# Stationary states

1. Dec 11, 2004

### George Isaac

First, read this; it's a two slit experiment carried out recently that is supposed to weaken the Copenhagen interpretation.

http://en.wikipedia.org/wiki/Double-slit_experiment#Shahriar_Afshar.27s_experiment

What are the stationary states physically? The problem with the stationary states is that they are stable i.e. if a particle is somehow in one of these states it remains in this state forever, if it does not interact with anything. I find it difficult to imagine a stable state that does not have the minimum energy possible, although such states are predicted by QM( I mean stable states having energies higher than the ground level). Second, what makes an electron attached to the atom? The smoke of mathematics involved in solving the Schrodinger equation for the hydrogen atom prevents me from seeing how does QM predict a stable atom? Also in books I am told that stationary states present no motion, so we have to build a superposition state to have time dependent expectations. Fine, let's pick an example to be specific, the harmonic oscilllator. Fine, then I find to my dismay that books compare the classical probability distribution for a classical "moving" harmonic oscillator to the stationary state probability distribution which comprises no motion. Also comparisons are carried out between the classical time averages and the expectation values for the same dynamical variables belonging to the stationary states.??????

2. Dec 11, 2004

### dextercioby

Textbook definiton:Stationary states of a QM system are the eigenvectors of the Hamiltonian operator assumed time independent in the Schroedinger picture.

Why do you find that a problem??The problems in QM appear when systems have other states than the stationary ones.

That's right.
That's your problem.This means your're running low with the imagination responsable part of your CPU.How's your geometry...??

Hmmmm...I don't know... Physics maybe...?? Quantum mechanics...????

AHA...So that's the reason... :surprised Geeeee...i find the Schroedinger eq. for the H atom really delightful... For now,i cannot see a better way to showing that the H atom is stable with the poor lectron is the 1s state without using a bit of Mr.Erwin Schroedinger's tricks... :tongue2:
Anyone who could come up with another method is invited to state and prove it.

U mean time evolution.That simply follows from the definiton.

Yap,that's a way to do it.
Since both classically and quantum,the probability distribution is time independent,why the hell not??By motion for the second part (the quantum) u imply time dependence again... :yuck: I resent this synonimy...It actually does not exist...But if u want to use it,be my guest... :tongue2:

Yes,why not??What's wrong with explaining why the quantum case is entirely different than the classical one and showing that exploring all possible reasons...??

Yap,u did better than the statistical part... Though i strongly suggest going through the mathematics involved by the Schroedinger's equation,since it's the easiest u can find in all QM.

Daniel.

3. Dec 11, 2004

### cjellison

Stationary States and Uncertainty

I have a question related to stationary states and the uncertainty principle. Stationary states have definite energy. That is, the standard deviation in the energy is zero.

In regards to the uncertainty principle, the usual interpretation is that the particle (or whatever you're thinking about) takes an infinite time to change its state---or that it doesn't change at all. I understand this interpretation, but I am caught up on the mathematics of it.

$$\Delta E \Delta t \geq \frac{\hbar}{2}$$

Clearly, as $\Delta E \rightarrow 0$, we must have $\Delta t \rightarrow \infty$. However, we do not say that the standard deviation for stationary states is approaching zero. We say that it is zero. Mathematically, I can't 'solve' for $\Delta t$ since I would be dividing by 0.

So what is the deal? Is the uncertainty principle "defined" by the qualitative definition that says you cannot simultaneously know such and such? If so, does that mean the mathematical formulation is 'useful' only in situations where both uncertainties are nonzero?

4. Dec 11, 2004

### dextercioby

Congratulations!!!U have touched a very sensitive point in the interpretation of QM.The principles of QM lead us to two things:
1)The uncertainty relations,among which the one including energy and time is very important.
2)The definiton of stationary states and hence their properties.

The issue is that these 2 practically do not match each other,though awkwardly they have been deduced starting from the same point:the postulates.Think about the more famous analogy:the uncertainty position-momentum.Apparently,there is the same problem,as one can imagine eigenstates of momentum with definite momentum (the de Broglie waves for which the momentum is uniquely determined).The problem there is solvable.(See below).

The question raises as follows:does QM (in the Copenhagen-standard-textbook formulation) has an answer..??The answer is:I certainly does.

