Glenn Rowe said:
I've got a question about the identification of SU(2) with O(3) in Ryder's QFT book (2nd edition) pages 34 - 35.
The other posts on this topic I could find don't seem to address this question, so here goes.
He derives the matrix in eqn 2.47:
$$H=
\left[\begin{array}{cc}
-\xi_{1}\xi_{2} & \xi_{1}^{2}\\
-\xi_{2}^{2} & \xi_{1}\xi_{2}
\end{array}\right]$$
He then constructs another matrix ##h## from the position vector and the Pauli spin matrices (eqn 2.49):
$$
h=\boldsymbol{\sigma}\cdot\mathbf{r}=\left[\begin{array}{cc}
z & x-iy\\
x+iy & -z
\end{array}\right]$$
We can identify ##h## with ##H## if we take
$$
\begin{eqnarray}
x & = & \frac{1}{2}\left(\xi_{2}^{2}-\xi_{1}^{2}\right)\\
y & = & \frac{1}{2i}\left(\xi_{1}^{2}+\xi_{2}^{2}\right)\\
z & = & \xi_{1}\xi_{2}
\end{eqnarray}
$$
He says that both ##H## and ##h## transform according to ##H^{\prime}=UHU^{\dagger}## and ##h^{\prime}=UhU^{\dagger}## which seems OK so far. However, he then says that if ##U## belongs to SU(2) and therefore has determinant 1, we can take the determinant of ##h^{\prime}=UhU^{\dagger}## to say that ##x^{\prime2}+y^{\prime2}+z^{\prime2}=x^{2}+y^{2}+z^{2}##.
This seems reasonable, except that when you actually evaluate ##x^{2}+y^{2}+z^{2}## using equations (1) to (3) you get ##x^{2}+y^{2}+z^{2}=0## identically. This also follows by taking the determinant of ##H##.
I guess my question is: how can you get around the fact that this equivalence seems to apply only to a single point at the origin to say anything useful about the relation of SU(2) to rotations?
Thanks for any comments.
Okay, let me clarify the thing for you. But first I must tell you that Ryder’s treatment is
correct. He just stops short of explaining the small details. I suppose, he assumes that you know what he is doing.
1) For any
real or
complex 3-vector \vec{\pi} = (\pi_{1},\pi_{2},\pi_{3}), we can define the matrix \Pi = \vec{\sigma} \cdot \vec{\pi}. So, when Ryder identifies his H with h = \vec{\sigma} \cdot \vec{r}, he is telling you that \vec{r} is no longer a position vector in the
real Euclidean space E^{3}. Rather, in that identification, \vec{r} is an
isotropic vector (i.e.,
complex null cone) in a 3-dimensional
complex Euclidean space. Isotropic vector is a vector of length
zero, i.e., orthogonal to itself: \vec{r} \cdot \vec{r} = | \vec{r}|^{2} = x^{2}+y^{2}+z^{2} = 0 . \ \ \ \ (1)
2) In \mathbb{R}^{3}, eq(1) has one and only one
trivial solution, \vec{r} = (0,0,0), but in \mathbb{C}^{3} it has
infinitely many solutions.
3) As we will see below, isotropic vector is an essential object for defining spinors. In fact in 3D,
isotropic vector and its associated spinor are two faces of the same coin.
4) There are many (equivalent) ways to define spinors, I will explain few of them for you, but the easiest one is the following. A spinor \Psi = \begin{pmatrix} u \\ d \end{pmatrix} is defined to be a
non-trivial solution to the following equation
\Pi \Psi \equiv \begin{pmatrix} \pi_{3} & \pi_{1} - i \pi_{2} \\ \pi_{1} + i \pi_{2} & - \pi_{3} \end{pmatrix} \begin{pmatrix} u \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} .Linear algebra tells us that a non-trivial solution (to the above matrix equation)
exists if and only if \det \Pi = 0. This means that \vec{\pi} = (\pi_{1},\pi_{2},\pi_{3}) is an
isotropic vector. So, given the isotropic vector (\pi_{1},\pi_{2},\pi_{3}), an associated spinor can be defined: Just write \pi_{3}^{2} = - (\pi_{1} - i \pi_{2})(\pi_{1} + i \pi_{2}) , then set u^{2} = - (\pi_{1} - i \pi_{2}) , \ \mbox{and} \ \ d^{2} = \pi_{1} + i \pi_{2} . Conversely, given the two complex numbers u and d (i.e., spinor), the components of
two isotropic vectors can be calculated from
\pi_{1} = \frac{1}{2} (d^{2} - u^{2}) , \ \pi_{2} = \frac{1}{2i} (d^{2} + u^{2}) , \ \pi_{3} = \pm u \ d . One of the two signs of \pi_{3} can be chosen arbitrarily (plane symmetry), I will choose the plus sign.
