# Energy eigenstates

1. Sep 30, 2008

### llamascience

Solving the time-independent Schrodinger equation gives the wavefunction for an energy eigenstate i.e. definite energy, so by the E, t uncertainty principle the uncertainty in time would be in a way "infinite". Is this what gives it the time independence? If so, how is this state physically significant?

2. Oct 1, 2008

### muppet

This energy time relation is usually understod to mean something quite different to the position-momentum uncertainty relation. It's usually understood to be physically meaningful only in the context of a state which has a finite (and hence in practice, short) duration. (Wiki "uncertainty principle" or this thread for a longer discussion on the relation between energy and measurement- the thread links to a paper written by a member of this forum who claims that the Energy-time relation is actually a "myth"). So for a particle in a box (say) which just stays there, the energy-time relation isn't relevant, and has nothing to do with the time-independence of of the solution (which is really only time-independent because the method of separation of variables has been employed to that specific end).

3. Oct 2, 2008

### DeShark

The time-independence of the solution is not time-independent because the method of separation of variables has been employed. It is more that the method of SoV has been employed *because* the potential function (i.e. the box) is time independent.

The energy-time relation *is* relevant. The energy-time relation concerns itself with the precision of the measurement of energy based on the amount of time it is measured for. If the energy of the particle is changing in time, then one small time interval after the energy was measured, it may have changed its energy. Thus at a time $$t+\delta T$$ after the measurement, the uncertainty in the energy is $$\delta E$$ which has a maximum value based on the energy-time relation.

Additionally, the energy-time relation can be useful for working out decay times. A state which exists for a short amount of time cannot have a definite state of energy. To quote wikipedia "In order to have a definite energy, the frequency of the state needs to be accurately defined, and this requires the state to hang around for many cycles, the reciprocal of the required accuracy." This is the important factor here, because the particle in a box is completely stable, has a definite state of energy and therefore will exist in that state for a definite amount of time. That is, the state is time independent.

In an undisturbed atom, the potential is constant and thus there would be no change of state from the stationary states calculated in the schroedinger equation. However, there are always fluctuations in the electric potential in a vacuum and it is this purtubative effect which causes decays. Increasing entropy says that they will decay releasing a photon, but it is possible that they can be excited by the vacuum fluctuations also.

4. Oct 2, 2008

### Redbelly98

Staff Emeritus
Another significance of energy eigenstates is that they form a complete set of basis functions, so that any arbitrary state of the system can be written in terms of them.

Moreover, the time evolution of an arbitrary state can be calculated based on the evolution of the component eigenstates.

5. Oct 2, 2008

### muppet

Sorry- the point I was trying to make was that the whole solution to the Schroedinger equation is not time-independent (energy eigenstate multiplied by a complex exponential in t), but the energy eigenstates are constant in time for the reason you specified, and this is what the OP wanted to know.

I'm not sure that's right. If the energy of the particle is changing with time, the particle is surely less likely to be in the measured energy eigenstate as time progresses. As you write below, the important thing is the duration of the particle possessing a given energy:

The only thing I'd clear up here is a linguistic caveat: that the particle posesses a definite amount of energy and will therefore exist in that state for an indefinite amount of time. So yes llamascience, ignore my first post- you can think of it like that.

6. Oct 3, 2008

### koolmodee

Truly stationary states are seldom found in nature, thus they are not very physically. They would imply no external pertubations, no interaction and as you correctly point out they must thus have existed for an infinite time. I think stable elementary particles count as being in true stationary states.

7. Oct 7, 2008

### reilly

By definition, a solution of the time independent SE is just that. Consider an isolated hydrogen atom, in its ground state. Taking isolated to mean no interactions, implies that the atom will stay in the ground state forever -- this could almost certainly apply to very sparse hydrogen gas in the Universe.

Now lets toss in another H atom, far away from the first one. These two atoms will interact with Coulomb forces, and will, in fact have a joint ground state. They will make transitions to the overall ground state primarily by means of photon emission. Once in the ground state, always in the ground state -- we are talking overall isolation. But, the emitted photons count as well, so the ground state involves atoms + photons, and, given isolation, that ground state will be stable.

A major reason we use free particle states in scattering is the assumption that interactions are negligible when the scattering system is widely dispersed -- in and out states and all that. (See most any discussion of scattering theory.)

In QFT and many-body theory, distinctions between time independent and time dependent approaches become blurred by the use of the Interaction Representation, in which the unperturbed states carry the time dependence generated by the time dependent Schrodinger EQ. with out benefit of interactions.

Indeed, the theory does require that an eigenstate of the Hamiltonian goes on forever without change. That translates, practically, into lifetimes of perhaps years or centuries for atoms in a sparse gas in interstellar space.

You can get a good idea of how all this works by investigating the basics of scattering theory, in and out states and the Interaction Rep in particular.

Good question.

Regards,
Reilly Atkinson