What do the uncertainty relations tell us??U cannot measure either the energy of an eigenstate of the Hamiltonian,or the time the system stays in that eigenstate with arbitary precision,as long as u do those measurements simultaneously.That is,if u measure the energy of the system in one the eigenstate of the time independent Hamiltonian,u cannot make any estimation upon the imprecision with which u could be measuring the time the system spends in that eigenstate.Otherwise,the uncertainty principle cannot prevent us from measuring the energy in the stationary eigenstate.It just ensures that we cannot measure the time spent in that eigenstate.And yes,if u want to be strict and you wouldn't wanna enter conflict with mathematics,we cannot equal uncertainties with zero,but instead ought to speak about limits when those quantities go to zero.
The situation is a bit tricky though,as the hamiltonian is not a bounded operator (it comprises the term p squared),so we cannot speak about normalized eigenstates in the continuous spectrum.Therefore,though we can say that the scaterring states of the H atom (which is still time independent) satisfy the H atom specral equation,they cannot describe physical states,as they are not normalized,as their norm is found to be infinite.That would contradict the first principle who says that the physical states must have finite norm.So this part should be clear.
For the bound states,which,for example in the case of the H atom,are of finite norm,and the spectrum of the H atom is purely discrete,i guess the only problem is that for any eigenstate,the uncertainty in the energy would be zero.Exactly zero.So,by trying to find the uncertainty in time,u'd have to devide a finite number (larger or equal to hbar/2) through zero.If u can see this not mathematically rigurous,the uncertainty in time could be found +infinty.Else,i think u should put a limit there.In general,physicists don't fancy too much about dividing through zero,as one of them invented the Delta Dirac "function" and mathematicians had no problem with it.Moreover,they took it and developed distribution theory.Why would they have a problem here??

Daniel.

PS.I'll think about it deeper,though i cannot realize what should be the problem.Dividing through zero...??
The case x-p_x is simpler,as the two operators do not have discrete spectrum,so what i said about unbounded operators and there connection with the physical states applies fully.

Last edited: Dec 11, 2004
5. Dec 11, 2004

### caribou

Hmmm... now that this experiment is taking up 1/3 of the Wikipedia entry on the double-slit experiment, I'm definitely tempted have a word with the guy behind it all.

6. Dec 12, 2004

### George Isaac

don't need one

I know this. This is not what I meant. I know the math. You're turning things around.

My geometry is fine and it's not about imagination of shapes in space.
How can we have a stable state not having the least possible energy? Shouldn't stable systems be that way? Imagine an experiment where we put an electron in one of its stable states not having the minimum possible energy. It should remain there forever shoudn't it? That is ofcourse if it does not interact with anything. This is what I find confusing.

I will not comment. I don't think you get what I mean.

Mathematics must be followed by physical interpretations. What do you think?

I solved the hydrogen atom problem and much more difficult problems that do not admit a seperation of variables solution so I know when is the mathematics simple and when it is not. Unfortunately for you I seek a physical picture after doing the mathematics. Regarding the statistical part, well you could have told me where to read this stuff instead of being sarcastic although I doubt that you fully understood what I meant.

7. Dec 12, 2004

### dextercioby

I don't see why,but let's say you're right.

Good for you.
The theory of QM says that way.If u won't accept it,that's your problem.

Weird,really weird.What's confusing about it??You're tempted to making a comparison with the classical case,in which the only stable state is the one with minimum energy.Quantum phyiscs is enetirely diffrerent in this respect.The notion of state is entirely different.
If u have this problem (though you may not,so i'd be wrong then),it means that the famous "tunnel effect" is a real enigma.Because i think that tunnel effect is a bit more difficult than the H atom.

It may be the case.Sometimes i have my way of seeing things.We're different individuals,we can see the same problem in different manners.

I think that's the difference between mathematics and theoretical physics.And it's obvious.

That,my friend,is selfcontradiction.Weren't the one that said something like "the smoke od mathematics" in connection with the H atom problem??Or am i mistaking you with someone else??

I don't get the part with "unfortunately for you".But i seek a physical interpretation too,as i imagine myself a physicst,not a mathematician.
Your problem apprently resides with the physical interpretation of the mathematical results in QM.You won't accept it.As i said before,that's your problem.

