# Natural line width and energy uncertainty question

1. Jan 29, 2005

### anonymous299792458

Lets take an electron which is "in orbit" around the nucleus. As far as I know, its (ANY conceivable) wavefunction can be represented as a superposition of energy eigenfunctions, which correspond to the discrete eigenvalues of the electron's enrergy. What I do not understand is where the natural line width of spectral lines comes from. I've heard that it's a consequence of the energy time uncertainty relation. I think I understand the position-momentum uncertainty relation, as both x and p take on a continuous spectrum of values. The energy levels in an atom are, on the other hand, DISCRETE. My understanding of this is that an electron can be in one energy level, or another, or in a superposition of several. Natural line width seems to imply that energy can have a CONTINUOS range of values. As I have mentioned previously, I do not quite get how the energy-time uncertainty relation applies to the case of DISCRETE energy values. Could someone explain this?
Now, it is usually said that an electron drops from one enegy level to another and emits a photon. But, due to the time-energy relation, doesn't the electron have to be in a superposition of at least a couple energy levels unless we wait for an infinite amount of time? So, the difference in energy eingenvalues of the electron has some EXACT value, and the electron is in a superposition of these eigenstates, although the energy of the emitted photon does not exactly correspond to this energy difference????????
After the emission, the atom (electron) is only APPROXIMATELY in its ground state, right (i.e. the COEFFICIENT of the higher energy level eigenfunction is not exactly 0)?????? Otherwise, wouldn't the time-energy uncertainty relation be violated? I'm sort of confused.

Last edited: Jan 29, 2005
2. Jan 29, 2005

### Kane O'Donnell

Think of it like this - the excited wavefunction "shape" only exists for a finite period of time. As such, instead of having a definite energy, it has a range of energies. The size of this range is related to the length of time the excited state exists and is given by the time-energy uncertainty relation.

The "definite energy level" wavefunctions are the idealised case, when there is no external electric field interacting and the electron is in the same state for all of time.

This is the same kind of situation you get when you use pulsed lasers - the frequency bandwidth of the pulse is high in order to have a pulse that is short in space and time. The really, really narrow linewidth lasers are all continuous wave lasers, not pulsed ones.

Cheerio,

Kane

3. Jan 29, 2005

### anonymous299792458

Well, I realize that an arbitrary wavefunction does not have a definite energy. But I thought that it could ALWAYS be represented as a superposition of energy eigenfunctions which DO have definite, DISCRETE energy values. Hence, it seems that the range of energies is discrete, not continuous as you imply. I do not understand where the continuous range comes from. Maybe external influences have something to do with it???

4. Jan 29, 2005

### Kane O'Donnell

Yes! This is what I was saying - the 'external influence' is the fact that we're in the real world and idealised situations don't exist. They *almost* do (otherwise we wouldn't see such sharp lines to a good approximation) but if you look close enough you get perturbations on the system.

Ok. Suppose you have a *real* Hydrogen atom in it's ground state, and suppose that it gets excited by a photon to the n=2 state. Now, in a ideal situation, the electron's wavefunction is now in the n=2 energy eigenfunction, ok? In the idealised case (ie if a Hydrogen atom was *alone* in the universe) stationary states are not even superpositions - they are the pure eigenfunctions.

However, if you go and ask the question "How long will the n=2 state last" for such an isolated system, you will get "forever" as the answer. Why? Because the treatment of the hydrogen atom in such an isolated sense does not allow for a photon to carry away the energy that would let the electron jump back down, so our model is unable to cope with ideas like transitions. It does explain what shapes the wavefunctions are, and so forth. To calculate transition times and so forth, you need to add the interaction with the electric field.

You can, to a good approximation, just add a classical electromagnetic field. Now you can calculate the probability that an electron will jump from n=2 to n=1. Notice before that because the eigenfunctions are orthogonal, the probability of transition is zero, since the transition probability is defined as $$\langle 1 | 2 \rangle$$, where the 1 and 2 are just labels to represent the n=1 and n=2 states.

Now, since we can jump from state to state, we get the kind of effects I described in my first post. Before, when our states lasted forever, the uncertainty principle said the variation in energy can be zero. Now, since $$\Delta t$$ is finite, we must have a finite (but very small) $$\Delta E$$ for the energy level. Now at each instant in time the n=2 state can have energy $$E_2 \pm \Delta E$$ - the exact value *when the electron jumps* is just a matter of probability, and hence for a large number of transitions (ie over gazillions of atoms all doing much the same thing) we see a finite linewidth.

So, the natural linewidth is a result of *perturbations* to the isolated wavefunctions due to the interaction between the atomic system and the electromagnetic field. What I was trying to point out in the last post was that this "ideal to perturbed" broadening thing happens *all the time* in applications, for example, in laser theory, in digital signaling, in gas-beam velocity distributions, etc. All kinds of places. In general, whenever two properties of a system are related by a Fourier transform (position/momentum, frequency/time, energy/time, etc) then you will see these effects.

5. Jan 29, 2005

### anonymous299792458

So in an IDEALIZED case, the electrons wavefunction will be strictly in the n=2 state? Doesn't this violate the time-energy uncertainty relation? Assuming the transition (WE ARE STILL ASSUMING THE IDEALIZED CASE) from n=1 to n=2 happened 5 seconds ago, wouldn't the energy uncertainty be AT LEAST planck's constant divided by 5? Or, in other words, wouldn't the electron HAVE TO BE in a SUPERPOSITION of n=1 and n=2? After all, the energy-time uncertainty realation STILL HOLDS in this idealized case, right???

6. Jan 29, 2005

### Kane O'Donnell

Yes, it still holds, but the point is, you can't consider *transitions* in the idealised case. That is, you can look at the n = 1 state, and look at the n = 2 state, but you can't model a transition between them - the model is *insufficient*. Therefore, we can't talk about transitions happening 5 seconds ago or a billion years ago.

It is a very instructive exercise to actually do the calculation - calculate <E> and <E^2>, then calculate:

$$(\Delta E)^2 = \langle E^2 \rangle - \langle E \rangle^2$$​

You will find that for any stationary state, $$\Delta E = 0$$. Then obviously from the energy-time uncertainty relation you get that $$\Delta t = \infty$$, ie, the state will last forever and has lasted forever.

The uncertainty relation isn't violated because in the idealised case we can only calculate the difference between energy levels and the uncertainty in energy is 0. If you add in a classical electromagnetic field, you can consider states with finite lifetimes and hence with slightly uncertain energies.

Two different models, two different uses.

Cheerio,

Kane