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First, I shall say that I am familiar with the intuitive idea that a spinor is like a vector (or tensor) that only transforms "up to a sign" when acted on by the rotation group. I have even rotated a plate on my palm to explain this to my fiancee! I have also looked at spinors as mathematical objects, such as the 2d subspace of complex 3-space such that $X·X = 0$, and feel that I understand this well also.

I am confused by spinors in physics. Are they still vectors that are isotropic (inner product with themselves is 0)? In what vector space? Normally states are vectors in infinite dimensional spaces! Every attempt to find literature which pins down specifically what a spinor is (in physical terms) seems to assume that one is already well acquainted with the idea.

Take, for example, the Dirac equation. I can see that the solutions are four-component wavefunctions, which then splits into two parts. Is this a spinor? Why? What vector space do these solutions live in? I believe I've heard that the answer has something to do with representation theory, perhaps of the Poincare group? I am also familiar with the basics there, so don't hesitate to explain in terms of representations.

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There are two different notions of a "vector". First is the definition used by mathematicians according to which any element of a vector space is a vector. Another definition is what is often used by physicists according to which a vector is that which behaves as a vector under rotations. Spinors are of course vectors in mathematical sense however they are not vectors in physics sense. – user10001 Aug 20 '13 at 15:33
The short answer is that spinorial representations are special representations of $SO(p,q)$. To understand their construction ,Clifford algebras are the most direct way. The even part of the Clifford algebras,(roughly) corresponds to spinorial representations.( A little introduction in this article, p. $1$ to $5$). Spinors are objects upon which spinor representations (which are matrix) act. – Trimok Aug 20 '13 at 15:59
Related: – Qmechanic Aug 20 '13 at 16:24
Let me elaborate on Trimok's comment. If you have a Lie group $G$ and you want to find the projective representations of it, you can go and find linear representations of its universal covering group $\tilde G$. Here $G$ and $\tilde G$ are locally diffeomorphic (isomorphic Lie Algebras), but globally distinct ($G = \tilde G/\pi_1(G)$). In the case of $G=SO(p,q)$, the projective representations which are NOT linear (so they are strictly projective) are called spinor representations! This is what Trimok calls special representations of $SO(p,q)$. – Heidar Aug 20 '13 at 21:56
The universal cover of $G=SO(p,q)$ is called $\tilde G=\text{spin}(p,q)$. These spin-groups can be constructed using Clifford algebras and they are therefore useful in constructing spinor (projective) representations of $SO(p,q)$. As an simple example take $G=SO(3)=SU(2)/\mathbb Z_2$, it has only integer spin representations. To find the projective representations we can use $\tilde G = \text{spin}(3) = SU(2)$. This has integer spin representations AND half-integer spin (spinor) representations. – Heidar Aug 20 '13 at 21:57

2 Answers 2

up vote 10 down vote accepted

At the risk of telling you how to "suck eggs" (your level in these things is not altogether clear), here goes.


The essential ingredients to this explanation are:

  1. A physical "system" which evolves in and whose "events" happen in some space $\mathcal{U}$ (ordinary Euclidean 3-space or Minkowsky spacetime, for example); in physics this space is always a linear (wontedly it's Minkowsky spacetime) space wherein stuff happens: let's call $\mathcal{U}$ the "scene" of where stuff we want to talk about happens;

  2. A connected Lie group $\mathfrak{G}$ which represents the co-ordinate transformations the a system can undergo: in physics these are all linear transformations $\mathcal{U}\to\mathcal{U}$ of the scene $\mathcal{U}$. Wontedly in physics, $\mathfrak{G} = SO^+(1,3)$ (the identity connected component of the Lorentz group comprising all rotations and boosts of "physical space", sometimes called the "proper, orthochronous Lorentz group" (proper = unimodular determinant =1, i.e. "does not invert space" and orthochronous = does not switch the time direction) or the Poincaré group;

  3. A cover of $\mathfrak{G}$; this is almost always (I've never seen it not so) the universal cover $\tilde{\mathfrak{G}}$ of $\mathfrak{G}$ as explained in my article "Lie Group Homotopy and Global Topology" on my website here;

