# Heisenberg's matrix mechanics.

1. Feb 26, 2004

### BlackBaron

How do you use Heisenberg's matrix mechanics to predict the energy levels of the Hydrogen atom?
Can anyone give me a reference (preferably something in the web) on that?

Is Heisenberg-Born-Jordan original paper available in the web? (preferably translated to english, spanish or italian).

Last edited: Feb 26, 2004
2. Feb 27, 2004

### 1100f

The energy level of the hydrogen atom are used by using the lentz vector.
When you look at the kepler problem, since the potential (1/r) is a central potential, angular momentum is conserved. However, if you look at the classical solution of the bound states, the trajectory consists in an ellipse. The axis of this ellipse is conserved (corrections due to general relativity, for example, give a perturbation that make a precession of the ellipse). It is found that the Lenz vector is a conserved quantity. The Lentz vector is given by $$\vec{A} = {\vec{r}\times\vec{J}\over 2} + {\vec{r}\over r}$$
(This is the vector for m = 1 and coupling constant = 1).

It is easily shown (not complicated but long calculation) that the commutation relations of the lenz vector with angular momentum are those of a vector operator (i.e. $$[J_i,A_j] = \epsilon_{ijk}A_k$$, and the relation of the components of the Lenz vector are:
$$[A_i,A_j] = 2i\epsilon_{ijk}J_kH$$ where H is the Hamiltonian. Since we are trying to find the states that are eigenstates of the Hamiltonian, we can replace in the comm. rel. the Hamiltonian by its eigenvalues. So that for each energy level, the comm. rel. between the angular momentum and the Lentz vector are closed and they form the generators of the so(4) algebra. One of the casimirs of the representations is identically 0, from the other you can find the values of the energy.

If you want, I can further develope the algebra for the eigenvalues of the energy.

Hope I helped.

Last edited: Feb 27, 2004
3. Feb 27, 2004

### BlackBaron

Yep, you helped a lot.
I'll start working on it to see if what you tell me is enough.
If need anymore help, I'll let you know.

Thanks a lot