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Let's define $P_k \in \{ I, X, Y, Z \}^{\otimes n}$ and called each of these $P_k$ as a Pauli string (or word) then given that $$U = \sum_{k=1}^L c_kP_k $$ with the following conditions:

  1. $\sum_{k=1}^L |c_k|^2 = 1 $
  2. $Im(c_i \cdot c_j) = 0$
  3. $\{P_i, P_j\} = 2 \delta_{i,j}$ where $\{ \}$ indicates the anti-commuting relationship. In other words, the Pauli strings $P_i$ in the sum that made up $U$ are anti-commute.

With the above three conditions, $U$ is a unitary operator and thus can be implemented as a quantum circuit.

In the following paper https://arxiv.org/pdf/1907.09040.pdf at the end of page 3, it shows that $U$ can be implemented as the product of $2L-1$ exponents of these Pauli strings (or words) in the following way:

$$ U = \prod_{k = 1}^L e^{i \theta_k P_k/2} \prod_{k = L}^1 e^{i \theta_k P_k/2} $$

where $\theta_k = \arcsin\bigg(\dfrac{c_k}{\sum_{j=1}^k c_j^2 } \bigg) $. Is there a neat proof of this fact or is it just full out algebraic manipulation which gets really messy? That is, is there an intuitive way to see why this is true? Since the authors seem to indicate this is a pretty trivial fact.


Bonus: What is the best implementation for $U$ in term of the quantum circuit outside of the proposed method above?

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  • $\begingroup$ You've missed out an arcsin from your definition of $\theta_k$. $\endgroup$
    – DaftWullie
    Mar 28 at 7:07

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You can prove this by induction. Start with the simplest non-trivial sequence $$ e^{i\theta_2 P_2/2}e^{i\theta_2 P_1}e^{i\theta_2 P_2/2}=e^{i\theta_2 P_2/2}(I\cos\theta_1+i\sin\theta_1 P_1)e^{i\theta_2 P_2/2}. $$ Now, $$ e^{i\theta_2 P_2/2}Ie^{i\theta_2 P_2/2}=e^{i\theta_2 P_2} $$ and, due to the anti-commutation, $$ e^{i\theta_2 P_2/2}P_1e^{i\theta_2 P_2/2}=P_1e^{-i\theta_2 P_2/2}e^{i\theta_2 P_2/2}=P_1. $$ Thus, the full sequence becomes $$ I\cos\theta_1\cos\theta_2+i\cos\theta_1\sin\theta_2 P_2+i\sin\theta_1P_1. $$

Now if we pre- and post-multiply by $e^{i\theta_3 P_3/2}$, then the anti-commutation will basically leave the Pauli-terms $P_1$ and $P_2$ unchanged, replacing $I$ with $e^{i\theta_3 P_3}$. And this sequence will keep repeating. You would be able to solve exactly for all the $\theta_i$ at that point, but assuming that the $\sin\theta_i$ are all small, you can approximate $\cos\theta_i\approx 1$ (I'm not convinced this is a good approximation once you have a large number of terms), which leaves you setting $\sin\theta_i=c_i$ up to normalisation.

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  • $\begingroup$ Thank you for your answer! I suppose the authors worked out a few terms first and that's how they got the proposed circuit construction as well. I didn't see the proof of it so I thought maybe this is really trivial and one can see it from intuition... and maybe it is for most people but I am just too slow. $\endgroup$
    – KAJ226
    Mar 28 at 14:03
  • $\begingroup$ A little late getting back since I just get around to try and implement this now... but what about the Identity operator and the imaginary coefficient floating around? If I have $U = c_1P_1 + c_2P_2$ then this does not give me back exactly $U$, does it? $\endgroup$
    – KAJ226
    Apr 2 at 16:18
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    $\begingroup$ If $\sum_i|c_i|^2=1$, then it will automatically come out that the coefficient of the identity (if you solve the exact problem, not the approximation) is 0, as required. $\endgroup$
    – DaftWullie
    Apr 4 at 6:46
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    $\begingroup$ I'm not sure how to deal with the complex values $\endgroup$
    – DaftWullie
    Apr 4 at 6:53
  • $\begingroup$ Thank you, @DaftWullie! Best wishes. $\endgroup$
    – KAJ226
    Apr 4 at 13:31

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