  4. A vector space $\mathcal{V}$ which can be, for example, a quantum state space, possibly infinite dimensional and its group $GL(\mathcal{V})$ of linear endomorphisms, i.e. bijetive, linear maps $\phi:\mathcal{V}\to\mathcal{V}$. Informally, $GL(\mathcal{V})$ is the group of invertible matrices acting on $\mathcal{V}$. Most importantly: this space is different from the physical "scene" $\mathcal{U}$. The scene is $\mathcal{U}$ spacetime all around us, the state space $\mathcal{V}$ is a Hilbert space of quantum states. And, actually, although we talk about a "linear" state space $\mathcal{V}$, we're a bit sloppy: sure, all quantum states are linear superpositions of the basis for $\mathcal{V}$ but they are always of unit magnitude: the probabilities of a measurement's "collapsing" the state into one of the basis vectors must all sum up to one - "we have to end up in some state". So, if we're being precise, we take heed that we are actually talking about the unit sphere within $\mathcal{V}$ as the state of quantum states. This state space is very different in character from spacetime, where there is no obligation for 4-positions of events to be unit magnitude;

  5. Representations $\rho : \mathfrak{G}\to GL(\mathcal{V})$, $\tilde{\rho}:\tilde{\mathfrak{G}}\to GL(\mathcal{V})$ of both $\mathfrak{G}$ and its cover $\tilde{\mathfrak{G}}$, respectively. Recall that a representation of a Lie group $\mathfrak{G}$ is a homomorphism from from $\mathfrak{G}$ to $GL(\mathcal{V})$, i.e. a transformation that "respects the group product" so that, given $\gamma,\,\zeta\in\mathfrak{G}$, we have $\rho(\gamma\,\zeta)=\rho(\gamma)\,\rho(\zeta)$. And, as discussed above, the linear transformations of the form $\rho(\gamma),\,\tilde{\rho}(\tilde{\gamma}) \in GL(\mathcal{V})$ for $\gamma \in\mathfrak{G}$ and $\tilde{\gamma} \in\tilde{\mathfrak{G}}$ must be unitary so that the transformed quantum states stay normalised. So we can see that $GL(\mathcal{V})$ is very different from $\mathfrak{G}$ or $\tilde{\mathfrak{G}}$: Lorentz boosts most assuredly are not unitary! We say that $\mathfrak{G}$ or $\tilde{\mathfrak{G}}$ "act on the state space $\mathcal{V}$ through the respresentations $\rho$, $\tilde{\rho}$".

I'm using here more of the mathematician's description of a representation, because here (i'm not always so fussed) I believe it is clearer than the physicists because we need to take heed that there are two different classes of representations in our discussion: those whereby the group of co-ordinate transformations $\mathcal{G}$ act on the state space $\mathcal{V}$ and those whereby its cover $\tilde{\mathcal{G}}$ acts on $\mathcal{V}$.

Baking Instructions: Wigner's Theorem and Why Covers Are Interesting

Why are we interested in covers at all? After all, elements of $\tilde{G}$ are not the "physical" co-ordinate transformation. This is where we meet our baking oven for our ingedients: Wigner's theorem. Clearly, when our scene $\mathcal{U}$ undergoes a co-ordinate transformation, then the transformations wrought on the quantum state has to preserve inner products in the quantum state space so that the state stays properly normalised. From this assumption alone, i.e. one does NOT have to assume linearity, Wigner proved that the when the scene $\mathcal{U}$ undergoes a "symmetry" (a Lorentz transformation), the state space must undergo a "projective homomorphism" $\sigma$, i.e. if $\gamma,\,\zeta$ are two Lorentz transformations, then the state space transformation corresponding to their product is:

