## Physics Questions People Ask Fermilab

**Radiation Direction**

**Hello,**

What is the direction of radiation emitted from an atom ?

Thanks,

*Bob Patton*

Bob,

Your question has always been an area of research whenever new processes are found and is a very valid one. The answer depends on the specific process in question. For one thing, a coordinate system must be chosen. If the atom is moving, then you, might use the momentum vector of the atom as the direction of reference. Usually with atoms though, the problem is solved in the rest frame. If no external field such as laboratory magnetic field is present, then there is no preferred coodrinate frame so just pick one.

The next thing is that the radiation is emitted randomly. This means you cannot predict the angular direction of any specific photon emitted. However, there is an overall probability distribution that predicts where most of the photons will be emitted over time. By the way, a photon is a quantum of radiation, a piece of radiation which is emitted by the atom in a bunch which acts in some ways like a particle. The energy of the photon is equal to the difference in energy of the energy levels involved in the transition giving rise to the radiation. So when I say a specific process, I usually mean a specific set of energy levels. Obviously, if you know the direction of emission, on average, and the energy, then in majority of the cases, you can build a detector to measure the radiation you want to observe.

As far as places to look. I would start with a book or article on atomic physics such as one you could buy in Borders or Waldenbooks. Fermilab's technical documents involve high energy particle angular distributions and the are considerably more complicated. One can search through recent articles (preprints) on the HEP SLAC server also.

http://www-spires.slac.stanford.edu/find/hep

*Glenn Blanford
Fermilab Public Affairs*

Dear Glenn,

Thank you for your reply to my question about the direction of visable radiation emitted from an excited atom changing to a lower energy state.

You indicated that this radiation is random. Does this mean that there is no specific direction and that the radiation is symmetrical? If it is symmetrical, then that is the answer to my question. However, I am interested in the direction of a spedific event, not a statistical average. Is this an unreasonable inquiry? If all of these events result in an average that is symmetrical, what then is the nature of a single event? By "the nature", I mean what is the mechanism.

You indicate that there is an overall probability distribution that predicts where most of the photons will be emitted over time.

That's great! How do I find out about this? It would be very helpful.

I have gone thru most of the library and book store texts you suggested, and find no answers. I searched the HEP SLAC server with negative results.

Thank you again for your help. However, I remain unsatisfied with an answer to my original questions. If you do not want to follow-up on these complicated questions, I'll understand.

Thank you again,

*Bob Patton*

Hi,

Breifly, it depends on the states involved in the transition. If you have taken a class in chemistry, you know that the atom has major levels defined by the principal quantum number, n, which are separated in energy by a fair amount (Energy is proportional to -1/n^2). Each of these levels is really broken down to sublevels, some which are slightly different in energy from one another. For example, the n = 2 level has the 2S sublevel as well as the 2P level. The S and P designations refer to the amount of orbital angular momentum the electron(s) has. An S state has 0 ang. mom. and the distribution of where it is in space as it orbits the nucleus is spherical. The P state has 1 unit of ang. mom. Its distribution in space is not spherical. So when a transition takes place between levels, one must take into account the where the electron could have been in the initial state and where it can possibly go in the final state. In other words, what the starting and ending distributions are like.

If you really want the specific distributions of photons in terms of a probability distribution over the emission angles, the best thing to do is to define what the quantum numbers of the initial and final states are and then look up in an atomic physics textbook. The majority of transitions that one would be concerned with are electric dipole transitions because they have the largest rates for happening. These occur between the S and P states. For example, in hydrogen, the 2P -> 1S transition would have an angular distribution P(w) = Constant * (sin(w))^2 where w is the angle between the direction you are looking from and the electric dipole moment direction. This creates a radiation pattern. If you take a cross section using a plane which includes the dipole vector and an axis perpendicular to it, the distribution has two lobes (searchlight type lobes) on either side of the dipole vector (perpendicular to it). The dipole vector is basically centered on the atom. Of course, unless your atoms are all moving in a certain direction such as in a beam or are polarized, their dipole vectors will have random orientations. The type of state will determine the orientation of the dipole vector if it has one but with respect to some axis in the atom. If that axis is randomly distributed because you have no control over the atoms direction (such as in a gas), then your radiation pattern is likewise random (ie. spherical).

The reason a laser works like it does is that even though the atoms inside radiate spherically, mirrors on both ends of the laser cavity keep only the photons moving in a particular direction bouncing back and forth. The rest are lost.

Hope this helps.

*Glenn Blanford
Fermilab Public Affairs
blanford@fnal.gov*

Dear Glenn,

Thank you again for your replies. I am beginning to get an appreciation for the complexity of atomic visible light radiation direction. Your patience and toleration to my questions are greatly appreciated.

From what you have described, it is necessary to define the transition states involved in light radiation from atoms. To be specific, I would select the transition of an electron of a hydrogen atom from the 1px state to the 1s state. where the position of the electron is or could have been at each end of the transition is, to me, unimportant. Select any appropriate or useful point for each, and proceed to calculate or describe the direction of resulting visible radiation. Is it spherically symmetrical from the atom? If not, what is it's distribution?

