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Quantum Mechanics Problem - [Rotating Molecule]

  1. Nov 25, 2004 #1

    AKG

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    Consider the molecule CN, which may be described by a dumbbell consisting of two masses [itex]M_1[/itex] and [itex]M_2[/itex] attached by a rigid rod of length [itex]a[/itex]. The dumbbell rotates in a plane about an axis going through the center of mass and perpendicular to it.
    • Write down the Hamiltonian that describes the motion.
    • What is the energy spectrum?
    • Write down an expression for the difference in energy between the ground state and the first excited state in terms of the masses and [itex]a[/itex].
    Now, am I right in saying that the effective mass is:

    [tex]\mu = \frac{M_1 + M_2}{M_1M_2}[/tex]

    and treating the problem as that of a point mass, [itex]\mu[/itex], travelling in a planar circular orbit or radius [itex]a[/itex]? Given this, I believe I would have:

    [tex]H = \frac{p^2}{2\mu} + \frac{1}{2}(\mu a^2)\omega ^2[/tex]

    Now, my book isn't clear on what the "energy spectrum" specifically is, but does it have to do with the spectral decomposition of a (linear) operator? What exactly am I to do for (b)?

    My book has some stuff on harmonic oscillator, where the energy eigenvalues are given:

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

    If I were to find the energy difference between the ground state and first eigenstate, would that simply be:

    [tex]E_1 - E_0 = \hbar \omega[/tex]

    If I can find the energy eigenvalues for my Hamiltonian (since this isn't a harmonic oscillator in this problem) is that all I need to do for (c): express [itex]E_1 - E_0[/itex]?

    Thanks.
     
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  3. Nov 26, 2004 #2

    Dr Transport

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    If the molecule is only rotating, the only term in the Hamiltonian is [tex]\frac{1}{2} I \omega^{2} [/tex], no potential energy term. Form there calculate the enegy spectrum and then you can find the difference between the first excited and ground state energies.
     
  4. Nov 27, 2004 #3

    AKG

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    How do I find the energy spectrum?

    The section in the text that this question is from is "Linear Momentum" and my TA said that we should use linear momentum to solve this question. So, I can do (a):

    [tex]H = \frac{\mathbf{L}^2}{2I} = \frac{\mathbf{L}^2}{2\mu a^2}[/tex]

    I'm not sure how to get (b). Let's say that [itex]|x_n\rangle[/itex] were an eigenvector of [itex]\mathbf{L}[/itex] corresponding to eigenvalue [itex]\lambda _n[/itex]. Then:

    [tex]H|x_n\rangle = \frac{1}{2\mu a^2}\mathbf{L}\left (\mathbf{L} (x_n)\right ) = \frac{1}{2\mu a^2}\mathbf{L}\left (\lambda _nx_n\right ) = \frac{\lambda _n ^2}{2\mu a^2}|x_n \rangle[/tex]

    So, for all n, the eigenvalues of H are in the form:

    [tex]\frac{\lambda _n ^2}{2 \mu a^2}[/tex]

    The answer for part (c) becomes:

    [tex]\frac{\lambda _1 ^2 - \lambda _0 ^2}{2\mu a^2}[/tex]

    is that correct?

    Now, there is still a big problem! I don't know the [itex]\lambda _n[/itex], i.e. the eigenvalues of [itex]\mathbf{L}[/itex], or equivalently, the [itex]\lambda _n ^2[/itex], i.e. the eigenvalues of [itex]\mathbf{L}^2[/itex]. My book talks more about [itex]\mathbf{L}^2[/itex] than it does about [itex]\mathbf{L}[/itex], so perhaps there is some way to find the eigenvalues for this operator, but how do I find them? What are they?
     
  5. Nov 27, 2004 #4

    dextercioby

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    [tex]L^2 [/tex] is the same operator that appears in the Schroedinger equation for the hydrogen atom/hydrogenoid ion.So,from solving that problem,u'll be able to find the spectrum for [tex]L^2 [/tex].From then on,point c) is very simple.The spectrum of [tex]L^2 [/tex] is well known [tex] \hbar^2 l(l+1) [/tex],from the theory of angular momentum in QM,or from solving the hydrogen atom/hydrogenoid ion using Schroedinger's equations as indicated above.
     
  6. Nov 27, 2004 #5

    Dr Transport

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    So for the lowest energy state, [tex] l = 1 [/tex], the next highest energy state has [tex] l = 2 [/tex] so [tex] E_{1} - E_{0} \propto 2(2+1) - 1(1+1) = 6 - 2 = 4 [/tex] with the other factors......
     
  7. Nov 27, 2004 #6

    AKG

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    Okay, so the spectrum for [itex]\mathbf{L}^2[/itex] is simply [itex]2\hbar ^2, 6\hbar ^2, 12\hbar ^2, \dots , n(n + 1)\hbar ^2, \dots[/itex], i.e. [itex]l[/itex] takes on all natural values?
     
  8. Nov 27, 2004 #7

    Dr Transport

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    Exactly.....
     
  9. Nov 27, 2004 #8

    AKG

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    Thanks Dr Transport and dextercioby.
     
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