Confocal excitation and detection point spread function ($H_a(\mathbf{r})=E(\mathbf{r})D_a(\mathbf{r})$)

For an isolated fluorescent molecule at displacement $\mathbf{r}$ from the scanning optical focus, write $E(\mathbf{r})$ for relative excitation light intensity and $D_a(\mathbf{r})$ for the fraction of its collected emission passed by a centred detection pinhole of radius $a$. The expected response, apart from a common brightness and collection factor, isHere $\mathbf{u}$ is position in the detection-pinhole plane and $p_{\mathrm{em}}$ is the emitted light intensity distribution there conditional on collection, with units of inverse area. This point spread function normalization retains throughput: $D_a$ is a fraction, not a curve rescaled to have peak one. The product expresses excitation followed by detection; it is an intensity/probability product, not multiplication of coherent wave amplitudes from different molecules. ^[ref]

Assume linear, unsaturated fluorescence, independent emitters whose light intensities add, a stationary specimen, fixed collection efficiency and quantum efficiency, and negligible changes from scattering, absorption or bleaching during a scan. Different excitation and emission wavelengths generally give different excitation and detection responses. For a sufficiently small pinhole, $D_a$ is approximately its area times $p_{\mathrm{em}}$ at the centre; this is the familiar excitation-times-detection point spread function. It is not universally the square of one point spread function. A spatially invariant response further allows the image to be expressed as a convolution with the emitter distribution; depth-dependent optical aberrations can invalidate that approximation.

For an explicitly synthetic axial example, put $t=z/z_0$ and choose an on-axis excitation $E(t)=1/(1+t^2)$, inspired by a Gaussian beam. Stipulate a circular Gaussian function for the detection-plane light intensity, with standard deviation $s(t)=s_0\sqrt{1+t^2}$. In dimensionless coordinates $\mathbf{v}=\mathbf{u}/s_0$, put $\alpha =a/s_0$. ThenThe expression for $D_{\alpha }$ follows by integrating $p$ over a disk using polar coordinates. The assumptions specify a toy model with matched axial scales; its parameters are not a calibration for a real objective optical lens. It omits Airy disk rings, vector diffraction, and unequal excitation and emission wavelengths.

For a vanishingly small pinhole, the peak-normalized synthetic response tends to $(1+t^2{)}^{-2}$, whose full width at half maximum is $2\sqrt{\sqrt{2}-1}{z}_0\simeq 1.2872z_0$. With all collected emission accepted it tends to $(1+t^2{)}^{-1}$, of width $2z_0$. Meanwhile $H_{\alpha }(0)=1-e^{-{\alpha }^2/2}\to 0$ as the pinhole closes. The narrower normalized curve therefore accompanies a vanishing detected signal; peak normalization must not conceal that loss.