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Thermal States of Light

  1. Nov 5, 2007 #1
    The Hamiltonian (ignores vacuum energy), [tex] H = \hbar\omega_{p}a_{p}^+a_{p} [/tex], represents some cavity at temperature T. For simplicity assume the cavity only supports a single mode.
    [tex] H = \hbar\omega_{m} a_{m}^+ a_{m} [/tex]

    1) Given that in thermal equilibrium the probability of a system to have energy E is proportional to ~exp(-E/kT), give an expression for [tex]\rho_{m}(n) [/tex] where:
    [tex] \hat{\rho}_{m} = \sum_{n}\rho_{m}(n) | n>_{m} _{m}<n| [/tex]
    The constant is independent of n
    2) Find this constant by imposing condition [tex] Tr( \rho_{m}) = 1 [/tex]

    For the first part since the constant is independent of n and the Hamilitonian is governed by [tex] \hbar\omega a^+a[/tex] where [tex]a^+a[/tex] equals n so in order to isolate the n component from the Hamiltonian and write an expression [tex]\rho[/tex] I think we should start by evaluating <m|[tex]\rho[/tex]|n> where we will get a summation expression that needs to be evaluated. This is what I think we should do and I tried it but it is not taking me anywhere..I am afraid that I don't even know what I am doing. I just need a small hint to get started and then I will be all set...
  2. jcsd
  3. Nov 5, 2007 #2


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    You know that the probability that the system has energy E is proportional to exp(-E/kT).
    What are possible values for E and what is the probability of finding the system to have energy a particular energy E? (In terms of rho(n))
  4. Nov 6, 2007 #3
    Thanks Galileo, I believe I have gotten the answer since E (non-vacuum state) is:

    [tex] \hbar\omega n [/tex]

    We could just assume that [tex] \rho (n) = C*exp(- \hbar\omega n [/tex]) and then it is easy street from there on...
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