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Harmonic Oscillator

  1. Feb 16, 2012 #1
    Problem:
    Consider a harmonic oscillator of mass m undergoing harmonic motion in two dimensions x and y. The potential energy is given by
    V(x,y) = (1/2)kxx2 + (1/2)kyy2.
    (a) Write down the expression for the Hamiltonian operator for such a system.
    (b) What is the general expression for the allowable energy levels of the two-dimensional harmonic oscillator?
    (c) What is the energy of the ground state (the lowest energy state)?

    Hint: The Hamiltonian operator can be written as a sum of operators.

    Now I'm a bit lost on how to write the expression for the Hamiltonian.
    Is the Hamiltonian simply H = - h2/2m d2/dx2 + V(x,y) [where V(x,y) is given above]?
    Then with that Hamiltonian, solving the Schrodinger eqn is pretty straightforward to get H*psi = E*psi, now I'm a bit lost here as well to solve for the general expression for the allowable energy levels?
     
  2. jcsd
  3. Feb 16, 2012 #2

    vela

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    Almost. You need to include a term for the kinetic energy due to movement in the y-direction.

    Have you solved the one-dimensional harmonic oscillator already?
     
  4. Feb 18, 2012 #3
    I can represent the Hamiltonian as a sum of operators like this?
    \hat{H} = \hat{H_x} + \hat{H_y}
     
  5. Feb 18, 2012 #4

    vela

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    Yes, depending on what you mean by Hx and Hy.
     
  6. Feb 18, 2012 #5
    I get this, this is the general expresion of Hamiltonian Operator for the Quantum Harmonic Oscillator ??

    [-ħ/2m (d^2 Ψ_x)/(dx^2 )+1/2 k_x x^2 Ψ_x ]+[-ħ/2m (d^2 Ψ_y)/(dy^2 )+1/2 k_y y^2 Ψ_y ]= EΨ_x Ψ_y
     
  7. Feb 18, 2012 #6

    vela

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    Part (a) is simply asking you for the operator ##\hat{H}##. The wave function doesn't appear in that expression. You wrote in your first post
    $$\hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x,y)$$ which isn't correct, but it's essentially the type of answer you want to give for (a). You just need to correct it, which I think you know how to do.

    The Schrodinger equation says what happens when you apply that operator to a wave function:
    $$\hat{H}\psi(x,y) = \hat{H}_x \psi(x,y) + \hat{H}_y \psi(x,y) = E \psi(x,y)$$ Note that the two pieces act on the same function. If you assume the solution has the form ##\psi(x,y) = \psi_x(x)\psi_y(y)##, you'll get something similar to what you have. Your expression isn't quite correct. You're getting there though.
     
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