# Can a spinor be defined as any quantity which transforms linearly under Lorentz transformations?

Recently I’ve come across a few papers from China (e.g. Xiang-Yao Wu et al., arXiv:1212.4028v1 14 Dec 2012) that make the following statement:

...any quantity which transforms linearly under Lorentz transformations is a spinor.

It’s my understanding that e.g. a 4-momentum vector also transforms linearly under a Lorentz transformation.

Is the first statement simply false, or should one take it to be true in the sense that a 4-vector is capable of being written in spinor notation?

Perhaps the first statement might be a confusion between Lorentz transformations and spin matrices? In the chapter on spinors in Misner, Thorne and Wheeler’s Gravitation (p. 1148) they show that while a vector transforms under a spin matrix (aka rotation operator / quaternion / spinor transformation) as:

$$X \to X' = RXR^*,$$

a quantity that transforms as

$$ξ \to Rξ'$$

is known as a spinor.

UPDATE

After further web-searching, I’ve come across references making statements that seem to throw some light on the issue

1. Andrew Steane’s recent and very readable (to such tyros as myself) “An introduction to spinors” (http://arxiv.org/abs/1312.3824 13 Dec 2013), in which he writes (p.1, 2nd para): “… One could say that a spinor is the most basic sort of mathematical object that can be Lorentz-transformed.” (But see (3), here below).

2. I’ve also now traced back the original quotation, repeated word-for-word, through a number of earlier papers (Chinese and Russian) to: V. V. Varlamov arXiv:math-ph/0310051v1 (2003), in which he cites – as do all the later papers – one of the earliest writers on spinors, B. L. van der Waerden, Nachr. d. Ces. d. Wiss. Gottingen, 100 (1929). Varlamov also wrote a densely mathematical and well-referenced paper “Clifford Algebras and Lorentz Group” (math-ph/0108022, 2001), which inclines me to give more credence to the original statement, even though it was parroted by a number of later authors.

3. However, it appears that Dirac himself suggested an even more general entity than the spinor: “A new kind of quantity with components which transform linearly under Lorentz transformations must be introduced, and I call it an expansor. It is rather more general than a tensor or a spinor in that the number of its components is infinite, but enumerable.” P. 1205, section 1946:1 DEVELOPMENTS IN QUANTUM ELECTRODYNAMICS (p. 21 of the section). The Collected Works of P. A. M. Dirac: 1924-48: 1924-1948 By (author) P. A. M. Dirac, Volume editor Richard Henry Dalitz. Cambridge University Press, 26 Oct 1995 - Science - 1310 pages.

Further, on p. 1163 of the above, Dirac notes that “the present theory of expansors applies, of course, only to integral spins, but probably it will be possible to set up a corresponding theory of two-valued representations of the Lorentz group, which will apply to half odd integral spins.”

These two-valued entities were subsequently supplied by Harish-Chandra, who called them “expinors” (“Infinite Irreducible Representations of the Lorentz Group”, Published 1 May 1947 doi:10.1098/rspa.1947.0047Proc. R. Soc. Lond. A 1 May 1947 vol. 189 no. 1018 372-401).

I haven’t access to van der Waerden’s classic, so cannot check whether he made the claim referenced by Varlamov (but have little reason to doubt him).

So what can I conclude?

The subsequent work by Dirac and Harish-Chandra seems to invalidate the van der Waerden claim, as later cited by Varlamov, at least for infinite dimensional IRs of the Lorentz group.

So perhaps the queried definition does apply fairly generally, but not universally?

If so, it would be good to have an expert clarify the distinction.

PS Gratitude to Qmechanics for tidying up the original posting.

• If you want to update the question, click the "edit" button right under it, rather than editing the answers. Keep in mind that often this isn't the best thing to do anyway. If you are expanding the scope and essentially asking a new question, just start a new question with a link back to this one if appropriate. If you are answering your own question, you should post it as a separate answer to this question (the answers need not be merged into one). – user10851 Feb 7 '14 at 19:15
• Thanks, Chris: As I'm adding more information and thereby refining the question, I'm neither answering the original question nor asking a new one - so I think it is still appropriate to use the edit button (I just used the wrong one first time). – iSeeker Feb 7 '14 at 20:10

In many contexts, we would like to determine how Lorentz transformations act on the mathematical objects that characterize a particular theory. In the case of classical, Lorentz-invariant field theories on Minkowski space for example, we need to specify how Lorentz transformations act on the fields of the theory. This leads naturally to determining how Lorentz transformations can act on both Minkowski space, and the target spaces of the fields. This, in turn, leads naturally to the notion of a finite-dimensional representation of the Lorentz group.

On the other hand, in quantum mechanics, and by extension in quantum field theory, we often want to specify how Lorentz transformations act on the Hilbert space of the theory. In this context, Wigner's theorem in symmetries in quantum mechanics demands that up to phase, Lorentz transformations as unitary or anti-unitary operators on the Hilbert space. In turn, the fact that these transformations are only defined up to phase implies that one, in general, needs to consider the projective representations of the Lorentz group in addition to its "ordinary" representations.

Now, it turns out that determining the projective representations of the Lorentz group $\mathrm{SO}(3,1)^+$ is equivalent to determining the ordinary representations of its universal cover, which is called $\mathrm{Spin}(3,1)$! It's a spin group! In fact, for any $p,q$, the group $\mathrm{SO}(p,q)^+$ of isometries of the space $\mathbb R^{p,q}$ has universal cover $\mathrm{Spin}(p,q)$.

