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From Cartan's Theory of Spinors, x1^2 + x2^2 + x3^2 = 0

  1. Jun 22, 2009 #1
    Let x1, x2, x3 be the components of a complex vector.

    If x1^2 + x2^2 + x3^2 = 0 Cartan calls this a isotropic vector.

    So if,

    x1 = a*exp(i*theta) then
    x1^2 = a^2*exp(2*i*theta) ?

    I think I'm being confused with what I read here,


    in particlular the definition of x*x = a^2 + b^2 + 2*i*ab

    Thanks for any help.
  2. jcsd
  3. Jun 22, 2009 #2
    The last equation seems to be wrong. With x defined as x=a+ib you get
    x^2 = x*x = (a+ib)*(a+ib) = a^2 + iab + iab + (ib)^2 = a^2 - b^2 + 2iab
  4. Jun 22, 2009 #3


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    Note in that definition, the vector x is expressed as a sum of a real and imaginary vector.
    [tex](\vec{a} + i\vec{b})\bullet (\vec{a}+i\vec{b}) = \vec{a}\bullet \vec{a} + \vec{b}\bullet\vec{b} + 2i\vec{a}\bullet \vec{b}[/tex]
    where [itex]\bullet[/itex] is the dot product.

    However I do not think your reference is the best way to understand spinors. The construction does not well generalize to higher dimensions. The best way to understand spinors is imnsho to consider first Clifford algebras (orthogonal algebras).

    You start by defining the cllifford algebra abstractly as the algebra generated by a set of vectors subject to the property:

    [tex] uv + vu = u\bullet v \mathbf{1}[/tex]
    where again the big dot is the dot product (the metric of the vectors).
    (Some conventions introduce either a factor of 2, -1, or -2 into the dot product on the rhs. This is just a matter of rescaling the magnitudes of the generators.)

    So if two vectors are orthogonal,
    [tex]u\bullet v = 0[/tex]
    then they anti-commute
    [tex]uv = -vu[/tex].

    These vectors form a graded (semi-graded) algebra with the identity being grade 0, the vectors, grade 1, and grade two elements are generators of rotations on the vector space.
    [tex] \sigma = uv-vu[/tex]
    generates a rotation in the plane spanned by u and v. Assume now that u and v were orthogonal and normalized vectors. This means in the algebra we can exponentiate:
    [tex] R(t) = \exp( t\cdot \sigma/2)[/tex]
    To yield a rotation operator which acts adjointly on the vector generators:
    [tex] w' = R(t) w R^{-1}(t)[/tex]
    yields the rotation of w in the u, v plane by angle t. (I may be off by a scaling factor on this angle depending on the above mentioned defining convention.)

    All the algebraic properties of the Clifford algebra are defined by its given relations so you can do all calculations without writing them down specifically as matrices. However if you write a matrix representation of the Clifford algebra then the "vectors" upon which these matrices act are in fact spinors.

    Examples, Hamilton's quaternions are a clifford algebra expressing rotation in 3 dimensions. So to are the complex algebra generated by the Pauli spin matrices. (Note there is a choice of sign convention in the dot product which gives two Clifford algebras for a given vector space.)

    Also the Dirac gamma matrices are the generators of the clifford algebra corresponding to lorentz transformations on Minkowski space-time. (There's a second all real representation yielding Majorana spinors).

    You have to do a bit more initial work playing with clifford algebras to get the feel for them but once you do, you will find that they are the best way to deal with and understand spinors.

    An interesting feature is as you increase the dimension of the underlying vector space you get a doubling of the dimension of the spinors (doubles for each two added dimensions) so after a while the spinor space is larger than the vector space.

    You also find a periodicity in the type of algebra (real, complex, quaternionic) you get when you consider clifford algebras of higher dimension and of various metric signatures.

    Porteus' book "Clifford Algebras and the Classical Groups" is the best reference I've found but may not be a good starting point.

    Take a look at http://www.av8n.com/physics/clifford-intro.htm" [Broken]
    On a quick glimpse it looks to be a good intro.
    Last edited by a moderator: May 4, 2017
  5. Jun 22, 2009 #4
    Back on track, thank you. There must have been a typo in the web page.
  6. Jun 22, 2009 #5
    Thank you, I think I'm back on track.

    If a complex number can represent a point in the phase space of 1D harmonic oscillator then I'm guessing a triple of complex numbers can represent a point in the phase space of a 3D harmonic oscillator. If we make the restriction X*X = 0 then spinors "pop" out. My hunch is that the restriction X*X = 0 means the phase space path lies in a plane?

    I think I have read that there is a quantum mechanical connection between the group SU(2) and a 2D harmonic oscillator. The long shot is that some form of restricted motion of a 3D harmonic oscillator at each point of space gives us spin?

    Thanks for your trouble, I will try and understand what you wrote.
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