$$\sigma(\gamma\,\zeta) = \pm \sigma(\gamma)\,\sigma(\zeta)\tag{1}$$

The fact that we don't get exactly a homomorphism is why we are interested in covers: the image of the representation $\tilde{\rho}(\tilde{\mathfrak{G}})$ (recall that this is a group of unitary operators in $GL(\mathcal{V})$ acting on the state space) of the cover $\tilde{\mathfrak{G}}$ contains both the transformations that fulfill the genuine $+$-sign homomorphism in (1) (which are simply unitary operators in the image $\rho(\mathfrak{G})$ of the co-ordinate transformation group $\mathfrak{G}$) AND those that flip the sign. So if we allow representations of the cover, we get every possible unitary transformation (even without an assumption of linearity - this automatically follows) that can be wrought on the state space $\mathcal{V}$ when the scene $\mathcal{U}$ is transformed.

Here's the punchline.

Quantum states that transform by the transformations belonging to the image $\rho(\mathfrak{G})$ of $\mathfrak{G}$ under the genuine homomorphism $\rho$ are called vectors.

Quantum states that transform by the transformations belonging to the image $\tilde{\rho}(\tilde{\mathfrak{G}})$ of the cover $\tilde{\mathfrak{G}}$ under the "projective homomorphism" $\tilde{\rho}$ are called spinors.

The above can be intuitively thought of as follows: in quantum mechanics, a global phase $e^{i\phi}$ multiplying a system's quantum state does not affect measurements we make on the system. So quantum systems "don't care whether a homomorphism is genuine or projective".

The universal (only) cover of the Lorentz group $SO(1,3)$ is the group $SL(2,\,\mathbb{C})$. So "spinors" transform by a representation of $SL(2,\,\mathbb{C})$. Vectors transform by a representation of $SO(1,3)$. The word "spinor" can be pretty vague in my experience: it can refer to the transformation in $SL(2,\,\mathbb{C})$ rather than the quantum state that is transformed by it, and people often speak of the unit quaternions as "spinors": Roger Penrose's "Road To Reality" Chapter 11 simply defines a spinor as something that takes a negative sign when rotated through $2\,\pi$ and comes back to its beginning point after a rotation through $4\,\pi$. This is actually a pretty good definition, for that is exactly how elements of a representation of $SL(2,\,\mathbb{C})$ act on the state space $\mathcal{V}$, and is the essential difference between how elements of a representation of $SO(1,3)$ act on state spaces.

Forget about "quantities with direction" as a vector's definition: in physics the word "vector" always talks about how something transforms when our scene $\mathcal{U}$ undergoes a symmetry. Remember this is pretty near to the word's literal meaning vehor (transliterated as vector) literally means "I am borne" or "I am carried" in Latin, so it's all about how the "vector" is borne, either by a transformation in physics or as a pathogen in biology (the word's original English meaning).

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Should there be bars in the second line of your boxed punchline? Also, does that comma in (1) belong there? I'm just trying to wrap my head around the notation. – Chris White May 10 '14 at 2:56
@ChrisWhite Right on both counts. Thanks for being my proofreader yet again! – WetSavannaAnimal aka Rod Vance May 15 '14 at 4:45

From the question I see that you are confused by the meaning of "Normally states are vectors in infinite dimensional spaces", not by spinors. Function is a good representation of a vector in infinite space. Let us consider function $\psi(\bf{r})$. This is a vector from infinite dimensional space. What will be with this function when we rotate the space? Not that imaginary infinite dimensional space, but real space which is $\bf{r}$ (the argument of that function). In the real space it is a constant (spin $=0$) because if you rotate that function, it may be done only by considering how function argument $\bf{r}$ is changed when you rotate the space.

It turns out, that there are more complicated cases. In some cases you have to consider not one function, but two $(\psi_{\uparrow}(\bf{r}),\psi_{\downarrow}(\bf{r}))$ or three $(\psi_{x}(\bf{r}),\psi_{y}(\bf{r}),\psi_{z}(\bf{r}))$ functions. After rotations it is not enough just to replace $\bf{r}$ with a new rotated $\bf{r}'$. The new (two or three) functions will be the linear combination of rotated original ones. The matrix which connect them is usually the matrix which describes rotation of spinors or vectors.

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