Do Transition Dipole Moments Have Different Signs? Resolving a Contradiction

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SUMMARY

The discussion centers on the properties of transition dipole moments, specifically whether they possess different signs when transitioning between quantum states. It is established that the transition dipole moments, denoted as μ12 and μ21, are equal (μ12 = μ21). The electric dipole operator, represented as \(\vec{r}\), has odd parity, resulting in zero diagonal elements for non-degenerate states. This leads to the conclusion that non-degenerate eigenstates can exhibit either even or odd parity, impacting the behavior of transition dipole moments.

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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.
 
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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.
 

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