# Transition integral from 1-D cartesian into 3-D polar coordinate system

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: I used the wrong operator in $\langle M|r|N\rangle$ . $r$ is not the correct operator to be used in the transition integral. Please refer to the other answers given here for the right operator.

$|N\rangle$ is state corresponding to $\Psi_n$ and $|M\rangle$ is the state corresponding to $\Psi_m$. Amplitude in this case is proportional to $\langle M|x|N \rangle$. In 3D polar coordinates (assuming energy is constant) the set of observables $\{r,\theta,\phi\}$ form a complete commuting set like $\{x, y, z\}$ did or $\{x\}$ in the 1D case.

Position eigenstates are now $|r,\theta,\phi\rangle$. So amplitude is now proportional to $\langle M|r|N\rangle$. One should also calculate $\langle M|θ|N \rangle$ and $\langle M|ϕ|N \rangle$ , however since the system is usually spherically symmetric the absolute value of the latter 2 have no meaning, as all angles are equally probable.

In 1D case, $x$ formed an complete commuting set all on its own. So $\int_{-\infty}^\infty |x \rangle\text dx\langle x|= \mathbb I$, which is the identity operator. Plugging in $\mathbb I$ in between $\langle M|x|N \rangle$ one gets the first formula you wrote.

To get the analog for polar co-ordinates one gets: $$\int_0^\infty\int_0^\pi\int_0^{2\pi} |r,\theta,\phi\rangle \,\rho \,\text dr\,\text d\theta\,\text d\phi\langle r,\theta,\phi|=\mathbb I,$$ where $\rho$ is a weight function equal to the determinant of the metric, so in this case $\rho= r^2\sin\theta$.

So, plugging in this operator in $\langle M|r|N\rangle$ one gets the required value.

Note however that only in spherically symmetric systems can one separate the angle integrals of $θ$ and $ϕ$ so your second formula is true only in such cases. The volume integral may be an area integral or even a discrete sum depending on what set of complete commuting observables you have taken. So one has to first know this set and from here he defines a Hilbert space of eigenkets of this commuting set. From this set of independent orthogonal eigenkets one defines the identity operator like above.

In this case you chose $\{x\}$ as a complete commuting set and then moved to $\{r,θ, ϕ\}$, so the 'volume' has a completely different meaning in both cases.

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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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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? – dj_mummy Sep 18 '13 at 14:40
You can't delete an accepted answer, I believe. – Emilio Pisanty Sep 18 '13 at 14:53