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I am currently working through some QFT derivations and running into conceptual problems. In particular, I am deriving the free field Hamiltonian of the form:

$H_{k} = \frac{1}{2} \int d^{3}k \hspace{0.2cm} \omega_{k} ( a^{+}(k)a(k) + a(k)a^{+}(k)) $

from the free field operator:

$\psi(\vec x, t) = \int{\frac{d^{3}k}{\sqrt{(2\pi)^{3}2\omega_{k}}} (a(k)e^{i\vec{k}\cdot\vec{x}-i\omega_{k}t}+a^{+}(k)e^{-i\vec{k}\cdot\vec{x}+i\omega_{k}t})}$

If the field is taken to be free (Klein-Gordon), I obtain the following form from the Hamiltonian density:

$H(\pi,\psi) = \frac{1}{2}(\pi^{2}(\vec{x},t)+|\nabla \psi(\vec{x},t)|^{2}+m^{2}\psi^{2}(\vec{x},t))$

The issue I am having is how to simplify the field, its time derivative $\pi(\vec{x},t)$ and the gradient to the Hamiltonian expression of the needed form. For example, in the term $|\nabla \psi(\vec{x},t)|^{2}$ I need to take the complex inner product of two vectors (of infinite dimensions). Can I do that under one integral sum (same variable $k$) or do I need to run two integrals? If the latter is the case, how do I simplify it? The same question applies to the other two terms. I see that the answer only has one sum running over the variable $k$, whereas the terms in the Klein-Gordon Hamiltonian expression each have more than one field (read integral).

Any help, advice and pain-saving tips are appreciated.

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You have two integrals, for example in $k$ and $k'$. If you do the math using the Canonical Commutation Relations, you will end up with some Dirac delta functions that will cancel out one of the integrals.

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  • $\begingroup$ I think I got it using your advice. What if the term has more than 3 fields? Do I use 3 integrals in that case and do a pairwise Dirac-delta trick? $\endgroup$ – ArtforLife Sep 11 '15 at 13:36
  • $\begingroup$ If I am not wrong, your Hamiltonians should have even powers of the fields. Whatever the numbers, you will have as many integrals as many fields are in the Hamiltonian, because you are using the Fourier expansion of every field. $\endgroup$ – Gian Abr Sep 11 '15 at 21:45

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