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I begin where Feynman et al. explain off-resonance absorption of light by molecules in volume III chapter 9 section 5 of [6]. They consider the case that the amplitude of the electric field is small and also the period of time during which photon absorption may happen $\mathrm{\Delta }t$ is small, so that ${\omega }_0\mathrm{\Delta }t\ll 1$, where ${\omega }_0$ is the resonant radial frequency of absorption for a molecule with an electric dipole moment. For this case, they derive the conditional probability that such a molecule will absorb a photon of monochromatic light over a time interval $\mathrm{\Delta }t$ is

The absorbed photon’s energy raises the molecule’s energy from $E_{\text{I}}$ to $E_{\text{II}}$, ${\omega }_0$ is the resonant absorption radial frequency, $ℐ$ is the spectral radiance, $\mathrm{\Delta }t$ is the time interval during which the absorption may happen, $\mu$ is the electric dipole moment of the molecule, $c$ is the speed of light in vacuum, $\mathrm{\hslash }$ is the reduced Planck constant, and ${ϵ}_0$ is the vacuum permittivity. $P_2(E_{\text{II}},\mathrm{\Delta }t|E_{\text{I}})$ is plotted in Fig. 3 for a fixed $\mathrm{\Delta }t$ and variable $\omega$.

Unlike monochromatic light, sunlight has an intensity $ℐ(\omega )$ per unit frequency interval, covering a broad range which includes ${\omega }_0$. Then the probability of absorption will become an integral over $\omega$:

The solar spectral radiance varies much more slowly than the sharp resonance term. The two functions in the integral might appear as shown in Fig. 4. Since the solar spectral radiance is relatively constant over the relevant range of integration, it can be taken out of the integral.

The mean energy absorbed over a time interval $\mathrm{\Delta }t$ is

or more explicitly,

This can be simplified by integrating over the interval where the probability peaks ${\omega }_0\pm 2\pi /\mathrm{\Delta }t$, making a change of variables $x=(\omega -{\omega }_0)\mathrm{\Delta }t/2$, and approximating $\int 𝑑x(x^{-2}{\sin }^{2}x)$ over the finite integration interval by its exact, closed form over $-\mathrm{\infty }$ to $\mathrm{\infty }$, which equals $\pi$. The error of this approximation can be estimated by complex contour integration in the upper-half complex-plane along a semicircle of radius $\pi$ with an indentation around the origin. The approximation is exact to first order in the radius of the contour. The expression for the mean energy absorbed becomes

The mean energy absorbed over a time interval $\mathrm{\Delta }t$ is inversely proportional to the half-width $\pi /\mathrm{\Delta }t$ of the transition probability distribution’s central peak at ${\omega }_0$. It is also directly proportional to the spectral radiance at the absorption peak.

The variance of the energy absorbed over a time interval $\mathrm{\Delta }t$ is

or more explicitly,

Following much the same procedure as for the mean energy absorbed leads to the simplified form

Recall that this analysis began by considering the case where the electric field is small and also the period of time $\mathrm{\Delta }t$ is small, so that ${\omega }_0\mathrm{\Delta }t\ll 1$, so the last term in square brackets is approximately equal to one, simplifying the variance to

The variance of the absorbed energy is directly proportional to the half-width $\pi /\mathrm{\Delta }t$ of the transition probability distribution’s central peak at ${\omega }_0$. It is also directly proportion to the solar spectral radiance at the absorption peak.

Now I will take the analysis from the quantum scale up to biological scale. After $N\gg 1$ absorption events, the cumulative energy absorbed will reach a value of $𝐄(N)=𝐄(0)+{\sum }_{j=1}^N\mathrm{\Delta }{E}_j$. The central limit theorem guarantees that the probability distribution of $\mathrm{\Delta }𝐄\equiv 𝐄(N)-𝐄(0)$ is Gaussian with a mean and variance

This is to be expected from a one-dimensional random walk, where the distance walked is analogous to the cumulative energy absorbed. This is why the noise I am considering here is generally referred to as ‘random-walk’ noise.

