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This is a sequel to How are two different registers being used as "control"?

I found the following quantum circuit given in Fig 5 (page 6) of the same paper i.e. Quantum Circuit Design for Solving Linear Systems of Equations (Cao et al.,2012).

enter image description here

In the above circuit $R_{zz}(\theta)$ is $\left(\begin{matrix}e^{i\theta} & 0 \\ 0 & e^{i\theta}\end{matrix}\right)$

As @DaftWullie mentions here:

Frankly, I've no chance of getting there because there's an earlier step that I don't understand: the output on registers $L, M$ after Figure 5. The circuit diagram and the claimed output don't match up (the claimed output being separable between the $L$ and $M$ registers, when qubit $l−1$ of register $L$ should be entangled with those of register M.

I understand that after the Walsh-Hadamard transforms the state of register $L$ is $$\frac{1}{\sqrt{2^l}}\sum_{s=0}^{2^l-1}|s\rangle$$

and that of register $M$ is

$$\frac{1}{\sqrt{2^m}}\sum_{p=0}^{2^m-1}|p\rangle$$

But after that, I'm not exactly sure how they're applying the $R_{zz}$ rotation gates to get to $$\sum_s\sum_p|p\rangle \exp(i p/2^m t_0)|s\rangle$$

Firstly, are all the $R_{zz}$ gates acting on a single qubit i.e. the $l-1$th qubit in the register $L$? (Seems so from the diagram, but I'm not sure).

Secondly, it would be very helpful if some can write down the steps for how're they're getting to $\sum_s\sum_p|p\rangle \exp(i p/2^m t_0)|s\rangle$ using the controlled rotation gates.

Sanchayan Dutta
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  • Your statement of $R_{zz}$ can’t be right! I think there are problems with the paper. Unless you get lucky and find someone on here who’s specifically interested in that paper (which is not me!) you’re probably better off directly contacting the authors. – DaftWullie Jul 02 '18 at 05:29
  • @DaftWullie By the way, could you just tell me why you thought that the $l-1$ th qubit should be entangled with Reg M? – Sanchayan Dutta Jul 02 '18 at 06:16
  • because there are lots of controlled gates acting from a superposition of states on register M onto qubit l-1. That generically is going to create entanglement, whereas the stated output can be created just by acting phase gates on the qubits of M and Not using register L at all. It probably is that phase which is created, but only if qubit $l-1$ is in the $|1\rangle$ state. – DaftWullie Jul 02 '18 at 06:27
  • @DaftWullie "whereas the stated output can be created just by acting phase gates on the qubits of M and Not using register L at all": How do you introduce the $p$ in $e^{ip/2^mt_0}$ without L? – Adrien Suau Jul 02 '18 at 07:43
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    @Nelimee That only depends on the index $p$, which is on the register $M$. It has nothing to do with $L$. – DaftWullie Jul 02 '18 at 07:51
  • is there are a reason why you are not linking to the latest version of the paper? Also, should this also be tagged with hhl-algorithm? – glS Jul 06 '18 at 12:30
  • @glS A lot of things have been changed in v3 of the paper. v2 deals with a $4\times 4$ system, whereas v3 deals with a $2\times 2$ system. As for the HHL tag, I forgot. Adding it now – Sanchayan Dutta Jul 06 '18 at 12:32
  • that's weird... if the authors themselves removed some details, I wouldn't trust those details to be correct – glS Jul 06 '18 at 12:34
  • @glS Nelimee spotted the mistakes in the paper and simulated the algorithm on QISKit after making some modifications. His code is indeed giving the correct outputs for the system of 4 simultaneous linear equations, which is given in the paper. From what he told me, the part I quoted in the question doesn't have any error. – Sanchayan Dutta Jul 06 '18 at 12:36
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    Those controlled $R_{zz}(\theta)$ operations don't make any sense. They're strictly more complicated than necessary, because they're equivalent to just applying an $R_z(\theta)$ gate to the control without involving the target. – Craig Gidney Jul 06 '18 at 15:47

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