{c} \rangle$ different from $\vert c_1, c_2, .., c_k, 0..(n-k)...0 \rangle$ ? I am not understanding what does 4.2 mean. $\endgroup$
    – user6818
    Commented Jul 21, 2014 at 17:50
  • $\begingroup$ Hi @user6818, you mean equation 4.3 not 4.2, right? The barred-c is different because it is an element of a smaller Hilbert space than the broader Hilbert space into which the right hand side vectors belong. Therefore it's an "encoding transformation", as explained in the text around the equation. Do you read the paper as well or are you just taking equations out of the context? $\endgroup$
    – Luboš Motl
    Commented Jul 22, 2014 at 6:30
  • $\begingroup$ @LubošMotl So on a generic vector what would the encoding transformation do? Is it just a trivial projection operator? $\endgroup$
    – user6818
    Commented Jul 24, 2014 at 20:52
  • $\begingroup$ The coding in 4.3 only defines it for special vectors that end with n-k zeroes. So it is legitimate to say that your question is meaningless. One doesn't have to define U_enc more generally. But yes, just ignoring the last n-k coordinates is the most natural generalization of a transformation that would act on the whole space $\endgroup$
    – Luboš Motl
    Commented Jul 25, 2014 at 10:34

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