# Why does the Förster resonance energy transfer scale as 1/r⁶?

The title more or less contains my question - why does the Förster resonance energy transfer (FRET) (photon?)-based energy transfer mechanism, which folks sometimes use as a spectroscopic ruler to measure the inter-particle spacing between dye pairs, exhibit $1/r^6$ scaling for energy transfer efficiency? Since we're in 3-space when performing a FRET measurement, I could understand some $1/r^3$ contribution due to us having an expanding sphere of radius $r$, but where does the rest of the distance-dependent efficiency dropoff come from?

• Perhaps this is a question best answered by the folks at physics.SE. I think the $r^{-6}$ law may come from the interaction of two electric dipoles, but I am not sure. – Nicolau Saker Neto May 16 '13 at 1:32
• A random article I pulled up on Google suggests that is the case. – Nicolau Saker Neto May 16 '13 at 1:57
• I think a dipole-dipole interaction energy is proportional to $r^{-3}$ and the rate constant of this energy transfer is proportional to the square of the interaction namely $r^{-6}$. Therefore from the quantum yield of the rate constants this gives you the desired factor. Thats my understanding but I'm not an expert on this field by any stretch of the imagination!! :) – AngusTheMan Oct 17 '14 at 12:00
• The interaction is via induced dipole moment. The example most familiar to the chemist would be London dispersion forces, which indeed show a 1/r^6 dependence. – Abel Friedman Oct 17 '14 at 16:23

This coupling is strongest if the corresponding optical transitions in both molecules are allowed for electric dipole radiation. Then these transitions are coupled not only to the radiation field but also to each other. Naturally, the interaction energy is of a dipole-dipole nature, depending on an inverse proportionality to the third power of the molecular distance. The probability of energy transfer is then proportional to the square of this interaction energy and decreases, therefore, with the sixth power of the distance.

This paper does not give a detailed derivation though. But in his seminal paper from 1948 Förster goes through the derivation in detail (unfortunately it is not in English but in German and the condition of the text is not very good).

The Forster (or resonance or dipole-dipole) energy transfer is a non-radiative transition between an electronically excited molecule and an acceptor. Non-radiative really does mean that no photons are involved even though the excited donor looses its electronically excited state energy and this is transferred to the acceptor which becomes electronically excited. The transfer is represented as

$$\ce{D^* + A -> D + A^*}$$

The figure below shows the orbital changes in Forster dipole-dipole transfer (left) and for comparison by the exchange interaction (right).

The $R^{-6}$ distance dependence is due to the coulomb interaction which is expanded into a point multipole series. Generally the first term is a dipole-dipole term representing the interaction between the transition dipole moments of both molecules, and the squares of these quantities are proportional the the oscillator strengths of the ground state to excited state transitions, i.e. the absorption and emission spectra as vibrations are also involved.

If higher order multipole contributions may be neglected then the interaction matrix elements are proportional to $R^{-3}$ where R is the separation of the centres of the molecules. As the transfer probability is proportional to the square of the matrix element, the distance dependence is $R^{-6}$.

For Forster transfer to occur the separation between the two molecules must not be too small (otherwise exchange transfer occurs instead), and the interaction energy must be very small, less than the width of a typical vibronic transition (otherwise excitons are formed between the molecules). Because the molecules are not close to one another there is effectively no orbital overlap.

The calculation of the the constant starts with time dependent perturbation theory and given by the Golden Rule equation. In short this can be made into the product of the square of the electronic interaction energy, Franck-Condon factors for electronic transitions and density of states in the acceptor all averaged over a thermal distribution of energies. It has the form

$$W_{ab} = \frac{2\pi}{\hbar}\sum_{v’}\sum_{v’’}P_{Dv’} \left| \left \langle \psi_{Dv’}\middle |\, \hat H\, \middle|\, \psi_{Av’’} \right \rangle \right|^2 \delta (E_{Dv’}-E_{Av’’})$$

where $P_{Dv’}$ represents a Boltzmann factor over the vibrational levels of the donor molecule D and $\hat H$ is the interaction energy between donor and acceptor. $\psi_{Dv’}$ is the state in which the donor is electronically excited and the acceptor A is not and $\psi_{Av’’}$ is the state when the acceptor is excited and the donor is not. The delta function represents the density of states and also ensures energy conservation.

In using the adiabatic approximation to evaluate this expression the vibrational and electronic wavefunctions are approximated as the product $\psi = \theta \phi$ where $\theta$ are wavefunctions nuclear motion and $\phi$ electronic. In this case

$$\left \langle \psi_{Dv’}\middle |\, \hat H\, \middle|\, \psi_{Av’’} \right \rangle = \left \langle \theta_{Dv’} \middle | H_{DA} \middle | \theta_{Av’’} \right\rangle$$

and

$$H_{DA} = \left \langle \phi_D \middle | \hat H \middle | \phi_A \right \rangle$$

The interaction matrix element is

$$H_{DA} \approx \left\langle \phi_D \middle | \sum _i \sum_j \frac{e^2}{n r_{ij}} \middle | \phi_A \right \rangle \approx \frac{1}{n^2R^3} \left[\vec \mu_{D^*} \cdot \vec \mu_{A^*} -\frac{3}{R^2}(\vec R\cdot \vec \mu_{D^*})(\vec R\cdot \vec \mu_{A^*})\right]$$

The distance dependence is thus seen to be $R^{-6}$ in the final rate constant and the remaining term gives the orientational factor $\chi$ and for random motion in solution $\chi^2$ averages to $2/3$. The $\mu$ are the respective electronic transition dipole moments and n is the refractive index (relative permittivity) of the medium the molecules are in.

Solving the for the rate constant involves replacing the delta function with its integral representation and then noticing that the functions produced are the fourier transforms of the absorption and fluorescence spectra. The final result adding constants to put the absorption spectrum in extinction coefficient units ($\pu {dm^3 mol^{-1} cm^{-1}}$) and frequencies in wavenumbers is

$$k(R)= \alpha \chi^2k_f\frac{1}{R^6}\int \frac{F_{\mathrm{donor}}(\nu)\epsilon_{\mathrm{acc}}(\nu)}{\nu^4}\,\mathrm{d}\nu = \chi^2k_f\left(\frac{R_0}{R}\right)^6$$ where $$\alpha=\frac{9000\ln (10)}{128\pi^5n^4N} ~~~ \text{ and }~~~ \chi =\vec \mu_D \cdot \vec\mu_A - 3(\vec \mu_D \cdot\vec R)(\vec \mu_A \cdot \vec R)$$

The integral represents the overlap of the (normalised) fluorescence spectrum of the donor with the absorption spectrum of the acceptor, and the constant $R_0$ represents this integral and has values of a few nanometres, far larger than molecular diameters. The larger the overlap the greater the rate constant, this in effect an energy conservation term, the more vibrational levels of the donor that are equal to to exceed those of the acceptor the larger the rate constant. The orientational term $\chi^2$ can take values between $0$ and $4$ and $k_f$ is the radiative lifetime of the donor molecule. Notice that all the terms in the rate constant can be determined experimentally.

The importance of Forster transfer cannot be understated as it is the primary mechanism by which energy moves among the antenna molecules in photosynthetic organisms.