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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
    [tex]E=(n+\frac{1}{2})\hbar\omega[/tex]

    3. The attempt at a solution
    https://www.physicsforums.com/attachment.php?attachmentid=23025&stc=1&d=1263339354
    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.

    http://131.104.156.23/Lectures/CHEM_207/CHEM_207_Pictures/p75a_72gif

    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

    Mark44

    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)

    [tex]
    E^\prime_1=E_3=\left(3+\frac{1}{2}\right)\hbar\omega=\frac{7}{2}\hbar\omega
    [/tex]

    and you should also write

    [tex]
    E^\prime_0=E_1=\left(1+\frac{1}{2}\right)\hbar\omega=\frac{3}{2}\hbar\omega
    [/tex]

    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

    [tex]
    E^\prime_n=\left(2n+\frac{3}{2}\right)\hbar\omega
    [/tex]
     
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