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In vibrational spectroscopy only transitions between neighboring vibrational states ($\Delta \nu = \pm 1$, $\nu$ being the vibrational quantum number) are allowed within the harmonic approximation. Where does this selection rule come from?

I have some ideas but haven't come to a satisfactory conclusion. I guess it has something to do with the transition dipole moment $P$ having to be nonzero. This is given by

\begin{equation} P = \left\langle \psi^{\text{final}} \big| \hat{\mu} \big| \psi^{\text{initial}} \right\rangle \end{equation}

where $\hat{\mu}$ is the dipole moment operator and $\psi^{\text{final}}$ and $\psi^{\text{initial}}$ are the wave functions of the final and the initial state, respectively. Since the wave functions of the harmonic oscillator are even functions for $\nu = 2n$ and odd functions for $\nu = 2n + 1$ (with $n = 0, 1, \dots$) and $\hat{\mu}$ transforms as a vector (and so can be considered an uneven function) the integral for $P$ vanishes if $\psi^{\text{final}}$ and $\psi^{\text{initial}}$ are both even (or uneven).

So, if my ideas about the problem are correct (please correct me if they aren't) I would expect a selection rule $\Delta \nu = \pm 1, \pm 3, \pm 5, \dots$. But why is it only $\Delta \nu = \pm 1$?

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2 Answers

up vote 3 down vote accepted

Parity is not sufficient to calculate this integral. The eigenfunctions of harmonic oscillator are mutually orthogonal meaning that $\left\langle \psi^n \big| \psi^m\right\rangle=0$ for every $n\neq m$ even if both are even.

Now, the important part of harmonic wavefunctions is Hermite polynomials. They have a particular recursive property that $$ xHe_n(x)=He_{n+1}(x)+nHe_{n-1}(x) $$ As you said, the dipole moment operator behaves as $x$ and, since polynomials are orthogonal, you get nonzero integrals only for $\Delta\nu=\pm1$.

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Just to add to gigacyan's answer, the harmonic oscillator Hamiltonian may be written in terms of raising and lowering operators: \begin{eqnarray} \hat{H}\psi&=&-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega^2x^2\psi\nonumber\\ &=&(a^{\dagger}a+\frac{1}{2})\omega\psi \end{eqnarray} where \begin{equation} a=\sqrt{\frac{m\omega}{2}}\hat{x}+\frac{i\hat{p}}{\sqrt{2m\omega}} \end{equation} and \begin{equation} a^{\dagger}=\sqrt{\frac{m\omega}{2}}\hat{x}-\frac{i\hat{p}}{\sqrt{2m\omega}} \end{equation} so \begin{equation} \hat{x}=\frac{1}{\sqrt{m\omega}}(a+a^{\dagger}) \end{equation}

In other words, the $\hat{x}$ contains a single raising and lowering operator, so facilitates only transitions with $\Delta\nu=\pm1$.

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Thanks, that clarifies gigacyan's answer even further for me. –  Philipp Feb 22 '12 at 15:07
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