Recently I've read a little about the description of particles with spin in the book Quantum Mechanics by Cohen-Tannoudji. Although I yet didn't fully study the subject, I've read one interesting part in which the author considers the description of such particle being given by the state space $\mathcal{E}=\mathcal{E}_{S}\otimes \mathcal{E}_{\mathbf{r}}$ being $\mathcal{E}_S$ the spin state space generated by $\{|+\rangle, |-\rangle\}$ and $\mathcal{E}_{\mathbf{r}}$ the usual state space of a particle without spin.
In that setting we can consider the representation $\{|-\rangle\otimes| \mathbf{r}\rangle, |+\rangle\otimes |\mathbf{r}\rangle\}$. With this, denoting $|\mathbf{r}, -\rangle = |-\rangle\otimes| \mathbf{r}\rangle$ and $|\mathbf{r}, +\rangle =|+\rangle\otimes |\mathbf{r}\rangle$ we find that if $|\varphi\rangle \in \mathcal{E}$ we can express it as
$$|\varphi\rangle = \int_{\mathbb{R}^3}(\varphi_-(\mathbf{r})|\mathbf{r}, -\rangle + \varphi_+(\mathbf{r})|\mathbf{r},+\rangle )d^3\mathbf{r}.$$
The author then states the following:
In order to characterize completely the state of an electron, it is therefore necessary to specify two functions of the space variables $x,y,z$:
$$\varphi_{+}(\mathbf{r})=\langle \mathbf{r}, + | \varphi \rangle$$
$$\varphi_{-}(\mathbf{r})=\langle \mathbf{r}, – | \varphi \rangle$$These two functions are often written in the form of a two-component spinor, which we shall write $[\varphi](\mathbf{r})$:
$$[\varphi](\mathbf{r}) = \begin{pmatrix}\varphi_{+}(\mathbf{r}) \\ \varphi_{-}(\mathbf{r})\end{pmatrix}$$
The main point is that it seems here that a spinor is just an element of $L^2(\mathbb{R}^3)\times L^2(\mathbb{R}^3)$.
Now, some time ago, before reading this I tried to learn what a spinor is, and looking on the internet I've found the Wikipedia page on which many things are said. I couldn't locate at the time any direct definition on the page.
On the page many things are said, considering Clifford algebras, spin groups and many other things. It highly depends on Clifford Algebras (which although I know the definition I haven't had the time yet to fully study) and IMHO it doesn't relate immediately to the idea of spin found in quantum mechanics.
On the other hand the idea of spinor introduced by Cohen is a thousand times clearer, simpler and much more connected to the idea of spin. I believe the relationship to rotations could be made even clearer seeing as the spin operators are generators of rotations in $\mathbb{R}^3$ and that the elements $|+\rangle$ and $|-\rangle$ are its eigenvectors.
My question here is: what is the relation between these two points of view about spinors? What is the relation between Cohen's idea of just picking those functions together and the quite complex algebraic construction? How can we connect those two viewpoints?
Best Answer
You are a bit confused by the wording in Cohen-Tannoudji et al. Namely, it is not the function $[\varphi]: \mathbf R^3 \to \mathbf R^2$ that is called a spinor, it is its value at a particular point, the two numbers $[\varphi](\mathbf r)\in\mathbf C^2$. So a spinor is not even an element of $L^2(\mathbf R^3)\times L^2(\mathbf R^3)\equiv L^2(\mathbf R^3)\otimes\mathbf C^2$: it is just an element of $\mathbf C^2$! It’s even simpler than you think. The function $[\varphi]$ is then properly called a spinor-valued wavefunction, as opposed to the more common scalar-valued wavefunctions.
However, in quantum mechanics, people still frequently call these objects spinors and not spinor-valued wavefunctions. It’s shorter and more convenient, but you have to keep in mind that it’s no more than an abuse of terminology. Otherwise, it will bite you later in field theory, where you will have spinor fields—spinors specified at every point of space(time)—in the same sense that the vector potential $\mathbf A(\mathbf r)$ is a vector field. The field theory doesn’t even have to be quantum: classical will do.
Now that we understand that a spinor is two numbers, why all the mathematical fuss? There is a good reason for it. It is indeed two numbers, but not just two numbers; thinking of it as two numbers will not help you understand what’s going on. For example, a vector is three numbers, but it’s not the three numbers you picture when you talk about vectors: it’s arrows in (Euclidean) space. Arrows make sense whether you choose to draw a coordinate system in the space or not. And, for example, associated with each arrow is a length, a number that also makes sense independent of any coordinate system you may or may not have.
But if you have one, $K$, you can describe any vector by three numbers: for example, $(x\;y\;z)$. If you have one more, $K'$, you can get three different numbers, $(x'\;y'\;z')$. Some numbers are more equal than others:* “length of $\mathbf v$” makes sense in the absence of a coordinate system, but “the $x$ coordinate of $\mathbf v$” doesn’t. However, once you have the coordinates of a vector in a particular system $K$, you can say what they will be in any other $K'$, consistently: that is, transforming from $K$ to $K'$ and then from $K'$ to $K''$ is the same thing as transforming from $K$ to $K''$ directly. And the transformation is linear.
