Spontaneous and stimulated emission in Planck's radiation law

[dbabic]

Hello,

Einstein introduced stimulated emission (along with spontaneous emission and absorption) to derive Planck's radiation law using his A and B coefficients in his 1917 paper. My question is, is it possible to separate the Planck radiation spectrum into a fraction that is spontaneous emission and a fraction that is stimulated emission? Is this even a good question? I took a simple simple approach and got the attached graph. Does anyone have an opinion on this?

Thank you...

 

[vanhees71]

How did you define the separation in these two contributions? It's not clear to me, how to make such a split from the standard derivation of the Planck spectrum from QED.

 

[dbabic]

Einstein's derivation assumes that all the generated EM radiation (spontaneous + stimulated) must be balanced by what's absorbed:

$$A_{21} N_2 + B_{21} N_2 \rho _{EM} (\omega ) = B_{12} N_1 \rho _{EM} (\omega )$$

where the first term is spontaneous (SPE) and the second stimulated emission (STE) and on the right is absorption. I evaluate the portion of the left side that is due to spontaneous emission. If you just divide the first term on the left side with the entire left side you get that the portion due to spontaneous emission is equal to

SPE contribution = $1 - \exp( -\hbar\omega/kT )$
STE contribution = $\exp( -\hbar\omega/kT )$

Now, I simply take the radiation law and multiply it by the fractions shown above to get the curve for SPE and STE.

The interesting thing is that the fraction of stimulated emission coincides exactly with Wien's distribution, which is the starting point of Einstein's arguments in the 1917 paper .

 

[dbabic]

Correction: the fraction of SPONTANEOUS emission coincides exactly with Wien's distribution.

 

[berkeman]

@vanhees71 and @dbabic -- are you in agreement after the updates? We received a report that there may be confusion in this thread after some edits...

Thanks.

 

[vanhees71]

Ok, let's do the derivation a la Einstein. It's a kinetic argument for the occupation numbers of a two-level system due to thermal radiation, i.e., the two-level system is supposed to be in thermal equilibrium at temperature $T$. Let $N_1$ and $N_2$ be the occupation numbers of the lower and upper level with energy difference $E_2-E_1=\hbar \omega$:
$$\dot{N}_1=A_{21} N_2+B_{21} N_2 I-B_{12} N_1 I$$
$$\dot{N}_2=-A_{21} N_2-B_{21}N_2+B_{12} N_1 \rho=-\dot{N}_1.$$
Here $A_{21}$ is the transition rate for spontaneous emission, $B_{21}$ the rate for spontaneous emission, and $B_{12}$ for absorption, $I$ is the intensity (energy density) of the radiation.

In thermal equilibrium one has $N_2/N_1=\exp[-\hbar \omega/(kT)]$ and $\dot{N}_1=\dot{N}_2=0$. From this one gets
$$(A_{21}+B_{21} I) \exp[-\hbar \omega/(kT)]=B_{12} I .$$
This gives
$$I=\frac{A_{21}}{B_{12} \exp[+\hbar \omega/(k T)]-B_{12}}.$$
On the other hand from equilibrium thermal QFT we can derive Planck's radiation Law
$$I = \frac{\hbar \omega^3}{\pi^2 c^3} \frac{1}{\exp[(\hbar \omega)/(k T)]-1}.$$
From this it follows
$$B_{12}=B_{21}, \quad \frac{A_{21}}{B_{12}}=\frac{\hbar \omega^3}{\pi^3 c^3}. \qquad (1)$$
So the spontaneous emission fraction is
$$\frac{A_{21} N_2}{B_{12} N_1 I}=\frac{\hbar \omega^3}{\pi^3 c^3}(1-\exp[-\hbar \omega/(k T)])$$
and of the induced emission
$$\frac{B_{21} I \exp[-\hbar \omega/(k T)]}{B_{12} I}=\exp[-\hbar \omega/(k T)].$$
So it's, of course, the induced emission part which coincides with Wien's radiation law.

The relations between the coefficients (1) can be directly derived from QFT too.