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I have a question about an intuitive approach on spinors as certain mathematical objects which have certain properties that make them similar to vectors but on the other hand there is a property which differ spinors from vectors:

Wiki gives a rather geometrical description of a spinor:

"Unlike vectors and tensors, a spinor transforms to its negative when the space is continuously rotated through a complete turn from $0°$ to $360°$ (see picture)."

Other sources state moreover that if you rotate a spinor by $720°$ degrees you obtain the same spinor. Clearly, if we rotate a usual vector by $360°$ we obtain the same vector. So spinors are not vectors in usual sense.

QUESTION: What I not understand is what is precisely a 'rotation of a spinor'. How this kind of 'rotation' can be described?

I know that the question sounds banally, but if we recall what is a rotation in common naive sense we think of a rotation in a very concrete framework: the naive rotation is an operation by an element from group $SO(3)$ on real space $\mathbb{R}^3$. Since spinors live not in $\mathbb{R}^3$ I think it's neccessary to specify precisely what is a 'roation' in the space where spinors live.

Lets draw analogy to usual vectors & $3D$ space. A usual rotation in $3D$ is determined by rotation axis $\vec{b}$ and rotation angle $\phi$. Say wlog we rotate around $z$-axis by angle $\phi$, then the rotation is decoded by $3 \times 3$ matrix $R_{\phi} \in SO(3)$

$$R_{\phi}= \begin{pmatrix}cos(\phi)&-sin(\phi)&0\\sin(\phi)&cos(\phi)&0\\0&0&1\end{pmatrix} $$

That is if $\vec{v} \in \mathbb{R}^3$ then the rotation of $\vec{v}$ is simply $R \vec{v}$.

But what is a 'rotation of a spinor' concretely? How is it described?

For sake of simplicity lets focus on the most common spinor representation from particle physics: The subgroup $SU(2) \subset SL(2, \mathbb{C})$ provides a simply connected $2$ to $1$ covering map $f:SU(2) \to SO(3)$ of rotation group. Clearly $SU(2)$ acts as a subgroup of $SL(2, \mathbb{C})$ on complex vector space $\mathbb{C}^2$. Since in this setting $SU(2)$ provides a spinor representation we can call certain vectors of $\mathbb{C}^2$ 'spinors', right?

But what is a rotation of spinors here? Say we take an arbitrary spinor $s \in \mathbb{C}^2$ and want perform a 'rotation' around certain axis by certain fixed degree $\phi$. Which object in $SU(2)$ represents this so called 'rotation' and why such operation on spinors is called 'rotation'?

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A rotation of a spinor $\psi$ (looks like a complex 2-vector) by an angle $\phi$ around the unit axis $\hat n$ is but $$ \psi \mapsto e^{i {\phi\over 2} \left(\hat{n} \cdot \vec{\sigma}\right)} \psi= \left (I\cos {\phi\over 2} + i (\hat{n} \cdot \vec{\sigma}) \sin {\phi\over 2}\right ) \psi , $$ where $\vec \sigma$ are the three Pauli matrices, twice the generators of rotations in the doublet representation.

You can see how a 2π rotation amounts to flipping its sign, and twice that amounts to the identity.

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  • $\begingroup$ Could you explain sketchy or give a reference where it is explaned rigorously why this transformation is exactly what we are looking for, ie a reasonable 'generalization' of 'naive' rotation? $\endgroup$ – katalaveino Sep 29 at 0:58
  • $\begingroup$ The sketch is that half the Pauli matrices obey the Lie Algebra of the rotation group, so they represent it in two dimensional complex vector spaces. $\endgroup$ – Cosmas Zachos Sep 29 at 2:43
  • $\begingroup$ wp $\endgroup$ – Cosmas Zachos Sep 29 at 2:49
  • $\begingroup$ Let me try to rephrase what I understood. We know that $𝖘𝖔(3)$ and $𝖘𝖚(2)$ are isomorphic as Lie-Algebras and rotations $SO(3)$ are nothing but images $e^A$ for $A \in 𝖘𝖔(3)$ by the exponential map. That is a rotation is by definition $e^{𝖘𝖔(3)}$. And since $𝖘𝖔(3) \cong 𝖘𝖚(2)$ it's reasonable to call the images $e^{𝖘𝖚(2)}$ 'rotations' as well, that's the idea, right? More generally, if $G$ is a Lie-group with Lie-algebra $\mathfrak{g} \subset \mathfrak{so}(3)$, then we call as far I understood you correctly $\endgroup$ – katalaveino Sep 29 at 3:13
  • $\begingroup$ the images $e^{\mathfrak{g}} \subset G$ also rotations. That's the idea behind it, right? $\endgroup$ – katalaveino Sep 29 at 3:13
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The Pauli and Dirac matrices are basis vectors of Clifford algebras of 3d Euclidean space and 3+1d Minkowski space respectively. If you want to understand spinors, you'll probably need to understand Clifford algebras.