We know the reason for this, of course: these are just arrows in space, so consistency is obvious. However, suppose we forgot about arrows and keep just the coordinates. We could then infer the existence of the geometric arrows from the fact that we have those algebraic transformation rules. Now generalize and say that any consistent set of rules defines a geometric object. (Keep linearity for convenience.) What other objects can we think of? Well, the six numbers $(x_1\;y_1\;z_1\;x_2\;y_2\;z_2)$ obtained from any two vectors $\mathbf v_1$ and $\mathbf v_2$, but this is kind of obvious, so we’d like to consider things that are not built from simpler ones like that. A rotation looks nice: it makes geometric sense, and at the same time it’s described by (nine) numbers conveniently arranged into a matrix. (It transforms like $A' = SAS^{-1}$ when $v' = Sv$, of course.) Turns out there’s a convenient way to talk about all these well-behaved objects.
What does this have to do with quantum mechanics? Everything. If the state of a particle at each point $\mathbf r$ in space is described by a complex number $\psi(\mathbf r)$, this number is of course doesn’t change under a coordinate transformation. The Hilbert space is $\mathscr H = L^2(\mathbf R^3)$. But what if there’s more than one “internal” degree of freedom at each point? Say $2s+1$ of them? Then $\mathscr H = L^2(\mathbf R^3)\otimes\mathscr I$ and there’s the “internal” space $\mathscr I =\mathbf C^{2s+1}$. They have to (1) mix with one another under a coordinate transformation; otherwise it’s just several simple particles lumped together, not one complex one. Mix (2) consistently, (3) linearly, and (4) the norm of the wavefunction doesn’t change. Here you go, an irreducible(1) unitary(4) linear(3) representation(2) of the rotation group $\mathrm{SO}(3)$ on $\mathscr I$. Representation theory tells you there is one such representation for $s = 0, 1, 2,\ldots$ and so on—but not for $s=1/2$! Where are the spinors?
It turns out we have missed a subtle but critical point. You don’t suddenly take away your axes and put them back differently: you change them continuously. The internal degrees of freedom $\mathscr I$ need to transform in a certain way given a process of rotation, not just its “endpoints”. But if I deform this process while keeping the endpoints fixed, the result had better stay the same. Now can all the rotation processes with fixed endpoints be transformed into one another? If yes, then we haven’t gained anything. But it turns out that the answer is no: not doing anything is the same thing as rotating twice around a fixed axis, but not the same thing as rotating once.
Why twice? Easy. Rotate (continuously) 360° around the axis in question, and then 360° around the axis pointing in the opposite direction: this is manifestly the identity. Rotate 360° around the given axis and then 360° around it again: this is the 720° rotation. But the two processes can be deformed into one another by moving the axis of the second 360° rotation!
The talk about processes strongly suggest a picture of integrating along a “curve in rotation space”. The relevant mathematics is the theory of Lie algebras. And what you need is not a representation of the group $\mathrm{SO}(3)$ of “big” rotations, but a representation of the algebra $\mathfrak{so}(3)$ of “infinitesimal” ones. [This is what physicists mean when they talk about “(infinitesimal) generators of a group”.] And, as we have seen, there’s more of them. More precisely, apart from ones with $s = 0,1,2,\ldots$ you also get one for each of $s = 1/2,3/2,\ldots$. It also turns out the operators of (conserved) angular momentum are closely connected with the Lie algebra of the associated symmetry—which is the rotations we have been talking about all along.
In a relativistic theory, the relevant symmetry is the bigger Lorentz algebra $\mathfrak{so}(3,1)$, so relativistic spinors have four components, just like four-vectors. In eleven dimensions, vectors have eleven components but spinors have thirty two. So they are not always “smaller”.
Finally, now that we have recovered spinors, what’s the matter with Clifford algebras and spin groups? Remember two inequivalent rotation processes between each pair of endpoints in $\mathrm{SO}(3)$? Turns out there’s always two of them in $\mathrm{SO}(n)$ for $n\ge 3$, so we can look at how these processes compose modulo the equivalences. (To compose two processes, do the first one and then the second.☺) The result (for $n\geq 3$, and similarly for the Lorentz groups) is exactly $\mathrm{Spin}(n)$. I’m not aware of a simple motivation† for the connection with Clifford algebras [edits welcome!], but you’ll get there once you understand the Lie algebra picture. The Clifford algebras themselves are quite simple objects, in fact, and can easily be motivated by taking the square root of the Laplacian; it’s just not clear why you get spinors this way.
* Animal Farm, of course.
† More precusely, I don’t know why the quadratic elements of the Clifford algebra form a representation of the corresponding orthogonal algebra. Apart from “just calculate it”, that is.