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The hamiltonian of the quantized Klein-Gordon field $\phi(\textbf{x},t)$ can be writting using the creation and annihilation operators: $$\hat{H} = \frac{1}{2} \int d^{3}\textbf{p} \ \omega_{p} (\hat{a}^{\dagger}_{\textbf{p}}\hat{a}_{\textbf{p}} + \hat{a}_{\textbf{p}}\hat{a}^{\dagger}_{\textbf{p}}),$$

where $\omega_{p} = \sqrt{\textbf{p}^{2} + m^{2}}$ (in natural units). I read that, if we use the commutation relation $[\hat{a}_{\textbf{p}}, \hat{a}^{\dagger}_{\textbf{p}'}] = \delta(\textbf{p} - \textbf{p}')$, so the expection values of this hamiltonian with respect to any state will be infinity. Then, we should adote periodic boundary conditions and quantize the momentum. But why? What's the reason for this? I think it must be simple, but I can't see why.

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    $\begingroup$ The main reason is that the world is quantum but you are starting from the classical theory to guess the right Hamiltonian. So, the trick is to put the creation operators always on the right. $\endgroup$ – Jon Jun 12 at 20:18
  • $\begingroup$ It was not exactly my doubt. I would like to know why the expection values will be infinity. I'm sorry, my question was confused. $\endgroup$ – AlfredV Jun 12 at 20:23
  • $\begingroup$ The reason is that, when you move the creation operator to the right with the commutation rule, you are left with a constant that, when integrated, yields an infinity. Remember that the the Hamiltonian must have the form $a^\dagger a$ to be diagonalized. $\endgroup$ – Jon Jun 12 at 21:40

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