The Vacuum Field and Einstein’s Stimulated Emission Theory

This block failed to display. Double-click this text to correct any errors. Your changes are saved.

Einstein’s paper on stimulated emission of a two state system leaves us with two clues. The first clue is that blackbody radiation has a statistically equal ability to create excitation and stimulated emission that is proportional to the spectral intensity. In other words, the cross-section for excitation,\(B_{12}\), from a lower state to an upper state is equal to the cross section for stimulated emission,\(B_{21}\), from an upper state to a lower state. Clue 2 says that the spontaneous emission coefficient \(A_{21}\) is related to the stimulated emission coefficient by \[A_{21}=\frac{B_{21} 8 \pi h \nu^3}{c^2}=\frac{1}{\tau_{decay}}\] These clues can be used to find a deeper connection between the vacuum field and stimulated emission. To find the power spectral density that the vacuum electromagnetic contains, look to “Random Electronics” by T.W. Marshall [4]. Marshall describes a distribution function \(Q(x_R,p_R,t)\) where \(X_R\) and \(P_R\) satisfy the equation \[\ddot{x}+\frac{2e^2\omega^2}{3mc^3} \dot{x}+\omega^2x=\frac{e}{m}E(t)\] with the initial values of \(x_R=\dot{X_r}=0\) at \(t=0\) and \(p_r=m\dot{x_R}\). After analyzing the expectation values for \(x_R^2\), Marshall arrives at the conclusion that at \(t=\infty \), the distribution function \[Q(x_R,p_R,\infty)=\frac{1}{\pi \alpha}e^{(\frac{ m\omega x_R^2}{\alpha}-\frac{p_R^2}{m\omega \alpha})}\] where \(\alpha=\frac{\pi e^2}{\gamma m \omega} I(\frac{\omega}{2 \pi})\) where \(\gamma=\frac{e^2 \omega^2}{3 m c^3}\) Marshall explains that the distribution given by \(Q(x_R,p_R,\infty)\) is an ensemble that is identical to that of a quantum harmonic oscillator. This is in agreement with the second quantization given before as long as \(\alpha=\hbar\). Then solving for \[I(\frac{\omega}{2 \pi})=\frac{\omega^3 \hbar}{3 \pi c^3}\] \[I(\nu)=\frac{4 \pi h \nu^3}{3 c^3}\] This result is the spectral field density of the vacuum field density, denoted \(I_{vac}(\nu)\). There is a small difference between Hutchin’s definition of \(I_{vac}=\frac{4 \pi h \nu^3}{c^2}\) and the one given above. The units given by Marshall are in \(joules/(cm)^3/Hz\) , to get the units in terms of watts, multiply by c. Then the 3 given my Masrhall, I believe is from getting the spectral field density for all dimensions (x,y,z), where Hutchin is considering the spectral field density along one direction, in agreement with Faria et al’s given by their equation seven [5]. It should be noted that when integrated \(I_{vac}(\nu)\) over \(4 \pi\) steradians, the result is the power field density.

Experiments have found that wavelengths longer 10 nm, the vacuum field density is about, \(1.8 x 10^{12} watt/(cm)^2\), this is incredibly large. This result leads to pressure that can bring two parallel plates together or force them about, this is the Casimir effect. One should be concerned with the inverse relationship with distance, \(P_{vac}=\frac{-\hbar c \pi^2}{240 d^4}\). At the small scales that we work with in laser systems and micro or nano electronics, this effect can lead to problems. This reinforces the mystery as to why the world around us isn’t ionized with such great vacuum field power. The fields are in fact there, since they can be measured, but there must be some property of the vacuum fields that don’t allow ionization.

Return to clue 2 which, in Einstein’s words, says spontaneous decay of any atom is identical to stimulated emission from a background spectral intensity.

## Share on Social Media