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Confusion in bose statistics

  1. Sep 10, 2009 #1
    Hello all,
    we have the B-E function
    <n>=1/(exp(x)-1), where x is hv/kT.
    What is <n> actually? because i ready everywhere that it is average no. of phonons!! but it also probability..?
    Please explain clearly all terms involved in the above relation..
    thanks
     
  2. jcsd
  3. Sep 10, 2009 #2

    f95toli

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    What you have written isn't anything "physical" as such.
    It is just a factor (a distribution function) that appears when dealing with systems that obey BE statistics.

    Do you know what a Boltzmann distribution is?
     
  4. Sep 11, 2009 #3
    Okay..
    so <n> is a probability of what???
     
  5. Sep 11, 2009 #4

    alxm

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    <n> would be the expectation value of n, which typically would be a number of something.
     
  6. Sep 11, 2009 #5
    So <n> is the expectation value of n. And n is the no. of phonons that occupy a certain state (here it is hv). Also hv is energy of that certain occupied phonon??
    correct???
     
  7. Sep 11, 2009 #6

    f95toli

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    Almost.
    Note what I wrote above: The expression you've written above is a distribution.
    (another well-known example of a distribution would e.g be a Gaussian ("Bell-curve")).
    Now, distributions can be USED to calculate the probabilities/expectation values but they ARE not probabilities themselves. It is important to understand the difference.
    Consider this: If you put numbers into the expression above you will just get a result (lets say 0.23) which in itself does not mean anything (0.23 phonons in which volume?)
    Note that what I am writing here about distributions has nothing as such to do with with physics; it is basic statistics.

    If you look closely at the expression you've written above you should notice that it is not normalized in any way. The reason why this is important is because in order to answer a question like "how many phonons are there in a particular state" where you need an actual number on phonons you ALSO have to know how many phonons there are in ANY state in your sample (or whatever volume you are considering).
    However, in most cases we are usually more interested in the probability to find a phonon in a given state but even in order to answer that question you first have to normalize your expression. To do this we use the fact that we know that the probability to find a phonon in SOME state with SOME energy is equal to 1 (because the phonon has to be somewhere in the system). Hence, in order to find the normalization constant you can integrate the expression over ALL possible energies and set the result to one. Once you have done that you can use another integral+said constant to get the probability to find a phonon in a particular energy interval.


    If this is confusing (and I am sure it is) you would be better off starting with something simpler. Take a two-level system (i.e a system which can only be in one of two possible states) which obeys some statistics (if you want to start with something simple a Boltzmann distribution) and where the two levels are separated by some energy deltaE. Calculate the probability of finding the system in the upper state at a given temperature T.

    Your question and most similar question are just slightly more complicated (because you need solve more tricky integrals) variations of this problem; so if you understand this the rest should be easy.

    btw, in a system with a continuum of states the probability to find a phonon with an energy of exactly hv is always zero. For the same reason that the probability of finding a golf ball exactly 4.21 cm from the hole is zero:
    When dealing with non-discrete systems (like energy, distance etc) we must always consider INTERVALLS (the probability of finding the ball in the intervall 4.21+-0.05 cm is non-zero).
    This is why you need to use integrals in the case of the phonon distribution (whereas in the example with the two-level system -which is discrete- you can just use sums).
     
    Last edited: Sep 11, 2009
  8. Sep 15, 2009 #7
    hi, thanks...now i could understand...[still a little confusion remains..but will solve it]
    thanks for your time.
     
  9. Sep 15, 2009 #8
    one final question..
    so <n> is just a probability that a single particle will have energy hv? [i guess correct!].
    Now i notice in some books..that they substitute the n in E=hv(1/2+n) by <n>..
    How?
    thanks
     
  10. Sep 15, 2009 #9
    <n> is the thermal expectation value of the operator n. So that substitution gives the thermally averaged value for the energy.
     
  11. Sep 16, 2009 #10
    Hi Kanato,
    so after substituting <n> we get thermally averaged value for energy. then how to call the energy E = hv(n+1/2)?
    thanks again
     
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