I understood what u allowed me to understand.I frankly doubt that anyone else would have understood differently,as it was obvious you could't make a distinction between a photon and a phonon,to give the most resounding example.
Bernard Diu,Landau&Lifschitz,Kerson Huang could be the guys who could come up with all the answers regarding SM.

Daniel.

8. Dec 12, 2004

### George Isaac

Fed up

This is directed to Daniel.

Why are you being sarcastic? Who do you think you are anyway? When I say mathematics smoke, I mean(and obviously you never get anything I say right) the physics gets lost when you solve lengthy equations, not that the equations are difficult, it's that you keep much attention of the maths negelcting any physical explanations. And I am not contradicting myself. You are a wierd person and I can tell you are just a scientific balloon very empty from the inside. How can people stand you anyway?

9. Dec 12, 2004

### RedX

Sorry, don't have time to see the Wikipedia entry. Wikipedia is good, by the way, but I think a good book in QM is better for you.

You have to admit that if you find your system in an excited energy state, and your system is isolated, then it has to stay in that excited energy state. It's stable. Now a system is never purely isolated as it can emit a photon. So what will happen is that a photon will be emitted causing the system to be dropped to the ground state. The reason this happens has to do with statistical physics and not quantum mechanics. Well it has to do with quantum mechanics but statistical physics explains it simply. The reason the system prefers the ground state is so that the much larger environment can have more energy, and it'll have a lot more entropy with the addition of that energy.

The Coloumb potential causes attraction between electron and nucleus.

When the smoke clears and you solve for your "radial function times the radius", then you see there is a centrifugal-type barrier/potential which keeps the electron from being in the nucleus for nonzero orbital angular momentum states.

I think what your book means when they say stationary states have no motion is that the expectation values of stationary states don't change (not only the expectation values don't change but the probability distributions don't change as the imaginary exponential time-dependence cancel), and that includes momentum, and if your state is bounded , then the expectation value of the momentum has to be zero, for otherwise the particle would drift to infinity as it can't stop or change course as the expectation value of the momentum can't change.

For the classical oscillator the average position is zero too, just as in the quantum case. Having an expectation value of zero for the position in QM doesn't mean necessarily you'll find that the particle is always at the origin, but if you make a series of measurements on an ensemble, then you'll find the average is at the origin.

10. Dec 13, 2004

### Kane O'Donnell

George Isaac:

I think the question has already been answered, but in short, the 'stable' states of the quantum mechanical Hydrogen atom are only stable up to the approximation of ignoring the vacuum fluctuations. So, for example, if |i> and |j> are different stationary states you can show that <i|j> = 0, ie the probability of measuring a state |i> to 'be' in a state |j> is zero and vice versa.

However, if you use add in a classical electromagnetic field A and treat it as an operator on say |j>, it is possible that <i|Aj> is non-zero, that is, there is a probability of transition in the presence of an em field. To make this much more rigorous (it's fairly dodgy the way I've said it) takes a lot of work and requires quantum electrodynamics, but the basic outcome is to a first approximation the same. QED also answers the question of *why* transitions can happen even if there is no 'external' em field to start with (this is where the vacuum fluctuations come in).

The minimum energy configuration is the most 'stable' in the sense that electrons will tend to be in low-energy states around an atom for example, but the point is, for observed conservation of energy to hold you need to *lose* energy to drop states, and the ordinary Hamiltonian for say the Hydrogen atom doesn't take into account the interaction between an electron and the electromagnetic field (except of course for the Coulomb potential).

Cheerio,

Kane O'Donnell

11. Dec 15, 2004

### sifeddin

Would you give a link to more elaborate document or lecture or even online course where we can elaborate on the subject of vacuum fluctuations in QFT?

yeah,I know this from the classical point of view: to be more stable go to the trough of the potential; where the force will be zero, but this is QM right? could you give us some QM reason?

Would that be the answer or you may know some one to tell us a more general Hamiltonian? Would that someone be you? (dextercioby do not answer this!) This is so much related to what is called "spontanous emission" is there a QM way to treat it? I read one in an appendix of Coldren and Corzine's Diode Lasers and Photonic Integrated Circuits" but it involves some "random virtual" EM field. Have you got something more solid?
I believe George got a fix on that question that he will be delighted to have it answered .

Last edited: Dec 15, 2004