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When an atom has an electron in an excited energy level and it transitions to a lower level it emits a photon. What direction is it likely to emit the photon in? Are all directions equally likely, even toward the nucleus?

Alex
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    Have a look, might be along the same lines: http://physics.stackexchange.com/q/22001/ actually, how would you explain a mirror, or Snell's law, in the above case? –  Sep 06 '15 at 17:24
  • Take a look at the dipole emission pattern. This is a good approximation for luminescent molecules (and probably for atoms too). You end up with a $sin^2(\theta)$ profile if memory serves. – boyfarrell Sep 10 '15 at 18:30

2 Answers2

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In the general case, "no" as there is a angular momentum transfer involved (meaning there are preferred directions relative the original and/or final angular momentum of the atom).

That said, for most matter at room temperature the atoms have random orientation so you can treat the answer as "yes" for experimental purposes.

Now, I see that you are wondering about the nucleus in the question. There are two thing to keep in mind here.

  • First you may be thinking that the photon is emitted by the electron and that the electron is on a particular side of the atom at the time of emission. (That is, you may have some version of the almost completely incorrect Rutherford/Bohr/de Broglie atom in your head.)

    It is the system of nucleus and electron(s) that undergoes a change and emits a photon, and even from the beginning asking "from what part of the atom is the photon released?" has no precise answer.

  • Second the nucleus is also a quantum system and can't absorb arbitrary amounts of energy in internal changes (it wouldn't be able to absorb it in arbitrary translational motion either because that would violate the conservation of momentum---it drags the electron with it, after all). The nuclear energy splittings are generally too large to absorb an atomic photon.

dmckee --- ex-moderator kitten
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    "'no' as there is a angular momentum transfer involved". I think that this is wrong. Yes, there is angular momentum transfer involved, but that would correspond to the polarization of the photon emitted, but not the direction of it's propagation. – aquirdturtle Sep 06 '15 at 17:40
  • @aquirdturtle Photons don't have degrees of polarization. Each one is circularly polarized either parallel or anti-parallel to their direction of motion. This is a feature of their being massless (which also precludes them from exhibiting the third state that would be expected of a spin=1 particle). You can make experimental use of this limit by building polarized targets. – dmckee --- ex-moderator kitten Sep 06 '15 at 17:45
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    "Each one is circularly polarized either parallel or anti-parallel to their direction of motion." It's better to think about it relative to the atom. Relative to the axis of the atom you have $\sigma+$, $\sigma-$, or $\pi$ polarization. These different configurations have different directions of angular momentum relative to the atom, allowing the photon emitted to carry away angular momentum from the particle via it's polarization orientation, not it's direction of propagation. – aquirdturtle Sep 06 '15 at 17:50
  • @aquirdturtle True , the wavefunction of the photon has a component of polarization but the measurement of a single photon can only show + or - spin . The polarization is a mathematical construct that adds up in an emergent em beam, not measurable on a single photon. Going away from the atom it takes spin + or -1 away in its direction of motion and leaves the atom minus that spin. – anna v Sep 06 '15 at 18:02
  • @annav I'm sorry but I think you're entirely wrong on two points. Photons are spin-1 particles and can have 3 orientations, m_s = {-1,0,1}, not just + or -. That corresponds to the three polariztations I mentioned before. Secondly, the polarization of an individual photon has very real experimental implications, the effects of which can be seen in photonic Hong-Ou-Mandel experiments. – aquirdturtle Sep 06 '15 at 18:06
  • @aquirdturtle Photons have zero mass , are elementary particles they have spin one but only two components of spin +/-1 because of the zero mass. I agree that in the bulk of electromagnetic wave composed by many photons the polarization is there. I am speaking of individual photon scatterings.http://physics.stackexchange.com/questions/46643/why-is-the-s-z-0-state-forbidden-for-photons I can see that the collective crystal+photon ( was not aware of the effect) state could also depend on the polarization component of the photon wave function, since again it is a one to many solution. – anna v Sep 06 '15 at 18:34
  • @annav hmm, interesting. I take back my point about 3 polarizations; $\pi$ polarization is just an equal superposition of $\sigma+$ and $\sigma-$. However, my point about individual photon spins having real experimental consequences still stands. Look up the HOM effect. Also still stands that it isn't the direction of propagation of a photon that carries away angular momentum, but the (restricted) spin-orientation. – aquirdturtle Sep 06 '15 at 18:42
  • @aquirdturtle But the spin of a single photon is either along its direction of motion or against it. The wavefunction which will give finally this spin has many components, but in two particle scatterings the spin is either + or - in the photon's direction of motion. – anna v Sep 06 '15 at 18:44
  • @annav I agree. I don't see how this changes my points. – aquirdturtle Sep 06 '15 at 18:46
  • I'm not in a place where I can dig up references, but experimental particle physics relies on the anisotropic emission effect at times. Admittedly usually in a nuclear context. – dmckee --- ex-moderator kitten Sep 06 '15 at 19:12
  • @dmckee Interesting. I'd love if you could post some references at some point; I still don't see how the direction of a particle's leaving an atom would correspond to any angular momentum exchange at all. Admittedly I'm thinking more in a context of lasers and stuff. – aquirdturtle Sep 06 '15 at 19:49
  • @aquirdturtle I also agree that in complex systems, emergent electromagnetic beams and interactions with crystals the difference between polarization and spin is accessible to measurement. The question is about one atom and one photon though. I was not aware of the HOM effect. Thanks for bringing it to our notice. It is a higher order effect where virtual exchanges enter (as in the photon photon scattering) and there all aspects are involved.https://en.wikipedia.org/wiki/Hong%E2%80%93Ou%E2%80%93Mandel_effect – anna v Sep 07 '15 at 03:15
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    @aquirdturtle this might help , http://feynmanlectures.caltech.edu/III_18.html – anna v Sep 07 '15 at 03:31
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It depends on the mechanism by which the photon is emitted.

  • Stimulated Emission: Yes, the emitted photon will inherit the characteristics of the photon that stimulated it, including its propagation direction. That's how lasers get coherent light.
  • Spontaneous Emission: No, this should be random orientation. Another way to think about spontaneous emission is as stimulated emission where the stimulating photons are virtual photons, who have random orientation themselves.
aquirdturtle
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