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|    sci.physics.research    |    Current physics research. (Moderated)    |    17,516 messages    |
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|    Message 16,442 of 17,516    |
|    Lawrence Crowell to ben...@hotmail.com    |
|    Re: Preparation of electron spin directi    |
|    28 Feb 19 21:49:42    |
   
   From: goldenfieldquaternions@gmail.com   
      
   On Tuesday, February 26, 2019 at 4:18:54 PM UTC-6, ben...@hotmail.com wrote:   
   > Sorry, if that garbling of text was my fault.   
   > The a* in the last line of the quotation should have been -a.   
   >   
   > To put the question another way: if an electron with a hypothetical   
   > exact spin vector p is subsequently prepared by presenting it to   
   > to a detector with vector a (Alice's detector), then does the   
   > electron exact spin axis p become changed to be exactly aligned   
   > with vector a (or vector -a, whichever is appropriate)? IMO the   
   > electron spin vector after preparation is either along exact vector   
   > p or along exact vector -p. The outcome of p or -p alignment depends   
   > on which of p and -p is nearest in alignment to vector a.   
      
   To prepare an electron spin state you first generate an electron beam.   
   This can be done by heating a cathode and using a positive charged grid   
   to move the electrons and then accelerate with RF cavities. To prepare   
   the spin state one can have this electron beam pass through a highly   
   asymmetric magnetic field. The spinning electron with its magnetic moment   
   has a magnetic field. This will then force the electrons to be aligned or   
   anti-aligned. Then with magnetic means one controls one of these split   
   beams with spin up or down.   
      
   This is the Stern-Gerlach apparatus. This is usually discussed with respect   
   to measurement of spin. However, the preparation of a quantum state is really   
   much the same as measurement. The alignment of the electron spins in principle   
   can be made as exact as possible. For the region of magnetic field having   
   the size d and the electrons with momentum vector k the angle uncertainty of   
   the beam or cone of the electron beam one prepares is an angle @ ~ (kd)^{-1}.   
   So for E ~ 1ev then k ~ 10^{7}cm^{-1} and if the apparatus magnetic port is   
   d ~ 1cm then the angle uncertainty is about 10^{-7}rad. The uncertainty in   
   the direction of the spins of the electrons will have a comparable value.   
      
   LC   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   
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