# Hydrogen atom in ground state: some puzzle

1. Jun 27, 2011

### M@2

I noticed many PF threads mention ground state of Hydrogen atom.
At the same time it is two body problem considered to be solved by separation of variables.

It is true, of course, that we can find basis wave functions (solutions of Shroedinger equation). But why does anybody think, that ground solution is the member of this basis functions? Try find Google for theorem. You find nothing exact.

Let us take "usual ground state" wavefunction of H atom with total momentum of atom P=0

Psi (R, r) level 1, L=0 (s orbital)
Psi is the solution of Shroedinger equation where variables separated, where R is proton coordinate, and r is electron coordinate.

Consider new function:
Psi+=Psi(R, r) + Psi(R-x, r-x)
where x is vector with |x| ~ 4* (atomic radius)

Every physisist who deals with double quantum dot knows, that atom levels be repelled and Psi+ as trial variational function, gives lower energy, than initial Psi(R, r).

2. Jun 27, 2011

### LostConjugate

Is the lower energy outside the bounds of uncertainty?

3. Jun 27, 2011

### SpectraCat

That's not correct. Only states with the same symmetry repel each other. Furthermore, if your trial basis function contains the ground state, then it will collapse to the ground state when you apply the variational method. As specified, your Psi(R,r) is the ground state of the H-atom Hamiltonian. Unless you are changing the Hamiltonian somehow (and you didn't say that you were), your variational result will just end up as Psi(R,r).

4. Jun 27, 2011

### M@2

I know, how to calculate ground state energy exactly for simple hamiltonians, but it is limit of some sequence of values.
Ground state is EXACT function with exact value. Calculate it even numerically is rather difficult.

The main point in this thread is that separation of variables may not give ground state.
Thow it gives the basis set of solutions.
Separation of variables can not give All solutions. All solutions are overcrowded.

5. Jun 27, 2011

### M@2

I gave trial wavefunction exactly. I hope You can use theory of perturbation for degenerate levels for H atom (atom!!!). Though it is better to begin with quantum well atom.

Trial function is composed of initial "supposed ground" state and shifted (-x) "supposed ground" state (though as we see it is not ground state at all.

BTW where don't You see same symmetry? We in fact try symmetrical superposition of H ATOM wave functions (not only electron) where wave functions are shifted by distance "x". It is almost evident that there be repellion of levels of ATOM (not ATOMS). We have one proton and one electron!

Collapse? I don't see any collapse.

Psi(R,r) is exact solution of Hamiltonian with momentum of H atom equal to zero.Shifted Psi(R-x,r-x), where x is any vector is also exact solution of Hamiltonian with momentum of H atom equal to zero.

So we have infinite number of Psi with the same energy. We combine two of them in one trial function and calculate energy of trial function and energy of trial function occurs lower, than is the energy of "supposed ground state" Psi(R,r).

Last edited: Jun 27, 2011
6. Jun 27, 2011

### SpectraCat

If you think so, then please run through the calculation and post your results here. Like I said, if you are using the normal H-atom Hamiltonian, then you will end up back where you started, with the function you call Psi(R,r). That is, you will end up there if you allow the coefficients on your two terms in your trial state to vary. If you don't allow them to vary, then you will end up with a higher energy state.

How do I know this? Because the solutions to the non-relativistic H-atom Hamiltonian are EXACT! It is one of the only problems that can be solved exactly, so it tends to stick in one's head. Now, if you want to add some other terms to the Hamiltonian, then perhaps you can find a lower energy solution, but I am talking about the standard H-atom Hamiltonian (in atomic units):

$\hat{H}=-\frac{\nabla^2_R}{2m_p} - \frac{\nabla^2_r}{2} - \frac{1}{|\vec{R}-\vec{r}|}$

(where mp is the proton mass in atomic units, and R and r are the position vectors for the proton and electron, as you specified)

If you stick to that, then you can't improve on the ground state function found in the exact treatment of the problem. That's why we call it exact ...

