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Uncertainty principle, relating the uncertainty in position to the uncertainty

  1. Mar 22, 2007 #1
    Prove the uncertainty principle, relating the uncertainty in position (A=x) to the uncertainty in energy ([tex]B=p^2/(2m + V)[/tex]):

    [tex]\sigma x\sigma H \geq \hbar/2m |<P>|[/tex]

    For stationary states this doesn't tell you much -- why not??
     
  2. jcsd
  3. Mar 22, 2007 #2
    What have you done so far? Do you know the generalized uncertainty relation? Read the guidelines to this forum, please, and we'll be able to help more.
     
  4. Oct 5, 2009 #3


    solution-
    [x,p2/2m+V]=1/2m[x, p2]+[x,V];
    
    [x, p2]= xp2 − p2x = xp2 − pxp + pxp − p2x = [x, p]p + p[x, p].

    using the equation [x,p]=ih{this is known as canonocal commutation relation}
    
    [x, p2]= ihp + pih = 2ihp. and And [x, V ] = 0,
    so [x,p2/2m+ V]=1/2m(2ihp) = ihp/m

    The generalized uncertainty principle says, in this case,

    σ2xσ2H≥{(1/2i)(ih/m)<p>}^2={h/2m<p>}^2⇒ σxσH ≥h/2m|<p>|. QED

    For stationary states σH = 0 and p = 0, so it just says 0 ≥ 0.
     
  5. Oct 5, 2009 #4
    for reference u can use {griffiths_d.j._introduction_to_quantum_mechanics__2ed.}
     
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