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Identical particles

  1. Jul 19, 2007 #1
    1. The problem statement, all variables and given/known data

    There are identical particles in a harmonic potential [tex]V(x)=\frac{1}{2}m\omega^{2}x^{2}[/tex]. The number of particles is 2N, where N is a positive integer.

    2. Relevant equations

    a) What is the system's Hamilton's operator for bosons in the second quantization? How about the fermions?

    b) What is the energy of the ground level, if the particles are i) S = 0 bosons, ii) S = 1/2 fermions?

    c) Write the state vector of the ground level for S = 0 bosons and S = 1/2 fermions using creation and annihilation operators to vacuum.

    3. The attempt at a solution

    a) I have no idea... :cry:

    b) Maybe I have to solve the Schrödinger's Equation, and using boudary conditions (?) solve the equation for energy E. There are 2N bosons all in the ground state, because Pauli's Exclusion Principle does not hold for bosons. So, I just put n = 1 to the E-equation and multiply it by 2N. Right? :uhh:

    How about fermions? The Pauli Exclusion Principle holds for them, at least... :confused:

    c) I have no idea... :cry:
    Last edited: Jul 19, 2007
  2. jcsd
  3. Jul 20, 2007 #2
    You look very confused, have you comprehended your notes completely, before trying the qeustions?

    to answer this question, you need to know what are energy levels of harmonic oscillator, the fact that no two fermions can occupy one energy level, and all bosons can occupy one energy level.

    then you need to know how to use Fock space, the difference with Hilbert space (tensor products of hilbert spaces). A ket in Fock space contains numbers that tell you how many particles occupy what energy level.

    Then you need to know the creation and annihilations operators, Fermionic and Bosonic, a creation operator labeled by an energy level creates a particle at at energy level. What commutation, anti-commutation relations these operators satisfy.

    e.g. one particle at energy level 1, another at energy level 2
    the Hamiltonian should be E1 a1^dagger a1 + E2 a2^dagger a2
    Last edited: Jul 20, 2007
  4. Jul 21, 2007 #3


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    (b) What is the ground state energy of a single simple harmonic oscillator? [You might want to memorize this expression. Memorize the expression for all the higher energy levels too.] Now, the bosonic case, if there are 2N particles all of which are in the single-particle ground state then they all have an energy equal to the ground state energy of a single simple harmonic oscillator. Hence what is the total energy of the system?
  5. Jul 21, 2007 #4


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    (a) your teacher should have given you a general form of the hamiltonian "in second quantization" for a given potential V. Something along the lines of
    H=\int d^3x\psi^{\dagger}(\vec x)\left(
    \frac{-\nabla^2\hbar^2}{2m}+V(\vec x)
    \right)\psi(\vec x)
    where psi and psi-dagger are particle annihilation and creation fields, respectively.
  6. Jul 21, 2007 #5


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    (c) the creation and annihilation operators mentioned about can be expanded in terms of operators which create particles not at a given position, but rather in a given mode of the single-particle Hamiltonian. These mode-creation and annihilation operators are what you use to do this problem.
  7. Jul 22, 2007 #6
    The ground state energy of a simple harmonic oscillator is [tex]\frac{1}{2}\hbar\omega[/tex].

    The energy levels of a simple harmonic oscillator are given by
    [tex](n+\frac{1}{2})\hbar\omega,\qquad n=0,1,2,...[/tex]

    a) for bosons: [tex]H=2NE_{1}a_{1}^{\dagger}a_{1}[/tex]?
    b) for S = 0 bosons: [tex]E=2N\cdot\frac{1}{2}\hbar\omega = N\hbar\omega[/tex]?
    Last edited: Jul 22, 2007
  8. Jul 22, 2007 #7


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    (b) correct.
    (a) incorrect.
  9. Jul 23, 2007 #8
    b) For S=1/2 fermions: There are two fermions in the ground state. Right?
    So, the energy is [tex]E=2\cdot\frac{1}{2}\hbar\omega = \hbar\omega[/tex]?
  10. Jul 23, 2007 #9


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    ...that's the energy of those 2. There are 2N-2 more that you have to take into account. The next 2 may not be in the lowest state, but you can put them in the next lowest state with energy [tex]3/2\hbar\omega[/tex].

    Those next 2 together contribute an energy [tex]3\hbar\omega[/tex].

    So, for the first 4 the energy is [tex]\hbar\omega + 3\hbar\omega[/tex], but there are still 2N-4 more to take into account.

    Now just keep on filling up the states until you have put all 2N fermions into the lowest N energy levels.
  11. Jul 23, 2007 #10
    Yes. Yes, of course! Thanks! :)
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