So, in general (if I understand this correctly):

  • Force particles behave differently than matter particles under rotation
  • The matter particles need a 720° rotation to put them back into their initial state (the force ones only 360°)
  • Certain objects called Spinors can be described as a columns matrix of complex numbers that tells you how matter particles (like fermions) transform

My question is can someone help me get a basic intuitive (yet still somewhat rig.) mathematical description of how the 360° rotation occures for force particles (I guess using Gauge Fields?) and then the basic math for Spinor rotations (and it's Spin Group "upper-level abstraction" if you will).

A followup question would be:

  • Does there exist a formal meaure for symmetry comparing the two different mathematical rotations, in that for example the Spinor one would be "less effective" in terms of additional rotations.

I'm sorry if this question is stupid, I do not have a physics background.

  • 1
    $\begingroup$ I'm not exactly sure what you want here as an answer that's not "you need to read several textbooks on quantum mechanics/QFT to really understand what's going on", but do any of physics.stackexchange.com/a/677753/50583, physics.stackexchange.com/a/167470/50583, physics.stackexchange.com/a/322017/50583 help? These are answers of mine that are about how the rotation is "implemented" $\endgroup$
    – ACuriousMind
    May 4 at 12:09
  • $\begingroup$ If the question is "why are fermions spinors and (gauge) bosons not?", then the answer to that is the spin-statistics theorem, for which we have an entire tag spin-statistics. $\endgroup$
    – ACuriousMind
    May 4 at 12:09
  • $\begingroup$ Thank you very much for pointing me to those threads, this is exactly what I was looking for as a sort-of jumping up point for further reading. "why are fermions spinors and (gauge) bosons not?" This is somewhat of a specific binary of the last question, where I was basically asking if symmetry has been formalized in a way. $\endgroup$
    – blazg_7S
    May 4 at 12:17
  • 2
    $\begingroup$ "Has symmetry been formalized ?" I guess a starting point would be the theory of representations of Lie groups and Lie algebras, and in particular of the rotation group $SO(3)$ $\endgroup$ May 4 at 12:37
  • $\begingroup$ Google "An introduction to spinors Andrew M. Steane" and this may give you an idea why spinors are needed. $\endgroup$
    – user226006
    May 4 at 15:40

1 Answer 1


Consider a point in space, and the effect a rotation would have on that point. Because the point has no internal structure (in other words, it is characterized entirely by its spatial location), a rotation of the universe would simply move that point from one location to another.

From there, imagine a single real number (i.e. a scalar) attached to that point in space. A real number has no internal structure either, so a rotation of the universe would simply move that number to a new location.

Next imagine an arrow (i.e. a vector) attached to that point in space. If we perform a rotation, the arrow moves to a different spatial location just as before - but in addition to that, the arrow itself is rotated. Unlike a point, an arrow is characterized not only by its spatial location and magnitude but also by the direction in which it points, and so a rotation will affect both its location (its "external" state) and its direction (its "internal" state).

Finally, imagine a pair of complex numbers $\pmatrix{\alpha,\beta}$ (i.e. a 2-spinor) attached to a point in space. If we perform a rotation, then the pair will of course be moved to a different spatial location as before, but now what will happen to its internal state? How do we implement a rotation on a pair of complex numbers?

To answer this, we must dive into what's called the representation theory of the rotation group. In essence, this subject is concerned with how we might take a collection of objects which possess a group structure and implement them as transformations on some vector space ($\mathbb R$ for scalars, $\mathbb R^3$ for vectors, $\mathbb C^2$ for 2-spinors, etc) in a way which is compatible with that group structure. A representation is called faithful if different elements of the group correspond to different transformations, and irreducible if there are no non-trivial subspaces of the vector space which are left invariant by every transformation (an example of the latter would be an implementation of rotations on $\mathbb R^4$ in which we left the first number alone and then rotated the other three just like we would rotate a vector in $\mathbb R^3$). In this language, our question becomes the following: Can we find a faithful, irreducible representation of the 3D rotation group $\mathrm{SO}(3)$ which acts on the vector space $\mathbb C^2$?

It turns out that there is no true, non-trivial representation of $\mathrm{SO}(3)$ which acts on $\mathbb C^2$. However, there does exist a projective representation of $\mathrm{SO}(3)$ on $\mathbb C^2$, which implements the elements of $\mathrm{SO}(3)$ as transformations on $\mathbb C^2$ in a way which respects the group structure possibly up to multiplication by a phase factor. For technical reasons, it turns out that this is exactly what we want in quantum mechanics, and that restricting our attention to true representations of the group would be too restrictive.

As an example, imagine that we perform a sequence of small rotations which add up to $360^\circ$. A true representation respects the structure of the rotation group insofar as a sequence of small rotations adding up to $2\pi$ is the same as performing a single $2\pi$ rotation all at once; however, a $2\pi$ rotation is exactly the same as no rotation at all (they correspond to the same element of $\mathrm{SO}(3)$) and so such a sequence returns our system to its original state.

On the other hand, in a projective representation this is no longer true; a sequence of small rotations which add up to some angle $\theta$ is not equivalent to a single rotation by $\theta$, but rather a rotation by $\theta$ and multiplication by a phase factor. To be concrete, if we understand our elements of $\mathbb C^2$ to correspond to the intrinsic spin of a particle as expressed in the $\hat z$-basis and we perform a rotation by an angle $\theta\in[0,2\pi)$ around the $\hat z$-axis, we obtain $$\pmatrix{\alpha\\\beta} \mapsto e^{i\theta/2}\pmatrix{1&0\\ 0& e^{-i\theta}}\pmatrix{\alpha\\ \beta}$$

Note that $\pmatrix{\alpha\\\beta}\mapsto \pmatrix{1&0\\0&e^{-i\theta}}\pmatrix{\alpha\\\beta}$ would correspond to a true representation, in which gradually increasing $\theta$ from $0$ to $2\pi$ would bring us back to where we started. However, this representation is not irreducible, because it leaves the first component alone (i.e. the subspace $(\alpha,0)$ is invariant). The addition of the phase factor $e^{i\theta/2}$ turns the representation from a reducible true representation into an irreducible projective one. As a result, we observe that increasing $\theta$ from $0$ to $2\pi$ brings us back to our original vector multiplied by the phase factor $e^{i(2\pi)/2} = e^{i\pi}=-1$.

The take-away is that if we want to implement rotations on 2-spinors in a non-trivial way, we must resort to a projective representation in which e.g. performing $N$ rotations of an angle $\theta/N$ each is not the same as a rotation by an angle $\theta$; it will generally come with an additional phase factor which in this case is $-1$ for $\theta=2\pi$.

It turns out more generally that a representation (or projective representation) of $\mathrm{SO}(3)$ can be classified by a number $J$ which is either an integer or a half-integer, and which is related to the total angular momentum of the particle under consideration. When $J$ is an integer, then the representation turns out to be a true representation and so in particular a $2\pi$ rotation is equivalent to no rotation at all. When $J$ is a half-integer, then the representation turns out to be projective, and in particular a $2\pi$ rotation about any axis is equivalent to multiplication by $-1$.

  • $\begingroup$ What a lovely answer. Thank you. $\endgroup$
    – user226006
    May 4 at 18:03
  • $\begingroup$ Thank you very much for a lucid answer, now I have more than enough material before I post here again. $\endgroup$
    – blazg_7S
    May 5 at 10:24

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