13
$\begingroup$

When a single molecule absorbs or emits light, it does so perpendicular to the direction of the respective transition dipole. In principle, the directions of the dipoles for absorption and emission can be different, due to different electronic configurations of the ground and excited states of the molecule, but in practice the difference is very small. Photons are thus preferentially reemitted into the same plane in which absorption took place.

A large angle between these two dipoles would probably require a significant change in the molecular structure when the molecule is excited. I have heard of some fluorescent proteins which exhibit large angles between absorption and emission directions, but I do not know any details. (Edit: I found this interesting paper, which reports negative fluorescence anisotropy in YFP and a large angle between the absorption and emission dipoles.)

I would be interested in specific compounds which have a large angle between the two transition dipoles. Particularly, are there any simple compounds (less than ~100 atoms) that show such a behavior?

I have read about twisted intra-molecular charge transfer (TICT), in which a segment of the molecule rotates by ~90 degrees upon excitation. I could imagine that in such a case the emission might take place in a plane perpendicular to the original plane of absorption, but I haven't found any confirmation of this either.


Update: to illustrate, I have created a small sketch that outlines what I have in mind. The molecule I drew is just for illustration and does not necessarily show that behavior; I also don't know how exactly such a restructuring process actually takes place (the one I drew would correspond to a rotation of the rings on the right around the single bond). I further assumed that the dipole directions coincide roughly with the positions of the two rings, which might not actually be the case.

That said, the process I have in mind looks as follows:

  • A photon with a polarization vector parallel to the dipole direction arrives (its direction of propagation is necessarily perpendicular to the dipole direction)
  • The photon is absorbed and the molecule is left in an excited state
  • In the excited state some internal rearrangement of the molecular structure takes place, such that the dipole direction changes
  • A fluorescence photon is reemitted perpendicular to the new dipole axis. The photon has been effectively "redirected" (and shifted to higher wavelengths)

Change of molecular structure (potentially) leads to different fluorescence emission direction

$\endgroup$
  • $\begingroup$ "Very" different may be the issue here, but did you take a look at optical rotation? You can find all sizes of compounds that are optically active. $\endgroup$ – TAR86 Feb 20 '18 at 5:03
  • 1
    $\begingroup$ Not necessarily. Actually the molecule will finally decay to the original one else is photochemistry Like a state for the emission is formed in the gap. Emission could be even more red shifted. But this is what seems to me. I am not sure/ don't know the situation you ask for. ... $\endgroup$ – Alchimista Feb 22 '18 at 15:21
  • 3
    $\begingroup$ I do share your interest and I actually calculate spectra of compounds with this property. I never realized that the direction of emission dipole could be of any importance. I will dig through my data to find this out. Meanwhile, have a look at the most conscious list of fluorescent molecules under a name I would never guess: chem.ucla.edu/~craigim/pdfmanuals/catalogs/… $\endgroup$ – ssavec Feb 22 '18 at 18:43
  • 2
    $\begingroup$ OK, then I would suggest to take a look at molecular motors, some are driven by light absorption and I suspect that some of those also emit after rearranging. $\endgroup$ – TAR86 Feb 22 '18 at 18:54
  • 2
    $\begingroup$ Great first question! $\endgroup$ – hBy2Py Feb 22 '18 at 18:59
1
$\begingroup$

In part of the naphthalene molecule you draw you indicate a transition moment. This is the direction for absorption into the lowest excited state $S_1$. As naphthalene is highly symmetric the fluorescence is also emitted from a transition moment aligned in the same direction. The second excited state $S_2$ has a transition dipole set perpendicularly to the long axis and so to the $S_1$ direction. Again excitation and emission would be from similarly aligned transition dipoles if emission was from $S_2$. However, in solution $S_2$ decays very rapidly ( a few picoseconds) to $S_1$ thus the observed absorption and emission dipoles would be at right angles to one another.

The wavefunction (molecule orbital) in an excited state is different to that in the ground state since an electron is excited to an anti-bonding orbital. This means that bond lengths and angles can change a little. If the molecule has polarisable substituents that are asymmetrical placed then the excited state dipole may differ in direction from that of the ground state. However, in molecules such as rose bengal, rhodamine 6G and many dicarbocyanines, which are not strictly symmetric ($C_1$ point group), the absorption and emission dipoles are parallel as determined by experiment. (Presumably this is because substituents are not sufficiently close in energy to affect the electronic state in a significant way)

How can we know what the angle between absorption and emission is? The probability of absorption is proportional to the cosine squared of the angle $\theta$ between the electric field direction of the light and the transition moment direction, $\approx (\bar e \cdot \bar\mu)^2= e^2\mu^2\cos^2(\theta)$. So we can describe the molecules are being photo-selected, the probability of absorption being greatest when $\theta=0$, i.e. absorption and excitation directions are parallel.

If we use polarised light to excite the molecules, and assume that it is polarised along the $z$ direction, then the emission intensity observed through a polariser parallel to the $z$ (or vertical) direction is $I_v\approx (\bar e_z \cdot \bar\mu_z)^2$ and observed at right angles to this ($x$ or horizontal direction) $I_h\approx (\bar e_z \cdot \bar\mu_x)^2$ where $\bar e_z$ is a unit vector aligned along the electric field and $\bar \mu_z$ the unit vector component of the fluorescence dipole along the $z$ and $\bar \mu_x$ along the $x$ direction. The amount of polarisation of the fluorescence is usually measured via the anisotropy which is defined as

$$r=\frac{I_v-I_h}{I_v+2I_h}$$

where the denominator is the total emission intensity.

In fluid solution molecules undergo rotational diffusion and if the molecule is regarded as a sphere, in fact not such a bad approximation in many cases then, the anisotropy becomes $\displaystyle r(t)=r_0e^{-6Dt}$ where $D$ is the rotational diffusion constant. The anisotropy can be measured by recording the fluorescence decay such as by time-correlated single photon counting. Taking the decay back to time zero produces $r_0$, this is the anisotropy that would be observed if the molecules were in rigid solution and effectively measures the angle between absorption and emission dipoles.

Now this relates directly to your question. The value of $r_0$ can be also be calculated as well as being measured and, again, considering the angle between absorption and emission dipoles $r_0=(3\cos^2(\alpha)-1)/5$ where $\alpha$ is the angle between absorption and emission dipoles. When the angle is zero then $r_0=0.4$ which is easily measured and when at right angles $r_0=-0.2$ so this limits the range of anisotropy.

(references. Text. M. Daune. Molecular Biophysics publ. OUP 1999. Some original literature; T Tao, Biopolymers v8,p609, 1969, Y. Yguerabide Meth. Enzymology, v26, p498, 172, Chuang & Eisenthal, J. Chem. Phys. v57, p5094, 1972)

$\endgroup$
  • $\begingroup$ Thanks for this detailed review. Do you happen to know any substances for which $\alpha$ is large, i.e. $r_0$ is very small or negative? $\endgroup$ – Sentry Feb 25 '18 at 12:50

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.