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Einstein A & B Coefficients

  1. Jun 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2016-06-03 at 17.23.13.png

    2. Relevant equations


    3. The attempt at a solution

    Very confused by this problem. For one thing, it doesn't specify if there is or isn't any light present to drive the stimulated emission/absorbtion. I guess there's no reason to assume that there is no light - but since the question is asking about lifetimes, that would seem more sensible...[plus introducing an unspecified ##\rho(\omega_{0})## seems odd]

    Assuming there is no light involved, then this is just a three level system, with three sets of coupled differential equations describing the behaviour - let ##N_{i}## be the number of atoms in state ##i##:
    (1) $$ \frac{dN_{A}}{dt} = -(A_{ac}+A_{ab})N_{A} $$
    (2) $$ \frac{dN_{B}}{dt} = A_{ab}(N_{A} -N_{B}) $$
    (3) $$ \frac{dN_{C}}{dt} = A_{ac}N_{A}+A_{bc}N_{B} $$
    [Though,one of them is made redundant by the fact that total particle number must be constant.]

    We can easily solve the first equation:
    $$ N_{A} = N_{A0}exp\bigg[ - (A_{ac}+A_{ab})t) \bigg] $$
    By substituting this into the equation (2), we can then solve for ##N_{B}##:

    $$ \frac{dN_{B}}{dt} +N_{B}A_{ab} = A_{ab}N_{A0}exp\bigg[ - (A_{ac}+A_{ab})t) \bigg] $$

    So by using the standard method of integrating factors:
    $$ N_{B} = Cexp(-A_{ab}t)-\frac{A_{ab}}{A_{ac}}N_{A0}exp(-A_{ac}+A_{ab}t) $$
    But how am I meant to define a lifetime for that?, Assuming ##N_{B}(t=0)=0##, i.e:
    $$ N_{B} =\frac{A_{ab}}{A_{ac}}N_{A0}exp(-A_{ab}t)\bigg[1-exp(-A_{ac}t)\bigg]$$
    Do we take the dominant exponential to define the lifetime?
     
  2. jcsd
  3. Jun 6, 2016 #2

    Simon Bridge

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    You are only told to consider the decays a-c and b-c.
    The presence of photons is not relevant to the lifetime calculation for a state ... check the definition of "lifetime" in your notes.
     
  4. Jun 6, 2016 #3
    Ah,okay, so lifetimes are generally defined to only involve the Einstein ##A## coefficients.
    So I can just ignore completely ##\psi_{A} \rightarrow \psi_{B}##? I wasn't sure the question implied that.... I guess if it does:

    $$ \frac{dN_{A}}{dt} = -A_{ac}N_{A} $$
    $$ \frac{dN_{B}}{dt} = -A_{bc} N_{B} $$

    Implying that ## \lambda_{a} = A_{ac}## ##\lambda_{b} = A_{bc}##. We know that ## \frac{1}{\tau} = \lambda## where ##\tau ## is the lifetime, so:
    $$ \frac{\tau_{A}}{\tau_{C}} = 2 $$

    and then the information about the matrix element is completely redundant? It's that simple??
    [Edit : I mean, it tells us the same thing]
    Thanks!
     
  5. Jun 6, 2016 #4

    Simon Bridge

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    Well done.
     
  6. Jun 6, 2016 #5
    Oh dear, that's embarrassing.. Ah well, many thanks for the help, much appreciated!
     
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