Two Level System with thermal population

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
ktolaus
Messages
1
Reaction score
0
Hello,

the task is the following:

Consider a two-level system with thermal population.

a) Show that the rate equation for the state [itex]N_2[/itex] is the following:

[itex]\frac{dN_2}{dt}=w(N_1-N_2)-\frac{N_2-N_2^e}{\tau}[/itex]

[itex]w=B_{12}\rho({\nu}) \,\,\, N_2=N_2^p+N_2^e\,\,\, N_1=N-N_2[/itex]

[itex]N_2^p[/itex] is the portion of the total population caused by pumping.

b) Show, that using such a system it is impossible to build a CW Laser.

c) Calculate the progress of [itex]N_2(t)[/itex] when the system is stimulated by a constant monochromatic signal with spectral density I and frequency [itex]h\nu=(E_2-E_1)[/itex]. For t=0 only thermal population exist.

My ideas are the following:

a) [itex]\frac{dN_2}{dt}[/itex] is just the sum of absorption, stimulated emission and spontaneous emission. Spontaneous emission occurs only from the the "pumped portion":
[itex]\frac{dN_2}{dt}=B_{12}\rho({\nu})N_1-B_{21}\rho({\nu})N_2-A_{21}N_2^p[/itex]
[itex]\frac{dN_2}{dt}=B_{12}\rho({\nu})N_1-B_{12}\rho({\nu})N_2-A_{21}(N_2-N_2^e)[/itex]
using [itex]A_{21}=\frac{1}{\tau}[/itex] leads to:
[itex]\frac{dN_2}{dt}=w(N_1-N_2)-\frac{N_2-N_2^e}{\tau}[/itex]

b) I'm not sure, but I think CW is only possible if [itex]\frac{d^2N_2}{dt^2}=0[/itex]:
[itex]\frac{d^2N_2}{dt^2}=B_{12}\rho(\frac{dN_1}{dt}-\frac{dN_2}{dt})-\frac{1}{\tau}\frac{dN_2}{dt}+\frac{1}{\tau}\frac{dN_2^e}{dt}[/itex]
[itex]N_2^e[/itex] should be independent on time. Using [itex]\frac{dN_1}{dt}=-\frac{dN_2}{dt}[/itex] leads to:
[itex]0=-2B_{12}\rho\frac{dN_2}{dt}-\frac{1}{\tau}\frac{dN_2}{dt}[/itex]

This can't be 0 since the left term depends on the frequency but the right term doesn't.

c) I tried several things but none of them were promising. The problem is: [itex]N_1[/itex] depends on the time.
I really would appreciate it if you gave me a hint.

Sorry for my english, but it's not my mother tongue.
 
Physics news on Phys.org
Thanks and best regardsTo solve part c), you can use the equation from part a):\frac{dN_2}{dt}=w(N_1-N_2)-\frac{N_2-N_2^e}{\tau}This equation can be written as: \frac{dN_2}{dt}=wN_1 - (w+\frac{1}{\tau})N_2 + \frac{N_2^e}{\tau}Now, we can use the initial conditions given in the question: N_2(0)=N_2^e and \frac{dN_2}{dt}(0)=0We can also assume that N_1 is constant over time. Using these conditions and the equation above, we can solve for N_2 as a function of time: N_2(t)=N_2^e + (N_1-N_2^e)e^{-(w+\frac{1}{\tau})t}This is the solution for the population of state N_2 when stimulated by a constant monochromatic signal.