Solving phi-fourth theory using Fourier analysis

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The equation of motion for phi-fourth theory is nonlinear due to the presence of the cubic term on the right-hand side, making it unsuitable for straightforward Fourier analysis. Attempting to apply Fourier transforms results in a complex triple integral, complicating the problem rather than simplifying it. This transformation leads to a nonlinear integral equation, which does not provide a direct solution. Ultimately, perturbation theory is necessary to approach the solution of the equation. Thus, Fourier analysis is not an effective method for solving the phi-fourth theory equation.
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The equation of motion of ##\phi^4## theory is ##(\partial^{2}+m^{2})\phi = -\frac{\lambda}{3!}\phi^{3}##.

Why can't this equation be solved using Fourier analysis? Can't we simply write the equation in Fourier space and take it from there?
 
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failexam said:
The equation of motion of ##\phi^4## theory is ##(\partial^{2}+m^{2})\phi = -\frac{\lambda}{3!}\phi^{3}##.

Why can't this equation be solved using Fourier analysis? Can't we simply write the equation in Fourier space and take it from there?

Well, the right-hand side of the equation is nonlinear, so performing a Fourier transform will produce a mess---a triple integral. You end up replacing a nonlinear differential equation by a nonlinear integral equation. That might be a good first step, but it doesn't solve the equation. You would have to resort to perturbation theory in any case.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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