Fluorophores and Old Aged Particles

The fundamental particles of particle physics are memoryless insofar that they have no internal state that tells them when they are old and when to die. A newly created muon is exactly as likely to decay in the next infinitessimal time interval $\mathrm{d} \,t$ as one that somehow has lingered for ten mean muon lifetimes. Therefore their lifetimes are exponentially distributed, and this statistical distribution is the fundamental hallmark of memorylessness. I illustrate this point both by showing the equivalence of the exponential lifetime distribution and memorylessness and by giving a counter-example of a non-fundamental particle whose lifetime is not exponentially distributed. This latter behaviour is witnessed in some fluorophores.

The Exponential Lifetime Distribution

Firstly, it should be emphasised that exponential distribution is the unique distribution that is memoryless in the sense spoken about above. In other words, to experimentally verify that a particle doesn’t “heed its age”, you look for the exponential distribution. If you see it, then you are observing memorylessness. But it’s stronger than this: if you don’t see an exponential distribution, you know there is some memory of age present.

To understand this uniqueness, we encode the memorylessness condition into the basic probability la:

$$p(A,\,B) = p(A) \, p(B|A)$$

That is, suppose after time $\delta$ you observe that your particle has not decayed (event A). If $f(t)$ is the propability distribution of lifetimes, then the probability the particle has lasted at least this long is $1-\int_0^\delta f(u)du$. The a priori probability distribution function that the particle will last until time $t+\delta$ and then the probability for decay in the time interval $dt$ is $f(t+\delta) dt$. Now this is events $B$ and $A$ observed together, but it is the same as simply plain old $p(B)$ since the particle cannot last until time $t + \delta$ without living till $\delta$ first! Hence the conditional probability density function is:

$$p(B|A) = \frac{f(t+\delta)\,\mathrm{d}t}{1-\int_0^\delta f(u)du}$$

But this must be the same as the unconditional probability density that the particle lasts a further time $t$ measured from any time, by assumption of memorylessness. Thus we must have:

$$\left(1 – \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta)$$

for all values of $\delta$. Letting $\delta\rightarrow 0$, we get the differential equation:

$$f^\prime(t) = – f(0) f(t)$$

whose unique solution is $f(t) = \frac{1}{\tau}\exp\left(-\frac{t}{\tau}\right)$. You can readily check that this function fulfills the general functional equation:

$$\left(1 – \int_0^\delta f(u)du\right)\,f(t) = f(t+\delta)$$

for any $\delta > 0$ as well. So memorylessness implies the exponential distribution. Conversely, the exponential distribution implies $f^\prime(t)\, \mathrm{d} \,t = – f(0) f(t) \,\mathrm{d} \,t$, which means the proportion $\mathrm{d}_t N(t) \,\mathrm{d} \,t/ N(t)$ of a large population $N$ that decays in time $\mathrm{d} \,t$ is $- f(0) N(t) \,\mathrm{d} \,t/ N(t)$ and is thus independent of $t$, so we have memorylessness.

This argument can be reproduced for discrete probability distributions as well. Again, the exponential distribution implies and is implied by memorylessness: the distribution of the number $n$ of coin tosses before a head shows up is $p(n) = 2^{-n}$ and shows that “a coin has no memory“.

Old Aged Fluorophores

There are “particles” that do remember their age, although they’re not fundamental particles. But they illustrate thconcepts above by showing what fundamental particles would need if they were to remember their age. If we think of an excited fluorophore (instead of a quantum field in a raised state), then fluorophores generally undergo one or more changes of state in their fluorescence process. We can think of this as a psuedoparticle – a quantum superposition of free photons and raised matter states – in the same way as a polariton is thought of as a pseudoparticle. Real fluorophores are more complicated – the quantum superposition involves states other than simply the primary excited state and photon, so there is an internal state to record the “particle’s” “age”. I have drawn below a schematic diagram of the energy levels for something like fluorescein below. The fluorophore generally gets raised to a higher level than it will fluoresce from, and thus undergoes a series of decays between these higher states before dropping back to the ground state (or, more often, something in a band just above the ground state). So the total fluorescence lifetime is the sum of several, memoryless pdfs: the total pdf – being the pdf of the sum of exponential distributions – is the convolution of all the individual exponential distributions.

FluoreceinJablonsky

Figure 1: Fluorescein Jablonsky Diagram

If there is one dominant higher energy state with lifetime $\tau_1$ and the main fluorescence transition has lifetime $\tau_2$, then the probability density function for the overall lifetime is $\frac{1}{\tau_1\,\tau_s}\int_0^t e^{-\frac{u}{\tau_1}} e^{-\frac{t-u}{\tau_2}} du =\frac{e^{-\frac{t}{\tau_1}}-e^{-\frac{t}{\tau_2}}}{\tau_1-\tau_2}$ and I have drawn a sample function of this kind for $\tau_1 = 1$ unit and $\tau_2 = 10$ units. Mostly, raised fluorescence states look like memoryless particles in practice because the higher states are so shortlived compared with the lowest singlet state, but there are some for which the behaviour below is quite observable: i.e. there is a time throughout which an excited population is quiet, then the fluorescence comes with a rush, then goes quiet again.

DoubleFluorescenceRiseAndFall
Figure 2: Composite State Behaviour Shown as Non-exponential Lifetime Distribution