Viewed on energy-scales $\gg \mathrm{\hslash }{\omega }_0$, $\mathrm{\Delta }𝐄(N)$ looks like a continuous random process. At this energy-scale, the energy absorbed over the time interval $\mathrm{\Delta }\tau \equiv N\mathrm{\Delta }t$, defined as $\mathrm{\Delta }𝐄(\tau )\equiv 𝐄(\tau +\mathrm{\Delta }\tau )-𝐄(\tau )$, has a mean and variance of

So, by Eqs. 6 and 10, the mean energy absorbed and its variance over a time $\mathrm{\Delta }\tau$ are

and

The mean power $\overline{W}$ absorbed over a time interval $\mathrm{\Delta }\tau$ is related to the mean energy absorbed by

and the absorption power variance ${\sigma }_W^2$ is

The signal-to-noise ratio (SNR) of a Gaussian signal is the ratio of its mean squared and its variance (a unitless or dimensionless quantity). The SNR of the absorbed power is then

The signal-to-noise ratio of light power absorption ${\text{SNR}}_W$ takes on its extreme values where its derivative with respect to ${\omega }_0$ is zero, which is where

The spectral density of the environment at and above sea level can be approximated by Planck’s law. This approximation ignores (1) the absorption spectrum of the Earth’s atmosphere and (2) the reflectance spectrum of the Moon, making no distinction between nocturnal and diurnal organisms. According to Planck’s distribution law, the spectral radiance $ℐ(\omega )$ of a black-body at a temperature $T$ is

where $\mathrm{\hslash }$ is the reduced Planck constant, $c$ is the speed of light in vacuum, and $k_B$ is the Boltzmann constant.

Using this functional form for $ℐ(\omega )$, the left hand side of Eq. 18 is proportional up to a constant scalar multiple to

where $z\equiv \mathrm{\hslash }\omega /(5k_{B}T)$. Writing the expression this way makes clear where the singularities of the derivative of ${\text{SNR}}_W$ are located: where the denominators $e^{5z}-1$ and $z-1$ equal zero, which are at $z=0$ and $z=1$. The zeros of Eq. 20 can be found by analyzing its Laurent series expansions about these singularities. These will be all of the zeros by the uniqueness theorem of Laurent series within their annular domain of convergence and by the fundamental theorem of algebra.

An analytic function $f(z)$ has a zero of order $k$ at $z_0$ if

and $g(z_0)\ne 0$.

Expanding the exponential functions about $z=0$ in terms of powers of $z$, the right hand side of Eq. 20 can be written as

where

so $g_0(0)=4/25$, satisfying the condition $g_0(0)\ne 0$.

Expanding the powers of $z$ and the exponential functions about $z=1$ in terms of powers of $(z-1)$, the right hand side of Eq. 20 can be written as

where

so $g_1(1)=-1/25$, satisfying the condition $g_1(1)\ne 0$.

So, the zeros of the first derivative of ${\text{SNR}}_W$ with respect to peak absorption radial frequency are indeed at $z=0$ and $z=1$ ($\omega =0$ and $\omega =5k_{B}T/\mathrm{\hslash }$). The higher-order (or general) derivative test can be used to determine whether either of these radial frequencies maximize ${\text{SNR}}_W$.

Let all the derivatives of $f$ at $z_0$ be zero up to and including the $n$-th derivative, but with the $(n+1)$-th derivative being non-zero:

The higher-order derivative test guarantees that if $n$ is odd and , then $z_0$ is a local maximum; and if $n$ is odd and $f^{(n+1)}(z_0)>0$, then $z_0$ is a local minimum.

By Eqs. 22 and 24, the first non-zero derivative at the zeros is

Recalling that $f_{z_0}$ is the first derivative of ${\text{SNR}}_W$ near $z_0$, the value of $n$ in the higher-order derivative test needs to be incremented by one more than is written in Eq. 27, and so is equal to 3, which is odd. So, according to the test, since $g_0(0)>0$ by Eq. 23, $z=0$ is a local minimum; and since by Eq. 25, $z=1$ is a local maximum. These are the global minimum and maximum since these are the only zeros of the derivative of ${\text{SNR}}_W$, as explained previously.