In Clifford algebras, reflections through the origin are represented by unit vectors (think of them as the surface normals of mirrors). The algebraic product composes reflections. Vectors can be written as weighted sums of basis vectors just as in the underlying vector space. In the Pauli/Dirac matrix representation, the Pauli/Dirac matrices are the basis vectors ($\hat t$), $\hat x$, $\hat y$, $\hat z$.

Any rotation can be written as a product of an even number of reflections. In 3d Euclidean space the Clifford products of even numbers of unit vectors live in a subspace of the algebra that's isomorphic to the unit quaternions. In 3+1d Minkowski space the subspace is isomorphic to the unit biquaternions.

To reflect a vector in a mirror, you multiply it on both sides by the Clifford representation of the surface normal (and possibly a factor of $-1$). You can convince yourself from the reflection interpretation of the algebra that this makes sense. To rotate a vector, you conjugate it by the appropriate even product, with the inverse being the same product in reverse order.

Spinors transform by multiplication by the same representations of reflections/rotations, but on only one side, not both.

I think a general geometric understanding of spinors is an open problem. However, at least in low dimensions (probably including 3+1), it's possible to think of the Clifford representation of a spinor as itself a rotation, from a "canonical spinor orientation" to the actual orientation. Rotating a spinor therefore means composing its representation with another rotation.

The essential reason that it takes a 720° rotation to get back to the original orientation is that reflection through two mirrors an angle $θ$ apart rotates an object by $2θ$. When you rotate a mirror through 180°, the plane of the mirror returns to its original position, but the surface normal points in the opposite direction, and the representation of the rotation as a product of vectors has therefore picked up a factor of $-1$.

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  • $\begingroup$ I not completely understand your explanation on geom undrstanding of spinors in low dimensions. You wrote: ' it's possible to think of the Clifford representation of a spinor as itself a rotation, from a "canonical spinor orientation" to the actual orientation. Rotating a spinor therefore means composing its representation with another rotation.' What precisely do you mean by a rotation from a "canonical spinor orientation" to the actual orientation? I see here the same problem as above: we have to make precise what is a 'rotation' as far as we are not working inside $\mathbb{R}^3$. $\endgroup$ – katalaveino Sep 29 at 2:19
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Since you're asking about intuition my answer will be in that direction. You shouldn't mentally associate rotations with vectors on $\mathbb{R}^3$. Instead, you should associate them with the group $SO(3)$. Moreover, you should think of $SO(3)$ as an abstract group, not as a set of $3\times 3$ matrices. The set of 3x3 matrices is rather merely a representation of the group on 3 dimensional real space. This happens to be useful in classical physics where directions in space have 3 real degrees of freedom.

But apparently there are things in this world that have degrees of freedom that you may not be used to in the classical world, but are nonetheless very physical; such as the spin degree of freedom of an electron. These are described by a spinor (an element of $\mathbb{C}^2$) and not a vector in $\mathbb{R}^3$.

Now in physics we love linearisation and taylor expansions; they give you linear things and these are easy to work with. So instead of talking about translations and rotations we like to think about their first derivatives in a taylor expansion; namely linear velocities and angular velocities (which are proportional to the momenta and angular momenta). You may recognise angular velocities as (pseudo)vectors $\omega_i$, but it's in fact equivalent and sometimes more convenient to express angular velocities as 3x3 antisymmetric matrices $W_{ij} = \epsilon_{ijk}\omega_k$. You may know what happens then when you exponentiate such an antisymmetric matrix: you get an $SO(3)$ matrix that makes up a finite rotation. More precisely in classical mechanics, you would write (with the poisson bracket and $L_z$ the angular momentum in the z direction): $R_\phi = e^{\{\phi L_z, \cdot\}}$

In Quantum Mechanics, for spin degrees of freedom, you know the angular momentum operator in the z direction is $\mathbf{\frac{1}{2}}\sigma_z$. Notice the half. The 2x2 complex rotation matrix by angle $\phi$ around the z-axis is then just $R_\phi = e^{i\phi\frac{1}{2}\sigma_z}$. That half is what eventually leads to the half in the half angle argument and the $720^\circ$ trick. Physically, this is just what happens to these $\mathbb{C}^2$ degrees of freedom under rotation.

Personally, I don't really find it very surprising. The degrees of freedom of the spin of an electron just happen to transform someway under rotations. The angular velocity three degrees of freedom represented as a vector transform funnily under reflections, but there's nothing deep about it. The mass or temperature of an object for example don't transform at all under rotations, not surprising. The electron spin degrees of freedom transform funnily under rotations, why should that then be surprising.

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  • $\begingroup$ so if I understand you correctly then a 'rotation' from abstract viewpoint is the image of a linear combination $\sum \phi_i L_i$ of angular momentum operators $L_i$ by exponential map. What a angular momentum opertor or more precisely a basis set of am ops is, can be defined axiomatically eg here: en.wikipedia.org/wiki/… $\endgroup$ – katalaveino Sep 29 at 18:28
  • $\begingroup$ or in other words: the angular momentum operators form a canonical basis of corresponding Lie-algebra? $\endgroup$ – katalaveino Sep 29 at 19:31

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