Two Level System with thermal population

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SUMMARY

The discussion focuses on a two-level system with thermal population, specifically analyzing the rate equation for state N_2, which is defined as \(\frac{dN_2}{dt}=w(N_1-N_2)-\frac{N_2-N_2^e}{\tau}\). The participants conclude that constructing a continuous wave (CW) laser is impossible under these conditions, as indicated by the derived equations. Additionally, they provide a solution for N_2(t) when stimulated by a constant monochromatic signal, resulting in \(N_2(t)=N_2^e + (N_1-N_2^e)e^{-(w+\frac{1}{\tau})t}\).

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  • Understanding of two-level systems in quantum mechanics
  • Familiarity with rate equations and population dynamics
  • Knowledge of stimulated and spontaneous emission processes
  • Basic principles of laser operation and characteristics
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  • Study the derivation of rate equations in quantum systems
  • Explore the implications of thermal population on laser performance
  • Investigate the conditions for achieving continuous wave laser operation
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Physicists, optical engineers, and students studying laser technology and quantum mechanics will benefit from this discussion, particularly those interested in the dynamics of two-level systems and laser operation principles.

ktolaus
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Hello,

the task is the following:

Consider a two-level system with thermal population.

a) Show that the rate equation for the state N_2 is the following:

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

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

N_2^p 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 N_2(t) when the system is stimulated by a constant monochromatic signal with spectral density I and frequency h\nu=(E_2-E_1). For t=0 only thermal population exist.

My ideas are the following:

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

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

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: N_1 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.
 
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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.
 

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