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

  1. Aug 22, 2013 #1
    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.
     
    Last edited: Aug 22, 2013
  2. jcsd
  3. Aug 22, 2013 #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##.
     
  4. Aug 22, 2013 #3
    Thanks for your excellent expplanation. You teach me how to use parity.
     
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