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Rotation Spectra; energies

  1. Mar 19, 2014 #1
    I don't really understand the explanation given in Binney's text about:

    htinb4.png

    Hamiltonian is given by:

    [tex]H = \frac{\hbar^2}{2} \left( \frac{J_x^2}{I_x} + \frac{J_y^2}{I_y} + \frac{J_z^2}{I_z} \right)[/tex]

    Orient axes such that ##I_x = I_y = I##.

    [tex]H = \frac{\hbar^2}{2} \left( \frac{J^2}{I} + J_z^2(\frac{1}{I_z} - \frac{1}{I})\right)[/tex]

    Energy is given by:

    [tex]E_{jm} = \frac{\hbar^2}{2} \left[ \frac{j(j+1)}{I} + m^2(\frac{1}{I_z} - \frac{1}{I}) \right][/tex]

    We are only interested in states:

    [tex]E_{jm} = \frac{\hbar^2}{2I} j(j+1) [/tex]

    Emitted energy and frequency are:

    [tex]\Delta E_p =\pm (E_j - E_{j-1}) = \pm j\frac{\hbar^2}{I}[/tex]
    [tex]v_j = j\frac{\hbar}{2\pi I}[/tex]



    Let's try to analyze the explanation here.

    1. Yes, energy, Jz and J2 share the same eigenstates ##|j, m>##.

    2. <J2> = j(j+1) : Yes, since that is the eigenvalue and eigenvalue correspond to real observables.

    3. Why do low lying states with ##m = 0## and ##j~O(1)## lead to: ## j(j+1) >> j ##? Firstly, doesn't low lying states correspond to a low ##j##? And what does m have to do with anything? ##m## was defined as the eigenvalue of Ji and ##j = m_{max}##

    The rest of the argument doesn't make any sense at all..
     
    Last edited: Mar 19, 2014
  2. jcsd
  3. Mar 19, 2014 #2

    Bill_K

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    He didn't say j(j+1) >> j. He said "significantly larger".

    It would help to know what system this discussion is about, what the upper and lower states are, what νj is, and what is contained in Eq.(7.24).
     
  4. Mar 19, 2014 #3
    It's about the energy levels in rotation spectra of diatomic molecules - specifically CO molecule in this case.

    The problem is I'm not sure what he is referring to. This is taken from Binney's book, pg 140:

    25tz4hy.png
     
  5. Mar 20, 2014 #4

    DrDu

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    Much more importantly, he said larger than ##j^2##, not j.
     
  6. Mar 20, 2014 #5
    Ok besides that, I don't get the rest of the argument about rotation frequencies at all!
     
  7. Mar 20, 2014 #6

    DrDu

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    Classically, you have two ways to determine the rotation frequency: Either from the energy of the state using: ##\omega=\sqrt(2IE)## or measuring the frequency of the emitted radiation, ##\omega_\mathrm{transition}##. In QM, the energy is quantized, so ω, as determined from E will depend on J. He is saying ##\omega(j-1)<\omega_\mathrm{transition}<\omega(j)##, where the transition is from j to j-1. In the limit of large j, all thre values will converge to the same classical value.
     
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