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Is the de Broglie wavelenght incongruent with particle in a box calculation?

  1. Jun 14, 2012 #1
    Dear Physics Forum,
    I think I am confused in applying the correspondence principle in QM, using the particle in a box and de Broglie wavelength calculations for massive (macroscopic) objects.

    Why is the calculated energy of a massive object so small in the particle in a box

    E = h2/8mL2

    but when the de Broglie wavelength of a massive object is calculated,

    λ = h/p

    it is very small, meaning its frequency and so energy are high?

    And if the energy level spacing is always n2, why is it said that the spacing becomes continuous, not discrete, for massive objects?

    Thanks for your help, Mark
     
  2. jcsd
  3. Jun 14, 2012 #2
    The energy of the particle in a box is calculated by plugging in the wavefunction of the particle into the Schrodinger Equation. Since the potential energy inside the "box" is zero, the Schrodinger equation takes the form [tex] \frac {d^{2} \psi} {dx^{2}} + \frac {8 \pi m} {h^{2}} E \psi = 0 [/tex] The wavefunction of the particle is given by [tex]\psi = \sqrt{\frac{2}{L}} + sin \left ( \frac{n \pi x}{a} \right )[/tex] Substituting this into 1 gives you the equation for the energy levels of the particle that you posted (except that there should be an n2 next to h2).

    When people say that it becomes continuous, it means that the difference in energy levels is so small that it appears to be continuous, as in classical mechanics.
     
  4. Jun 14, 2012 #3

    mfb

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    If you combine your first post in a different way, you answer your own questions:

    For macroscopic objects in macroscopic boxes, the energy levels are very close to each other. Therefore, any "macroscopic energy" is at very high n, where the relative distance between the levels is really small (relative distance of n^2 and (n+1)^2 is ~2/n and therefore falling for large n).
    As the energy is large, the De-Broglie wavelength is very small and smaller than usual box sizes.
     
  5. Jun 14, 2012 #4

    jtbell

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    This is for the ground state, n = 1. For any n you have

    E = n2h2/8mL2

    For a non-relativistic particle with no potential energy (only kinetic energy), E = p2/2m (which you can verify from the usual classical formulas for momentum and kinetic energy).

    1. Use de Broglie's formula λ = h/p to substitute for p and get a formula that gives E in terms of λ.

    2. Recall that for the particle in a box, the length of the box is an integer multiple of half the wavelength: L = nλ/2. Use this to substitute for λ in the equation you found in step 1. What do you get for E?
     
  6. Jun 18, 2012 #5
    Thanks very much for the help, I see the point that the relative spacing decreases with n, MarkM. I also did the substitutions you suggested jtbell, and did recreate the particle in the box equation from E = p2/2m. I guess that surprised me, as the energy in E= p2/2m increases with mass, but in E=n2h2/8mL2 the energy appears to decrease with mass. The math works out, but why (I apologize for not using the squared superscript properly, but nonetheless)? Can you help?

    Thanks very much, Mark
     
  7. Jun 18, 2012 #6

    mfb

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    The typical n increases with the mass. Write n as something like n=c*m, and you get E proportional to m for the quantum-mechanical treatment, too.
     
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