Acting the Hamiltonian operator on a function

Question

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 $$ H(x) \phi(x) = \left[ \int d^{3} \vec{x} \ h(x) \right] \phi(x) $$

Answer 1

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?

Answer 2

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.

Answer 3

The function h(x) under the integral is not an hamiltonian but an Hamiltonian density, and Doug is right.