Null vectors and spinors

My recent reading on the topic of spin and angular momentum in quantum mechanics led me to the concept of a spinor. It turns out that spinors are fearsomely nasty objects to wrap one’s head around in full generality, requiring Clifford algebras and other ingredients, although the three-dimensional case is quite accessible and is described below. The (slightly unhelpful) heuristic is that spinors behave like vectors except that they change sign under rotation through an angle of 2π — a somewhat confusing property that will be made clearer later.

First, some general notions: let V be a vector space over a field K, equipped with a bilinear map b: V × VK and hence a quadratic form q: VK given by q(v) ≔ b(v, v) for all v ∈ V. A vector vV is called a null vector or isotropic if q(v) = 0. Recall that the standard Euclidean bilinear form on ℝn is

\displaystyle b \bigl( (x_{1}, \dots, x_{n}) , (x'_{1}, \dots, x'_{n}) \bigr) \equiv x \cdot x' := \sum_{j = 1}^{n} x_{j} x'_{j} .

This bilinear form has no non-trivial null vectors (i.e. the only null vector is the zero vector), but two close relatives of the Euclidean bilinear form do have interesting null vectors.

[Somehow, WordPress deleted large chunks of this post. Apologies! If the text below differs from what you saw earlier, then blame the post-deletion restoration effort.]

Example 1 (Hyperbolic Space). A first example of a space with non-trivial null vectors is ℝ1+n equipped with the Lorentz bilinear form

\displaystyle b_{\mathrm{L}} \bigl( (t, x_{1}, \dots, x_{n}) , (t', x'_{1}, \dots, x'_{n}) \bigr) := - t t' + \sum_{j = 1}^{n} x_{j} x'_{j} ,


\displaystyle b_{\mathrm{L}} \bigl( (t, x) , (t', x') \bigr) := - t t' + x \cdot x' .

With the addition of a few factors of c (the speed of light), this bilinear form bL is the one used to describe positions (t, x) in spacetime in Einstein’s special theory of relativity; it is particularly useful in Minkowski diagrams, in which time t is usually drawn as the vertical axis and space x as the horizontal axis. A vector v = (t, x) ∈ ℝ1+n is called

  • time-like if bL(v, v) < 0;
  • light-like (or null) if bL(v, v) = 0;
  • space-like if bL(v, v) > 0.

As the name suggests, a light-like vector is one along which a ray of light will travel; the set of null vectors is the light cone through the origin of ℝ1+n. For a mere mortal like myself, who is constrained to travel at less than the speed of light, the vector joining any two points in spacetime on my worldline (i.e. when and where in ℝ1+n I have been) must be a time-like vector. On the other hand, when the difference between two spacetime locations is space-like, it really makes sense to say that they are in different places (spaces), rather than just being one place that has moved to the other at less than light speed (e.g. the Earth orbiting the Sun).

Example 2 (Simple Spinors). Another simple example of a vector space with non-trivial null vectors comes from applying the Euclidean bilinear form to complex n-space ℂn. Often, ℂn is equipped with the sesquilinear form

\displaystyle \bigl\langle (x_{1}, \dots, x_{n}) , (x'_{1}, \dots, x'_{n}) \bigr\rangle := \sum_{j = 1}^{n} \overline{x_{j}} x'_{j} ,

which defines an inner product on ℂn (with no non-trivial null vectors), and so its square root is a norm on ℂn. However, the standard Euclidean form on ℂn does have non-trivial null vectors. Indeed, write x ∈ ℂn in terms of real and imaginary parts as x = a + ib with a, b ∈ ℝn. Then

\displaystyle x \cdot x = (a + ib) \cdot (a + ib) = a \cdot a - b \cdot b + 2 i a \cdot b .

Thus, for example, if a and b are orthogonal unit vectors, then x = a + ib will be null. Indeed, x will be null if and only if its real and imaginary parts are orthogonal and have equal magnitude.

It is, then, perhaps not surprising that in the case n = 3 the set of all null vectors x in ℂ3 can be described by a “vector” z = (z1, z2) ∈ ℂ2. It is this representative z that is called a spinor, and the reason for the quotation marks around “vector” is that it does not obey the same transformation rules as vectors. The components of a null vector x and its spinor representation z are related by the following over-determined system of equations:

\displaystyle x_{1} = z_{1}^{2} - z_{2}^{2} ,
\displaystyle x_{2} = i (z_{1}^{2} + z_{2}^{2}) ,
\displaystyle x_{3} = - 2 z_{1} z_{2} ,
\displaystyle x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 0 .

Solutions to this system are given by

\displaystyle z_{1} = \pm \sqrt{\frac{x_{1} - i x_{2}}{2}} ,
\displaystyle z_{2} = \pm i \sqrt{\frac{x_{1} + i x_{2}}{2}} ,

with either of the two choices of sign (plus or minus) producing a solution. In this way, a spinor may be viewed as a null vector together with a choice of sign.

To see that a spinor does not transform in the same way as a vector, consider the effect of the rotation map Rθ that rotates each of the three components of a vector x by an angle θ ∈ ℝ about the origin in the complex plane ℂ, i.e.

\displaystyle R_{\theta} \colon x \mapsto e^{i \theta} x .

(Note that Rθ only maps null vectors to null vectors if θ is an integer multiple of π2.) What is the effect of Rθ on the spinor representation z of x? A quick application of the above formulae shows that

\displaystyle R_{\theta} \colon z \mapsto e^{i \theta / 2} z .

In particular, as claimed at the top of the page, the full rotation R2π takes x to itself but takes z to −z. (Mathematical pedants will notice that, because of the branch cut discontinuity in the complex square root, it is impossible to choose a sign consistently so that the spinor Rθz changes continuously as a function of arbitrary θ and x. However, the rotations do act unambiguously — by a fractional linear transformation — on the ratio z1:z2, and this suffices for our purposes.)


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s