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I always see the form of the field operators derived by, in the case of a scalar spin 0 particle, imposing the field commutation relations on the classical field solutions of the Klein Gordon equation and interpreting te Fourier coefficients of the field, Fourier expanded in momentum space, as operators which will then turn out to be annihilation and creation operators for particles in momentum eigenstates.

This procedure also works for the photon field and the electron field since both the Wave equation and the Dirac equation, like the Klein Gordon equation, are linear equations that can also be Fourier expanded in momentum space.

When considering interactions the equations are no longer linear but in this case the ingenious "interaction picture" saves the day. The field operators in the hamiltonian operator that is used to calculate the time evolution of the quantum states are just the free-field operators.

I am however wondering if it is possible to derive the exact form of the field operators in a more direct way, that could also work for nonlinear equations that do not have a free field version and a corresponding interaction picture.

Maybe we could use the Heisenberg equation of motion in combination with the commutation relations?

Let's look at scalar field theory:

The hamiltonian is,

$H=\int\frac{1}{2}\pi'^{2}+\frac{1}{2}(\triangledown\phi')^{2}+\frac{1}{2}m^{2}\phi'^{2}dx'^{3}\tag{1}$

Indicating the primed symbols as the ones over which we intgrate, the Heisenberg equation of motion in natural units is:

$\frac{\mathrm{d} \phi }{\mathrm{d} t}=i\left[ \int\frac{1}{2}\pi'^{2}+\frac{1}{2}(\triangledown\phi')^{2}+\frac{1}{2}m^{2}\phi'^{2}dx'^{3},\phi \right ]\tag{2}$

Using the equal time commutation relations this finally gives me:

$\frac{\mathrm{d} \phi}{\mathrm{d} t}=\pi\tag{3}$

so

$\frac{\mathrm{d} \phi}{\mathrm{d} t}=\frac{\partial \phi}{\partial t}\tag{4}$

It seems to be saying the the field operator doesn't explicitly depend on $x$, $y$ and $z$. I don't know where to go from here though.

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  • $\begingroup$ Qmechanic's answer is correct on the informal level, but there are caveats. Interaction picture doesn't exist nonperturbatively due to Haag's theorem. Defining these operators (or, more accurately, operator-valued distributions) in general is a very hard problem. There exist examples of well-defined QFTs in spacetimes of dimension 2 and 3, but afaik all known 4d QFTs are either non-interacting, or only valid as a perturbative asymptotic series. It has long been conjectured that Yang-Mills theory for compact semisimple gauge groups is well-defined in 4d, but it remains to be proven. $\endgroup$ – Prof. Legolasov Sep 20 at 5:57
  • $\begingroup$ @Prof.Legolasov: Right. I should probably have included some QFT disclaimers :) $\endgroup$ – Qmechanic Sep 20 at 12:32
  • $\begingroup$ Yeah I did take Qmechanic's hints into account, which lead me to a more sensible equation than I had initially. Maybe my question is unclear, if so I'll redo it, but I'm trying to ask if its possible to figure out how the field operator acts on for example a one particle state in a momentum eigenstate, without using the ansatz of just writing down the normal solution to the wave equation and interpreting the Fourier coefficients as annihilation and creation operators because this approach is limited to linear equations... $\endgroup$ – Stijn Boshoven Sep 20 at 20:11
  • $\begingroup$ It looks to me like all the Heisenberg equation of motion is telling me is that the field operator does not explicitly depend on spatial variables which is correct, but not enough information $\endgroup$ – Stijn Boshoven Sep 20 at 20:11
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Hints:

  1. The Hamiltonian $H(t)=\int d^3x~{\cal H}({\bf x},t)$ is a spatial integral of the Hamiltonian density ${\cal H}({\bf x},t)$.

  2. The Hamiltonian density ${\cal H}$ of a real scalar Klein-Gordon field is ${\cal H}~=~\frac{1}{2}\pi^2 + \frac{1}{2}(\nabla\phi)^2 + \frac{1}{2}m^2\phi^2$.

  3. The equal-time CCR reads $[\phi({\bf x},t),\pi({\bf y},t)]~=~i\hbar {\bf 1}\delta^3({\bf x}-{\bf y}).$

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  • $\begingroup$ Thanks so much, I'll see if I can figure it out now, gimme a sec. $\endgroup$ – Stijn Boshoven Sep 18 at 17:54

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