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Bosons and statistical physics

  1. Mar 15, 2009 #1
    Hello,

    I am stuck on the first part of this question. There are several parts that follow that depend on this bit, and I know I can do them if I can just work this out. Any help would be gratefully received.

    1. The problem statement, all variables and given/known data

    Consider an isolated system of N identical spin-0 bosons inside a container where the allowed energy levels are non-degenerate and evenly spaced. Let η be the spacing between energy levels, and let q be the number of energy units (each of size η in excess of the ground-state energy). Assume that q<N.

    (a) Draw diagrams representing all allowed system states from q = 0 up to q = 6. Let
    each column represent a different system state, and each row a single-particle state. Use
    numbers to indicate the number of bosons occupying each level.


    2. Relevant equations

    None

    3. The attempt at a solution

    I have no idea how to draw such diagrams. Since I don't know the number of particles, N, how can I say what number are ocuppying each level? And N could 1000 or more, does this mean I should draw 1000 diagrams?? Surely not!

    Thanks for reading.
     
  2. jcsd
  3. Mar 15, 2009 #2

    CompuChip

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    Even if N = 1000, you only have a few available diagrams. For example, if q = 1, then you have only one energy unit to distribute over N bosons. So N - 1 of them must be in the ground state and 1 is in the first excited state (with energy η above the ground state). So you only get one diagram!

    In general, I think, you have q diagrams. So to do q = 0 to q = 6 you "only" have to draw 21 diagrams.
     
  4. Mar 15, 2009 #3
    Thankyou for your help CompuChip!

    So, is the system state the total amount of energy, ie q? And then the particle state is the energy level the particle is occupying?

    Please see attached, is this the sort of thing they are looking for?

    Edit: Sorry about the attachment, I didn't realise my contract had expired, you can still see most of it though. I get 28 different states, rather than 21. Have I done somewhat wrong?
     

    Attached Files:

    Last edited: Mar 15, 2009
  5. Mar 16, 2009 #4

    CompuChip

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    Technically, the system state is given by the occupation numbers of each energy level.
    So in the first completely visible row (:tongue:) you could specify the complete state of the system as (N-1, 0, 1, 0, 0, 0, ....).
    Since the bosons indiscernable (i.e. any two bosons are interchangeable) and the energy levels are non-degenerate, this completely specifies the system. Note that this would not be true, for example, if the bosons had labels (a, b, c, ...) tagged onto them, because then the system where boson a is in the ground state and boson b is in the first excited state would be different than the system you get by interchanging a and b. Also them being bosons is crucial: if they were spin-1/2 fermions you could never have N - 1 fermions in a single state for N > 2.

    As for the table, it looks correct, except that I think you should transpose it to match the convention in the question. So each column would specify one system state (for example: the state (N-1, 0, 1, .....) with q = 2) and the rows would correspond to the energy levels (i.e. single particle states): E = E0, E = E0 + η, E = E0 + 2η, etc.

    And that you get 21 is correct, I need to learn basic arithmetic. If my argument was correct, you get q possibilities for each q, so you will get 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21 in total. (Can you think up for yourself why this should be so?)
     
  6. Mar 16, 2009 #5
    I seem to be getting 28 (or 29 if you count the ground) system states in total. I can't see why there would be only be q possibilities (for each q)? It's true if q=0,1,2 or 3 but when q=4 I get 5 possibilities. Have I gone wrong?

    I have altered the table as suggested (see atached).
     

    Attached Files:

    Last edited: Mar 16, 2009
  7. Mar 17, 2009 #6

    CompuChip

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    Ah, you seem to be right, for q = 4:
    (..., 0, 0, 0, 1)
    (..., 1, 0, 1, 0)
    (..., 0, 2, 0, 0)
    (..., 2, 1, 0, 0)
    (..., 4, 0, 0, 0)
    [the dots stand for: put in the ground state the number of bosons needed to get N in total :smile:]

    My argument was as follows: suppose that we have N(q) systems with energy level q (i.e. with energy q.η above the ground state). Then for energy level (q + 1), by taking one boson from the ground state to the first excited state, we can make a "good" system with energy level (q + 1) from any system with energy level q. Then we only haven't counted the system with 1 particle in the highest state, (N-1, 0, 0, ..., 0, 1, 0, 0, ...). So we get N(q) + 1.
    What I hadn't realized is that states like (N-2, 0, 2, 0, 0) cannot be obtained in this way, which occur because 2 divides (q = 4).
    So N(q) = q, as I claimed before is not true. Would have been too easy, anyway :tongue:

    And yes, you have to count the ground state (it is the only one with q = 0, and they request you to include it).
     
  8. Mar 17, 2009 #7
    Hey CompuChip,

    Could I be really cheeky and ask another question :shy:? It is one of the follow ups to to the previous question,

    (b) Compute the occupancy of each energy level, for q = 6. Draw a graph of the occupancy
    as a function of the energy of the level. (You may wish to let the occupancy of the
    ground level fall outside the area of your graph.)

    I thought I could do this question, but I am lost as to how to compute the occupancy of each energy level. I have the formula's,

    Occupancy,

    [tex] \overline{n}=\sum{n*P(n)}[/tex]

    where,

    [tex]P(n)=\frac{1}{Z}e^{-n(\epsilon-\mu)/kT}[/tex]

    [tex]Z=\sum{e^{-n(\epsilon-\mu)/kT}[/tex]

    and T=temperature
    k=Boltzmans constant
    mu=chemical potential
    epsilon=energy
    Z=Grand partition function

    For instance take total energy=6*eta. From my chart, where the rows represent the particle state and the columns the system state,

    q=1 (0 0 1 0 0 2 2 3 4 6)
    q=2 (0 1 0 0 3 0 2 0 1 0)
    q=3 (0 0 0 2 0 0 0 1 0 0)
    q=4 (0 1 0 0 0 1 0 0 0 0)
    q=5 (0 0 1 0 0 0 0 0 0 0)
    q=6 (1 0 0 0 0 0 0 0 0 0)

    How would I find the probability that a particle is in energy level q=1? My problem is that I have no idea what my 'n' is in the equations.
     
  9. Mar 18, 2009 #8

    CompuChip

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    IIRC correctly from my thermodynamics course, you have to sum over each energy level and n will be the number of particles in it. So writing it more explicitly,

    [tex]
    Z = \sum_i \exp\left[ - n_i \left( \epsilon_i - \mu \right) / k_\mathrm{B} T \right]
    [/tex]
    where i runs over all (available) energy levels, ni is the number of particles in it, [itex]\epsilon_i[/itex] is the energy of a particle in the energy level and all other variables as you said.
     
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