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

• bobydbcn
In summary, the two energy level atom's Hamiltonian includes ground and excited states with corresponding eigenvalues. When there is an interaction between the atom and electric field, the Hamiltonian of interaction is represented as the product of the electric dipole operator and the electric field. This interaction only exists in the off-diagonal elements due to the odd parity of the electric dipole operator. Parity is a property of wave functions and can be used to determine the value of certain operators.

#### 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.

Last edited:
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##.

eternalserv and bobydbcn
The_Duck said:
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.