The unique representation of Heisenberg Group. How can we construct the unique-up-to-isomorphism irreducible representation of Heisenberg Group.

The link you've provided in the comments on Dietrich Burde's answer illuminates the question quite significantly.  The definition of Heisenberg Group there is $H(V)=V\oplus U(1)$ for some symplectic vector space $(V,\omega)$.  Here $\omega$ is a nondegenerate, skew-symmetric, bilinear form.  The group law is:
$$(v,z)\cdot (w,u)=(v+w,\exp(\frac{i}{2}\omega(v,w))zu)$$
Next, choose a Lagrangian subspace $L\subset V$, which is a maximal isotropic subspace with respect to the form $\omega$.  With such a choice, we can define the Hilbert space $\mathcal{H}_L$ as done so in the link (the completion of a space of smooth, $L^2$ functions satisfying some translation identity).  It is on this Hilbert space that we define the action of $H(V)$ to be:
$$(v,z)\cdot \phi(x)=z\phi(x-v)\exp(\frac{i}{2}\omega(v,x))$$
Notice that group elements $(0,z)$ act via scalar multiplication.  This representation can also be viewed as being induced from the irreducible $1$-dimensional representation of the subgroup $L\oplus U(1)$ via projection onto $U(1)$.  I'm not exactly sure how the equivalence works.
Now, as claimed in the link, it seems that any representation of $H(V)$ in which $U(1)$ just acts as scalar multiplication is equivalent to the one constructed here, and so up to isomorphism, our representation is independent of our choice of Lagrangian, $L\subset V$.
Not sure if this is any more helpful.  As others have noted, it would be faster in the future if you would give more context for your questions.  As it was stated, it was very difficult to see exactly what you were asking.
