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Heinsenberg UC and n=0 state?

  1. Mar 13, 2006 #1
    How would you go about this question?

    Show that by allowing the state n=0 for a particle in a 1D box will violoate the uncertainty principle, delta(x)delta(p)>=h(bar)/2

    I have tried to substitute all sorts of different relationships but do seem to get anywhere. I have showed that E=0 for a ground state electron but can't seem to relate it to the uncertainty principle.
     
  2. jcsd
  3. Mar 13, 2006 #2

    dextercioby

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    You just have to compute the uncertainties for each variable, x and p_{x}, knowing the wavefunction.

    Daniel.
     
  4. Mar 13, 2006 #3

    If we allow n=0, the wavefunction will cease to exist hence the particle will cease to exist. Hence momentum and position of a particle does not exist or could you say 0. Hence any change in the n=0 state, the particle will continue to cease to exist. In this way the HU principle will not be satisfied. But my argument is pretty vague. Is there a quantifiable way to express this?
     
  5. Mar 14, 2006 #4

    dextercioby

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    What's the energy spectrum for the particle...?

    Daniel.
     
  6. Mar 16, 2006 #5
    What do you mean by the energy spectrum?

    Do you mean the energy levels?
    At n=0, Energy=0.
    At other levels, Energy=n^2(pie)^2(hbar)^2/(2mL^2)

    But how does it relate to the UC?
     
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