# Energy from single point rather than expectation integral in Full CI calculation?

1. Oct 10, 2007

### reddorange

energy from single point rather than expectation integral in Full CI calculation??

Hi, i guess this is kind of a stupid line of thought...

if you get a wavefunction, say from a Hartree-Fock calculation, you can find your energy by calculating the expectation value of the hamiltonian.

well actually for any operator, observables are the eigenvalues of the operator. my question is, if you have a wavefunction (or any eigenvector), why don't you just calculate the operator at a single point, then divide out the eigenvector part to get E?

For instance, if you have Psi, why solve the integral <Psi | H | Psi >? Why not do something like H(x=0)Psi(x=0) = E Psi(x=0), and solve for E at some convenient point?

I guess in CI you just get E from a diagonalization, maybe it's a moot point there?

But in in Hartree Fock you have your orbitals, and you go through this business of calculating an expectation value and get overcount terms and such.

in general, a molecule is a system of particles (electrons). but you are solving a wave equation for them. it also seems weird that whatever the position of the electrons, the hamiltonian operating on those electrons (parameterized with positions of nuclei, or whatever) gives the exact same energy...if your electrons are all bunched up beside each other, shouldn't the energy be really high at that point?

2. Oct 10, 2007

### reddorange

i mean, it's a WAVE-function, but it's a function of the positions of the electrons!

3. Oct 10, 2007

### quetzalcoatl9

ok, slow down. first of all, what you are asking has nothing to do with CI or HF, but QM in general.

the energy is not defined at a point, but rather over all of space as the expectation value of the Hamiltonian. what makes you think otherwise?

as far as the energy being a function of the electronic coordinates, consider the following. it was proven by Hohenberg and Kohn that is you define an electron density function as

$$\phi(r_1) = \int dr_2 \int dr_3 ...\int dr_N \psi^*(r_1, r_2, r_3,...,r_N) \psi(r_1, r_2, r_3, ..., r_N)$$

then there exists a unique functional of $$\phi$$ that yields the ground state energy, i.e.:

$$E[\phi] = E_0$$

this leads to DFT (since you seem to be interested in numerical solutions).

4. Oct 10, 2007

### reddorange

thanks, quetzal.

but if you look at just the time-independent equation, HPsi = EPsi, doesn't this hold for every point along Psi? This is why you can write the expectation value as <Psi|H|Psi> = E<Psi|Psi>, ie E is constant for every volume element.

5. Oct 11, 2007

### quetzalcoatl9

because that would not be a unique solution, ie.

what exactly is H(x=0)? remember what you are doing, you are acting an operator (which maps a function onto another function) on an eigenfunction.

for example, consider the free-particle hamiltonian - you cannot act the derivative operator on a point (you agree that Psi(x=0) would be a point?), it must act on a function.

perhaps what you mean (and what i was trying to get at with the DFT example) is some sort of fictious local energy function that is useful for the computation. another example is via diffusion quantum monte carlo (see equation 2.58):

http://www.tcm.phy.cam.ac.uk/~ajw29/thesis/node27.html#SECTION00551000000000000000