The Appendix to a recent paper Graph States as a Resource for Quantum Metrology states:

We model an $n$ qubit graph state $G$ undergoing iid dephasing via $$G \to G^{\text{dephasing}} = \sum_{\vec{k}}p^{k}(1-p)^{n-k}Z_{\vec{k}}GZ_{\vec{k}}$$ where $p$ is the probability that a qubit undergoes a phase flip. This effectively maps the graph state onto the orthonormal basis $\{Z_{\vec{k}}|G\rangle\}_{\vec{k}}$.

Question: Is it clear why this expression is given in this form (in particular what does the $Z_{\vec{k}}$ denote and how do we know that $\{Z_{\vec{k}}|G\rangle\}_{\vec{k}}$ is an orthonormal basis)?

These might be trivial questions but I am new to this area hence some of the conventions I am still unfamiliar with.


1 Answer 1


Imgine that there is a bit string $\vec{k}\in\{0,1\}^n$. We use this to specify sites (bit value 1) where an error has occurred, and sites (bit value 0) where no error has occurred. The number of 1s in the bit string is $k$. The probability of this particular error arising is then $p^k(1-p)^{n-k}$ because there are $k$ sites with an error and $n-k$ sites with no error. The error is decribed as $Z_{\vec{k}}$, i.e. specifying which qubits (bit values 1) the Pauli $Z$s are applied to.

As for why $Z_{\vec{k}}|G\rangle$ are all orthogonal, consider the inner product of two: $$ \langle G|Z_{\vec{k}}Z_{\vec{l}}|G\rangle $$ Now, we can write the graph state as $U|+\rangle^{\otimes n}$, where $U$ corresponds to the set of controlled-phase gaes that entangle the system. Phase gates commute with $U$, so $U^\dagger Z_{\vec{k}}=Z_{\vec{k}}U^\dagger$, so this means that the inner product is $$ \langle+|^{\otimes n}Z_{\vec{k}}Z_{\vec{l}}|+\rangle^{\otimes n}. $$ Provided $\vec{l}\neq\vec{k}$, this is clearly 0 because there is at least one site where there is an inner product $\langle+|-\rangle$.

  • $\begingroup$ Why is it that for dephasing and quantum erasures of graph states, the error is described by applying Pauli $Z$ operators to the graph state? $\endgroup$
    – John Doe
    Commented Nov 6, 2019 at 15:13
  • 1
    $\begingroup$ Dephasing is Z errors. Erasure is basically setting a qubit state to 0, which we can write as $I+Z$. $\endgroup$
    – DaftWullie
    Commented Nov 6, 2019 at 17:11
  • $\begingroup$ @DaftWullie Still not very understand, does partial trace out the erasure part is the same as what you said, i.e., $I + Z$ ? $\endgroup$
    – narip
    Commented Apr 11, 2021 at 0:51

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