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|    sci.physics.research    |    Current physics research. (Moderated)    |    17,516 messages    |
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|    Message 16,516 of 17,516    |
|    Lawrence Crowell to Jos Bergervoet    |
|    Re: Stern-Gerlach question.    |
|    15 Jun 19 19:18:56    |
   
   From: goldenfieldquaternions@gmail.com   
      
   On Friday, June 14, 2019 at 4:27:19 PM UTC-5, Jos Bergervoet wrote:   
   > Consider a Stern-Gerlach device using a field in the z-direction.   
   >   
   > Entering is a single-electron Gaussian wave packet with spin   
   > Psi(x) ( |u> + |d> ) / sqrt(2)   
   > where the initial spin is clearly in x-direction.   
   >   
   > We know the apparatus will split the trajectory of the components   
   > and due to the non-uniform grading of the field two separate wave   
   > packets will come out, moving in slightly different directions.   
   >   
   > But before the separation is complete, isn't there fast precession   
   > of the spin around the z-axis? And then wouldn't the electron radiate   
   > energy, so fall down to the energy level of the |d> spin which in   
   > the field is lower than the |u> spin?   
   >   
   > So my question is: do we expect more down-spin to come out than up-   
   > spin? (When the input components both have exactly equal amplitude,   
   > I mean!) Or is there a reason for this mechanism to be insignificant?   
   >   
   > --   
   > Jos   
      
   The frequency of the electron spin resonance (esr) is f = g*mu*B/h,   
   which is usually in the radio frequency (MHz) range. The electron   
   beam passing through the highly asymmetric magnetic field does so   
   in a time T = v/d, for d the "aperture." or scale of the magnetic   
   region, of the device. So one makes the Stern-Gerlach apparatus   
   such that d/v << f if you want to avoid this problem. So if the   
   aperture is 1cm then v >~ 10^6cm/sec should to the trick. An electron   
   in a 1 volt potential will then have eV energy or 1.6x10^{-19}j.   
   Equating that to E = 1/2mv^2, m =~ 10^{-30}kg then v^2 =~   
   2x10^{11)m^2/s^2 or about v = 4.5x10^5m/s. So this should avoid the   
   problem of esr spin flipping.   
      
   Of course the electron spin resonance is done usually in a Helmholz   
   coil with a uniform magnetic field. The Stern-Gerlach apparatus has   
   a highly divergent magnetic field, which is opposite of what is   
   standard in esr. So to fully address this question might take a bit   
   more analysis.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   
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