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[tex]\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3}(a_p e^{-i\omega_p t + i\mathbf{x}\cdot \mathbf{p}}+a_p^\ast e^{i\omega_p t - i\mathbf{x}\cdot \mathbf{p}})[/tex]

Now to get this I expandand [itex]\Box \ \phi = (\partial_t^2 - \nabla^2)\phi[/itex] and took the Fourier transform so that we get the equation [itex](\partial_t^2 + \omega_p^2)\hat{\phi}(\mathbf{p},t)=0[/itex] where [itex]\omega_p = |p|[/itex].

This equation now has [itex]\mathbf{p}[/itex] just as a parameter and we can easily solve it since it is just the simple harmonic oscilator equation of motion. We end up with

[tex]\hat{\phi}(\mathbf{p},t)= a_p e^{-i\omega_p t}+b_p e^{i\omega_p t}[/tex]

If we now use the Fourier inversion formula we have that

[tex]\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3}\hat{\phi}(\mathbf{p},t)e^{i\mathbf{p}\cdot \mathbf{x}}=\int \dfrac{d^3\mathbf{p}}{(2\pi)^3}(a_p e^{-i\omega_p t}+b_p e^{i\omega_p t})e^{i\mathbf{p}\cdot \mathbf{x}}.[/tex]

This is almost the result, but we need, however, to ensure [itex]\phi[/itex] is a real field. For that we need to apply the reality condition to the Fourier transform:

[tex]\hat{\phi}(\mathbf{p},-t)=\hat{\phi}^\ast(\mathbf{p},t)[/tex]

This provides us with

[tex]a_p e^{i\omega_p t} + b_p e^{-i\omega_p t} = a_p^\ast e^{i\omega_p t} + b_p^\ast e^{-i\omega_p t}[/tex]

Now I'm stuck here. I'm not getting how from this condition I can arrive naturally at the textbook answer. How can I proceed this to arrive at the same solution that the textbook presents?