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A Probability of photon emission from quantum dot

  1. Sep 7, 2016 #1
    Hi guys, I am looking for a formula which I am sure exits but I cannot locate it.


    The problem is that a quantum dot absorbs a photon of wavelength λ0(dot is semiconductor or could be any other material). Assuming that it reemits a photon, what is the probability that this emitted photon will have a wavelength λ. Of course if λ<λ0 then the probability is zero, but what kind of distribution does it have for λ>λ0 ?

    I am sure this is a very standard problem and the formula exits, I would be glad if someone can help me find out a reference for this.

    Thanks
     
  2. jcsd
  3. Sep 12, 2016 #2

    Ygggdrasil

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    Not sure if it applies to semiconductors as well, but for molecules, one would apply the Franck-Condon principle.
     
  4. Sep 12, 2016 #3

    f95toli

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    I suspect the answer will depend on many factors.
    In the simplest case you could presumably think of this as a two-level problem where you have a ground state and an excited state with some lifetime T. The lineshape should therefore -in most cases- be Lorentzian so that should also be the energy distribution of the photons.
    If you are talking about a real dot and need to start worrying about conservation rules etc it will get messy quite quickly, but the above should be a good approximation.
     
  5. Nov 11, 2016 #4
    I would think that the distribution would be a Lorentzian. And I will elaborate further on the probability of its emission
     
  6. Nov 11, 2016 #5
    The distribution should be a Lorentzian. The probabil
    The spectrum should be a Lorentzian, meaning that it is uniformely distributed. For the wavelength, wouldn't it be related to the probability of obtaining certain eigenvalues? If this statement is correct, you would have to solve the Schrodinger equation.



    meanthatitiu
     
  7. Nov 11, 2016 #6
    I think the above statement is way more complex than needs to describe the wavelength propability of the quantum dot. Wouldn't it be ideal to just use the Schrodinger equation?
     
  8. Nov 12, 2016 #7

    Ygggdrasil

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    What do you mean by uniformly distributed? A Lorentzian distribution is very different from a uniform distribution.

    Also, see https://en.wikipedia.org/wiki/Quantum_dot#Optical_properties for graphs of the emission spectra of real quantum dots. They are not Lorentzian.
     
  9. Nov 14, 2016 #8
    You are right. I got confused with something being homogeneously broadened (Lorentzian) as normally distributed. The spectra is a gaussian, which is normally distributed.
     
  10. Nov 14, 2016 #9

    Ygggdrasil

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    They are not exactly Gaussian either as you can see some appear to be asymmetric. The structure of the spectra are likely defined by the how much the wavefunction of the conduction band overlaps with the wavefunction of the different bands in the valence band. A more detailed answer would require a more detailed understanding of solid-state physics, and I don't remember much from when I studied that.
     
  11. Nov 14, 2016 #10

    blue_leaf77

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    I haven't checked myself, but to say whether the spectra are Gaussian, Lorentzian or neither one should look at the spectrum as a function of frequency not wavelength which is used in the Wikipedia link.
     
  12. Nov 15, 2016 #11
    I obtain the Gaussian curve info from here (number 7). Let me know what you think about it.
     

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  13. Nov 15, 2016 #12
    What's the reasoning behind it?
     
  14. Nov 15, 2016 #13

    blue_leaf77

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    A custom. Probably that's because the spectrum is the Fourier transform of field in time domain. This means if you apply FT on ##E(t)## you will get ##\tilde{E}(\omega)##, the spectrum. Of course you can simply use the relation ##\omega = 2\pi/\lambda## to get ##\tilde{E}(\lambda)## but this function will generally have different form than ##\tilde{E}(\omega)##. In some cases where the bandwidth is much smaller than the central wavelength, Gaussian ##\tilde{E}(\omega)## will approximately result in Gaussian ##\tilde{E}(\lambda)##.
     
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