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Application of the harmonic oscillator to an arbitrary number of free bosons

  1. Jan 18, 2009 #1
    Aplication of the harmonic oscillator to an arbitrary number of free bosons

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

    I will like to know if my answer to this problem is correct and If not, what I am missing.

    Given a system with an arbitrary number of free bosons, where the hamiltonian for one particle with two independent states is given by the equation:

    [tex]
    H = :(\frac{1}{2}\pi _1^2 + \frac{1}{2}\pi _2^2 + 2\phi _1^2 + \frac{1}{2}\phi _2^2):
    [/tex]

    Calculate the expected value for the Hamiltonian in the state:

    [tex]\left| {\Psi \rangle } \right. = \frac{1}{2}\left| {1,0\rangle + } \right.\frac{{\sqrt 3 }}{2}\left| {2,1\rangle } \right.
    [/tex]


    2. Relevant equations



    3. The attempt at a solution

    The Hamitonian correspond to two non coupled harmonic oscillators with energies [tex]E_{1}=2; E_{2}=1[/tex] in natural units.
    Since the Hamiltonian is already diagonalized, we proceed to the quantization introducing the ladders operators:

    [tex]
    \eqalign{
    & a = \frac{1}
    {{\sqrt 2 }}\left[ {\sqrt 2 \phi _1^{} + \frac{i}
    {{\sqrt 2 }}\pi _1 } \right] \cr
    & b = \frac{1}
    {{\sqrt 2 }}\left[ {\phi _2 + i\pi _1 } \right] \cr}
    [/tex]

    we obtain for the Hamiltonian:

    [tex]
    \eqalign{
    & H = :(aa^ + )\,2(aa^ + )^t \, + (b\,b^ + )(b\,b^ + )^t : \cr
    & \cr
    & H = 2\,a^ + a + b^ + b\,\,; \cr
    & \cr}
    [/tex]

    Solving the independent Schrödinger equation, we obtain that the energy of the system equals

    [tex]E = 2n_1 + n_2 [/tex], where n1 and n2 indicate the number of particles for the two monoparticular states.

    To calculate the value of H in the given state:
    [tex]
    \left\langle H \right\rangle = \left\langle {\Psi H\Psi } \right\rangle = \sum\limits_{ij} {\left| {C_{ninj} } \right|} ^2 E_{ninj} = \frac{1}
    {4}E_1 + \frac{3}
    {4}(2E_1 + E_2 ) = \frac{2}
    {4} + \frac{3}
    {4}(4 + 1) = \frac{{17}}
    {4}
    [/tex]
     
    Last edited: Jan 18, 2009
  2. jcsd
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