De Broglie and experiment have told us the wavelength of a particle with momentum p, ?=h/p but what if the particle is rolling up a hill: it will eventually turn around and roll down again. Suppose the hill has a very gentle slope, so that the - PowerPoint PPT Presentation

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De Broglie and experiment have told us the wavelength of a particle with momentum p, ?=h/p but what if the particle is rolling up a hill: it will eventually turn around and roll down again. Suppose the hill has a very gentle slope, so that the

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... h/p, then the de Broglie formula must give the wavelength pretty accurately. ... and there is a smallest momentum h/(2L) which we call the 'zero point energy' E ... – PowerPoint PPT presentation

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Title: De Broglie and experiment have told us the wavelength of a particle with momentum p, ?=h/p but what if the particle is rolling up a hill: it will eventually turn around and roll down again. Suppose the hill has a very gentle slope, so that the


1
  • De Broglie and experiment have told us the
    wavelength of a particle with momentum p, ?h/p
    but what if the particle is rolling up a hill
    it will eventually turn around and roll down
    again. Suppose the hill has a very gentle slope,
    so that the momentum is not changing much in a
    distance h/p, then the de Broglie formula must
    give the wavelength pretty accurately. As the
    particle rolls up the hill, p goes down and the
    wavelength goes up. We see that the wavelength
    depends on the local potential energy--- we are
    going to need a wave equation that contains the
    potential energy U(x)

2
  • But let us put off worrying about that equation
    and see what we can do with our knowledge of the
    wavelength of a free particle-- one moving in a
    region where U(x) is constant.
  • If the particle has to be held in the box, it has
    to have a rigid wall, so U(x) has to go to
    infinity outside, then we should keep it inside...

box
U(x)
L
0
x
3
  • Inside the box, the particle is free, so the
    wavelength is h/p, ?(x) sin kx, but the
    probability has to be zero outside the box-a
    boundary condition!! Prob (x) ?(x)2 0 at
    edges of the box, and our practice with strings
    stands us in good stead here, and we see that sin
    k0 0 (no information) and at the other end sin
    kL 0 -gt kL n? -gt
  • k n?/L or 2 ?/? n ?/L or ? 2L/n
  • much like the string. But this now gives us the
    momentum!! From ? h/p, p (nh)/(2L)
  • and there is a smallest momentum h/(2L) which we
    call the zero point energy Ep²/2m(n²h²)/(8mL²)


4
  • There are lots of other problems we can do by the
    locally free approximation, such as the ball
    rolling up the hill we mentioned,
  • We do have to face up to the problem of the wave
    equation for the general U(x) cant quite be
    deduced but Schrödinger didnt have much trouble
    guessing it,

5
  • Try separation of space and time variables,
    because we want at first the stationary states
    which do not change in time

6
  • Example of use of the wave equation tunneling
  • the bump in the middle, of thickness a, we call a
    barrier. When 0ltxlta the energy E is less that
    U(x) U? and the time-independent Schrödinger
    equation reads

U(x)
U?
a
7
  • Match the wavefunctions at the boundaries
  • and after the amplitudes are calculated, there is
    found a transmission coefficient
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