Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 4 down vote accepted

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?

share|cite|improve this answer
Yeah, that summation part makes a lot of sense. Thanks. Also, my mistake for H(x) instead of H(t). – johndmalcolm Aug 20 '12 at 13:40

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.