Subject: Positives eigenvalues ?

From: [email-masked]
Date: Tue, 28 Jul 1998 09:24:27 +0200 (DFT)

Dears CCLers,

I have a very basic question about quantum mechanics using finite 
basis set.

1) Let's start with the continuous form of the Hamiltonian (H). For simplicity,
let's take the simple system, the Hydrogen atom.

H = -(1/2)D + V, where D is the Laplacian, V the electrostatic potential 
(V = -1/r ; r : distance between the nucleus and the electron ) in atomic
units.

We have then the Schrodinger equation to solve : H Y = E Y
where Y is the wave function of the system and E the energy associated to Y.
There is an infinity of bound states solutions, called 1s, 2s, 2p ,....
All these solutions have an associated negative energy, which are accumulating
under zero. [E = -(1/2)(Z^2/n^2)]
However, when V is zero, the solutions of the Schrodinger equation form a
continuum with positive energies.
The essemble composed of the discrete negatives eigenvalues and the positive
continuum is called the spectrum of the Hamiltonian.

2) Now let's use a basis set to solve the previous problem. All states Y 
can be written as a linear combination of the basis functions. Then, we use
the variational principle (orthogonalisation + diagonalization) to find
the best set of coefficient which gives the best (lower) energy. We are
then focusing on the ground state.
With any basis set of N functions, one finds only one negative eigenvalue
(corresponding to the 1s state) and (N-1) positives eigenvalues.

And now here is my question : (at least!)

WHAT ARE THESE POSITIVES EIGENVALUES ? 
   1) Are they some values of the positive continuum of the spectrum ? 

 I personally don't think so since V is not equal to zero.

   2) Or do they correspond to other bound states that are so badly 
      represented, because of the variationnal principle, that they are 
      positives ? 

 That is what I believe but have no proof.

   3) Or are the zeros different in the two representations ?

 I don't think so because the zeroes mean the same : the atom is ionized.

   4) Anything else ?

I would be very gratefull if someone can point me some references dealing
with this, or some textbook, or just teach me.
Please reply to me directly, I will summarize to the list after.

PS : I deliberatly choose the hydrogen atom since the Hamiltonian is the
same as the Fockian, and people can perform calculations easily with any
quantum chemical package with various basis set.

                                      ...Xav

---------------------------------------------------------------------------
Ast. Pr. Xavier Assfeld                          [email-masked]
Laboratoire de Chimie theorique                  (T) 33 3 83 91 21 49
Universite Henri Poincare                        (F) 33 3 83 91 25 30
F-54506 Nancy BP 239                             http://www.lctn.u-nancy.fr

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From: "Hans Joergen Aa. Jensen" <[email-masked]>
Date: Tue, 28 Jul 1998 11:31:39 +0200

Dear Xav,

You don't explain how you get your positive eigenvalues.
However, from what you write I would guess you have
obtained them with an ROHF program, specifying just one
electron. You see, the Fock matrix is NOT identical to
the Hamiltonian matrix (bare nucleus matrix) for a one-
electron system. The Hamiltonian matrix has as expected
a number of negative eigenvalues corresponding to 1s, 
2s, 2p, etc., the number and accuracy depending on the
quality of the basis set. However, the Hamiltonian matrix
correspond to the zero-electron Fock matrix, not the 
one-electron Fock matrix. The eigenvalue of the occupied
orbital is identical to the value from the Hamiltonian matrix.
The other eigenvalues, the virtual eigenvalues of the 
one-electron Fock matrix, correspond to approximate electron
affinities (Koopmans' theorem), that is to forming the H-
system.

  -- Hans Joergen.