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I What is a spinor?

  1. Oct 13, 2016 #1
    Okay, I have read on spinors here and there but I really don't understand geometrically or intuitively what it is. Can someone please explain it to me and how/when it is used? Thanks!
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  3. Oct 13, 2016 #2


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    A spinor is a vector in a representation space of ##SU(2)##. Mathematically, we are led naturally to consider representations of ##SU(2)## because of the physical irrelevance of global phase factors in front of the wave function.

    A representation of a group ##G## is a pair ##(V,\rho)##, where V is a vector space, called the representation space, and ##\rho:G\rightarrow GL(V)## maps group elements into the set of operators on ##V## in such a way that ##\rho(g_1 g_2) = \rho(g_1) \rho(g_2)## and ##\rho(e) = \mathrm{id}##. If ##G=SU(2)##, the group of unitary 2x2 matrices with unit determinant, you would call elements of ##V## spinors. For instance, ##V = \mathbb C^2## and ##\rho(g)=g## would be the spin-##\frac{1}{2}## representation of ##SU(2)## and any vector in ##V## would be called a spinor.
  4. Oct 13, 2016 #3
    Hmm alright, do you think there would be a more geometric explanation of what is going on? Thinking in this rigorous analytic fashion is useful but I don't think it really captures the essence of the structure.
  5. Oct 13, 2016 #4


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    You can check out the Penrose flag representation of a spinor, but I don't think it is very helpful.

    Well, spinors are algebraic objects. I think what I wrote captures the essence quite exactly. However, the key to understanding why we are interested in such objects is the relation between the ##SO(3)## and ##SU(2)## groups. I guarantee that if you understand this, spinors will feel very natural. It is nicely described here: http://www.damtp.cam.ac.uk/user/examples/D18S.pdf
  6. Oct 13, 2016 #5
  7. Oct 13, 2016 #6
    Isn't it correct to say: "a rank 1 spinor is a quaternion"? I know it's almost correct. Surely this is the simplest intuitive picture. Then you can move on to higher dimensional spinors.
  8. Oct 13, 2016 #7
    I do understand quaternions, how would one extend spinors to higher ranks?
  9. Oct 13, 2016 #8
    Well, first you should investigate the relation between quaternions and spinors. For one thing, Hamilton used (mostly) real representation of quaternions (4 real numbers). Spinors, however, always seem to use complex representation (2 complex numbers), because it's more suitable for QM math. You may not be so familiar with that representation.

    The most intuitive quaternion representation is Hamilton's exponential form (basically the Penrose-Rindler "flag" picture) so keep that in mind as well. The half-angle rotation and double cover are trivial from that point of view.

    The most important aspect of spinors which you didn't learn in quaternion math class is their invariance under Lorentz transformations. You should understand how that works.

    Quaternions gives you all the relevant math but the way that's used in physics is not obvious.

    Assuming you're comfortable with rank 1 spinors, you're ready to go to higher dimensions. AFAIK the easiest route is Clifford algebras. Quaternions, and spinors, are equivalent to a couple flavors of Clifford algebras (real, and complex). Once you see how they work in those low-dimension algebras it's easy to extend it to higher dimensions.
    Last edited: Oct 13, 2016
  10. Oct 14, 2016 #9
    Spinor is a geometrical object that has magnitude, direction and a special property describing parity of number of rotations.

    There is a nice construction of spinors I personally like, involving "half derivative". Half derivative takes a scalar produces an object that has to be differentiated again to yield the full derivative, that means gradient, a vector. The object produced halfway is a spinor.
  11. Oct 14, 2016 #10


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    Hmm. This might be a terminology confusion on my part. I thought that a spinor was represented as a two-component column matrix, while quaternions can be represented as 2x2 square matrices. So a spinor is not a quaternion, as I understand it.
  12. Oct 14, 2016 #11
    Ooohh that does sound interesting. I do know how to take half derivatives and such. Would I just have to take half partial derivatives with respect to each variable to create like a half-gradient? But then how would this exhibit the properties of a spinor like parity of rotation?
  13. Oct 14, 2016 #12
    Yes. You can also clearly construct pseudovectors this way, but only in 3d space.

    There are two sets of spinor coordinates for each direction. When you are creating a spinor of unit magnitude pointing in particular direction, there are two ways to do it.

