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Reduced Mass

  1. Sep 2, 2010 #1
    If the electron of the h-atom is not moving in a classical orbit (like a circular orbit) why is the reduced mass used in Schrodinger's equation?
     
  2. jcsd
  3. Sep 2, 2010 #2

    alxm

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    Because, even though you don't have motion in the classical sense, the motion of the nucleus and electron is still correlated.
    The potential [tex]\frac{1}{|\mathbf{R}-\mathbf{r}|}[/tex] is dependent on both coordinates; the equation in ordinary coordinates is not separable.
    You have to switch to reduced-mass coordinates to be able to make it separable and soluble.

    In this sense, it's no different than a classical problem.
     
  4. Sep 2, 2010 #3
    I dont know but have you seen this
    http://www.sciencedirect.com/scienc...be8d63904a0f7703962a3faf050dd121&searchtype=a

     
  5. Sep 2, 2010 #4

    alxm

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    What about it? (Doesn't seem like terribly interesting/important paper)

    But if you want to look at H2+ from a coordinate-transform point of view, then it has no analytical solution using any coordinate transform. However, if you assume fixed nuclei (Born-Oppenheimer approximation), it becomes separable in prolate spheroidal coordinates. This transform was in fact being used to try to to solve H2+ even before quantum theory, or rather in the days of the 'old' quantum theory.
     
  6. Sep 2, 2010 #5
    [tex]\mu=\frac{m_e m_p}{m_e + m_p}[/tex]

    When the reduced mass is effective in the two-body problem, the next two conditions must be satisfied.

    1. The two particles are moving by the attractive force between them.
    (If there is no force between them, the electron is moving "independently" of the proton,
    so [tex]m_p[/tex] of the reduced mass becomes meaningless.)

    2. The electron needs to be moving.
    (If the electron is not moving, the center of mass is not changing,
    so we don't need to use the reduced mass to correct its motion.)

    As shown in this site,
    I wonder why the ordinary books or sites about QM don't try to explain the reduced mass from the QM viewpoint ?
    (Why they don't show the detailed explanation about it (from not classical viewpoint) ?)
    We should accept this conception naturally?
     
  7. Sep 2, 2010 #6
    If you have ever measured the transition energies of electrons in both hydrogen and deuterium gas using a diffraction grating, you will see the reduced mass shift.

    Hydrogen 3d->2p red line is 656.2852 nanometers

    Deuterium 3d->2p is measurably different at 656.1065 nanometers

    The fractional difference is 1 part in 3672. Can you relate this to the proton to electron mass ratio?

    Bob S
     
  8. Sep 2, 2010 #7
    We use reduced mass since the proton would be moving too, and so the potential energy depends on both the position of the proton and electron. This is near impossible to solve, and moreover, we usually aren't interested in the proton.

    If we redefine the problem in terms of the separation between the proton and electron and in terms of the location of the center of mass, this equation separates nicely. The schrodinger equation for the center of mass coordinate, with mass M_p+M_e, gives a free wave equation. The equation for the separation vector gives the normal hydrogen wavefunction that describes a fictitious particle of mass mu moving around in a potential field with a fictitious positive unmoving charge at the origin. The solution of this equation gives the energies of the hydrogen atom.
     
  9. Sep 4, 2010 #8
    Thanks for the replies.

    It appears that the reduced mass is necessary in predicting hydrogen's electronic states, but what about the angular momentum of the ground state of the H-atom? Doesn't this type of motion being discussed imply nonzero angular momentum?
     
  10. Sep 4, 2010 #9

    tom.stoer

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    There is a mathematical rigorous approach to see how the reduced mass appears.

    One starts with a two-particle system with masses m1 and m2 and quantizes this two-particle system. The Hilbert space is spanned by the states |p1, p2>. Then one defines a unitary transformation (QM analogue of a canonical transformation) which generates the new momentum operators p and P; the states Hilbert space is now spanned by the states |p, P>

    The Hamiltonian splits into two parts H[P] and H[p, V(x)]

    The part of the Hamiltonian H[P] (with P = total momentum) commutes with H[p, V(x)]; therefore one can impose the condition P|phys> = 0 which is the QM analogue of selecting the frame with vanishing momentum P.

    The coefficient in p2/2m is just the reduced mass. It is not introduced by hand but generated by the unitary transformation.
     
    Last edited: Sep 4, 2010
  11. Sep 4, 2010 #10
    Actually not. Consider two masses, m and M, connected by a linear spring with a spring constant k. Assuming no angular momentum and no damping, so that the motion is only axial, what is the natural resonance frequency ω0 of this linear harmonic oscillator? It should be (I am guessing)

    ω0 = sqrt[k(M+m)/Mm]

    Please check the equation. Note presence of reduced mass in equation. Also confirm that this equation has units of s-1 (i.e., radians per second).

    Bob S
     
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