# Wave function of Stationary State

1. Oct 23, 2009

### I_am_learning

I learned (University Physics, 9th Edition, Extended Version) that the wave function of a particle having a definite energy is independent of time. This means the probability Density of the particle don't change with time, i.e. If a particle is 90% likely to be found some where now, There is always 90% chance of finding the particle at the same place.

What I am confused on is in this Scenario---
Suppose a particle is moving in freely in free space with certain constant velocity therefore having constant Energy.
Then How is it possible that the particle is always most likely to be found at its starting position.?
My institution Tells me that the probability density should constantly change with time as the particle moves.

2. Oct 23, 2009

### kanato

A particle in free space that has a definite energy, and for a free particle that also means a precisely defined momentum, that is $$\Delta p = 0$$. So the uncertainty principle tells us that uncertainty in position must be infinite; the particle cannot be specifically located, so it has the same probability of being located anywhere.

3. Oct 23, 2009

### sokrates

You said it yourself: Definite velocity (momentum) and a starting position...

What starting position?

If you have a certain velocity you don't have a starting position because you could be anywhere.

You cannot assume a starting position as well as a definite energy. If your particle is in a "starting position", then we would have to argue that the electron must somehow be localized, but then you wouldn't have a pure momentum state.

4. Oct 23, 2009

### I_am_learning

I quite got where I was slipping. But one question still remains.
For an electron revolving round the nucleus in stationary state, It can't be true that the uncertainty in position is infinite. We at least Know that the electron is somewhere around the nucleus, It obviously can't be in the another corner of the solar system.

5. Oct 23, 2009

### sokrates

By making similar arguments, it can't be true that the electron is revolving with a "pure" momentum state. Because the levels are always "broadened", they are not "real" discrete states (delta functions).

Remember all these are really toy examples, and do not faithfully represent reality.

6. Oct 23, 2009

### I_am_learning

The electrons are in stationary State, Thats for sure, Right?
Then The wave function is time independent, Right?
And don't this means that there is the equal probability of finding the electron anywhere?

7. Oct 23, 2009

### zonde

Are you sure about what you imply?
To me it seems that "the wave function of a particle having a definite energy is independent of time" means that sum of potential energy and kinetic energy is independent of time while separately potential energy and kinetic energy can vary with time.

8. Oct 23, 2009

### nnnm4

I think the most basic objection to an argument that involved a free particle of definite momentum is that a free particle with definite momentum simply DOES NOT EXIST. The stationary states (definite momentum) of a free particle are not normalizable and hence, are unphysical.

9. Oct 23, 2009

### zonde

A side question.
Does it mean that free particle (photon) with definite velocity does not exist?

10. Oct 23, 2009

### mikeph

I think what the original poster is asking in another way is,

If you have a particle in an infinite space with no potential, and perform an energy measurement of that particle, do you not collapse the wavefunction into a single energy eigenstate which is time-independent?

Is the solution that the energy of the particle gives you no information about the position or momentum?

Also, I am not certain if the problem is well-defined- how can you normalise a wavefunction which is defined over an infinite space, and by symmetry, transnationally invariant? (but non-zero!)

11. Oct 23, 2009

### sokrates

No, that's not right.

An electron in an excited state will relax, right? If the excited level was really a stationary state, then there would be no relaxation to the ground state. An excited electron would stay in that energy forever.

One-particle Schrodinger equation does not capture this, there's always a slight broadening in the levels.

These are toy examples. Don't take them seriously. Or study many-body perturbation theory.

12. Oct 23, 2009

### sokrates

Not at all... It just means that because you know the velocity exactly

you have lost all the information regarding the position. Exact frequencies are both theoretically and experimentally possible (,say, for a free electron in vacuum) - it is the coupling to the environment that corrupts pure states. This "environment effect" is usually much more stronger. It "broadens" the energy levels and introduces effects like relaxation etc..

Last edited: Oct 23, 2009
13. Oct 23, 2009

### xepma

The case of the the electron orbitting a nucleus is different from the free particle case. The reason is that in this case a definite energy state does not correspond to a definite momentum state (in the case of a free particle it does). So it is in fact possible for the particle to sit in a stationary state (an energy eigenstate). And in this state there is both an uncertainty in the position and the momentum (but it doesnt stretch out to infinity).

The relaxation process is a different process. For that to take place you need some way to divert the energy away from the nucleus/electron system. For that to take place you need some sort of interaction process. In fact, what happens is that the electron interacts with the QED vacuum -- in some sense a virtual photon can kick the electron to a lower energy state. As a result the electron emits a photon.

But strictly speaking, in the absence of interactions quantum mechanics does not predict any relaxation and the energy eigenstates become truly stationary. But you are right that this is just a toy model.

14. Oct 23, 2009

### victorphy

NO,the equal proabability of finding the electron anywhere means the electron is in a momentum eigenstate.However an electron revolving round the nucleus is effected by a potential,so its momentum doesn't commute with the Hamiltonnian.That means although the electron is just in stationary state,it's not in a momentum eigenstate.

15. Oct 23, 2009

### sokrates

What do you mean the relaxation process is different? After interacting with the vacuum, electron still -relaxes- right? I'd still call it "relaxation" even if the excited atom falls back on a lower sate in vacuum.

Relaxation is pretty loose terminology so I couldn't figure whether you are saying anything different.

16. Oct 23, 2009

### I_am_learning

Seems Like I have to do further Drilling into the subject. I quite don't know whats meant by momentum eigenstate

17. Oct 23, 2009

### I_am_learning

I think you would be able answer this for me.
Suppose I go into a large vacant space (to get rid of all environment influences). Then I project single electron with an electron gun.
--Do this electron now have a definite velocity/momentum or not?
--What is the wave function of this electron like?
(Since the electron always moves in zero potential region, I guess that its Kinetic Energy (whatever that may be) remains constant. Since the only energy involved is the kinetic Energy, I find it sensible to say that the Total Energy of the electron also remains constant. Then there should be no objection in saying the the electron is in Stationary State, is there? Correct me where I am wrong.)

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