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

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.
  • #1
bobydbcn
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the two energy level atom, whose hamiltonian takes the form
[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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  • #2
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##.
 
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Likes eternalserv and bobydbcn
  • #3
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.
 

1. Why are the off-diagonal elements of the electric dipole moment matrix zero?

The off-diagonal elements of the electric dipole moment matrix are zero because the electric dipole moment is a vector quantity that describes the separation of positive and negative charges within a system. As a vector quantity, the direction of the electric dipole moment is important, but the magnitude of the vector is not. This means that the off-diagonal elements, which represent the orientation of the electric dipole moment, are not crucial and can be set to zero without changing the overall description of the system.

2. Can the off-diagonal elements of the electric dipole moment matrix ever be non-zero?

Yes, the off-diagonal elements of the electric dipole moment matrix can be non-zero if the system possesses anisotropy, meaning it has a preferred direction. In this case, the off-diagonal elements would represent the orientation of the electric dipole moment in relation to the preferred direction of the system.

3. How does the symmetry of a system affect the off-diagonal elements of the electric dipole moment matrix?

The symmetry of a system plays a crucial role in determining whether the off-diagonal elements of the electric dipole moment matrix are zero or non-zero. If a system possesses perfect symmetry, such as a spherical or cubic shape, the off-diagonal elements will be zero. On the other hand, if a system has an irregular or asymmetric shape, the off-diagonal elements may be non-zero.

4. Are the off-diagonal elements of the electric dipole moment matrix always zero in a molecule?

No, the off-diagonal elements of the electric dipole moment matrix can be non-zero in a molecule if the molecule has a permanent dipole moment. This means that there is a separation of positive and negative charges within the molecule, resulting in a non-zero electric dipole moment. In this case, the off-diagonal elements would represent the orientation of the dipole moment in relation to the molecular axes.

5. How do the off-diagonal elements of the electric dipole moment matrix relate to the polarizability of a system?

The off-diagonal elements of the electric dipole moment matrix are related to the polarizability of a system through the dielectric constant. The off-diagonal elements represent the orientation of the electric dipole moment, while the polarizability represents the ease with which the system can be polarized. A higher polarizability means a higher off-diagonal element in the electric dipole moment matrix, indicating a stronger response to an external electric field.

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