    Also, I wanted to clearly distinguish classical and quantum spinors. All geometrical objects come in two variants: classical and quantum. Speaking a bit informally, classical spinors are anticommutative:

    a b + b a = 0

    while quantum spinors are non-anti-commutative:

    a b + b a = 1

    By taking half-derivative, you get geometric algebra, that is classical spinors. You then have to perform a quantum deformation to obtain Clifford algebra, that is quantum spinors.
  14. Oct 14, 2016 #13


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    Could you go through a demonstration of how half-derivatives lead to spinors? I've never heard that before.

    A terminology nitpick: the phrase "geometric algebra" as used by Hestenes is basically Clifford algebra. The case with anticommuting variables is (I think) called Grassman algebra.
  15. Oct 14, 2016 #14
    It's confusing. That's why I told @Aakash Lakshmanan he needs to apply his quaternion knowledge to understand rank 1 spinors before going ahead to higher dimensions: the correspondence is not entirely clear.

    First, consider David Hestenes http://geocalc.clas.asu.edu/pdf-preAdobe8/VeSp&ComUpdated.pdf:

    But he also says "Some readers may want to say, rather, that we have here two isomorphic algebras, but there is no call for any such distinction. ... The identification of quaternions with spinors is fully justified not only because they have equivalent algebraic properties, but more important, because they have the same geometric significance." So ... are they the same or not? Many other references, for instance https://en.wikipedia.org/wiki/Spinor#Three_dimensions and http://math.gcsu.edu/~jeb/Math2263Files/HamiltonsQuaternions.pdf, also say they are. But there may be caveats.

    Being mathematics, we know exactly what quaternions are. But spinors are physics, therefore there are conflicting views. (At least that's how it appears to a mathematician :-)

    Similar confusion happens with the terms Clifford, Grassman, Geometric algebras. From https://www.physicsforums.com/insights/interview-mathematician-david-hestenes/:

    Reading between the lines, we see other physicists prefer different terminology, but they simply don't understand the subject as well as he does. He also includes Grassman Algebra in his GA, I think.

    Terminology can lead to endless debates so let's not worry about it too much. Still the question remains, what is the relation between quaternions and spinors?

    Hamilton defines a quaternion as 4 real numbers with their associated axes, 1, I, j, and k: a + bi + cj + dk. It's an extension of the concept of complex numbers, with the multiplication rules i^2 = j^2 = k^2 = -1. One more rule is needed, ijk = -1, to complete quaternionic algebra definition. So a quaternion is a real 4-vector with special multiplication rules.

    But it can also be represented by a pair of complex numbers: (a + bi) + j*(c + di). Since a spinor is a pair of complex numbers, this quaternionic representation may be a spinor. See below.

    Quaternions are also representable by 2x2 matrices of complex numbers - as you (@stevendaryl) mention. That's 8 real numbers but pairs of them are equivalent, so there are only 4 reals (as required). The Pauli matrices are such matrices, forming a basis. Those matrices are, however, not called spinors. Instead, the vectors they operate on (usually considered column matrices, multiplying to the right) are rank-1 spinors (most physicists seem to agree, I think). Being a pair of complex numbers they're formally equivalent to the above-mentioned quaternion representation as two complex numbers. Lubos Motl said they are indeed the same (can't find the ref), but it's not clear to me.

    Bottom line, spinors are (almost, anyway) quaternions. If you understand quaternions then you already know spinor math. Unfortunately the exact connection is confusing.

    It's well known that spinors can be thought of as "the square root of a vector". Quaternions exhibit that ansatz clearly. I don't doubt the same idea can be expressed with half-derivatives but don't see how at the moment.
  16. Oct 18, 2016 #15
    Let's have 3-dimensional Euclidean space. Let's define 3 scalar fields x, y and z that are linear in one dimension and constant in other dimensions.

    We can define 3 "spinor differentials" that are anticommutative. They are elements of Grassman algebra. Let's call them sx, sy and sz.
    Let s be the half-differential operator. We can postulate its linearity, anticommutativity and the differential product rule:

    s(a F) = a sF
    s(F + G) = sF + sG
    s(F G) = sF G + f sG
    sF sG = -sG sF

    F and G are fields, a is a scalar. Note that this is noncommutative algebra, so order of multiplication is important, especially in the product rule. The unit spinors sx, sy and sz are half-differentials of the scalar fields x, y and z respectively.

