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Excited state of an atom

  1. May 5, 2008 #1
    How long an atom remains excited?
    How can we prolong the excited state of an atom? Whether its possible.
  2. jcsd
  3. May 5, 2008 #2


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    Taking only optical transitions into account, there are only two ways for the atom to emit a photon and get back into the ground state: stimulated emission and spontaneous emission.

    In stimulated emission the transition back into the ground state is triggered by another photon of the same energy as the transition back to the ground state. The incident photon "arrives" at the atom and two photons of identical energy and direction come out. In this case the lifetime is in general rather short and depends on the external conditions as we need to shine light on the atom.

    Spontaneous emission is a similar process, in which vacuum fluctuations play the role of the triggering photon. In this case the lifetime is longer. The exact value of the spontaneous emission rate is usually calculated by using Fermi's golden rule and depends mainly on the energy of the transition and the mode spectrum of the em-field. Therefore the rate of spontaneous emission can be altered by changing the spectrum of the em-field, for example by placing the atom inside a resonant cavity, which enhances spontaneous emission, which is called the Purcell-effect.

    One should also note, that spontaneous emission is a phenomenon, which can only be described in terms of a quantized em-field. Classical physics and "ordinary" qm predict, that there is no spontaneous emission.
  4. May 5, 2008 #3
    Electron promotes to higher E-level by absorbing certain frequency, but system always tends to attain minimum energy; so the electron should return to ground state as soon as possible. Now what I really want to know is- what deter the electron from coming back?

    And Cthugha thx for your response.
  5. May 5, 2008 #4


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    deter ?

    Anyway, you have a transition matrix element ([tex] < final|O(E_{\gamma}) | final > ) that gives you the probability amplitude for a certain decay. So you see that it depends on the wave functions and the operator for that decay transition - i.e how much O|initial> overlaps with |final>.
    The operator here, O(E), depends on what multipolarity the transition requires - and the multipolarity depends on what total angular momenta J and parity the states has(and some more things). So perhaps <final, J = 2|O(E2)|initial, J = 4> is larger than <final, J = 2|O(E4)|initial, J = 4>, then E2 will dominate over E4 in this made up case.

    The same things you have in nuclear physics.
  6. May 5, 2008 #5


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    Does it really? You have two coupled systems: the atom and the electromagnetic field. According to the argument you mentioned both should tend to attaind minimum energy, but due to conservation of energy one of these systems has to be in an excited state. In usual thermodynamics sytems in equilibrium tend to distribute the energy evenly among all degrees of freedom.

    In the microscopic picture the number of possible states in phase space is rather analogous to the degrees of freedom in thermodynamics. This means, the atom has only one state (or in case of degeneracy only few states), which corresponds to the given energy. The em-field however has a huge number of states, which match the given energy. For example the components of the momentum vector do not need to have a fixed value, only the magnitude of the vector is fixed. Therefore there are a huge number of possibilities for the em-field to get from the vacuum state to the one-photon state.

    So the coupled system has lots of states with one photon and an atom in its ground state and few states with an excited atom and the em-field in its ground state. All possible states have the same final energy, but it is far more likely for the system to be in a state with one photon. However the system reaches a state of higher entropy when a photon is emitted. The usual notion of systems tending towards minimal energy is just a consequence of this behaviour.
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