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  About the correspondence between SO(3) and SU(2)

+ 1 like - 0 dislike

In David McMohans QFT Demystified book on p66 it is claimed that the correspondance between $SO(3)$ which rotates space vectors $\vec{x} = (x,y,z)^T$ and the group $SU(2)$ which acts on two-component spinors $\psi = (\alpha, \beta)^T$ can be seen by making use of the maping between space vectors and two-component spinors given by

\[x = \frac{1}{2}(\beta^2-\alpha^2); \quad y= -\frac{i}{2} (\beta^2+\alpha^2); \quad z = \alpha\beta \]

Where does this specific mapping come from? Is it unique or are there other maps between space vectors and spinors?

Then, the author goes forward and writes that the 3 parameters of $SU(2)$ can  be associated with angles of rotation of $SO(3)$ such that for an arbitrary angle $\alpha$ (not a spinor component now) a 3D rotation around the x axis can be written as 2x2 or 3x3 matrix as

\[U = \left( \begin{array}{cc} \cos(\alpha/2) & i\sin(\alpha/2) \\ i\sin(\alpha/2) & \cos(\alpha/2) \\ \end{array} \right) \quad \mathrm{or} \quad R_x(\omega) \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos\omega & \sin\omega \\ 0 & - \sin\omega & \cos\omega \\ \end{array} \right) \]

a 3D rotation around the y-axis as by the angle $\beta$ as

\[U = \left( \begin{array}{cc} \cos(\beta/2) & \sin(\beta/2) \\ -\sin(\beta/2) & \cos(\beta/2) \\ \end{array} \right) \quad \mathrm{or} \quad R_y(\phi) \left( \begin{array}{ccc} \cos\phi & 0 & \sin\phi \\ 0 & 1 & 0 \\ -\sin\phi & 0 & \cos\phi \\ \end{array} \right) \]

and a 3D rotation around the z-axis by the angle $\gamma$ as

\[U = \left( \begin{array}{cc} e^{i\gamma/2} & 0 \\ 0 & e^{-i\gamma/2} \\ \end{array} \right) \quad \mathrm{or} \quad R_z(\theta) \left( \begin{array}{ccc} \cos\theta & \cos\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \]

First of all, do the angles of the two representations correspond to each other too, such that $\alpha = \omega$, $\beta = \phi$, and $\gamma = \theta$?

And how can it be proven or seen that the specific 2x2 matrix of $SU(2)$ represent exactly the same 3D rotation as the 3x3 matrix of $SO(3)$ in each case?

asked Jan 6, 2015 in Chat by Dilaton (6,240 points) [ revision history ]
recategorized Jan 7, 2015 by Dilaton
Most voted comments show all comments
$x = \frac{1}{2}(\beta^2-\alpha^2); \quad y= -\frac{i}{2} (\beta^2+\alpha^2); \quad z = \alpha\beta$ is indeed strange, there's no guarantee that $x$, $y$ and $z$ are real if $\alpha$ and $\beta$ are arbitrary. To your second question, yes, at least if the convention for $SU(2)$ is for example $e^{i\omega\sigma_x/2}$(for rotation around x-axis), you can quickly check the result by applying the exponential of Pauli matrix listed here: http://en.wikipedia.org/wiki/Pauli_matrices#Exponential_of_a_Pauli_vector

A good reference may be Chapter 2 in Angular Momentum in Quantum Physics- Theory and Application. Starting from Section 4 discusses the problem: Cartan's definition of a spinor (he considered a rotation of an isotropic vector in $\mathbb{C}^3$).

Thanks @JiaYiyang and @L.Su, I now see that if anything the issue already comes up in basic QM, apart from the first subquestion. I should really not have posted that here so feel free to vote to close/delete or move to chat maybe.
Yes, I'd suggest to move to chat.
Lubosh Motl wrote on his blog that a square root of vector is a spinor. Maybe this can help establish a direct relationship of rotations in both cases.
Most recent comments show all comments
Well, actually after reading @L.Su's reference, I think your first question might be on-topic, Cartan's geometric construction is not typically introduced in undergrad courses, at least I was completely ignorant of it until just now.
Maybe I could extract and repost a version of it on the main page ...

1 Answer

+ 2 like - 0 dislike

An effective way to think about the rotation group and the double cover is in terms of the Clifford algebra. One way to construct a Clifford algebra is to introduce for vectors a multiplication that satisfies $uv+vu=2(u,v)=2g_{\alpha\beta}u^\alpha v^\beta$. Immediately, $u^2=(u,u)$. Whatever the metric, the Clifford algebra is isomorphic to a matrix algebra. For a Clifford algebra over the complex field, the (3,1) signature case is isomorphic to the Dirac matrix algebra $M_4(\mathbb{C})$, the (3,0) case is isomorphic to the matrix algebra $M_2(\mathbb{C})$, mapping any choice of three orthogonal 3-vectors to the three Pauli matrices. Because of the defining equation above, we have, for non-null vectors, $$uvu^{-1}=uv\frac{u}{(u,u)}=\left(2(u,v)-vu\right)\frac{u}{(u,u)}=-\left(v-2u\frac{(u,v)}{(u,u)}\right),$$ which is just $-(v\mbox{ reflected in the plane (or, rather, }n\!-\!1\mbox{-space) perpendicular to }u)$. With enough pairs of reflections we obtain any rotation or boost, and the sign doesn't matter because the reflections have to be taken in pairs to give a rotation.

Because Clifford algebras are isomorphic to matrix algebras, we can introduce actions of the Clifford algebras on left and right ideals, corresponding to actions of the matrix algebras on column and row vectors, which, when rendered into components, leads to relationships such as those you detail in your question. This is quite a broad brush, but on this view of the mathematics, much of the structure comes from the equation $uv+vu=2(u,v)$. The map from a Clifford algebra to a matrix algebra is not unique. There is a Wikipedia page, http://en.wikipedia.org/wiki/Clifford_algebra.

answered Jan 9, 2015 by Peter Morgan (1,230 points) [ revision history ]
edited Jan 9, 2015 by Peter Morgan

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