Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spin and isotropic vectors

  1. Apr 6, 2013 #1
    I've been trying to wrap my head around spinors, and I think I'm beginning to understand the mathematical formalism, but one thing in particular continues to confuse me. Whenever I read anything mathematically rigorous on spinors, the authors refer to spinors being associated with isotropic (zero length) vectors.

    Does anyone know how this squares with the use of spinors in QM and QFT? Spin vectors clearly have length, so why do we use this mathematical formalism developed for isotropic vectors?
     
  2. jcsd
  3. Apr 6, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

  4. Apr 6, 2013 #3
    See,

    http://www.sjsu.edu/faculty/watkins/spinor.htm

    I think Mr. Watkins makes a math error in the second paragraph.

    XdotX = a^2 - b^2 + 2iab and not a^2 + b^2 + 2iab

    Good luck.
     
  5. Apr 6, 2013 #4
    Thanks for the links (I've already seen both of these pages, they're pretty much lifted directly from Cartan's book on spinor theory), but maybe I didn't explain my confusion well enough. I understand the mathematical formalism, and I understand that you can associate an isotropic vector in 3-space with a two-component spinor. What I'm confused about is why spinors are useful if they are only connected with isotropic vectors. This seems at odds with what we normally use spinors for --- to represent spin angular momentum, which is not a quantity of zero magnitude. It just seems confusing to me. I would've thought that the 3d vector to which a spinor corresponds would be the actual angular momentum vector in 3-space, but apparently I'm not thinking about this correctly.
     
  6. Apr 6, 2013 #5

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    I always thought this business with "isotropic vectors" was thoroughly obtuse. As far as I can tell, isotropic vectors are not used in any other context, so it seems silly to worry about how spinors are related to them. This may be the way Cartan originally discovered spinors, but it is not the most straightforward way to visualize spinors.

    A spinor is a quantity that indicates a direction in space. This is, essentially, what it means to "transform under a representation of the rotation group", because the rotation group is the group that maps directions in space onto other directions in space. A direction in space is the same thing as a unit vector (note, I'm not talking about isotropic vectors here; just ordinary, real vectors), which in turn is the same thing as a point on the unit sphere. So, to each spinor must correspond a point on the unit sphere. (In fact, it turns out this map is 2-to-1; i.e. to each point on the sphere there correspond two spinors).

    To visualize this mapping, you need to understand the Riemann sphere (which is just the sphere with a complex structure). Take the ordinary unit sphere, and allow it to sit on top of the complex plane, such that its south pole is sitting at the origin. Now project rays down from the north pole. Each ray will intersect the sphere in one point, and the plane in another. Hence the collection of rays define a map from the complex numbers onto the sphere (the north pole itself corresponds to complex infinity). So the complex numbers can be used as "projective coordinates" on the sphere.

    Now you can ask, what happens when the sphere rotates? It turns out that every rotation of the sphere corresponds to a fractional linear transformation of the complex plane,

    [tex]z \rightarrow \frac{a z + b}{c z + d}[/tex]
    for some complex numbers ##a, b, c, d##. It will be instructive if you work out how these complex numbers relate the the Euler angles of the rotation.

    It turns out that fractional linear transformations are a group, and if we represent such a transformation as a matrix

    [tex]g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}[/tex]
    then the composition of two fractional linear transformations is given by matrix multiplication (check this!).

    The last step is to write ##z## as a ratio of two complex numbers

    [tex]z = \frac{\xi_1}{\xi_2}[/tex]
    Then the fractional linear transformation is given by

    [tex]\frac{\xi_1}{\xi_2} \rightarrow \frac{a \xi_1 + b\xi_2}{c\xi_1 + d\xi_2}[/tex]
    But now this looks a lot like the matrix ##g## multiplying a column vector! In fact, we've identified the vector space where ##g## acts, and this transformation is given by

    [tex]\begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix} \rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix}[/tex]
    So here's a "projective representation" of the rotation group. It's called "projective" because it relies on projecting the sphere onto a plane. You should be able to check that the matrices that correspond to rotations form the group SU(2). (There are also fractional linear transformations that correspond to other rigid motions of the sphere, such as translations in 3-dimensional space. Altogether, they will form the group SL(2,C).)

