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.

Do you have constraints on $\alpha,\beta$? – DaftWullie – 2020-03-19T10:25:00.677

hi ! As far as I know, they should be both $\frac{1}{\sqrt{2}}$ – Antoine Henry – 2020-03-19T12:53:03.707

1what 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 – glS – 2020-03-19T15:20:51.447

Is it supposed to be $|\alpha|^2$ rather than $|\alpha|2$? – DaftWullie – 2020-03-19T15:35:51.150

@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 – Antoine Henry – 2020-03-20T08:30:53.940

@AntoineHenry note that it is encouraged to edit your post to add all of this information. Comments are not made to stay, so whatever information is important to the post should be added in the question – glS – 2020-03-20T08:37:52.730