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Tiny part of derivation for the energy levels occupied by a free particle in a box.

  1. Mar 3, 2012 #1
    Deduce the formula giving the energy levels in terms of n for a free particle in a box of length l, using the fact that only stationary waves can be occupied by the particle.
    - considering stationary waves, λ = 2l/n. (1)

    1) Using E=hv/λ= hvn (2)
    And
    P = h/λ, v = h/m*λ, v = hn/2ml (3)

    Subbing this into (2), E = h^2n^2/4ml^2

    2) However using E=1/2mv^2, p=mv, and λ = h/mv,

    1/2mv^2= p^2/2m = h^2/λ^2*2m
    = h^2/2m * n^2/4l^2 ( applying (1) )
    = h^2n^2/8ml^2


    I’m confused as to why this is - is there something fundamentally wrong with trying to derive the energy levels for a particle in a box using 1/2mv^2=E and p=mv ?

    Thanks a lot anyone - greatly appreciated :D.
     
  2. jcsd
  3. Mar 3, 2012 #2
    Re: Tiny part of derivation for the energy levels occupied by a free particle in a bo

    what you've got there isn't [itex]E=\frac{1}{2}m v^2[/itex]
    check your working

    also 'only stationary waves can be occupied by the particle.' isn't specifically true, if you have a superposition of energy eigenstates then the wave certainly isn't stationary
     
  4. Mar 3, 2012 #3

    BruceW

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    Re: Tiny part of derivation for the energy levels occupied by a free particle in a bo

    Yes, this is right. But really the technical term is 'stationary state'.

    Where did this come from? Did you use E=hf and then substitute f=v/λ? This does not work in quantum mechanics.

    This is the right answer.
     
  5. Mar 4, 2012 #4
    Re: Tiny part of derivation for the energy levels occupied by a free particle in a bo

    Ahh, thank you. Why is this? Is c= alanda*f derived from : E^2=(mc^2)^2+(pc)^2 and E=hf and p=h/alanda
    Whereas for particles, due to their mass, once arranging all of the above, the product of alanda and f will not equal its speed.
     
    Last edited: Mar 4, 2012
  6. Mar 4, 2012 #5
    Re: Tiny part of derivation for the energy levels occupied by a free particle in a bo

    Because the wave function of the particle must be zero outside the box (the walls of the box are impregnable, and the particle cannot be found outside the box, so the probability for finding it must be zero everywhere). Since the wavefunction ought to be a continuous function (a postulate in Schroedinger's Wave Mechanics), it must be zero on the ends of the box as well.

    Inside the box, the particle is free and is described by a single wavelength standing wave.

    Since the wave has 2 nodes at the ends of the box, there must be an integer number of half-wavelengths of the standing wave accomodated on the length of the box, i.e.:

    [tex]
    n \, \frac{\lambda}{2} = l \Rightarrow \lambda = \frac{2 l}{n}, \ n = 1, 2, \ldots
    [/tex]
     
  7. Mar 4, 2012 #6

    BruceW

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    Re: Tiny part of derivation for the energy levels occupied by a free particle in a bo

    That's not really the reason. Remember, you used non-relativistic physics for your particle in a box question, and f=v/λ still did not work.

    The reason that it doesn't work in quantum mechanics is because 'frequency' and 'wavelength' don't have the same meaning as they do for classical waves (confusing, I know). In quantum mechanics, you should simply think of the 'frequency' as telling us about the energy of the particle, and the 'wavelength' as telling us about its momentum.

    Having said that, there can be a strong connection between the classical frequency of motion and the energy of the analogous quantum state. For example, in the quantum harmonic oscillator, the energy is quantised by hf (where f is the classical frequency of motion).

    So in summary, I'd say be careful with the quantum frequency, its proper definition is to do with the energy of the system, not the classical frequency.
     
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