7. Jun 27, 2011

### M@2

I feel i need explain what is it exact wave function.
If i say Psi(R,r) is exact stationary wave function it means:
H|Psi(R,r)> = E0|Psi(R,r)> plus boundary conditions for R, r
So is for shifted Psi(R-x,r-x):
H|Psi(R-x,r-x)> = E0|Psi(R-x,r-x)> plus boundary conditions for R-x, r-x

But!!!!
Psi+=Psi(R, r) + Psi(R-x, r-x) is not stationary function with E0!!!! We must see new combined boundary conditions!!!! It is stationary function with lower level E-

Psi(R, r) , Psi(R-x, r-x) are not orthogonal. We must find orthogonal basis in the space of nonorthogonal functions Psi(R, r) , Psi(R-x, r-x) for perturbation method.
Resulting basis is (if magnetic field is 0).
Psi(R, r) + Psi(R-x, r-x)
Psi(R, r) - Psi(R-x, r-x)

8. Jun 27, 2011

### M@2

I admit that in summing two "exact" wave functions is uncertainty. Boundary conditions!
But when we sum we get not exact wave function, but approximate! And its energy can be calculated (evaluated) and will be lower than for Psi.
See my post above.

I add: we have infinite number of Exact solutuons with the same energy.
For example H atom can be near Mars and near Moon at the same time (gravitation we ignore).
And wavefunctions near Mars and near Moon overlap!!!
Can Mars/Moon "LCAO" have lower energy? Why not? Relativistic effect we ignore for simplicity.

9. Jun 27, 2011

### SpectraCat

Your reasoning above is complete nonsense. You cannot lower the ground state energy by adding off-center basis functions to the exact ground state solution ... that should be utterly obvious. I can't explain it any better than I already have. Do you really think that this fairly trivial idea would have gone unnoticed for the better part of a century? I am sure you know that some pretty smart people worked on this problem before you did.

But go ahead, prove me wrong. Do your calculations and prove that you obtain a lower energy (you won't). If you are successful, then it is most definitely a publishable result ...

10. Jun 27, 2011

### SpectraCat

Ok .. now perhaps I see where the confusion lies. I had thought you were referring to the ground state of the electronic problem in the COM frame ... that is what we normally call "the ground state of the H-atom". However, you are thinking about the TOTAL energy of the H-atom, including translational degrees of freedom of the center of mass. In that case, I am not sure what the ground state wavefunction is. Can you have an H-atom with a stationary center of mass and zero kinetic energy? There is nothing mathematical that forbids it, although the translational wavefunction for the COM would be delocalized over all space, which is a non-physical result. How then do you define the "maximum delocalization" (or minimum momentum) for an H-atom, is it a de Broglie wavelength of 1 m?, or 1 km? .. I have no idea how to place a reasonable boundary on that quantity. If you can answer that question, and you consider an atom in the exact electronic ground state, then you will have found the "true" ground state TOTAL energy of the H-atom.

Last edited: Jun 27, 2011
11. Jun 27, 2011

### M@2

OK. My english is poor and i can't as a rule to explain my opinion clearly in english.
We can consider more simple artificial atom of two particles: qwantum well (string theory prototype of nuclear forces, if we take many shifted wavefunctions distributed along string).
For two shifted atomic wavefunctions we can calculate ground state exactly numerically.

But it is unpublished and so outside the rules of PF.

For polaron there was a paper in some sense similar to idear of the thread:
PHYSICAL REVIEW B 72, 224303 2005
Coherence of the lattice polarization in large-polaron motion
E. N. Myasnikov,1 A. E. Myasnikova,2 and F. V. Kusmartsev3
But it is not two body case.

12. Jun 27, 2011

### SpectraCat

A polaron is a completely different beast than an H-atom. For one thing .. a polaron is only a quasi-particle used to describe an excitation of an underlying medium. That is certainly not true for an H-atom in non-relativistic QM. So you can't just "shift" the H-atom nucleus and expect to lower the ground state electronic energy (assuming that is really what you are talking about).

You certainly can discuss and develop your ideas about the H-atom and perturbation theory on PF .. both of those concepts are well within the mainstream of QM. What you should *NOT* to do is claim that you know what the answer will be to your non-mainstream treatment of the problem when you have not shown a complete derivation.

According to mainstream QM, the exact ground state electronic energy for the non-relativistic H-atom is:

$$E_0=\frac{R_{\infty}}{1+\frac{m_e}{m_e+m_p}}$$

where $R_{\infty}$ is the Rydberg constant (~13.6 eV), me is the electron mass and mp is the proton mass.

Saying that you can get a lower value than that for the electronic energy is one of the more crackpot-ish claims that a person can make ... so you really need to be precise about what you are claiming.

Last edited: Jun 27, 2011