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Lets say we have an electron which can be in two states. Its wavefunction for two states is then $\Psi=A\Psi_n + B\Psi_m$, where $\Psi$ is time dependent wave function.

I know that the transition integral can be written in 1D cartesian coordinates like this - $\psi$ marks the time independent wavefunction:

$$\int\limits_{-\infty}^{\infty}x\,\overline\psi_{m}\psi_{n} dx.$$

How do I write the selection rule for the 3D polar coordinate system? I have been thinking about writing something like this:

$$\int\limits_{V}r\,\overline\psi_{m}\psi_{n} \,dV=\int\limits_{0}^{\infty}r^3\overline\psi_{m}\psi_{n} \,dr\int\limits_{0}^{\pi}\sin\theta \,d\theta\int\limits_{0}^{2\pi}d\phi.$$

Is this correct? Am I missing anything? I used the volume differential $dV=dr\cdot r\sin\theta d\theta \cdot rd\phi$. I am not sure if it is correct to just replace the $x$ with $r$. Should I instead replace $x$ with its polar analog which is $r\sin\theta\cos\phi$?

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EDIT: My answer was wrong, so,it was deleted. Please refer to the other answers here.

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The transition integral $$\int x \psi_m^\ast(x) \psi_n(x)\text dx$$ is, in perturbation theory, part of the transition rate from state $n$ to state $m$ caused by a weak perturbation $\hat V=\lambda \hat x$, which couples to the system via the dipole operator $\hat x$ with a small coupling constant $\lambda$. This is not the only possibility, and other choices for $\hat V$ will lead to other transition integrals, but this is probably the simplest possible case.

In three dimensions, an atom will mostly couple to an incoming electric field via a dipole coupling. This is called an E1 interaction, and there are further terms in the series (electric quadrupole, E2, magnetic dipole, M1, and so on) which refine the details of the interaction. For a dipole coupling, the interaction operator $\hat V$ is proportional to a component of the position vector, $x_m$. This is usually chosen to match the spherical harmonics, by choosing $$\left\{\begin{array}{lll} x_{-1}&=x-iy&\text{(for left circularly polarized light)}\\ x_0&=z&\text{(for linearly polarized light)}\\ x_{+1}&=x+iy&\text{(for right circularly polarized light)} \end{array}\right.$$

Thus the transition integrals are written in the form $$\int x_{m'} \psi_m^\ast(\mathbf r) \psi_n(\mathbf r)\text d\mathbf r,$$ where $m'=-1,0,1$.

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  • $\begingroup$ Uhmm.. I seem to have a problem. It seems I simply calculated the value < M |r| N > in my answer. It is wrong. I am trying to delete it. But it seems I can't do it. How do I do it? $\endgroup$ – dj_mummy Sep 18 '13 at 14:40

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