Decomposing a 4x4 unitary matrix

[CMJ96]

Homework Statement

I want to decompose the following matrix into a product of two level matrices $V_i$

$$ \begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & \frac{-\sqrt{3}}{2} & 0 & \frac{-1}{2} \\
\frac{\sqrt{3}}{2} & \frac{-1}{4} & 0 & \frac{\sqrt{3}}{4} \\
\frac{1}{2} & \frac{\sqrt{3}}{4} & 0 & \frac{-3}{4}
\end{bmatrix} $$

Homework Equations

I have only been given the 3x3 case, which I would like to extend to 4x4, in the 3x3 case the decomposition looks like
$$ U_3 U_2 U_1 U = I_n $$
Where $I_n$ is the identity matrix.
$$ U= V_1 V_2 V_3 $$
Where $V_i = U_i ^{\dagger}$

The Attempt at a Solution

I need to eliminate each entry below the $u_{i=j}$ terms (if that makes any sense).
Since $u_{2,1}$ is $0$ I can set $U_1 = I_n$, for $u_{3,1}$ I start to run into trouble, I know that for a 3x3 matrix I can eliminate this term by setting $U_2$ to
$$ \begin{bmatrix}
\frac{u_{1,1}^*}{n} & 0 & \frac{u_{3,1}^*}{n} \\
0 & 1 & 0 \\
\frac{u_{3,1}^*}{n}& 0 & -\frac{u_{1,1}^*}{n}
\end{bmatrix} $$
Where $n=\sqrt{u_{1,1}^2 + u_{3,1}^2 }$
I have attempted to expand this to a 4x4 matrix, and this is what I got
$$ \begin{bmatrix}
\frac{u_{1,1}}{n} & 0 & 0 & \frac{u_{4,1}}{n}\\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\frac{u_{4,1}}{n} & 0 & 0 & -\frac{u_{1,1}}{n}
\end{bmatrix} $$
Where $n= \sqrt{u_{1,1}^2+ u_{4,1}^2 }$
Is this along the right lines?

 

[StoneTemplePython]

I'd look for simple solutions here... since you know the 3x3 case, what about converting your problem into the 3x3 case?

i.e.

$\text{Original Matrix} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & \frac{-\sqrt{3}}{2} & 0 & \frac{-1}{2} \\ \frac{\sqrt{3}}{2} & \frac{-1}{4} & 0 & \frac{\sqrt{3}}{4} \\ \frac{1}{2} & \frac{\sqrt{3}}{4} & 0 & \frac{-3}{4} \end{bmatrix} = QP$

where

$Q = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{-\sqrt{3}}{2} & 0 & \frac{-1}{2} \\ 0& \frac{-1}{4} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{4} \\ 0& \frac{\sqrt{3}}{4} & \frac{1}{2} & \frac{-3}{4} \end{bmatrix}$

and $P$ is a well chosen permutation matrix. (Note all permutation matrices are orthogonal, which is a special kind of unitary).

This implies that $Q$ is unitary (why?). The reason this setup is nice is that you now have a separable blocked structure that conveniently has a submatrix $Q_s$ which is unitary and 3x3

$Q = \begin{bmatrix} 1 & \mathbf 0^* \\ \mathbf 0 & Q_s\\ \end{bmatrix}$

and hence you already know how to solve the problem
- - - -
If this approach is too far afield let me know. If you have questions about the approach, I'm happy to clarify. Finding nice blocked structures can simplify a lot of results about matrices.

 

[CMJ96]

I see, should I only apply my 3x3 method of decomposition to the $Q_s$ matrix?

 

[StoneTemplePython]

I see, would I only apply my 3x3 method of decomposition to the $Q_s$ matrix?

right. So if you find, for example: $Q_s = V_1V_2$
or something like that, then it could be written as

$\begin{bmatrix} 1 & \mathbf 0^* \\ \mathbf 0 & Q_s\\ \end{bmatrix} = \begin{bmatrix} 1 & \mathbf 0^* \\ \mathbf 0 & V_1V_2\\ \end{bmatrix}$

and if you carefully follow the blocked multiplication, you'll see
$\begin{bmatrix} 1 & \mathbf 0^* \\ \mathbf 0 & V_1V_2\\ \end{bmatrix}= \begin{bmatrix} 1 & \mathbf 0^* \\ \mathbf 0 & V_1\\ \end{bmatrix} \begin{bmatrix} 1 & \mathbf 0^* \\ \mathbf 0 & V_2\\ \end{bmatrix}$

 

[CMJ96]

Ahhh yes, I have applied my method to the 3x3 block in the $Q$ matrix, the $U_1$, $U_2$ and $U_3$ when multiplied together give $I_n$ which looks good, as it happens $U_1$ and $U_2$ don't change when taking $U_i^{\dagger}$ so this has simplified things a bit.

I probably should have mentioned this earlier, I have to use this decomposition to design and construct a quantum circuit that has to be implemented on IBM's quantum experience composer, will this method work fine with that?
In particular I have to give a realisation of each matrix V as controlled operations