Do Transition Dipole Moments Have Different Signs? Resolving a Contradiction

noi_tseuq
Messages
3
Reaction score
0
Hi,

Is dipole operator symmetric or antisimmetric? Or, in other words, do the transition dipole moments from state 1 to the state 2, and from the state 2 to the state 1 have different sign? I.e.

mu_12 = - mu_21.

As far as I understand, the diagonal elements for the dipole operator should be zero (since the transition dipole moments from state "n" to the same state "n" should ne zero). However, in this case the dipole moments of all excited states should be zeros! This does not look to be physical. How one can resolve this contradiction?

Thank you in advance.
 
Physics news on Phys.org
I assume you mean the electric dipole operator, which is \vec r.
This has odd parity, so that it is zero for any non-degenerate ground or excited state.
It is not zero for transitions from an excited state to a lower state of different parity.
In any event, mu_12=mu_21.
 
clem said:
I assume you mean the electric dipole operator, which is \vec r.
This has odd parity, so that it is zero for any non-degenerate ground or excited state.
It is not zero for transitions from an excited state to a lower state of different parity.
In any event, mu_12=mu_21.

Thank you for the answer. That what you say is an agreement with what I have. But I still does not understand the logic. You say that the dipole operator has odd party. I am not sure what exactly it means. Does it mean that the dipole operator is an odd function with respect to all Cartesian coordinates of the system? I also do not understand why in the case of the "odd parity" of the dipole operator the diagonal matrix elements should be zeros (for non-degenerate states)? Are wave-functions of non-degenerate states always even? If yes, why?

Thank you.
 
The "parity operation" is taking each Cartesian coordinate to its negative.
This changes \vec r to -{\vec r}.
A non-degenerate eigenstate has either even or odd parity, but then [tex|\psi|^2[/tex] will always have even parity since (-1)*(-1)=+1.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top