Spinor Intuition

Spinors are a hard mathematical object to gain geometric intuition for—they take two full rotations (not just one) to “reset”.

We can “prove” to our intuition that such objects exist with a visual demonstration:¹

However, it’s tricky to just watch this video and come away with a decent intuition for how these kinds of two-rotation-reset objects work. I prefer to use another trick: “induction” from a similar situation in a “lower dimension”.

Rotating a Spinor Around the \(x\) Axis

For the purposes of this demonstration, we’ll always be rotating around the \(x\) axis. I think it’s helpful to imagine the \(x\) axis as coming directly out of (and going directly into) the screen, so our imagined objects will be tracing out circles on the screen. (Incidentally, this is the default axis setup of Blender, which is why it’s become the default way I imagine 3d axes now.)

The formula for rotating a spinor² around the \(x\) axis³ by \(\theta\) is:

\[R_x(\theta) = \begin{pmatrix} \cos \frac{\theta}{2} & i \sin \frac{\theta}{2} \\ i \sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{pmatrix}\]

When we plug in \(\theta = 2 \pi\), a full rotation, we get:

\[\begin{split} R_x(2 \pi) &= \begin{pmatrix} \cos \frac{2 \pi}{2} & i \sin \frac{2 \pi}{2} \\ i \sin \frac{2 \pi}{2} & \cos \frac{2 \pi}{2} \end{pmatrix} \\ &= \begin{pmatrix} \cos \pi & i \sin \pi \\ i \sin \pi & \cos \pi \end{pmatrix} \\ &= \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \\ &= - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= -I_2 \end{split}\]

Instead of the \(2 \times 2\) identity matrix \(I_2\), we find that \(R_x(2\pi)\) is actually its negative. If we do another full rotation…

\[\begin{split} R_x(4 \pi) &= \begin{pmatrix} \cos \frac{4 \pi}{2} & i \sin \frac{4 \pi}{2} \\ i \sin \frac{4 \pi}{2} & \cos \frac{4 \pi}{2} \end{pmatrix} \\ &= \begin{pmatrix} \cos 2 \pi & i \sin 2 \pi \\ i \sin 2 \pi & \cos 2 \pi \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= I_2 \end{split}\]

…then we get back the identity transformation, and our spinor is back to where it started.

This is confusing behavior that’s hard to make sense of intuitively.

Induction by Analogy

When trying to understand something, one of the most powerful tricks is also one of the most obvious—simplify the thing you’re trying to understand, and then try again. This simplification step is one of building a situation that’s both:

1. Simpler than the original situation you’re trying to understand
2. A valid analogy for the situation you’re trying to understand

I can’t really explain the “induction” step—it’s something that sometimes just happens for me, when I’m trying to think of an equivalent, but simpler, situation. Often, it feels like more of a sudden leap. Probably a lot of the useful work is in trying to do the simplification yourself (in which case, this blog post won’t actually be very helpful to you, even though the process helped me).

For example, let’s try to build intuition for “a point in 100-dimensional space”. It’s easy enough for me to imagine a point in 3d space—something like a fly buzzing around the room does the trick. Trying to directly add on 97 spatial dimensions doesn’t work for me at all though.

Instead, we can try adding non-spatial dimensions, and a smaller number of them? Instead of a fly, let’s think of a grain of salt floating in space. Maybe it’s pink in the color dimension, and 100% salty in the saltiness dimension, for a total of five dimensions. I can do that imagining pretty easily.

This kind of thinking works for me for a few more dimensions (e.g. maybe our salt particle is 70°F in the temperature dimension), but pretty soon I start running out of properties. There’s no intuitive model of a 100-dimensional space directly available to me. The way I think of a 100-dimensional space is mostly by induction of an analogy to a much lower-dimensional space, maybe somewhere in the range from 3d to 7d.

Back to Spinors

So, what kind of simpler situation (with “lower dimensions”) can help explain spinors? What helped me was imagining the difference in rotational transformations not between spinors and vectors, but between vectors and scalars.

If we rotate a vector in 3d space around the \(x\) axis, it takes a full rotation of \(2 \pi\) to “reset”. The rotation matrix for this transformation is:

\[R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}\]

When we plug in \(2 \pi\) as our rotation, we should expect to get the identity matrix \(I_3\) back.

\[\begin{split} R_x(2 \pi) &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos 2 \pi & - \sin 2 \pi \\ 0 & \sin 2 \pi & \cos 2 \pi \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ &= I_3 \end{split}\]

Just as a spinor requires two full rotations, a vector requires a single full rotation to reset. There is a simpler object that doesn’t require any specific amount of rotation at all to reset: a scalar.

The \(x\) axis rotation matrix for a scalar looks like:

\[\begin{split} R_x(\theta) &= \begin{pmatrix}1\end{pmatrix} \\ &= I_1 \end{split} \]

That is, no matter what we plug in for \(\theta\), we get the original scalar back. One way to imagine this is with a property like charge: if we take a point charge (imagine an electron) and rotate it around the \(x\) axis, it doesn’t matter how much we rotate it. Its charge will remain the same.

This realization that there are some objects (scalars) invariant under rotations, and some objects (vectors) that vary unless at a multiple of a full rotation, helped me realize that there could be some objects (spinors) that vary unless at a multiple of two full rotations.

It’s not “rotating around a full circle” that maintains an object as it was⁴—scalars are maintained through a rotation of even one degree. Instead, it’s the transformation matrix canceling out to the identity that does that work. Spinors just happen to require two rotations for that canceling.