the two energy level atom, whose hamiltonian takes the form(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\hat{H_A}=E_g|g\rangle\langle g|+E_e|e\rangle\langle e|[/itex],

where [itex]E_g[/itex] as well as [itex] E_e [/itex] represent the eigenvalues and [itex] |g\rangle [/itex] as well as [itex] |e\rangle [/itex] denote the ground state and excited state.

There is an interaction between the atom and electric field. And the hamitonian of interaction reads

[itex] \hat{H}_{AL}=-\hat{\vec{p}}.\vec{E} [/itex],

where [itex]\hat{\vec{p}} [/itex] is an electir dipole operatior and [itex] \vec

{E}[/itex] is an electric field.

If [itex] \hat{H}_{AL}=-\hat{\vec{p}}.\vec{E} [/itex] is represented in the energy eigenstates, its diagonal elements is zero. I wonder why only its off-diagonal elements exist ? It is said because [itex] \hat{\vec{p}} [/itex] is an odd parity operator. I don't know about parity and don't know the causality between them.

Thanks. I don't know this parity part of quantum theory. I think maybe it is related with group theory.

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# Why the off-diagonal elements of electric dipole element are zero?

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