September 16, 2019, 08:42:18 PM
Forum Rules: Read This Before Posting

### Topic: Calculation of the emission intensities for a single fluorophore  (Read 226 times)

0 Members and 1 Guest are viewing this topic.

#### anndy

• Very New Member
• • Posts: 1
• Mole Snacks: +0/-0 ##### Calculation of the emission intensities for a single fluorophore
« on: September 02, 2019, 01:18:35 PM »
Hi all,
Can anyone explain to me how they finally obtain the equation for the vertical emission intensity and horizontal emission intensity? I lost somewhere as marked in the attachment. Thank you very much!

Bests,

Yao

#### Enthalpy ##### Re: Calculation of the emission intensities for a single fluorophore
« Reply #1 on: September 03, 2019, 06:13:43 AM »
If I interpret properly the linked diagram, the emitted electric field is along the axis of a molecule, so its components go like sin and cos. The power per unit of area is proportional to E×H, hence to E2 in a given medium and for plane waves, hence the sin2 and cos2. This is also consistent with the conservation of power: sin2 + cos2 = 1. The φ and θ only result from 3D rotations.

What I just wrote is a heresy in quantum mechanics. The electric field is just a statistical illusion that appears if enough photons are detected. A photon has no electric field. Proper wording would rather go like "the probability of detecting a photon with angle φ between the source and the detector's orientations varies like cos2φ". But as the received power depends on the mean number of detected photons, the probability of detecting a photon varies like E2.

Though, macroscopic quantities are often a means to find or double-check the probability calculations of QM, or provide one direction of understanding. After all, both must often meet if taking many particles. Sometimes it fails: for a hydrogen atom that de-excites from 3s to 2p, no electric field can represent the wave function ψ. As we're here, the wave function of a photon is a scalar ψ, it's not the electric field as many books allege.

A similar situation is for the "orientation" of proton magnetic moment in NMR. Probably everyone figures mentally the magnetic moment as a definite vector, despite this is not a measurable quantity for one proton, and not even a hidden entity inaccessible to measurements, it's only a statistics. Only probabilities for each component exist for one proton, and are treated as such by QM. A thought definite vector helps imagine how the probabilities of each component evolve in an external magnetic field.

#### Corribus

• Chemist
• Sr. Member
• • Posts: 2722
• Mole Snacks: +438/-20
• Gender: • A lover of spectroscopy and chocolate. ##### Re: Calculation of the emission intensities for a single fluorophore
« Reply #2 on: September 03, 2019, 11:30:30 AM »
The intensity of the spectroscopic transition is proportional to the square of the transition dipole integral. The latter basically relates the orientation of the electric field vector with the alignment vector of the shift in electron density (for electronic transitions)) between the two states involved in the transition. The vector relationship is a dot product. In that sense, it's basic trigonometry. The intensity is though proportional to the square of the transition moment, and this is the origin of the square in your equation.

As to why it's proportional to the square of the transition moment - that's a deeper question. It falls out of the fact that solving the problem of a spectroscopic transition requires application of the time dependent Schrodinger equation. I.e., it's not enough that the electric field vector has to interact successfully with the vector that describes the moment of electron density between the initial and final states, but it has to do so within a finite time window. Squaring the solution is necessary because of the imaginary component of the time-dependent Schrodinger equation.

Anyway, a crude and probably unsatisfying way to look at it is that quantum events are probabilistic. To get a successful transition, you have to have interaction between the electronic states and the field, or the field with the electronic states, at the same time. It's a 2D probability problem. So just like the probability of rolling two 3's simultaneously on two six-sided dice is (1/3)2, so the same occurs here. Dead spectroscopists everywhere are probably rolling over in their graves at that analogy but maybe it helps.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

#### Enthalpy ##### Re: Calculation of the emission intensities for a single fluorophore
« Reply #3 on: September 04, 2019, 03:55:00 AM »
Dead spectroscopists everywhere are probably rolling over in their graves at that analogy.

And how many of these spectroscopists used similar analogies all their career long to get some understanding of what's happening? I guess QFT would give a formal answer, but (1) it's abstruse (2) the heuristics that established its formalism must have relied on such analogies.

The φ and θ only result from 3D rotations.

Oops, I missed something. The molecule's axis isn't perpendicular to the direction of propagation. That's why there are φ and θ. Possibly it's not even a matter of detector orientation, I don't fully grasp the diagram.

Remembering that I built antennas during the Neolithic, the molecule radiates like a short dipole does, because it's nm long and the photon has µm wavelength. The E field radiation diagram is just a cos of the direction measured from the molecule's equator, and the power density a cos2. If the reference for the angles is θ away from the molecule's axis, the detector's direction is characterized by two angles, φ and θ, and the cos2 becomes an expression of both.