Emission line width and the uncertainty principle

[EnSlavingBlair]

Hey what's the relationship between the width of emission lines and the uncertainty principle? My lecturer mentioned it briefly but didn't go into it. I think I get it but would have trouble explaining it.

Cheers,
nSlavingBlair

 

[Bob S]

Delta E delta t = h/(2 pi) (where h = Plancks constant)

Writing E = h v and putting it in the above we get

Delta v delta t = 1/(2 pi) or

Delta v delta w = 1 (this is the electrical engineers version)

 

[EnSlavingBlair]

thanks Bob

and isn't it delta E delta t = h/(4 pi)? coz it's h-bar/2 and h-bar=h/(2 pi)?

 

[Bob S]

thanks Bob

and isn't it delta E delta t = h/(4 pi)? coz it's h-bar/2 and h-bar=h/(2 pi)?

I learned h-bar in school, but that was decades ago.

 

[f95toli]

Note that this does not really have anything to do with QM as such, that the bandwidth (=linewidth) of a system is given by (decay time)^-1 is just a basic result in in Fourier analysis (sometimes known as the mathematical UP).

 

[Bob S]

Note that this does not really have anything to do with QM as such, that the bandwidth (=linewidth) of a system is given by (decay time)^-1 is just a basic result in in Fourier analysis (sometimes known as the mathematical UP).

Then taking the Fourier analysis result and multiplying it by h-bar gets the Heizenberg Uncertainty Principle. Is this a proof of it?

 

[Meir Achuz]

Then taking the Fourier analysis result and multiplying it by h-bar gets the Heizenberg Uncertainty Principle. Is this a proof of it?

Yes, as in my post in this other thread:
"Re: Is it possible to derive HUP from SE?
The SE comes from saying that x is the Fourier transform variable for p.
Then the HUP is just the relation between widths for any Fourier transform."

 

[sokrates]

OR if you want to have a feel for it and understand it the careful wording is:

The lifetime of a state ( one over its decay rate) times the spread in its energy is hbar. An example: If you have a one-channel molecule and if make you perfect contacts to it (if you couple it perfectly to the reservoirs), the decay rate will be infinitely large. That means your electron will escape almost instantaneously to your contacts (and you can exactly define your electron's lifetime in that particular state), but then that one level will broaden out in infinite range and it will no longer be a sharp level. So you'll lose information in the density of states ( you had a delta function but now it's smeared out in the entire energy range). This is basically why people see no 'funny' quantum effects in field effect transistors eventhough the channel widths are on the order of 45 nms ( about 100 atoms) these days. Because they are making very well conducting contacts and the energy levels within the device broaden out.

The counterpart is that if you have a hydrogen atom that sits by itself (complete isolation from environment) the lifetime of the state goes to infinity ( decay rate goes to zero) because you have exact information on its energy.

 

[Meir Achuz]

The counterpart is that if you have a hydrogen atom that sits by itself (complete isolation from environment) the lifetime of the state goes to infinity ( decay rate goes to zero) because you have exact information on its energy.

If you are saying an isolated hydrogen atom in an excited state won't decay, you are wrong.
Maybe that's not what you mean.

 

[sokrates]

If you are saying an isolated hydrogen atom in an excited state won't decay, you are wrong.
Maybe that's not what you mean.

That's exactly what I mean. If you write the Schrödinger equation for a single particle (without taking the environment into account) , you'll never see the decay theoretically.

Again a simple question, yet full of pitfalls: Why does an atom emit light?
The decay totally follows from somewhere else. Let me first prove that SE does not capture the decay of an excited state:

In the eigenstate basis the Schrödinger equation neatly decouples into the following set of equations:

$$i\hbar\frac{d}{dt}\left\{\psi_n \right\}= [\epsilon_n]\left\{\psi_n\right\}$$

which can be solved trivially:

$$\psi_n(t)=\psi_n(0) e^{-i\epsilon_n t/\hbar}$$

then comes the coup de grace:

Let's look at the probability P for finding an electron in state n:

$$|\psi_n(t)|^2 = P_n(0)$$

where $$P_n(0)$$ is a constant and does not change with time.According to the Schrödinger Equation an excited electron stays there forever... Whatever it is that causes it to decay is clearly not above.

The missing part requires some many-particle physics knowledge which I'll skip here. So, yes, although not true, the decay never occurs according to the one-particle Schrödinger equation and thus according to HUP.

(Edit: Of course, the truth is if you start out in an excited state, you'll end up decaying to a lower level. But that will introduce a broadening as well (the excited state will not be as sharp anymore) to protect the validity of HUP ---- but I did not take that into consideration because the excited state is not a sharp level and it is NOT the counterpart of what I stated.)

I hope it's clear.

 

[Meir Achuz]

It isn't, unless you really deny QED. If you are saying an incomplete theory doesn't agree with reality... Anyway this is my last effort in this thread.

 

[Bob_for_short]

Sokrates is right, one-particle SE is not sufficient to make the atom radiate. This situation is quite similar to the Bohr model of atom where only discrete orbits are permitted.

To obtain the radiation, we have to introduce an interaction with always present quantized electromagnetic field - the term j*A_rad. It was Dirac who made it in the early QED. The quantized EMF is enormous set of degrees of freedom coupled permanently to charges but often considered perturbatively due to small coupling constant. The excited atoms give up their energy to these degrees of freedom, this happens within a finite interval of time which determines the line width.

Bob.

 

[sokrates]

It isn't, unless you really deny QED. If you are saying an incomplete theory doesn't agree with reality... Anyway this is my last effort in this thread.

I am not valiant enough to doubt the validity of either QED or QM. What I am saying is a well-established fact and I provided mathematical arguments, so I just suggest you go through the math and convince yourself.

"No, it ain't so" type of response (without providing any counter arguments) isn't really that good of a last effort. So in very elementary terms let me repeat:

If you are in the "single particle" Hamiltonian ------ No decays will be captured by the theory.
If you are in the "multi particle" Hamiltoinan ------- There will be broadening.

Thanks,

 

[sokrates]

The excited atoms give up their energy to these degrees of freedom, this happens within a finite interval of time which determines the line width.
Bob.

Exactly. You put it better than me.

 

[Bob S]

That's exactly [referring to "If you are saying an isolated hydrogen atom in an excited state won't decay, you are wrong.
Maybe that's not what you mean." That's exactly what I mean. If you write the Schrödinger equation for a single particle (without taking the environment into account) , you'll never see the decay theoretically.....I hope it's clear.

In Bethe and Salpeter "Quantum Mechanics of One and Two Electron Atoms" they calculate theoretically the transition probabilities (lifetimes) of all hydrogen levels up to the 6h state and tabulate them in a table on page 266.