Therefore,

determining all projective representations of the Lorentz group is equivalent to determining the ordinary representations of the corresponding spin group.

It is overwhelmingly likely, in my opinion, that this is what the authors are referring to in their quote because the "objects that transform linearly under Lorentz transformations" that we consider in physics are precisely those objects that transform under projective representations of the Lorentz group (ordinary representations are included as a subclass), and these are precisely those objects that transform under ordinary representations of spin groups, and such objects are called spinors.

By the way, you'll probably find the following related post illuminating.

https://physics.stackexchange.com/a/96060/19976

It's hard to know the intentions of the authors for certain, but here's some information that might help interpret what they're saying.

Recall that every representation of the Lorentz algebra $\mathfrak{so}(3,1)$ can be constructed from representations of $\mathrm{sl}(2,\mathbb C)$, the complexified angular momentum algebra (which is of course the algebra whose representations describe spin). The standard way of doing this is to note that if one complexifies the Lorentz algebra, when one finds that the complexification yields a direct sum of $\mathfrak{sl}(2,\mathbb C)$ with itself; \begin{align} \mathfrak{so}(3,1)_\mathbb C \approx \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C). \end{align} It follows that the representation theory of the Lorentz algebra reduces to the representation theory of the angular momentum algebra. In fact, every irreducible representation of the Lorentz algebra is essentially a tensor product of two irreducible representation of the angular momentum algebra, and these representations are often labeled by a pair $(s_1, s_1)$ of "spins" $s_1, s_1\in\{0,\frac{1}{2}, 1, \dots\}$. For example, the $(\frac{1}{2},0)$ representation is called the left-handed Weyl spinor representation, and the $(0,\frac{1}{2})$ is called the right-handed Weyl spinor representation.

The vector representation that you refer to, namely that standard representation that transforms a four-vector by a Lorentz transformation $\Lambda\in \mathrm{SO}(3,1)^+$ by mapping $V^\mu$ to $\Lambda^\mu_{\phantom\mu\nu} V^\nu$, corresponds to the $(\frac{1}{2}, \frac{1}{2})$ representation. So it is in a sense true as you say that "a 4-vector is capable of being written in spinor notation."

It's also the case that any finite-dimensional representation of the Lorentz algebra can be written as a direct sum of the irreducible representations $(s_1, s_2)$, so in a very real sense, all of the finite-dimensional objects that Lorentz-transform can be "built" out of "spin" representations, namely representations of the angular momentum algebra.

• Thanks, Josh - it's comforting that someone who's clearly more expert than myself sees no simple answer either. – iSeeker Feb 6 '14 at 23:43
• @user40110 Sure thing. Let me know if you determine something more convincing. Hopefully user V. Moretti will see this; I think he'd have something intelligent to say as well. – joshphysics Feb 7 '14 at 0:08
• Josh - Perhaps my update refines the question a bit? – iSeeker Feb 7 '14 at 20:13
• @user40110 I updated the answer; not sure why I didn't think of it before. – joshphysics Feb 7 '14 at 23:42
• Many thanks for your more complete answer and related post. I can see broadly what you're saying, but will have to chew on them for while. (I'm coming at the subject as a chemist fascinated by spinors but having to teach myself the necessary math background.) – iSeeker Feb 8 '14 at 13:40

This answer is coming several years late, but I think the intended meaning is actually quite simple. Start with the transformation of a Lorentz vector, $$V^\mu \to \Lambda^\mu_\nu V^\nu.$$ Then the transformation law of a rank $2$ tensor can be built by product, $$T^{\mu_1\mu_2} \to \Lambda^{\mu_1}_{\nu_1} \Lambda^{\mu_2}_{\nu_2} T^{\nu_1 \nu_2}$$ with similar expressions for tensors of arbitrary rank. Clearly, this transformation is perfectly linear in $T$, as it must be by the definition of a representation.

However, if you regard the matrix $\Lambda$ as a fundamental thing that specifies a Lorentz transformation, then you might say that the transformation of a rank $2$ tensor is quadratic, i.e. quadratic in $\Lambda$. More precisely, all the tensor representations can be built up by tensor products of the vector representation, so they are "higher order".

Now the quoted statement is just taking this one level further. When we consider projective representations of the Lorentz group, there are even smaller, more "elementary" representations which are the right-handed and left-handed Weyl spinors, which transform as $$\psi_\alpha \to M^\beta_\alpha \psi_\beta, \quad \bar{\chi}_{\dot{\alpha}} \to (M^*)^{\dot{\beta}}_{\dot{\alpha}} \,\bar{\chi}_{\dot{\beta}}.$$ Just as the rank $2$ tensor can be built from tensoring two Lorentz vectors, the Lorentz vector can be built by tensoring two spinors. Therefore we might say that the spinors are "linear" and the vector is "quadratic" (in $M$). Indeed you can prove that all projective representations of the Lorentz group can be found by tensoring the spinors.

The real issue here is the language barrier. It's not that people are blindly parroting each other, any more than you are parroting your textbook when you say "all representations are linear". It's just that they are using a word in a different way than you expect.

• This is yet another instance of the ambiguities of natural languages in "interaction" with formal languages; a frequent source of confusion for the amateur mathematician. I've seen it quite a lot in representation theory too. – iSeeker May 21 '18 at 16:03
• @iSeeker Yeah, I could write a good thousand words on how representation theory words differ between math and physics. For example, the words "real", "fundamental" and even "representation" itself... – knzhou May 21 '18 at 16:20