Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

1-D harmonic oscillator problem

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a particle of mass m moving in a one-dimensional potential,

    [tex]V(x)=\infty[/tex] for [tex]x\leq0[/tex]

    [tex]V(x)=\frac{1}{2}m{\omega^2}{x^2}[/tex] for [tex]x>0[/tex]

    This potential describes an elastic spring (with spring constant K = m[tex]\omega^2[/tex]) that can be extended but not compressed.

    By reference to the solution of the 1-D harmonic oscillator potential sketch and state the form of the valid eigenfunctions, and state the corresponding eigenvalues, for the ground and first excited state.

    2. Relevant equations

    3. The attempt at a solution
    Here is the potential sketch

    Since the potential V(x)=[tex]\infty[/tex] for [tex]x\leq0[/tex], The potential at x=0 is infinity. So the wave function at x=0 must equal 0. If it were anything else the particle would have infinite energy. Therefore eigenfunctions must pass through the origin.

    From the graph, n=1 and n=3 are the lowest eigenfunctions to pass through the origin.

    Ground state eigenfunction n=1 [tex]{E_1}=(1+\frac{1}{2})\hbar\omega=\frac{3}{2}\hbar\omega[/tex]

    1st excited state eigenfunction n=3 [tex]{E_2}=(3+\frac{1}{2})\hbar\omega=\frac{7}{2}\hbar\omega[/tex]

    Can someone tell me if this is right?
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 12, 2010 #2


    Staff: Mentor

    Please don't double post!
  4. Jan 12, 2010 #3
    I believe you mean [itex]E_3=\cdots[/itex] and not what you wrote. But otherwise, this is correct. This problem is a good example of http://en.wikipedia.org/wiki/Parity_(physics)#Quantum_mechanics".
    Last edited by a moderator: Apr 24, 2017
  5. Jan 12, 2010 #4
    No, i meant [tex]E_2[/tex] as i thought it is the 2nd energy level i.e. the 1st excited state, can you tell me why this its not [tex]E_2[/tex]
    Last edited by a moderator: Apr 24, 2017
  6. Jan 12, 2010 #5
    The reasoning is purely semantics. Since the formula is [itex]E_n=\left(n+\frac{1}{2}\right)\hbar\omega[/itex], if you have [itex]n=3[/itex] in the parenthesis, you should have [itex]n=3[/itex] as the subscript on [itex]E[/itex]. What you should say is (I made a mistake too)


    and you should also write


    because the first equation is the new first excited state and the second equation is the new ground state. If you calculate more levels, you should find that

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook