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3d harmonic oscillator

  1. Jul 11, 2016 #1
    1. The problem statement, all variables and given/known data

    An isotropic harmonic oscillator has the potential energy function U = 0.5 k (x²+y²+z²). (Isotropic means that the force constant is the same in all three coordinate directions.)

    (a) Show that for this potential, a solution to the three dimensional time-independent Schrodinger equation is given by ψ = ψ(x)ψ(y)ψ(z). In this expression, ψ(x) is a solution to the one-dimensional harmonic oscillator Schrödinger equation. The functions ψ(y) and ψ(z) are analogous one-dimensional wave functions for oscillations in the y- and z-directions. Find the energy associated with this ψ.

    (b) From your results in part (a) what are the ground-level and first-excited-level energies of the three dimensional isotropic oscillator?

    (c) Show that there is only one state (one set of quantum numbers) for the ground level but three states for the first excited level.

    2. Relevant equations


    3. The attempt at a solution

    (Sorry for my poor English.)

    I dont know even how to approach to this problem. I considered that a function of several variables can be written as a product of three functions of a single variable each one. I called these functions ψ(x), ψ(y), ψ(z). Then I wrote the 3d Schrod. Eq. substituting for ψ(x,y,z), and I divided both sides by that product:

    -(ħ²/2m) ((1 / ψ(x)) ∂² ψ(x) / ∂x² + ...) + U(x,y,z) = E

    I've read that the energy E cannot depend on the values of x, y, or z. Then I guess each term
    [ -(ħ²/2m) ((1 / ψ(x)) ∂² ψ(x) / ∂x² ] + [ -(ħ²/2m) ((1 / ψ(y)) ∂² ψ(y) / ∂y² + ...) ] + ... must be equal to a correspondent constant that doesnt depend on the direction. I called such constants Kx, Ky and Kz. So ψ(x)Kx + ψ(y)Ky + ψ(z)Kz + U(x,y,z) = E.

    But I'm not sure about my answer. In fact, I think it's wrong. :H
     
  2. jcsd
  3. Jul 11, 2016 #2

    BvU

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    Gold Member

    Hi Kent,

    Note that you don't have to find the solution any more, you just have to show that the given ##\Psi## is a solution
    So you assume ψ = ψ(x)ψ(y)ψ(z) With ##H\Psi(x) = E_x\Psi(x)## etc.
    and show that it satisfies ##H\Psi = E\Psi##, i.e. write out ##H\Psi##
     
  4. Jul 11, 2016 #3
    Hi,
    Thank you for your answer. After reading your hint I've tried this:

    I started supposing that ψ(x)ψ(y)ψ(z) satisfies the Schrödinger equation and can be written as a single wave function ψ.

    [ (-ħ²/2m) ψ(y)ψ(z) d² ψ(x) / dx² ] + [ (-ħ²/2m) ψ(x)ψ(z) d² ψ(y) / dy² ] + [ (-ħ²/2m) ψ(x)ψ(y) d² ψ(z) / dz² ] + 0.5 k'x² + 0.5 k'y² +0.5 k'z² = E ψ(x)ψ(y)ψ(z)
    I divided both sides by ψ(x)ψ(y)ψ(z) and added the terms as follows:

    (-ħ²/2mψ(x)) (d²ψ(x) / dx²) + 0.5k'x² = Ex, and so on for y and z.

    Then Ex + Ey + Ez = E, the total energy. So ψ(x)ψ(y)ψ(z) is a solution. Is it right?
     
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