 Gaussian pulse energy and peak power

For an isolated Gaussian pulse, the relation between laser pulse energy $U$ and peak power $P_{\mathrm{pk}}$ depends on the pulse-width convention. Here $P(t)$ is the total incident radiant power, the background is zero, and $\tau >0$ denotes the power full width at half maximum. Integrating $P$ over the entire pulse defines its positive energy $U>0$.

Pulse energy, peak power and average power resolve different information. Consider a Gaussian pulse withThe chosen pulse duration $\tau$ is the full width at half maximum of radiant power: $P(\pm \tau /2)=P_{\mathrm{pk}}/2$. It refers to radiant power, rather than the wave amplitude of the electric field. The Gaussian integral gives the relation between Gaussian pulse energy and peak power:The pulse energy shape factor is $\kappa =U/(P_{\mathrm{pk}}\tau )$. It equals $1.064467$ for this power profile; a rectangular pulse of duration $\tau$ has $\kappa =1$. Knowing only a width and an integrated energy therefore requires a pulse-shape assumption before inferring peak power.

An equivalent form makes the normalization transparent. Write the pulse centre as $t_0$, and let $\sigma$ denote the standard deviation of its energy-normalized pulse profile:The ratio $p(t)=P(t)/U$ is a probability density function in the mathematical sense: it is nonnegative, has units ${\mathrm{s}}^{-1}$, and integrates to one. Here it weights the pulse's energy over time. Its normal distribution profile has mean $t_0$ and variance ${\sigma }^2$. This gives $P_{\mathrm{pk}}=U/(\sqrt{2\pi }\sigma )$ without a separate area estimate.

For a stipulated $\sigma =2\mathrm{ns}$ and $U=1\mathrm{nJ}$, the power full width at half maximum is $4.70964\mathrm{ns}$ and the peak power is $0.19947\mathrm{W}$. Confusing the standard deviation with the full width at half maximum would insert the wrong duration into the conversion.

The Gaussian pulse field-envelope width is a second distinct convention. If a fixed spatial field profile gives $P(t)=K|A(t){|}^2$ for a constant $K>0$, the wave amplitude envelope obeysConsequently its amplitude full width at half maximum is ${\tau }_E=\sqrt{2}\tau$. Inserting ${\tau }_E$ into the formula for a power-width input underestimates the true peak power by a factor $1/\sqrt{2}$. State which measured waveform defines the width.

An electrical photodetector trace can also be broadened by its impulse response. The optical pulse-width inference therefore needs a characterized detector response.^[ref]