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QED vacuum density of states

  1. Jun 24, 2015 #1
    The Purcell effect is when an atom placed inside a high finesse cavity with a very small mode volume gets an increase in the spontaneous emission rate. I've tried to find correct explanation for this effect, but it seems hard to find, except that it comes from an increase in the vacuum density of states.

    Outside of the cavity, the vacuum density of states responsible for the spontaneous decay is given by:

    $$\rho_s = \frac{8 \pi f^2}{(c/n)^3}$$

    , whereas inside the cavity the vacuum density of states is instead given by:

    $$\rho_c = \frac{1}{\Delta f V}$$

    , where ##\Delta f## is the linewidth of the cavity and ##V## the mode volume. The formulas are taken from wikipedia and also mentioned in several articles, including the original by Purcell, although it does not contain any derivations. My question is now basically, how can these density of states be derived? I have tried to google and look through both wiki-like resources and papers, but have been unable to find a proper derivation. The closest thing I found was a derivation of the Casimir effect, which is certainly similar in nature, but not quite the same expressions.

    Do any of you the derivation for the vacuum density of states, or where I can find it?
  2. jcsd
  3. Jun 24, 2015 #2
    Give this a look, page 202,

    https://books.google.com/books?id=Q-4dIthPuL4C&pg=PA204&dq=The+Purcell+effect&hl=en&sa=X&ei=TFqLVfTmEYf0yAT-zYLYCQ&ved=0CCoQ6AEwAg#v=onepage&q=The Purcell effect&f=false



    Also see, page 126,

    https://books.google.com/books?id=I7sLAn4_ZVcC&pg=PA126&dq=The+Purcell+effect+cavity+qed&hl=en&sa=X&ei=F2GLVde9OYK9yQSX2YPYCg&ved=0CB4Q6AEwAA#v=onepage&q=The%20Purcell%20effect%20cavity%20qed&f=false [Broken]


    Last edited by a moderator: May 7, 2017
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