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

[bobydbcn]

the two energy level atom, whose hamiltonian takes the form
$\hat{H_A}=E_g|g\rangle\langle g|+E_e|e\rangle\langle e|$,
where $E_g$ as well as $E_e$ represent the eigenvalues and $|g\rangle$ as well as $|e\rangle$ denote the ground state and excited state.
There is an interaction between the atom and electric field. And the hamitonian of interaction reads
$\hat{H}_{AL}=-\hat{\vec{p}}.\vec{E}$,
where $\hat{\vec{p}}$ is an electir dipole operatior and $\vec {E}$ is an electric field.
If $\hat{H}_{AL}=-\hat{\vec{p}}.\vec{E}$ is represented in the energy eigenstates, its diagonal elements is zero. I wonder why only its off-diagonal elements exist ? It is said because $\hat{\vec{p}}$ 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.

 

[The_Duck]

Suppose you have a wave function $\psi(\vec{r})$ with the property that $\psi(\vec{r}) = \psi(-\vec{r})$. The wave function $\psi$ is said to have "positive parity." Meanwhile if $\psi(\vec{r}) = -\psi(-\vec{r})$ then $\psi$ has "negative parity." The standard set of hydrogen energy eigenstates, for instance, all have definite positive or negative parity. You should convince yourself that the parity of a hydrogen eigenstate with orbital angular momentum $\ell$ is $(-1)^\ell$.

Next you should convince yourself that if $\psi(\vec{r})$ has positive parity then $x\psi(\vec{r})$ has negative parity. Similarly if $\psi(\vec{r})$ has negative parity then $x\psi(\vec{r})$ has positive parity.

Next you should convince yourself that if $\psi$ has positive parity and $\phi$ has negative parity, then the inner product $\langle \psi | \phi \rangle = \int d^3 r \psi(\vec{r})^* \phi(\vec{r})$ is zero.

Finally you should put these together to argue that if $\psi$ has definite parity--either positive or negative--then $\langle \psi | \hat{x} | \psi \rangle = \langle \psi | \hat{y} | \psi \rangle = \langle \psi | \hat{z} | \psi \rangle = 0$. The operator $\hat{\vec{p}} \cdot \vec{E}$ is just a linear combination of $\hat{x}, \hat{y}, \hat{z}$, so this shows that $\langle \psi | \hat{\vec{p}} \cdot \vec{E} | \psi \rangle = 0$.

 

[bobydbcn]

Suppose you have a wave function $\psi(\vec{r})$ with the property that $\psi(\vec{r}) = \psi(-\vec{r})$. The wave function $\psi$ is said to have "positive parity." Meanwhile if $\psi(\vec{r}) = -\psi(-\vec{r})$ then $\psi$ has "negative parity." The standard set of hydrogen energy eigenstates, for instance, all have definite positive or negative parity. You should convince yourself that the parity of a hydrogen eigenstate with orbital angular momentum $\ell$ is $(-1)^\ell$.

Next you should convince yourself that if $\psi(\vec{r})$ has positive parity then $x\psi(\vec{r})$ has negative parity. Similarly if $\psi(\vec{r})$ has negative parity then $x\psi(\vec{r})$ has positive parity.

Next you should convince yourself that if $\psi$ has positive parity and $\phi$ has negative parity, then the inner product $\langle \psi | \phi \rangle = \int d^3 r \psi(\vec{r})^* \phi(\vec{r})$ is zero.

Finally you should put these together to argue that if $\psi$ has definite parity--either positive or negative--then $\langle \psi | \hat{x} | \psi \rangle = \langle \psi | \hat{y} | \psi \rangle = \langle \psi | \hat{z} | \psi \rangle = 0$. The operator $\hat{\vec{p}} \cdot \vec{E}$ is just a linear combination of $\hat{x}, \hat{y}, \hat{z}$, so this shows that $\langle \psi | \hat{\vec{p}} \cdot \vec{E} | \psi \rangle = 0$.

Thanks for your excellent expplanation. You teach me how to use parity.