Thanks again for your help.

Happy Holidays,

*Bob*

Dear Glenn,

Thank you again for your reply to my questions about the direction of visible radiation from an atom having undergone an energy transition. I think I am beginning to get the idea; but still can use some help in getting over some of the confusing bumps directly in front of me.

I don't have a supply of Physics Texts for reference; and those I have looked at in our local library have not been all that useful to me. If I did have a supply, I'm not sure I could use them effectively anyway.

Consider a H atom transition: (A) 2P(x) --> 1S and (B) 2P(y) --> !S Questions:

- Would the direction of radiation propagation be at 90 degrees to each other if each initial state were in the same xyz orientation?
- Would the direction of propagation from each be in two directions 180 degrees apart?
- Would the direction of propagation from each be 90 degrees from the direction of the electric dipole direction pointing ?
- If a 1S H atom would absorb visible radiation from a certain direction to yield a H atom in the 2P(x) state, would the electric dipole direction be at 90 degrees to the propagation of the radiation?

Thanks again,

*Bob*

Bob,

I will try to answer your latest questions as I believe you meant them.

- I think your first question has to be rephrased in terms of what states the H atom really exists w/o any external influences. These are the nlm states where n = 1,2,3,... is the principal quantum number, l = 0(S), 1(P), 2(D), .... is the angular momentum q.n., m = -l,...,+l (by integer steps, for a total of 2l+1 m-states for each l) is the z-component of l. Classically, l is the angular momentum vector which describes any rotation of the system.
So what we want to consider are electric dipole transitions:

- (2,P,-1) -> (1,S,0)=1S
- (2,P,+1) -> 1S
- (2,P,0) -> 1S

For the +-1 -> 0 transitions, the dipole vector lies in the x-y plane totally perpendicular to its direction previously. The photon is elliptically polarized depending on where k points. If k is along +(-)z, we get right(left) circular polarization. If k is in x-y plane, we get plane polarization. Anything in between is an elliptical combination. This is for the -1 -> 0. For the +1->0, reverse the definitions. I believe the angular distribution in this case goes like cos^2(theta)*sin^2(phi) where now theta is the angle between the z axis and k as it was before (but now the dipole is not along z so you get opposite trigonometric behavior). Phi is the angle between the dipole vector and k's projection into the x-y plane. So if k lies in x-y plane, you get sin^2(phi) dependence. If k is along z, you get no angular dependence since the photons only come out one (well really, two, up-down) way anyway.

So essentially, 2P+-1 -> 0 emits at 90 degrees wrt 2P0->0. You can see this from the cos^2(theta) vs. sin^2(theta). I think the problem is that you are expecting a specific direction. Each transition emits its maximum amount of photons in a plane perp. to the dipole. This is not a specific direction in space though. So the direction of these two planes for the two types of transitions are at 90 deg but you still get overlap of the two distributions.

I suppose you could make a superposition of two states, the dipoles from the states being perpendicular to each other. Then if you work out the radiation rate, you would probably get a distribution clustered around a single direction which is mutually perpendicular to both of these. I think this is right, but I haven't done the calculation to be sure. You could then construct two states with radiation direction perpendicular. I guess you could call these 2Px and 2Py.

- If you use the original two states, there is overlap of the angular distributions, so if you had two atoms in the same coordinate system to compare, you will get sometimes sets of photons with the same direction, sometimes 180 deg apart, but most times they will be at some arbitrary angle closer to 90 deg. If you use linear combinations, you can construct two atoms which have radiation
directions 180 apart.
- Yes, on average, you must build up the angular distribution. So over a successive number of emissions, one atom will emit no photons parallel to its dipole, a few near parallel, some more midway, a bunch closer to perpendicular, and the maximum at perpendicular. The specifics are in the angular distributions above. Quantum mechanics allows you to linearly add quantum states together so you could have state which is part 2P-1 and part 2P0 for example. But you have to get there somehow, you must prepare it somehow. I claim I can prepare 2P-1 and 2P0 because, they involve transitions from other stable states such as the ground states which occur in nature.
- Yes, the selection rules work in reverse as well. On average, the photon must have a polarization consistent with the emission rules given above for the states used above. So the ensuing dipole of your state would have to be at 90 deg to the photon's direction. On the other hand, this is only one photon. I claim that you cannot create a 2Px state just by hitting it with the right photon (direction,polarization), since the atom has no way of knowing that you want 2Px. It will think you want 2P+1 let's say. You already know there is overlap since we constructed the mixed state that way. On the other hand, let's say I hit my atom successively with many photons of a certain type, then what state does it relax into. That is a good question and I don't know offhand although, the laser experts must. I will get back to you.
Sincerely,

*Glenn Blanford*

Fermilab Public Affairs

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