Quantum Mechanics – Intuition About Analogy of Spinors in Clifford Algebra

clifford-algebraquantum mechanicsquantum-spinspinors

I've been studying the application of Clifford Algebra in quantum mechanics, more specifically in spin, but I'm stuck in a basic analogy that is part of the basics before starting to do things using Clifford Algebra. The point is that we have a spinor in spherical coordinates and then say that the equivalent representation of it in the way made below, but it looks like something taken out of nowhere, or simply established in a way that things could work. I am following three references about them:

Geometric Multiplication of Vectors An Introduction to Geometric Algebra in Physics (Compact Textbooks in Mathematics) by Miroslav Josipović;
Geometric Algebra for Physicists by Chris Doran, Anthony Lasenby;
William E. Baylis – Clifford (Geometric) Algebras With Applications to Physics, Mathematics, and Engineering – Birkhäuser Basel (1996)

I would like some intuition or explanation about why we do this identification $$\psi = a^0 + a^k \sigma_k$$
where $I=\sigma_1 \sigma_2 \sigma_3$ but I'd like more than because it works in that way, if possible. I searched a lot but I couldn't find anything.

I really appreciate your attention and help.

Best Answer

The capability of identifying a spinor with a rotor is a coincidence in 3D.

For a general dimension N, let's do an excise of counting

The degree of freedom of rotation (the rotor referenced in the OP): $\frac{N(N-1)}{2}$
the degree of freedom of a spinor (assuming it's the even part of the Clifford algebra):$2^{N-1}$

In 3D, both freedoms are 3, after you subtract the normalization factor freedom of a spinor. Therefore, there is a way to identify a spinor with a rotor.

In 4D, the degree of freedom of rotation is 6, while the degree of freedom of a spinor is 8. After subtracting the normalization factor freedom of a spinor, you still have one more degree of freedom ($6+1$) which can NOT be identified with a rotation. This fact has been noted early on by David Hestenes (please search his papers for more info).

In N>4 D, the situation is even worse. In summary, the gimmick cited in the OP only works in 3D.

Related Solutions

[Physics] Different definitions of spinors

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.}

quantum-mechanics – Understanding Rotations of Quantum States: $SO(3)$ versus $SU(2)$

On one hand, if $\vec{\alpha}\in \mathbb{R}^3$ denotes a 3D rotation vector$^1$, then the corresponding rotation matrix $$R(\vec{\alpha})~=~\exp(i\vec{\alpha}\cdot \vec{L})~\in~ SO(3)~ \subseteq ~{\rm Mat}_{3\times 3}(\mathbb{R}). \tag{1}$$ Here $iL_j\in so(3) \subseteq {\rm Mat}_{3\times 3}(\mathbb{R})$ are three $so(3)$ Lie algebra generators, defined as $$\begin{align} i(L_j)_{k\ell} ~=~& \epsilon_{jk\ell} ,\cr j,k,\ell~\in~& \{1,2,3\}, \cr \epsilon_{123}~=~&1,\tag{1'}\end{align}$$ and fulfilling the $so(3)$ Lie bracket relations $$\begin{align} [L_j, L_k] ~=~& i\sum_{\ell=1}^3\epsilon_{jk\ell} L_{\ell}, \cr j,k,\ell~\in~& \{1,2,3\}.\end{align} \tag{1"}$$
On the other hand, the corresponding $SU(2)$ matrix is$^2$ $$ X(\vec{\alpha})~=~\exp(\frac{i}{2}\vec{\alpha}\cdot \vec{\sigma})~\in~ SU(2)~ \subseteq ~{\rm Mat}_{2\times 2}(\mathbb{C}), \tag{2}$$ where $\sigma_{\ell}$ are the Pauli matrices.
The two matrix representations (1) & (2) are related via $$\begin{align} X(\vec{\alpha}) \sigma_k X(\vec{\alpha})^{-1}~=~& \sum_{j=1}^3\sigma_j R(\vec{\alpha})^j{}_k, \cr R(\vec{\alpha})^j{}_k~=~&\frac{1}{2}{\rm Tr}(\sigma^jX(\vec{\alpha}) \sigma_k X(\vec{\alpha})^{-1}), \cr j,k~\in~& \{1,2,3\}. \end{align}\tag{3}$$
The relation (3) exposes the fact that the adjoint representation $$ {\rm Ad}:~ SU(2) \quad\to\quad SO(su(2))~\cong~ SO(3),\tag{4} $$ given by $$ \begin{align} X\quad \mapsto&\quad{\rm Ad}(X)\sigma~:= ~X\sigma X^{-1}, \cr X~\in~& SU(2),\cr \sigma~\in~& su(2)\cr ~:=~&\{ \sigma\in {\rm Mat}_{2\times 2}(\mathbb{C})\mid \sigma^{\dagger}=\sigma~\wedge~{\rm tr}(\sigma)=0\}\cr ~=~&{\rm span}_{\mathbb{R}}\{\sigma_1,\sigma_2,\sigma_3\},\cr &||\sigma||^2 ~=~\det(\sigma),\cr &{\rm Ad}(\pm {\bf 1}_{2 \times 2})~=~{\bf 1}_{su(2)}, \end{align}\tag{4'} $$ is a surjective 2:1 Lie group homomorphism between $SU(2)$ and $SO(3)$, i.e. $$ SU(2)/\mathbb{Z}_2~\cong~SO(3). \tag{4"}$$
See also this related Phys.SE post, which explains the half-angle appearance in eq. (2).

References:

G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapters 3 & 6. The pdf file is available here.

$^1$ The direction of a rotation vector $\vec{\alpha}\in \mathbb{R}^3$ is parallel to the rotation axis, while the length $|\vec{\alpha}|$ denotes the counter-clockwise rotation angle (as seen from the tip of the rotation vector).

$^2$ The notation and conventions in this Phys.SE answer follows Ref. 1.

Best Answer

Related Solutions

[Physics] Different definitions of spinors

quantum-mechanics – Understanding Rotations of Quantum States: $SO(3)$ versus $SU(2)$

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