Acting the Hamiltonian operator on a function I have just a little confusion on some formalism in QM.  I have a Hamiltonian density function, $h(x)$, where the regular Hamiltonian is given by
$$
H(x) = \int d^{3} \vec{x} \ h(x)
$$
I'm wondering, if I am in a situation where I need to act the Hamiltonian on some function, would this function go inside the integral or after it?
$$
H(x) \phi(x) = \int d^{3}\vec{x} \ \left[h(x)\phi(x)\right]
$$
OR
$$
H(x) \phi(x) = \left[ \int d^{3} \vec{x} \ h(x) \right] \phi(x)
$$
 A: I think you are just mislabeling things and getting yourself confused. You wouldn't ever write anything like 
$  H(x) = \int d^3 x h(x)$
since you are performing a definite integral on the RHS of the equation with limits of $\pm \infty$ (an integral over all of space). 
What you might be thinking of for the Hamiltonian is 
$H(t) = \int d^3 x h(x,t)$ 
or a general operator in momentum space basis say:
$ \mathcal{O} (p) = \int d^4 x \mathcal{O} (x)$? 
At any rate, 
$H(x) \phi(x) = \int d^3 x h(x) \phi(x) $ 
doesn't make any sense because doing that is just like (if you were to consider it a summation) sticking terms into the summation:
$H \phi_i = ( \sum_i h_i ) \phi_i = ( \sum_i h_i  \phi_i) $
which, I think, clearly is wrong. Does this clear it up?
A: Adding on to what DJ wrote, the proper way to write the Hamiltonian acting on something else would be:
$$H(t)\phi(x) = \int d^3y\; h(y,t) \phi(x) $$
That is, we can still write it in terms of an integral, we just have to be careful about what is a dummy variable and what is a free variable. 
A: The function h(x) under the integral is not an hamiltonian but an Hamiltonian density, and Doug is right.
