Selection rules for atomic transitions

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SUMMARY

The selection rules for atomic transitions dictate that an electron can only transition in an atom if the change in orbital angular momentum, Δℓ, is ±1. This rule arises from the parity of electronic wavefunctions and the properties of the electric dipole operator, which has odd parity. Although the Schrödinger equation does not explicitly incorporate spin, the selection rules can still be derived mathematically, indicating that certain eigenstates are difficult to excite by external light waves. The interpretation of these rules often varies, but the underlying physics remains consistent.

PREREQUISITES
  • Quantum Mechanics fundamentals
  • Understanding of orbital angular momentum
  • Familiarity with electric dipole transitions
  • Basic knowledge of group theory
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  • Study the derivation of selection rules in quantum mechanics
  • Explore the role of parity in quantum transitions
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copernicus1
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Hi, I'm a little confused by the selection rules for atomic transitions. In pretty much any standard QM text, they derive the fact that an electron can't transition in an atom unless \Delta\ell=\pm1, i.e. unless the orbital angular momentum changes by one unit. Then the books go on to say that this can be interpreted as a consequence of the fact that the photon carries one unit of spin.

This seems strange to me, since the Schrödinger equation doesn't predict spin, and as far as I can tell, spin is not even put in by hand in these calculations. How do the transition rules somehow predict that the angular momentum must change by one unit when the whole formalism doesn't incorporate spin angular momentum?

It seems like someone working out this effect without any knowledge of spin would have to come up with some other interpretation.

Thanks!
 
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My understanding was that it's down to the parity of the original and new electronic wavefunctions. The electric dipole operator has odd parity (swap all the positions and it points in the opposite direction), and the transition probability is proportional to <new state|electric dipole operator|old state>, so for this integral to be nonzero the new and old states must have different parity values, i.e. <+|-|-> or <-|-|+>, otherwise the two wave-functions will cancel. States with the same l have the same parity, so transitions must involve the electron either gaining or losing one quanta of orbital angular momentum or the integral is zero and the transition is "forbidden".
 
Then the books go on to say that this can be interpreted as a consequence of the fact that the photon carries one unit of spin.

This seems strange to me, since the Schrödinger equation doesn't predict spin, and as far as I can tell, spin is not even put in by hand in these calculations. How do the transition rules somehow predict that the angular momentum must change by one unit when the whole formalism doesn't incorporate spin angular momentum?

It seems like someone working out this effect without any knowledge of spin would have to come up with some other interpretation.

The selection rule is just a rule of thumb; the radiation at frequency corresponding to "forbidden transition" has certain non-zero presence in the spectrum of the gas of atoms. The corresponding lines are just too faint when compared to regular ones.

You are right, there is never one correct interpretation. That's why the word "interpretation" - it is often very subjective view of what it all means and there are usually more than one.

Spins and photons indeed do not figure in the basic Schr. equation, but the selection rule still follow from the latter. This can be viewed as a purely mathematical fact, that certain eigenvibrations of Schr. equation are hard/impossible to excite by external light wave. This happens in simpler models too, for example a set of coupled harmonic oscillators can have eigen-modes which are hard to excite by external light wave.
 
Schrödinger eqn does not predict the nature of intrinsic spin,but it does predict orbital angular momentum.The change in total angular momentum is ±1 which can be derived from other means like invoking group theory.The result can be interpreted if we say photon has spin 1.By the way it is possible to prove using maxwell eqn that photon does have spin 1 character.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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