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If we look at canonical quantization approach to QFT, we see that particle numbers are accounted for by means of state vectors, while amplitudes sit in the corresponding field operators. Is it possible to have all of it in one place?

For example, for complex Klein-Gordon field $(\partial^2+m^2)\phi = 0$ is it possible to write something like $$\phi(x) = \sqrt{2} \frac{\exp[-i(px)]}{\sqrt{2E_\vec{p}V}} + \sqrt{1} \frac{\exp[-i(kx)]}{\sqrt{2E_\vec{k}V}}$$ for 2 particles with momentum $\vec{p}$ and one particle with momentum $\vec{k}$?

If we work out energy and momentum for this field configuration, we get:

$$E =\sum_{\vec{q}}E_{\vec{q}}(|a_{\vec{q}}|^2 + |b_{\vec{q}}|^2) = 2E_\vec{p} + E_\vec{k}$$ $$\vec{P} = \sum_{\vec{q}}\vec{q}(|a_{\vec{q}}|^2 + |b_{\vec{q}}|^2) = 2\vec{p} + \vec{k}$$

So, field's energy and momentum are consistent with our expectations. Why then we use creation/annihilation operators to account for chnage in particle numbers during reactions?

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  • $\begingroup$ "particle numbers are accounted for by means of state vectors, while amplitudes sit in the corresponding field operators" That doesn't sound right. It's no different from ordinary QM. In Schrodinger picture the state vector accounts for everything. In Heisenberg picture the field operators account for everything. $\endgroup$
    – knzhou
    Commented Nov 14, 2019 at 6:56
  • $\begingroup$ What you've done is neither like Schrodinger picture nor Heisenberg picture. For example, you've set the field operator equal to a number, which it isn't at all. $\endgroup$
    – knzhou
    Commented Nov 14, 2019 at 6:57
  • $\begingroup$ I'n not trying to make use of operators. Just want the field configuration for a couple of particles to be a regular function. $\endgroup$ Commented Nov 14, 2019 at 9:11

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