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I am currently studying the paper "Decomposition of unitary matrices and quantum gates (2012)" and referring to the textbook Quantum Computation and Quantum Information. Among the topics, I am particularly focused on understanding the decomposition of a $4 \times 4$ unitary matrix, but there are certain parts that I find challenging to grasp.

Source: arXiv:1210.7366

In the process of decomposing an arbitrary unitary matrix $U$ into multiple unitary matrices, I am unsure why we start by setting the entry in the third row and first column of $U_1U$ to zero. Additionally, I don't understand the reason behind the specific form of the $U_1$ matrix that is used to make the entry in the third row and first column zero.

To rephrase briefly, why do we start by setting the entry in the third row and first column of $U_1U$ to zero, and why is the form of $U_1$ chosen in such a way to achieve this?

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That paper appears to do their rotations in a very strange order. The method you're interested in is how to use Givens rotations to perform a QR decomposition (see, e.g. https://en.wikipedia.org/wiki/QR_decomposition#Using_Givens_rotations). The conventional ordering is that you start by setting the bottom-left element to zero, then work up the first column until all but the diagonal element are zero. Then you start from the bottom of the second column and work up, and so on.

Why do you choose $U_i$ a particular way? Consider $$ U_i=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos\theta & \sin\theta \\ 0 & 0 & -\sin\theta & \cos\theta \end{array}\right). $$ When you calculate $U_iU$, what's its effect? Notice, first, the the first two rows of the output are the same as the first two rows of $U$. What does the fourth row look like? $-\sin\theta r_3+\cos\theta r_4$ where $r_3,r_4$ are the bottom rows of $U$. So, the trick is that there will always be a linear combination that zeros the element we want, you just have to pick the correct value of $\theta$. The side-effect is that, because $U_1$ has to be unitary, the third row gets messed up a bit. But that'll get sorted in the next iteration.

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  • $\begingroup$ In the paper, they mention that the matrix in the solution was used to zero out the entry in the 3rd row and 1st column. However, based on what you said, does it mean that with the right choice of theta, we can zero out either the entry in the 3rd row and 1st column or the entry in the 4th row and 1st column? $\endgroup$
    – junghyunHa
    Nov 29, 2023 at 9:05
  • $\begingroup$ Yes, the standard ordering would be to zero out 4th row, 1st column, then 3rd row 1st column, then 2nd row 1st column. For some reason, the paper chooses to do it in a different order. I have no idea why. $\endgroup$
    – DaftWullie
    Nov 29, 2023 at 9:15
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Note the following: if we have a vector with 2 entries $\begin{bmatrix} a \\ b \end{bmatrix}$, then we can use a unitary matrix to change the second entry to $0$. Namely, set $r = \sqrt{|a|^2 + |b|^2}$ (the 2-norm of the vector), $c = \frac{a}{r}$, $s = \frac{b}{r}$, then $$\begin{bmatrix} \bar{c} & \bar{s} \\ -s & c \end{bmatrix} \begin{bmatrix} a \\ b\end{bmatrix} = \begin{bmatrix}r \\ 0\end{bmatrix}$$ and it is easy to check that the matrix is unitary.

This means that if we have a vector in $\mathbb{C}^d$, then we can set any entry to $0$ by only changing 2 entries — just put this small matrix into the rows we want to affect and the corresponding columns. And with repeated application of this we can first set the last coordinate to $0$, then the penultimate one, and so on, until only the first entry is nonzero. This technique is usually called Givens rotations.

Then we apply this to the first column of a matrix and then apply the same procedure to the block in lower right, and iterate.

In the paper they slightly modify the rotation matrix to have determinant $\mu$ instead of $1$, and choose a fixed order of rows and columns (this is why they first set the entry (3,1) to zero, their example uses order 1,2,4,3, so we first work on column 1 and set first entry 3 to zero, then entry 4, then entry 2), but the idea is the same.

What we have in the paper is a simple modification of the standard Givens rotation algorithm for QR decomposition, and I suggest that you only reference this paper mentioning this. The fact that this paper was published and does not mention Givens rotations is a shame for the current academic publication system.

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  • $\begingroup$ Thank you for your response! I'll study about QR decomposition. $\endgroup$
    – junghyunHa
    Nov 29, 2023 at 9:05

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