Converting back from $z$ to $\omega$ via $z=\mathrm{\hslash }\omega /5k_{B}T$, the analysis above shows that when the spectral density illuminating the environment at or above sea level is approximated by Planck’s law, the global maximum of the SNR is reached when a photoreceptor’s peak radial frequency ${\omega }_0=5k_{B}T/\mathrm{\hslash }$.

How does the attenuation of light by water affect the prediction of the optimal peak absorption radial frequency for environments beneath the surface of bodies of water? The diffuse attenuation coefficient for irradiance $\kappa (\omega )$ is the apparent optical property that provides the most direct measure of the penetration of radiant energy in water,

where $ℐ(\omega ,\delta )$ is the spectral irradiance at depth $\delta$ and $ℐ(\omega ,0)$ is the irradiance at the surface.

Raymond C. Smith and Karen S. Baker created a consistent set of data on the diffuse attenuation coefficient for irradiance of clear fresh- and salt-water environments for wavelengths in the range 200–800 nm [16]. Based on their data, the diffuse attenuation coefficient can be approximated in a neighborhood of its minimum value $\kappa ({\omega }_a)$ by a quadratic,

where the coefficient $a$ can be fit to their data. Such a fit is shown in Fig. 5. Fitting over all of their data, the coefficient of determination $R^2$ over the range 300–600 nm centered at 450 nm is 0.9961.

Using this approximation for $\kappa (\omega )$ and the attenuated spectral density in Eq. 28, the condition for maximizing the SNR in Eq. 18 is proportional up to a constant scalar multiple to

where

and $z_a$ is the scaled radial frequency where the diffuse attenuation coefficient $\kappa$ is locally minimal.

Compare Eqs. 30 and 20. Notice the additional term in the square brackets that introduces a factor of $\alpha$.

Since the zeros of Eq. 30 are sought, it can be simplified by ignoring the exponential decay factor on the far right and grouping factors to emphasize the change in the expression going from the above-sea-level analysis to the below-sea-level analysis,

Notice the term in square brackets in Eq. 32. Going from the above-sea-level analysis to the below-sea-level analysis, this term changes according to

because at sea level $\alpha =0$. To see how this change affects the zero of Eq. 32, the right hand side can be factored into powers of $(z-1)$ to analyze it in the neighborhood of the zero of the above-sea-level analysis, $z=1$. The expression becomes

Near $z=1$, this expression can be truncated to the first power in $(z-1)$, so the expression can be simplified to

The approximate change in the zero, $\mathrm{\Delta }{z}_0$, from $z_0=1$ to $z_0=1+\mathrm{\Delta }{z}_0$ is characterized by

Comparing this form to Eq. 35, the approximate change in the zero when going from the above sea level to the below sea level analysis is

The approximate change in the zero has the limiting values ${lim}_{\delta \to 0}\mathrm{\Delta }{z}_0=0$ (sea level) and ${lim}_{\delta \to \mathrm{\infty }}\mathrm{\Delta }{z}_0=-(z_a-1)/(z_a-2)$ (deep ocean). By the diffuse attenuation data of Smith and Baker [16], $z_a\approx 1.1079$, so ${lim}_{\delta \to \mathrm{\infty }}\mathrm{\Delta }{z}_0=0.1209$, which corresponds to a limiting shift of the peak absorption wavelength from $\approx 499$ nm at sea level to $\approx 445$ nm at extreme depths. Notice that the diffuse attenuation coefficient data of [16] is well approximated by a quadratic over this range as shown in Fig. 5 with a coefficient of determination $R^2$ of 0.9961, so the analysis is valid over the entire range of depths.

This predicted shift to shorter peak absorption wavelengths at lower and lower depths is consistent with what is observed in various species of fish [20]. Exceptions to this trend are deep sea fish that rely on bioluminescence to illuminate their environment, which have rhodopsins with peak absorption wavelengths of $\approx 526$ nm. Deep sea fish without bioluminescence have rhodopsins with peak absorption wavelengths in the range 479–486 nm, which are within the predicted range [20].