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Quantum mechanics, bidimensional harmonic oscillator

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    At t=0 the wave function of a two-dimensional isotropic harmonic oscilator is

    ψ(x,y,0)=A(4α^2 x^2+2αy+4α^2 xy-2) e^((-α^2 x^2)/2) e^((-α^2 y^2)/2)

    where A its the normalization constant
    In wich instant. Wich values of total energy can we find and which probability?
    2. Relevant equations
    None

    3. The attempt at a solution
    I dont know how to start it.
     
  2. jcsd
  3. Nov 1, 2015 #2

    blue_leaf77

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    Can you set up the Hamiltonian of an isotropic harmonic oscillator?
     
  4. Nov 1, 2015 #3
    I find that the hamiltonian of the isotropic harmonic oscillator is

    $$H=\frac{\hbar^2}{2m} \nabla^2+\frac{1}{2}m^2w^2p^2$$

    that's what you mean ?
     
    Last edited: Nov 1, 2015
  5. Nov 1, 2015 #4

    blue_leaf77

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    That does not seem to be what we are interested in. I just searched in Google and I can find quickly enough in the first few search results. What about 1D harmonic oscillator, do you know how the Hamiltonian looks like?
     
  6. Nov 1, 2015 #5
    No, i don´t know, I'm really cinfused about it
     
  7. Nov 1, 2015 #6

    DrClaude

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    Staff: Mentor

    Please use ##\#\# \: \#\### delimiters for inline LaTeX equations and ##\$\$ \: \$\$## for displayed equations. I've edited your post accordingly.
     
  8. Nov 1, 2015 #7

    blue_leaf77

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    The Hamiltonian for 1D oscillator reads
    $$
    H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2
    $$
    For 2D harmonic oscillator, you clearly need to add two more terms identical to those two above only that the variable is now ##y##. If you are still confused, go back to Google.
     
  9. Nov 1, 2015 #8
    Developing the Laplacian for a 2D harmonic oscilator, i´ll be



    $$H=\frac{\hbar^2}{2m} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2)+\frac{1}{2}m^2w^2(x^2+y^2)$$
     
  10. Nov 1, 2015 #9

    blue_leaf77

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    Correct, except for the absence of a minus sign in front of ##\hbar^2/2m##. Now this equation is easy to solve because the variables ##x## and ##y## are separated. Suppose the solutions of the separate Hamiltonians
    $$
    H_1 = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2
    $$
    and
    $$
    H_2 = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial y^2} + \frac{1}{2}m\omega^2y^2
    $$
    are ##u_p(x)## and ##u_q(y)## respectively. What do you think the solution of the total Hamiltonian ##H = H_1+H_2## will look like in terms of ##u_p(x)## and ##u_q(y)##?
     
  11. Nov 1, 2015 #10
    I´ll be the superposition rigth?
    If ##U_m(x)## and ##U_n(y)## are the solutions, the total Hamiltonian is
    $$H= U_m(x) + U_n(y)$$
     
  12. Nov 1, 2015 #11

    blue_leaf77

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    No, its not a superposition. Ok let me rephrase in previous statement, I should have said "solve the Schroedinger equation", not "solve the Hamiltonian". What we want to solve is
    $$
    (H_1+H_2)\psi_{pq}(x,y) = E_{pq}\psi_{pq}(x,y)
    $$
    with the total energy ##E_{pq} = E_p+E_q##. Now I will give you a hint. In this problem, we have that
    $$
    H_1\psi_{pq}(x,y) = E_p \psi_{pq}(x,y) \\
    H_2\psi_{pq}(x,y) = E_q \psi_{pq}(x,y) \\
    H_1 u_{p}(x) = E_p u_{p}(x) \\
    H_2 u_{q}(x) = E_q u_{q}(x)
    $$
    Using these, you are ought to find ##\psi_{pq}(x,y) ## in terms of ##u_{p}(x)## and ##u_{q}(y) ##.
     
  13. Nov 1, 2015 #12
    A question, ih this correct?

    $$ H_1 \psi_{pq}(x,y)=H_1u_p(x)$$

    As ##H_1## only depends of X, and from your last post you say ##u_p(x)## was the ##H_1## solution, Is it true?
    Or i´m misreading
     
  14. Nov 1, 2015 #13

    blue_leaf77

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    No, it's not like that since it will imply that ##\psi_{pq}(x,y)=u_p(x)##.
    This problem actually belongs to those which can be solved by separation of variables method. I suggest you read http://tutorial.math.lamar.edu/Classes/DE/SeparationofVariables.aspx, especially Example 1.
     
  15. Nov 1, 2015 #14
  16. Nov 2, 2015 #15

    blue_leaf77

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    Yes, you can find what I asked at the end of page 2.
    Yes, that's how we should proceed. The goal at this step is just to make you aware of how the total eigenfunction looks like in terms of the individual oscillator eigenfunctions. In the next step, we are supposed to expand the given wavefunction ##\psi(x,y,0)## in terms of ##\psi_{pq}(x,y)##'s.
    $$
    \psi(x,y,0) = \sum_p \sum_q c_{pq} \psi_{pq}(x,y)
    $$
    with ##c_{pq}## a constant. In general, each sum above runs from ##0## to ##\infty##. However, if we can relate the appearance of ##\psi(x,y,0)## with the nature of the function ## \psi_{pq}(x,y)## for a given ##p## and ##q##, we should find that only a few constants ##c_{pq}## out of these infinite series are non-vanishing. So, your next task is to find out the functional form of ## \psi_{pq}(x,y)##. First, you have known how ## \psi_{pq}(x,y)## looks like in terms of ##u_p(x)## and ##u_q(y)##, right?
     
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