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Given that operator $S_M$, which consists entirely of $Y$ and $Z$ Pauli operators, is a stabilizer of some graph state $G$ i.e. the eigenvalue equation is given as $S_MG = G$.

In the paper 'Graph States as a Resource for Quantum Metrology' (page 3) it states that the expectation value is given by

\begin{align} \langle S_M \rangle &= \text{Tr}(e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i}S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}G) \\ &= \text{Tr}(e^{i \theta \sum_{i=0}^{n}X_i}G). \end{align}

It seems that they are working in the Heisenberg picture and the above equations imply that $$S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i} = (S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i})^{\dagger} = e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i^\dagger}S_{M}^{\dagger} = e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i}S_{M},$$ but in order to do this I assumed that $S_M e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}$ is Hermitian. We only know that $S_M$ is Hermitian and unitary (being Pauli operators) and $e^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}$ is unitary. What am I missing that allows the above simplification?

Thanks for any help.

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    $\begingroup$ Surely this doesn’t work? The trace is invariant under cyclic permutations so in the first equation you move the last term to the front, and combine it with the first term to give the identity? $\endgroup$ – DaftWullie Nov 15 '19 at 6:35
  • $\begingroup$ @DaftWullie There was a typo there, apologies. The last term is the density operator $G$ (graph state) in the first equation. $\endgroup$ – John Doe Nov 15 '19 at 6:39
  • $\begingroup$ And X_i are Pauli X matrices on qubit i? $\endgroup$ – DaftWullie Nov 15 '19 at 6:58
  • $\begingroup$ @DaftWullie That's correct. $\endgroup$ – John Doe Nov 15 '19 at 7:00
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The way that I approach the calculation is quite different. I think of $$ e^{i\frac{\theta}{2}\sum_n X_i}=\prod_ne^{i\frac{\theta}{2}X_i}. $$ So, this means that you can consider the term $$ e^{i\frac{\theta}{2}\sum_n X_i}S_Me^{-i\frac{\theta}{2}\sum_n X_i} $$ on a qubit-by-qubit basis. You'll have terms like $$ e^{i\frac{\theta}{2}X}Ze^{-i\frac{\theta}{2}X}, $$ where $Z$ might be replaced by $Y$ (doesn't make any difference to the forthcoming analysis). Recall that $e^{i\frac{\theta}{2}X}=\cos\frac{\theta}{2} I+i\sin\frac{\theta}{2} X$. This lets us use the anti-commutation of $Z$ with $X$: $$ Z(\cos\frac{\theta}{2} I-i\sin\frac{\theta}{2} X)=(\cos\frac{\theta}{2} I+i\sin\frac{\theta}{2} X)Z. $$ This means that our per-qubit term has become $$ e^{i\frac{\theta}{2}X}Ze^{-i\frac{\theta}{2}X}=e^{i\frac{\theta}{2}X}e^{i\frac{\theta}{2}X}Z=e^{i\theta X}Z. $$ So, overall we have $$ \text{Tr}\left(e^{i\theta\sum_n X_i}S_MG\right), $$ at which point we just use that $S_MG=G$.

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  • $\begingroup$ Thanks for your answer. In the same section in the paper they refer to the "$S_M$ basis". Would you interpret this as the qubit-by-qubit basis consisting of eigenstates of the respective $Z$ and $Y$ Pauli operators (which would depend on how $S_M$ is defined)? $\endgroup$ – John Doe Nov 15 '19 at 8:08
  • $\begingroup$ @JohnDoe Without going through the paper carefully for context, I have no idea. $\endgroup$ – DaftWullie Nov 15 '19 at 8:26
  • $\begingroup$ In the interest of calculating the variance $$\Delta S^2_M = \langle S_M^2 \rangle-\langle S_M \rangle^2$$ would I be correct in stating that based on the above reasoning $\langle S^2_M \rangle$ gives terms of the form $$e^{i\frac{\theta}{2}X}Z^2e^{-i\frac{\theta}{2}X} = I$$ hence $$\langle S^2_M \rangle = \text{Tr}(e^{i\frac{\theta}{2}\sum_{i=0}^{n}X_i}S_Me^{-i\frac{\theta}{2}\sum_{i=0}^{n}X_i}G) = \text{Tr}(G)?$$ $\endgroup$ – John Doe Nov 26 '19 at 10:35
  • $\begingroup$ @JohnDoe Looks about right from a brief glance $\endgroup$ – DaftWullie Nov 26 '19 at 11:51

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