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In page 32 of Peskin & Schroeder, it is written:

But if $j(x)$ is turned on for only a finite time, it is easiest to solve the problem using the field equation directly.

I am wondering:

  1. How is it possible to "solve" the field equation at all? I thought that once we make the quantization, the field equation is meaningless. It is only classically that it gives a field varying in time and space and quantum mechanically it would be what exactly? A differential equation for an operator?
  2. If I insisted to find the spectrum of the system using the methods used before, how would I go about it? I tried to compute the Hamiltonian by the definition: $$ H=\int d^3x[\pi(x)\partial_0\phi(x)-\mathcal{L(\phi)}] $$ where $\pi(x)=\partial^0\phi(x)$ and $\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2+j\phi$. Thus: $$ H=\int d^3x[\frac{1}{2}\pi^2+\frac{1}{2}(\nabla\phi)^2+\frac{1}{2}m^2\phi^2-j\phi] $$ If we denote by $H_0$ the free Hamiltonian, then we get: $$ H = H_0 -\int d^3xj(x)\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}(a_p e^{-i p_\alpha x^\alpha} +a_p^\dagger e^{i p_\alpha x^\alpha})$$ which clearly does not lead to the same expression between equations 2.64 and 2.65 (for one, the integration over $x$ is only 3-dimensional and we would need a 4-dimensional integral to get to $\tilde{j}(p)$.)
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  1. I think the usual way we approach this issue is to first obtain the field equations, in the classical manner, and obtain classical solutions by free wave expansion or Green's functions. For example in equation (2.55) and (2.60), you can see that the propagators, which are derived from quantum mechanics, have "classical" solutions, that is they satisfy the classical field equation (Klein-Gordon). So, your equation of motion will be eqn. (2.61), \begin{align} (\partial^2 + m^2)\phi(x) & = j(x) \end{align} Take the Fourier Transform of this to momentum space, and you obtain, \begin{align} (-p^2 + m^2) \tilde{\phi}(p) = \tilde{j}(p) \end{align} To solve this, you solve for the free Green's function, \begin{align} (-p^2 + m^2) \tilde{\phi}(p) = -i \end{align} Then after you obtain a solution for this $\tilde{\phi}_{\rm free}(p)$, do this: \begin{align} \phi(x) = \int \frac{d^4 p}{(2\pi)^4} \tilde{j}(p) \tilde{\phi}_{\rm free}(p). \end{align} Where is the quantization, you may ask. It's in the solution of the free Green's function.

  2. To explain the second part of your question, you need to include information about when your source was turned on. The $\phi(x)$ over time will also not be the one for the free field, since $\phi$ is a solution to the field equations, as we showed in 1. In other words, \begin{align} \phi(x) \neq \int \frac{d^3 p}{(2\pi)^3} a_{\vec{p}} e^{i p\cdot x} + a_{\vec{p}}^\dagger e^{-i p \cdot x} \end{align}.

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  • $\begingroup$ Thanks for your detailed answer. In point number 2, is there, then, a way to obtain the time dependent $\phi$ without using the field equation? Maybe the Heisenberg equation of motion.. $\endgroup$ – PPR Nov 11 '13 at 8:48
  • $\begingroup$ I believe you can, but the problem remains that you might not be able to solve for the Heisenberg equation of motion immediately, or that you will return to the same equation that you will have to solve classically. $\endgroup$ – tanzhongm Nov 18 '13 at 2:52

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