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I want to construct fields from unitary representation of the Poincaré group but I do not know how. In Weinberg book he proposed that the Hamiltonian should be of certain kind and from that he derived the fields is there any another way to that.

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This answer explains how exactly to construct field operator so that it represents one-particle state. If your question is about the first principle from which we naturally introduce field operators, then I"ll rewrite an answer.

First, read my answer on this question. Here I'll add some technical details for the most simple case - fields which realizes irreducible massive representations of the Poincare group with integer spin (the other cases are realized analogically, but harder).

As was claimed in the linked answer, the basis of getting relativistic wave equations (in principle this is exactly the way to construct relativistic wave field) is that subspace of massive unitary representations of the Poincare group with nonzero mass $m$ and spin $s$ is characterized by two Casimir operators, $$ \tag 1 \hat{P}_{\mu}\hat{P}^{\mu} = m^{2}\cdot\text{id}, \quad \hat{W}_{\mu}\hat{W}^{\mu} = -m^2s(s+1)\cdot \text{id} $$ For the representation of creation/destruction fields $\hat{P}_{\mu} = i\partial_{\mu}$, and the first operator is just Klein-Gordon differential equation: $$ (\partial^{2}+m^2)\hat{\Phi}_{A}^{\pm} = 0 $$ As for the second condition, the way to construct it in a form of differential equation is harder. For the most convenient way to introduce e need to look for the irreducible representations of the Lorentz group $(A/2, B/2)$, which is realized by completely symmetric spinors $\psi_{a_{1}...a_{A}\dot{b}_{1}...\dot{b}_{B}}$. It can be shown that corresponding field operator $\hat{\Psi}^{\pm}_{a_{1}...a_{A}\dot{b}_{1}...\dot{b}_{B}}$ with $(A+B)/2 = s$ realizes one-particle massive representation of the Poincare group (in a sense of $(1)$) if $$ \tag 2 \begin{cases} (\partial^{2} + m^{2})\hat{\Psi}^{\pm}_{a_{1}...a_{A}\dot{b}_{1}...\dot{b}_{B}} = 0\\ \partial^{a\dot{b}}\hat{\Psi}^{\pm}_{aa_{1}...a_{A-1}\dot{b}...\dot{b}_{B-1}} =0 \end{cases} $$ (as you know, massive unitary representations with spin S are characterized by the $SO(3)$ little group, co corresponding field operator must have $2S+1$ component).

It can be even shown that representations $(A, B)$ and $(A-1,B+1)$ are equivalent, and corresponding equivalence operator is given by the operator $\Delta_{a}^{\ \dot{a}} = \frac{1}{m}\delta_{a}^{\ \dot{a}}$. Thus for representation of one-particle state with integer spin $s = 2S$ we may conveniently use representations $\left( S, S\right)$ of the Lorentz group, which can be given in a vector form by the well-known isomorphism $$ \tag 2 V_{\mu_{1}...\mu_{S}} = \frac{1}{2^{s}}(\sigma_{\mu_{1}})^{a_{1}\dot{b}_{1}}...(\sigma_{\mu_{S}})^{a_{1}\dot{b}_{S}}V_{a_{1}...a_{S}\dot{b}_{1}...\dot{b}_{S}} $$ With $(3)$, $(2)$ is converted into $$ \begin{cases} (\partial^2 + m^2)\hat{\Phi}_{\mu_{1}...\mu_{S}} = 0 \\ \partial^{\mu}\hat{\Phi}_{\mu...\mu_{S-1}} = 0 \\ \hat{\Phi}_{\mu_{1}...\mu_{S}} = \hat{\Phi}_{(\mu_{1}...\mu_{S})} \\ g^{\mu \nu}\hat{\Phi}_{\mu \nu ... \mu_{S-2}} =0 \end{cases} $$ Here $(...)$ means symmetrization.

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