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Stern-Gerlach experiment

  1. Feb 15, 2009 #1
    Hi, I've been trying to understand the Stern-Gerlach experiment. I've been reading this book which explains how to find the amplitudes of measuring a particle to have a spin 1/2 lined up along some direction.

    Here is the book:

    http://www-thphys.physics.ox.ac.uk/people/JamesBinney/QBc6.pdf [Broken]

    It's page 107 that confused me: that first matrix equation. The matrix representation of the spin matrix along theta seems to be in the basis of eigenstates of the Sz matrix. But then the object (1, 0) would be the state of the particle having spin along the z axis, in which case the equation doesn't seem to make sense, as it would suggest that the state of having spin along the z axis is an eigenstate of the spin matrix along theta.

    Can anyone help here?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 15, 2009 #2
    Can You explain your trouble more explicitly?
    Which equations are initial and where is misunderstanding?
    (6.114)???
     
  4. Feb 15, 2009 #3

    Fredrik

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    The equation would imply that if it had been true for any values of a and b, or at least specifically for a=1,b=0, but it isn't. See (6.117).
     
  5. Feb 15, 2009 #4
    Thanks for replying

    Yes 6.114, I don't really get where this comes from. I think that

    |+,z> = a*|+, p> + b*|-,p>

    (where p is theta)

    But I think that the matrix representation of sigma.n in 6.114 is in the basis of eigenstates of Sz, i.e. of |+,z> and |-,z>. I think this because the paragraph before 6.114 references 6.111, which is in the basis of eigenkets of Sz.

    So if this is the case, is (1, 0) (a column vector), the representation of |+,z>? If this is the case then the left hand side of the equation surely reads:

    sigma.n [ a|+,z> + b|-,z>] which isn't the same as the right hand side of [ a|+,z> + b|-,z>] as |+,z> isn't an eigenket of sigma.n

    ???????ARGH????

    Thanks
     
  6. Feb 15, 2009 #5
    Sorry would you mind elaborating, I don't fully understand this. Sorry if I'm being very slow!
     
  7. Feb 15, 2009 #6

    Fredrik

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    What 6.114-6.117 says is that if (a,b) is an eigenvector, then a and b must be specifically those numbers specified by 6.117. So (1,0) is clearly not an eigenvector (of [itex]\hat n\cdot\vec\sigma[/itex]). (You said in #1 that the equation suggests that it is).

    What you said in #4 is correct until the last statement before the argh. The two sides of the equations are the same, if a and b are as in 6.117. You seem to be assuming that a linear combination of two eigenvectors with different eigenvalues can't be an eigenvector of some other operator. That assumption is wrong.
     
    Last edited: Feb 15, 2009
  8. Feb 15, 2009 #7
    Ah OK so what I said where the two sides is correct, but I was incorrect in saying that they can't equal?

    What is the reason for setting (a,b) as an eigenvector of sigma.n? (last question I promise!)

    Thanks
     
  9. Feb 15, 2009 #8

    Fredrik

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    If the word "where" is supposed to be "were", then yes. That's what the equation says, and the equality holds.

    Because he wants to find out which linear combinations of |z+> and |z-> are eigenvectors of [itex]\hat n\cdot\vec S[/itex]. He's not "setting" (a,b) as an eigenvector. He's determining what the eigenvectors are.
     
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