    Also note this algebra is not quaternions. In quaternions you have i2 = -1, here we have dx2 = 0. We have derived geometric algebra.

    A general spinor is a field of a form:

    S = a sx + b sy + c sz

    Of course a, b and c may depend on x, y and z. Let's name their half-differentials too.

    sa = ax sx + ay sy + az sz
    sb = bx sx + by sy + bz sz
    sc = cx sx + cy sy + cz sz

    Now let's take the half-differential once again.

    sS = s(a sx) + s(b sy) + s(c sz) = sa sx + a ssx + sb sy + b ssy + sc sz + c ssz =
    = (ax sx + ay sy + az sz) sx +
    + a ssx +
    + (bx sx + by sy + bz sz) sy +
    + b ssy +
    + (cx sx + cy sy + cz sz) sz +
    + c ssz =
    = a ssx + b ssy + c ssz +
    + 0 + ay sy sx + az sz sx +
    + bx sx sy + 0 + bz sz sy +
    + cx sx sz + cy sy sz + 0 =
    = a ssx + b ssy + c ssz +
    + (bx - ay) sx sy + (cy - bz) sy sz + (az - cx) sz sx

    Note that we have not said anything about the product of two spinors. The only thing we know that a quare of a spinor is always zero and multiplication is antisymmetrical. In particular, it is not true that sx sy = sz . So this is not a quaternion algebra. In fact, the product of two spinors is another object we may call a pseudovector.

    So the double half-derivative of a scalar field gave us another object. The first part, the sum
    V = a ssx + b ssy + c ssz
    is a familar covariant vector.
    The second part
    P = (bx - ay) sx sy + (cy - bz) sy sz + (az - cx) sz sx
    we may call a general pseudovector.

    We may now define the differential operator dF = s2F = s(sF).
    The differentials dx = ssx, dy = ssy, dz = ssz are unit vectors. They are commutative and they behave as vectors should do.

    This construction can be generalized to any number of dimensions, except pseudovectors have 3 components only in 3d space. Number of components of pseudovectors is always (d2 - d)/2. It is an important coincidence that pseudovectors and covariant vectors have the same number of components in 3d. This is also the reason why quaternion algebra describes some properties of 3d space.
    A different coincidence occurs in 7d space, where pseudovectors have 21 = 7 * 3 components. The related algebra describing some aspects of rotations in 7d space is octonions.

    I personally believe, but this is only my gut feeling, that this vector plus pseudovector result has something to do with the V-A weak interaction theory and the CP symmetry breaking. But this is only my personal opinion.
  17. Oct 18, 2016 #16
    Ooohhh I really like this formulation. This is basically just differential forms but replacing d with s. Because before applying d to 1-form yielded the curl and some d^2 components which were regarded to be 0 but in this formulation they have meaning. On the other hand, the pseudo vector forms this weird version of a curl 2-form. So the first part is simply a vector but the second is a bit confusing as to meaning and geometric interpretation. Could you elaborate a little on it?
  18. Oct 18, 2016 #17
    I sense that you might want to start with a very simplistic approach to grasping the general concept of a spinor. Consider how you might mathematically describe physical things that vary in complexity.

    First, what can be a mathematical description of something quite simple, the current temperature of one point in a room? Here one number would suffice, and that number—that mathematical description—would be a scalar. It tells “how much right here and now.”

    Next what about a boat moving in one direction along the surface of a calm lake? Here one number would not suffice—a scalar would not do—but two numbers are enough, one for direction and another foe speed. We call that set of numbers, obviously useful in navigation, a vector. It tells “how much which way.” In airplane navigation things are a bit more complicated (direction includes up and down), but the description is still a vector.

    But what if we tried to describe something even more complicated, such as a point in curved space-time? Of course that’s done in general relativity, and the set of numbers required in this case forms a tensor. It tells “how much which way where space and time are curved.”

    Finally, what if things are even more complicated, like various subatomic particles (which may have spin or charge) in space-time? This description arises in Dirac’s approach to quantum mechanics, and it is called a spinor.

    I stress that this a very watered-down informal approach, but the point should be clear: We categorize mathematical descriptions in physics according to complexity, and spinors describe things that are so complicated that scalars and vectors will not do.
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