2
$\begingroup$

Fermi's Golden Rule states that the rate of a transition of an electron going from the ground state $0$ into some state $n$, is directly proportional to the square of the first order perturbation $\hat U$ of the Hamiltonian $$\Gamma_{0n} = \frac {2\pi}{\hbar} \left|\right\langle{0} \vert \hat U \vert n\rangle \vert^2 \delta(\omega-\omega_{0n}) = \frac{\pi A^2_0}{2 \hbar c}|T_{0n}|^2 \delta(\omega-\omega_{0n}) $$

where $A$ is a vector potential representative of the incident electromagnetic field, $\omega$ is the frequency of the incident field and $|T_{0n}|^2$ represents the sum of the various transition moments (i.e. electric dipole, electric quadrupole, magnetic dipole etc.) $$T_{0n} = T_{0n} ^{(0)} + T_{0n} ^{(1)} + T_{0n} ^{(2)} + ... $$

The oscillator strength $f_{0n}$ is related to $|T_{0n}|^2$ and the absorption cross-section $\sigma_{0n}$ via: $$ f_{0n} = \frac{m_e c}{2\pi ^2 e ^2 \hbar} \sigma_{0n} = \frac{2m_e}{e^2 E_{0n}}|T_{0n}|^2$$

where $E_{0n}$ is the transition energy.

For a dipole transition we know that $T_{0n} ^{(0)}$ can be written as $$ T_{0n} ^{(0)} = \frac{e}{m_e} \sum_{i} \langle{0} \vert \hat p_i \cdot E \vert n\rangle = E \cdot \langle{0} |\hat \mu_i \vert n\rangle$$

where $\hat \mu_i$ is the electric dipole moment operator in the length representation and the summation index $i$ represents the current electron being considered $$ \hat \mu_i = e \sum_{i} \hat r_i $$

Given this, we can express the oscillator strength in terms of the individual components of the transition dipole moment as: $$ f_{0n,x} = \frac{2m_e}{e^2 E_{0n}}|E \cdot \langle{0} |\hat \mu_{i,x} \vert n\rangle|^2$$

$$ f_{0n,y} = \frac{2m_e}{e^2 E_{0n}}|E \cdot \langle{0} |\hat \mu_{i,y} \vert n\rangle|^2$$

$$ f_{0n,z} = \frac{2m_e}{e^2 E_{0n}}|E \cdot \langle{0} |\hat \mu_{i,z} \vert n\rangle|^2$$

My question is: How can we express the oscillator strength of a tensor quantity like the electric quadrupole in terms of each of the individual tensor elements? In other words, what would be the appropriate expressions to obtain $f_{xx},f_{xy},f_{xz},f_{yy},f_{yz},f_{zz}$ ?

Thanks for any help/guidance!

References:

  1. J. Chem. Phys. 137, 204106 (2012); https://doi.org/10.1063/1.4766359
$\endgroup$
1
$\begingroup$

I think that there is some confusion here in the post, so I went and skimmed your very helpfully placed reference. Just as a matter of notation, $\hat{\mu}$ is almost always the dipole or transition dipole operator. So everything you wrote for $\hat{mu}$ is correct, given that you consider it as the electric dipole operator, which is a vector. So summing over the 3 terms in your final set of calculations will give you the operator strength of the electric dipole transitions.
The electric quadrupole, which in your references is given as $\hat{Q}$ of course is a rank 2 tensor, and can't be expressed using those formulae. Instead of what you wrote, we can write the quadrupole operator in the length representation as

$\hat{Q_{\alpha \beta}} =e\sum_i {r_{i\alpha}r_{i\beta}}$ (Eq 30) in your reference.
This will give you 6 independent rate terms ($\Gamma$). Once you have those 6 terms, you can then find the oscillator strength of each term using the usual equation $f_{n,0}^{Q_{\alpha \beta}}=\frac{2m_E}{e^2E_{n,0}}|Q^{\alpha,\beta}_{n,0}|^2 $

So that's how you would do it for each tensor element $\alpha, \beta$ Before moving on, I do want to be clear that his paper, and a followup from Xiaosong Li's group (https://doi.org/10.1063/1.4937410 sorry it's behind a paywall, not sure if it's okay that I still post the link) is that truncating the oscillator strength at the electric quadrupole is not the right way to proceed. This is because you get an oscillator strength that depends on your coordinate system choice, a clear non-physicality. In this case, it can be remedied by including octopole-dipole coupling, magnetic dipoles, and magnetic quadrupoles. This completes a 2nd order perturbation approach which contains all of the terms that of 2nd order in the wavevector of the light or less. Just a head's up, but the formulae above are how one would calculate the quadrupole moment only oscillator strength.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.