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Help me get my head around spin

  1. Oct 8, 2012 #1
    I'm currently taking a first course in quantum mechanics and we just recently introduced generalized angular momentum.

    Now, take the special case of a spin-1/2 particle. If i understood this correctly there are two possibilities spin up: ms = 1/2 , and spin down ms = -1/2 , this comes from the fact that ms can take on the values +s,..-s where s is the spin quantum number.

    The eigenvalues of the SZ operator are ms[itex]\hbar[/itex], so these are the values the spin can take on when measured along the Z axis. Now this is what confuses me, what is meant by measuring the spin "along the Z axis" or along any axis for that matter, is this some sort of projection of the spin on the Z axis?

    What bothers me is i think of spin as some property of the particle, its either up or down, so whats with the measurement along some axis. Whats the purpose of this, cant we just say that its up or down and be done with it?

    Hope i made some sense of what confuses me, lol.
     
  2. jcsd
  3. Oct 8, 2012 #2

    jtbell

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    "Spin up" and "spin down" don't mean that the spin angular momentum is literally exactly along the +z or -z directions. They're just intended to convey the general orientation. For an electron, the magnitude of the spin angular momentum is

    $$S = \sqrt{s(s+1)} \hbar =\sqrt{\frac{1}{2} \left( \frac{1}{2} +1 \right)} \hbar = \frac{\sqrt{3}}{2} \hbar$$

    and the projection of that magnitude along the z-direction is

    $$S_z = \pm \frac{1}{2} \hbar$$

    so the vector ##\vec S## is clearly at an angle with respect to the z-direction.
     
  4. Oct 8, 2012 #3

    Simon Bridge

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    Yes ...

    In addition to jtbell;
    The "z-axis" has to be determined by something - in the Stern-Gerlach experiment, it is determined by the apparatus ... in an atom, by the orbital angular momentum.

    See also:
    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qangm.html
     
  5. Oct 8, 2012 #4
    Right, so that makes sense then. Thank you.

    So the choice of expressing an arbitrary spinor as a linear combination of the projections of spin up and down on the Z axis is just by convention?

    Now, this also confuses me a bit, say i want to express the projection of the spin down angular moment on the x axis , | βX > as such a linear combination, how would i go about doing that?
     
  6. Oct 8, 2012 #5
    Yes.

    First you need to write down the S_x operator. Then you need to find the eigenvectors of the S_x operator. You should find that there are two of them, and that they span the space of spin-1/2 spinors, so you can represent any spinor as a linear combination of these eigenvectors.
     
  7. Oct 8, 2012 #6
    Ok, things seem alot clearer now after this and after i did some exercises. Thanks alot.

    In the basis { |[itex]\alpha[/itex] z > , | β z >} the matrix representation of S_x is:
    Sx= ( [itex]\hbar[/itex] /2 ) σx. ( Pauli matrix)
    Which has the eigenvalues +- ( [itex]\hbar[/itex] /2 ).
    z> is the eigenvector corresponding to eigenvalue -( [itex]\hbar[/itex] /2 ).
    Using this we find |βx> = (1/√2)( |[itex]\alpha[/itex] z > - | β z > ).

    Sound about right?
     
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