5) Elie Cartan’s definition of spinors: Rotation requires an
oriented plane. In the real Euclidean space E^{3}, consider two
orthogonal vectors \vec{X} = (x_{1}, x_{3}, x_{3}) and \vec{Y} = (y_{1}, y_{2}, y_{3}) with the
same length: \vec{X} \cdot \vec{X} = \vec{Y} \cdot \vec{Y} , \ \ \ \vec{X} \cdot \vec{Y} = 0 . \ \ \ \ (2) Clearly these two vectors define a plane, and if we specify the “order” in which we pick these vectors, this “order” defines an orientation for the plane (i.e., direction of rotation). An algebraic representation of the “order”, in which the pair (\vec{X} , \vec{Y}) have been chosen, can be achieved by multiplying the components of the second vector by the complex number i. Thus, an isotropic vector \vec{\pi} is born with the following components
\pi_{j} = x_{j} + i y_{j} , \ \ j = 1, 2, 3 .
Indeed, with some easy algebra and using eq(2), we find \vec{\pi} \cdot \vec{\pi} = \vec{X} \cdot \vec{X} - \vec{Y} \cdot \vec{Y} + 2i \vec{X} \cdot \vec{Y} = 0 . Now that we have an isotropic vector \pi_{3}^{2} = - (\pi_{1} - i \pi_{2})(\pi_{1} + i \pi_{2}) , we can repeat our work in (4) and find
\pi_{1} = x_{1} + i y_{1} = \frac{1}{2} (d^{2} - u^{2}) ,\pi_{2} = x_{2} + i y_{2} = \frac{1}{2i} (d^{2} + u^{2}) ,\pi_{3} = x_{3} + i y_{3} = u \ d . Thus, the two complex numbers u and d (i.e., spinors) form a representation of the
two vectors \vec{X} = (x_{1}, x_{2}, x_{3}) and \vec{Y} = (y_{1}, y_{2}, y_{3}) as well as of the
orientation of the plane defined by those vectors.
6) Here I will describe a method for defining spinor, although inaccurate, it allows you to draw a "picture" of spinor. Again, let \vec{\pi} = (\pi_{1} , \pi_{2}, \pi_{3}) be an isotropic vector. Rewrite the equation \pi_{1}^{2} + \pi_{2}^{2}+ \pi_{3}^{2} = 0 as \frac{\pi_{1} - i \pi_{2}}{\pi_{3}} = \frac{- \pi_{3}}{\pi_{1} + i \pi_{2}} . \ \ \ \ \ \ (3) This equation represents a complex null cone. Let P be a plane
tangent to the cone (3) along the vector (\pi_{1} , \pi_{2}, \pi_{3}). It is given by P: \ \ u \pi_{1} + s \pi_{2} + d \pi_{3} = 0 \ \ \ \ (4) where (u,s,d) are coordinates of a current point in the plane P. Let P_{0} be another plane tangent to the null cone along an isotropic straight line l_{0}
perpendicular to the \pi_{3}-axis: l_{0} : \ \ \pi_{1}^{2} + \pi_{2}^{2} = 0, \ \ \Rightarrow \ \ \pi_{2} = i \pi_{1} . \ \ \ (5) Thus, the plane P_{0} is given by P_{0} : \ \ u \pi_{1} + i s \pi_{1} + 0 = 0 , \ \ \Rightarrow \ \ s = i \ u . \ \ \ (6) Let e = P \cap P_{0} be the line where the two planes
intersect. From (6) and (4), we find
e : \ \ u ( \pi_{1} + i \pi_{2} ) + d \pi_{3} = 0 . \ \ \ (7) From (7) and (3), we obtain \frac{u}{d} = \frac{- \pi_{3}}{\pi_{1} + i \pi_{2}} = \frac{\pi_{1} - i \pi_{2}}{\pi_{3}} , and, therefore \frac{u^{2}}{d^{2}} = \frac{- (\pi_{1} - i \pi_{2})}{(\pi_{1} + i \pi_{2})} . Thus, there corresponds to the isotropic vector \vec{\pi} an object of only two complex components (u , d) (i.e., a spinor) such that
u^{2} = - (\pi_{1} - i \pi_{2}) , \ \ d^{2} = \pi_{1} + i \pi_{2}, \ \ \mbox{and} \ \ u \ d = \pi_{3} .
Now,
draw the two intersecting planes and a cone between them. Hold the plane P_{0}
fixed and let the other plane P
rolls around the cone. You then can observe the following: if P rolls
once around the cone, the intersecting line e turns only 180 degree in the fixed plain P_{0}. Therefore, a
double rolling of P is necessary to bring the line e back to its
original position. This “pictures” for you the
double-valued spinor representation of the rotation group SO(3).
7) Finally, since you are studying QFT, you need to understand the concept of spinor and spinor space through group theory. Have a look at the following link where I used elementary concepts from group theory to answer similar questions.
SU(2) ~ O(3) identification
