[scientific]

Hi all,

I always had a hard time with molecular symbols, and I was instructed to consider those when answering the problem.

Which of the following electronic transitions is forbidden for a hydrogen-like atom?
a. 2p  3p
b. 2p  1s
c. 2p  3s
d. 2p  4s
e. 2p  3d

[Corribus]

Well, what do you think?

[scientific]

Honestly, I would answer (e) since a hydrogen-like atom promoting from a 2p orbital to a 3d sounds absurd, but this is panic reasoning.




I know this.....A hydrogen like atom assumedly has an inversion center.When considering molecular symbols, I know there's the spatial "direction," the u/g thing.  The Laporte rule says that any transition that keeps the  symmetry is forbidden. But I think that applies only within given subshells? Like a 2p  2p transition wouldn't be allowed, which isn't here at all. I mean, does that apply for (a), even though they are different energy levels?

edit: is a change in orbital degeneracy always allowed by this rule (not in general, but according to Laporte rule alone) since it changes symmetry (?), because if so, I would say (a). Or are these not related?

[Corribus]

Look up the Grotrian diagram for hydrogen. This wI'll help you understand the allowed transitions.  You are on the right track with thinking of selection rules.  The critical one here is that which requires the angular momentum to change during a transition.  This classical origin of this rule is that a photon haso an angular momentum, so to conserve momentum, the atomic system has to change its momentum by an equal amount when the photon is absorbed.

[Enthalpy]

"promoting from a 2p orbital to a 3d sounds absurd"

"Transitions" include photon absorptions too. Which is a good choice, because these follow the same selection rules as photon emissions. Logically enough since an absorption is an emission where the time runs backwards.

This is consistent with thermodynamics (fortunately). Since bodies emit EM radiation due to their temperature, the ease of emission and of absorption must be equal. If not, a colder body could heat a warmer one. Since we can make frequency filters, polarization filters... the eases of absorption and emission of one body at one temperature are equal at any frequency and any polarization.

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"a photon has an angular momentum"

Or it can be linearly polarized. But the orbitals too. The equivalent for 2p would be a peacock orbital (linear) versus a doughnut one (circular). And at least for the transitions to spherical orbitals, this corresponds to the linear or circular polarization of the involved photon.

Corribus' "angular momentum" was a neat, clear and understandable answer. Apologies for spoiling it, I just couldn't resist.

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During a transition, the electron's wavefunction is a linear combination of two orbitals - a legitimate wavefunction, solution of Schrödinger's equation since it is linear. When the combined orbitals have different energies, the linear combination is not a stationary solution (not an orbital): it wobbles at a frequency equal to the energy difference (put h adequately).

More details for a weighted sum of 1s and 2p there
http://www.chemicalforums.com/index.php?topic=82337.msg299165#msg299165

If the difference between the quantum numbers is 1, the wobble is dipolar, and the electron can radiate or absorb light having the wobble's frequency. With other differences, the wobble is spherical, quadrupolar or worse, and this movement can't radiate nor absorb light. That's a forbidden transition.

Transitions are still possible, but through indirect mechanisms that often involve more atoms.

[Corribus]

Of course the classical view is a simplification. It is not only angular momentum that must be conserved, but also parity - although these concepts are somewhat related. In either case, the photon angular momentum is independent of the light polarization, and cannot be equal to zero due to the relativistic nature of the photon. This is one interpretation of why it violates a selection rule for, say, the 2S photon of hydrogen to decay directly to the 1S state. The transition would emit a photon with a total angular momentum >0, but the momentum of the system wouldn't change. This violates the (classical) conservation of angular momentum.

Naturally the physics involved are quite a bit more complicated, but it's, as you say, a neat explanation that students can appreciate.

Significantly more detail can be found here for the interested reader:

https://www.eng.fsu.edu/~dommelen/quantum/style_a/consem.html

And particularly about the nonzero angular momentum of the photon:

https://www.eng.fsu.edu/~dommelen/quantum/style_a/phwav.html#sec:phwavsp

[Enthalpy]

I hope the nice explanation with angular momentum can be adapted a little bit for orbitals and photons of linear polarization, by noting that both can be written as a linear combination of circular polarizations. I haven't put time in it, but this looks like mere redaction.

In the linked text, I dislike the claim "the wave function of the photon is a four-di­men­sion­al vector"
https://www.eng.fsu.edu/~dommelen/quantum/style_a/phwav.html#sec:phwavsp
it is presently my understanding that all wave functions are complex scalars, including for the photon or for several particles. The photon's wave function depends on xyzt (for instance) AND on an orientation, for instance the detector's one.

In the same linked text, I dislike too that it implicitly favours circular polarizations over linear ones. "The spin can be +1 or -1" holds if the detector is circular. The photon does not choose necessarily such a circular polarization. If the detector is linear, the photon can choose a vertical or a horizontal one. Linear polarization are just as fundamental as circular, as weighed sums tell. EPR experiments rely on that and show that entangled photons pairs don't choose a circular polarization upon emission; they can choose a linear one and upon detection.

All texts (and my professors) used to say "the E field is the photon's wave function" but that's not general enough. For instance if a 3s->2p transition emits a photon in any direction, this can't be written as an electric field, but it can as a scalar ψ that depends on a polarization vector. Same for a positron-electron annihilation, where a scalar ψ can represent a photon's wave function, or even better, one wave function for both photons, but a vector E field can't represent an isotropic probability.