# Conservative Systems

1. Sep 29, 2005

### rlduncan

Could some comment on conservative or nonconservative systems in the context of the hydrogen atom and the total energy E, as given in Schrodinger's equation.

2. Sep 29, 2005

### inha

Could you be a bit more specific?

3. Sep 29, 2005

### David

It's not clear exactly what you mean by this but if you are just looking at a (non-interacting) hydrogen atom, then there is a well-defined Hamiltonian which gives a conserved energy.

If you are referring to something to do with Heisenberg's uncertainty principle, then it might get subtle and you'll have to be very clear in your question as to what you are asking.

4. Sep 29, 2005

### rlduncan

It appears that a fundamental condition of Schrodinger's equation is that the total energy of the system remains constant. This then implies that the kinetic energy and potential energy must be constant, a conservative system. In the wave equation y=Asin(kx), k is constant and is not a function of the x variable (lets assume one dimension). However, k=2*pi/lambda and if lambda is the debroglie wave and varies as a function of x, how is it possible to take a simple derivative as if k is a constant. Also for the h-atom the total enery varies with the kinetic energy and potential energy, a nonconservative system.

Last edited: Sep 29, 2005
5. Sep 29, 2005

### David

Although not conserved separately, I am sure you realize.

Where does this equation come from? It's not a wave equation in the usual sense (unless you mean that x is a time variable). It certainly doesn't have anything to do with a hydrogen atom. Or am I misinterpreting and is supposed to be the wavefunction? (If so, it's not a regular solution to the hydrogen atom.)

Why do you want to take a derivative specifically? I think the problem here is that we don't know what you are trying to do here - you need to tell us more about what your are doing and then how you are doing it for us to give you useful feedback.

6. Sep 29, 2005

### rlduncan

If you use an exponential function as a solution to the hydrogen atom the same problem arises, that is y=Aexp(ikx) where k is the wave number and is equal to 2pi/lambda. Usual derivatives of this equation treats k as a constant. Also k can be set equal to sqr(2mE/h-bar squared) which suggest the kinetic energy,E is a constant, a conservative system. In the case of the hydrogen atom, whether or not the system is conservative or nonconservative is relative to taking proper derivatives of the wave equation. A nonconservative system suggest that k varies.

7. Sep 30, 2005

### vanesch

Staff Emeritus
In classical systems, a conservative system is one that can be formulated using a Hamiltonian, and that's what's needed to even be able to conceive a quantum system. Maybe there are some tricks that can give you a kind of quantum model of a dissipative system, but I wouldn't know how to do that rigorously. Happily, all systems (when analyzed microscopically) are conservative and dissipation is only apparent (lack of taking into account certain degrees of freedom).
The hydrogen atom is a conservative system...

8. Sep 30, 2005

### rlduncan

Looking at the total energies of the hydrogen atom as given buy Bohr's equation which I assume are valid. For example, for n=1:KE=13.6, PE=-27.2, and E, the total energy sums to -13.6ev. For n=2:KE=3.4, PE=-6.8, and E=-3.4ev. The sum of the initial KE and PE does not equal the sum of the final KE and PE which I assume should be the same for a conservative system.

9. Sep 30, 2005

### SpaceTiger

Staff Emeritus
A hydrogen atom cannot make the transition from an n=1 state to an n=2 state without receiving energy from somewhere. This sort of excitation usually occurs via collision or absorption of a photon, both of which involve an energy exchange.

10. Sep 30, 2005

### rlduncan

I agree, but this seems to suggest a nonconservative system.

11. Sep 30, 2005

### SpaceTiger

Staff Emeritus
Why? The difference in energy is provided by the photon or particle. Those would have to be included in the system if you're going to talk about conservation of energy.

12. Sep 30, 2005

### rlduncan

I agree and that is my point. In Schrodinger's equation the total energy is the sum of KE and PE and when solved the total energy is assumed to to be a constant. The KE =E-V(r). However, according to Bohr equations it is not mathematically a constant unless you included the photon energy.

13. Sep 30, 2005

### vanesch

Staff Emeritus
That's because in that case, you'd have to include the photon in the Schroedinger equation of course ! (and you'd be doing QED)

14. Oct 1, 2005

### Juan R.

Exactly! The absorption and emision of energy cannot be modeled using the Schrodinger equation nor QED.

All that people done in QED is the introduction of ad hoc phenomenological equations (without rigorous theoretical basis) like the so-called fundamental master equation of particle physics.

One needs a theory of quantum dissipation, there are several available in literature. One of most advanced in developed by Brusshels school that describe in detail the absorption and emision of photons (whereas QED cannot do that with detail) one very popular is axiomatic Lindblad theory. One more advanced still is obtained in canonical science. In fact, the quantum dissipative theories and phenomenological equations derived during the last 50 years by at least five communities are special cases derived in special limits of the sophisticated canonical equation. For example, the equation used in particle physics arises after of three approximations: Hilbert space assumption, factorization of initial and final states, and absence of memory effects (or that is usually called the Markovian approximation).

In condensed matter chemistry, QFT is not often applicable and one could, in principle, use Brushels School theory, but that only work in certain limits, like the thermodynamic limit.

Last edited: Oct 1, 2005