# Complex Inner Product for Integral Expressions

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.

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.