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Standing waves corresponding to a confined particle

  1. Feb 24, 2015 #1
    I'm just beginning to learn about quantum mechanics. My background in physics is very limited (Physics 1,2,3 and thermo). Here are my questions: A wave packet (which represents a particle) can be formed from the superposition of waves over a range of continuous wavelengths (i.e. an integral of a say a sine wave with respect to wave number k over some range of k). The range of k corresponds to a range of wavelengths which represents the uncertainty in the wavelength and therefore the momentum of the particle. The width of the wave packet corresponds to the uncertainty in location. So far so good. But when considering the wave packet corresponding to a particle in an infinite potential well, the boundary conditions for the wave packet dictate that the wave can only have particular values for wavelength similar to standing waves on a string. But since now there is reference to a PARTICULAR wavelength for the wave representing the particle where is the uncertainty in wavelength? For example, for a particular quantum state the value of n dictates the wavelength of the standing wave (which is a definite value). The only thing I can think of to reconcile this is the fact that standing waves themselves are the superposition of waves of different frequencies so the uncertainty in wavelength comes from the variety of component waves making up the standing waves? And even though the wavelengths of the standing waves take on definite values, do the wavelengths of the component waves that make up the standing wave vary continuously (like the wave packet constructed from an integral) or can those only take on certain values as well? Because if that was the case then the wave packet couldn’t be localized because any finite combination of waves with discrete wavelengths will produce patterns that repeat between –Infinity and Infinity?
     
  2. jcsd
  3. Feb 24, 2015 #2

    Nugatory

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    Staff: Mentor

    The energy eigenstates are the standing sine waves ##\psi(x)=sin\frac{n\pi}{L}x## (to a normalization constant). These correspond to states of definite energy but maximum uncertainty in position; the particle in that state can be found anywhere in the box (except at the nodes where ##\psi(x)=0##). A state in which there is less uncertainty about the location of the particle (the wave packet, for example) can be constructed as a superposition of these states; but because it is a superposition of states with different ##n##, it is a also a state in which there is greater uncertainty about the energy.

    It's really not much different than what you've been doing, except that you are summing over discrete values of ##n## instead of integrating over a continuous range.
     
  4. Feb 24, 2015 #3

    jtbell

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    In order to have a completely definite wavelength and momentum, the wave has to be infinitely long. An energy eigenstate of the infinite square well is not such a wave. The wave function has the form of a wave only inside the well; outside the well it is zero.

    You can construct that energy eigenstate by adding an infinite number of infinitely-long waves with different wavelengths, in such a way that they cancel outside the well, leaving only a wave inside the well, by using the methods of Fourier analysis. The different wavelengths in the sum correspond to different momenta, giving an uncertainty in momentum.
     
    Last edited: Feb 25, 2015
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