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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.

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  • $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Jul 2, 2018 at 5:29
  • $\begingroup$ @DaftWullie By the way, could you just tell me why you thought that the $l-1$ th qubit should be entangled with Reg M? $\endgroup$
    – Sanchayan Dutta
    Jul 2, 2018 at 6:16
  • $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Jul 2, 2018 at 6:27
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    $\begingroup$ @Nelimee That only depends on the index $p$, which is on the register $M$. It has nothing to do with $L$. $\endgroup$
    – DaftWullie
    Jul 2, 2018 at 7:51
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    $\begingroup$ 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. $\endgroup$
    – Craig Gidney
    Jul 6, 2018 at 15:47

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