My goal is to prove that I can synthesise arbitrary unitary from two components.

In the end, I find a matrix with the form \begin{equation} \mathbf{W}_j=\begin{pmatrix} |\alpha|2\cos{(\phi_{k,2}+\phi_{k,1})}+i|\beta|2\cos{(\phi_{k,2}-\phi_{k,1})} & i|\alpha|2\sin{(\phi_{k,2}+\phi_{k,1})}+|\beta|2\sin{(\phi_{k,2}-\phi_{k,1})}\\ i|\alpha|2\sin{(\phi_{k,2}+\phi_{k,1})}+|\beta|2\sin{(\phi_{k,2}-\phi_{k,1})} & |\alpha|2\cos{(\phi_{k,2}+\phi_{k,1})}+i|\beta|2\cos{(\phi_{k,2}-\phi_{k,1})} \end{pmatrix} \end{equation} where $\phi_{k,1}$,$\phi_{k,2}$,$\alpha$ and $\beta$ are parameters.

Where can I begin to prove that it is an arbitrary unitary?

  • $\begingroup$ Do you have constraints on $\alpha,\beta$? $\endgroup$
    – DaftWullie
    Commented Mar 19, 2020 at 10:25
  • $\begingroup$ hi ! As far as I know, they should be both $\frac{1}{\sqrt{2}}$ $\endgroup$ Commented Mar 19, 2020 at 12:53
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    $\begingroup$ what you mean with "two components"? I don't see any "component" here. Also, you should clarify what the parameters mean, and which one in particular have fixed values $\endgroup$
    – glS
    Commented Mar 19, 2020 at 15:20
  • $\begingroup$ Is it supposed to be $|\alpha|^2$ rather than $|\alpha|2$? $\endgroup$
    – DaftWullie
    Commented Mar 19, 2020 at 15:35
  • $\begingroup$ @glS yeah i was not clear. Actually i want to prove that from two optical components (Pulse Shaper and Electro Optic modulator) i can create quantum gates, and doing a combinaison of those components give me this matrix, which in the end should synthesise gates $\endgroup$ Commented Mar 20, 2020 at 8:30

2 Answers 2


There are multiple ways to show that $W_j$ is not, in general, unitary. The easiest might be to look at the determinant. A basic property of unitary matrices is that their determinant has unit modulus. In this case the determinant reduces to $$\text{det}(W_j)=4 \, (\vert \alpha \vert^2 - \vert \beta \vert ^2 + i2 \, \vert \alpha \vert \vert \beta \vert \cos(2\phi_{k,2})).$$ There is nothing in the problem statement constraining this value to unit modulus, and I don't see any natural constraints that can be imposed to make it so.

There are many other ways to show that $W_j$ is not generally unitary. Hopefully it's clear that if $W_j$ is not generally unitary, it's certainly not an arbitrary unitary.


The straightforward method is to compute $ W W^\dagger = W^\dagger W = I $ and to get constraint over your parameters solving this system. I'm going to show you how to do it only for $ W W^\dagger = I$ but it should be very similar for $ W^\dagger W$. I assume here that all your parameters are real.

first some notation for ease of reading, let's pose :

$W = \begin{bmatrix} A & B \\ C & D \end{bmatrix} $, $W^\dagger = \begin{bmatrix} A^* & C^* \\ B^* & D^* \end{bmatrix}, I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

this allows us to see easily that : \begin{align} AA^* + BB^* &= 1 \\ CC^* + DD^* &= 1 \end{align}

using the property over product of complex conjuguates ($zz^* = |z|^2$), the computation for these two equalities is pretty straightforward.

\begin{align} |A|^2 + |B|^2 &= 1 \\ |C|^2 + |D|^2 &= 1 \end{align}

I let you the details of the computation (The trick is to remember that $cos^2(x) + sin^2(x)=1$) but in the end you should get :

$$ |\alpha|^2+|\beta|^2 = \frac{1}{4}$$

for the two equalities. That's your first constraint over your matrix for it to be unitary.

Then you have the two following equalities :

\begin{align} AC^* + BD^* &= 0 \\ CA^* + DB^* &= 0 \end{align}

Those ones are a bit more expensive in time to compute, and I'm going to do only the first one. With some observations you can see that 4 terms cancels each other, making the computation quite fast. At the end you should get the equality (only difficulty is to remember the following formula, $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$) :

$$4 |\alpha||\beta| \sin(2 \phi_2) = 0$$

which has multiple solutions :

$(\phi_2 = z \frac{\pi}{2}, \alpha \in \mathbb{R}, \beta \in \mathbb{R}, z \in \mathbb{Z}),(\phi_2 \in \mathbb{R} , \alpha \in \mathbb{R}, \beta = 0), (\phi_2 \in \mathbb{R} , \alpha =0, \beta \in \mathbb{R}) $

You can do the same for

$$ CA^* + DB^* = 0$$

and it should give you a new equality constraint over your parameters.

For now we have the following non-linear system of constraint :

\begin{align} |\alpha|^2+|\beta|^2 & = \frac{1}{4} \\ |\alpha||\beta| \sin(2 \phi_2) & = 0 \end{align}

You now need to complete this system by finding new constraint with the last equality I left aside and the ones you'll get by computing $W^\dagger W = I$. I believe you'll get pretty close constraints, maybe one more over $\phi_1$.

In the end you will get a non-linear system, it should not be too hard to solve by hand, if it is for you, you can use the very useful online systems of equations solver from Wolframalpha. The solutions of this non-linear system will tell you what range your parameters can take such that your matrix is unitary.

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    $\begingroup$ I don't think that's what the question is asking. I think the question wants to prove not only that it is a unitary, but for any arbitrary single-qubit unitary that you specify, that it can be written in this form by finding suitable values of the parameters. $\endgroup$
    – DaftWullie
    Commented Mar 19, 2020 at 15:41
  • $\begingroup$ I see it now. I hope my answer will still have some use somehow. $\endgroup$ Commented Mar 19, 2020 at 15:45
  • $\begingroup$ Ok thank you for your answer, it was actually the aim of my question (though it was quite unclear) I actually did kind of this method beforehand, and got to the same constraints, however, i was not sure that i was right Another question that might look silly : if i prove that my matrix is arbitrary, it does mean that i can synthesise any 2x2 matrix from it right ? I started a position as an intern in quantum information where i'm quite new, so i'm really glad that there is a blog like that to help me ! $\endgroup$ Commented Mar 20, 2020 at 8:28
  • $\begingroup$ you can't describe every 2x2 matrix with it , a simple example of matrix you can't describe is $\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}$. $\endgroup$ Commented Mar 20, 2020 at 8:39
  • $\begingroup$ @AntoineHenry You cannot, in fact, describe the vast majority of unitary matrices with it. For example, you should be able to see that within $SU(2) \subset U(2)$ this matrix can only describe a one dimensional subspace $SO(2) \subset SU(2)$. $\endgroup$ Commented Mar 20, 2020 at 22:08

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