# Always in an eigenstate?

1. Nov 7, 2007

### Artaxerxes

Let's consider the hydrogen atom with one electron. If we observe the energy the result is always one of the eigenvalues even if the statefunction is arbitrary. I accept that. But why is the atom always treated as if the electron is in one of the eigenstates? Why is the statefunction always supposed to be an eigenstate?

Is that a postulate?
Or is the electron not always supposed to be in an eigenstate?
Does an arbitrary statefunction evolve into an eigenstate?

2. Nov 7, 2007

### f95toli

Energy eigenstates are the only states that are stationary , i.e. they do not evolve with time(expect for an irrelevant phase factor). Hence, it is the only kind of state that is "stable".

Last edited: Nov 7, 2007
3. Nov 7, 2007

### blechman

In general, the state of an atom can be anything it wants to be! subject to boundary conditions/schrodinger's equation, etc. General states of Hydrogen take the form of a superposition of the eigenstates. When you observe the energy, you collapse the wavefunction, but besides that, there is nothing special about the eigenstates.

Eigenstates do not evolve in time, as f95toli points out (although the excited states actually can "decay" to the ground state by emitting photons), but a general superposition of eigenstates will evolve according to Schrodinger's equation.

4. Nov 8, 2007

### dextercioby

Every QM text shows (or at least it should) that there are some "nice" properties of the stationary states of a quantum system. A stationary state is, in the standard Dirac-von Neumann formulation, a solution of the spectral equation of a time-independent Hamiltonian. Since such a Hamiltonian is a constant of motion, these special properties follow from the general properties of constants of motion, also mentioned in texts.

That's why we concentrate on solving spectral equations for operators in QM. Time dependent Hamiltonians (in the Schrödinger picture) do not lead to anaytical solutions of the SE and in this case perturbative methods are applied.

5. Nov 8, 2007

### OOO

As f95toli has already emphasized, eigenstates of the Hamiltonian are the only stable ones. One can easily show that superpositions of two energy eigenstates result in a time-dependent dipole moment, and this obviously leads to their decay due to the (induced) emission of radiation. In a thermal environment both levels will be occupied according to the Boltzmann factors of their energies (in the limit T=0 only the lower level will be occupied in the long term).

The reverse process of the above would be to irradiate the atom (or whatever system you refer to) with a laser the frequency of which coincides with the transition frequency between the two levels. This will result in the buildup of equal occupation numbers of both levels (but you will never reach a higher occupation number in the upper level, unless you are using an additional laser pumping process for it !) due to induced absorbtion. This state formally corresponds to infinite temperature (as far as the Boltzmann factor of this transition is concerned). But this is obviously not a closed system.

Last edited: Nov 8, 2007