    Finally, notice that the map from spinor space onto the unit sphere is given by

    [tex]z = \frac{\xi_1}{\xi_2}[/tex]
    and hence the "magnitude" of the spinor is irrelevant. We can normalize it to 1. Furthermore, the overall sign is also irrelevant, which is why 2 spinors map to the same point on the sphere.

    Spinors do not have "magnitude" because they only indicate direction (via the map onto the unit sphere). In QM, this corresponds to the direction of the spin axis.
     
  7. Apr 6, 2013 #6
    Thanks this is very helpful!
     
  8. Apr 7, 2013 #7

    samalkhaiat

    User Avatar
    Science Advisor

    Spinors are tensors of half-integer rank. They can only be defined in a manifold where squares of segments can be found to vanish. That is, spinors can be introduced only in a manifold admitting of the existence of isotropic vectors. Space-time is such manifold.
    The transformation
    [tex]z \rightarrow \frac{ az + b }{ cz + d } , \ \ ad - cb = 1[/tex]
    is the most general conformal and orientation preserving transformation of the Riemann sphere to itself. It is called spin transformation provided that [itex]z[/itex] is related to the Minkowski null vectors through
    [tex]
    z = \frac{ \xi_{ 1 } }{ \xi_{ 2 } } = \frac{ x^{ 0 } + x^{ 3 } }{ x^{ 1 } - i x^{ 2 } } = \frac{ x^{ 1 } + i x^{ 2 } }{ x^{ 0 } - x^{ 3 } } ,
    [/tex]
    where
    [tex]
    x^{ 0 } = \frac{ 1 }{ \sqrt{ 2 } } \left( | \xi_{ 1 } |^{ 2 } + | \xi_{ 2 } |^{ 2 } \right), \ \ \ x^{ 3 } = \frac{ 1 }{ \sqrt{ 2 } } \left( | \xi_{ 1 } |^{ 2 } - | \xi_{ 2 } |^{ 2 } \right) ,
    [/tex]
    [tex]
    x^{ 1 } = \frac{ 1 }{ \sqrt{ 2 } } \left( \xi_{ 1 } \xi^{ * }_{ 2 } + \xi_{ 2 } \xi^{ * }_{ 1 } \right) , \ \ \ x^{ 2 } = \frac{ - i }{ \sqrt{ 2 } } \left( \xi_{ 1 } \xi^{ * }_{ 2 } - \xi_{ 2 } \xi^{ * }_{ 1 } \right) .
    [/tex]

    Sam
     
  9. Apr 12, 2013 #8
    Sam, is there possibly a sign wrong in your definition of [itex]x^2[/itex]? I don't have Ward's book on GR in front of me, but I just notice the spin-up eigenvector for the y axis gives a negative y component using your definition.
     
  10. Apr 12, 2013 #9

    samalkhaiat

    User Avatar
    Science Advisor

    I do not see how can a sign difference change any thing in [itex]( x^{ 0 })^{ 2 } - ( \vec{ x } )^{ 2 } = 0[/itex]?

    Sam
     
  11. Apr 12, 2013 #10
    Well I've checked Ward's book and he defines [itex]x^2[/itex] without the initial minus sign. I just expected the spin-up eigenstate of [itex]\sigma_2[/itex] would point along the positive y axis in Minkowski space. I know it doesn't change the length of the 4-vector.
     
  12. Apr 12, 2013 #11

    samalkhaiat

    User Avatar
    Science Advisor

    Oh, in that case, you can exchange the order of the spinor components. I have no idea about what Ward's text treatment is? and what signature he uses.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook