I Equivalence of SU(2) and O(3) in Ryder's QFT book

1. Aug 30, 2016

Glenn Rowe

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?

2. Aug 30, 2016

Demystifier

$U$ belongs to SU(2), but nobody said that $h$ belongs to SU(2). Think about the quantity $e^{h}=1+h+...$!

3. Aug 30, 2016

Glenn Rowe

I realize that $h$ isn't in SU(2) but I don't see how that helps. The point of Ryder's argument seems to be that because $\det h^{\prime}=\det h$ we can say that $x^{\prime2}+y^{\prime2}+z^{\prime2}=x^{2}+y^{2}+z^{2}$, but if $\det h = x^{2}+y^{2}+z^{2}$ is identically zero, that doesn't seem to mean anything.

4. Aug 30, 2016

Staff: Mentor

I haven't gone through the calculations, but what came to my mind is, that $\det h =0$ cannot occur at all (in linear groups).

5. Aug 30, 2016

vanhees71

These matrices are not part of the group but the Lie algebra. For SU(2) you have $U=\exp(-\mathrm{i} \vec{\sigma} \cdot \vec{\phi}/2)$. Since $\ln \det{U}=\mathrm{Tr}(-\mathrm{i} \vec{\sigma} \cdot \vec{\phi}/2]=0$ and $U^{\dagger}=U^{-1}$ you have $\vec{\sigma}^{\dagger}=\vec{\sigma}$, i.e., the Lie algebra su(2) consists of all hermitean traceless $2 \times 2$ matrices.

Further what's constructed here is a group homomorphism $\mathrm{SU}(2) \rightarrow \mathrm{SO}(3)$ by acting with the SU(2) matrices on the Lie-algebra elements, the adjoint representation of SU(2) on its Lie algebra (a very natural construct for Lie groups and their algebras) . It thus turns out that SO(3) is the adjoint representation of SU(2). It's of course not one-to-one, because to each SO(3) you can find two SU(2) matrices, leading to the same result, namely with $U$ also $-U$ is represented by the same SO(3) matrix. The kernel of the homomorphism is inded $\{1,-1 \} \subset \mathrm{SU}(2)$.

Another way to see that is that you need $|\vec{\phi}|=4 \pi$ to get $U=1$.

6. Sep 1, 2016

feld

I have been reading the same section in ryder's book recently.
Ryder identifies h with H.
Clearly both H and h are traceless, so the identification is OK so far.
Ryder states that h is hermitian, but this is assuming x, y, z are real.
However it may readily be checked that H is not hermitian, except trivially.
If we are to continue with the identification of h and H, h must be nonhermitian, i.e. x, y, z cannot be all real.
So the interpretation of rotations in 3-space breaks down.
I find it better to instead consider the matrix $K=\xi \xi^\dagger$ and to identify it with
$$k=t+\boldsymbol{\sigma}\cdot\mathbf{r}=\left[\begin{array}{cc}t+z & x-iy\\ x+iy & t-z\end{array}\right]$$
Then setting $\xi_1 = A + iB$, $\xi_2 = C + iD$, where A, B, C, D are real, we find
\begin{align} x &= AC + BD \\ y &= AD - BC \\ z &= {\frac 1 2}(A^2 + B^2 - C^2 - D^2) \\ t &= {\frac 1 2}(A^2 + B^2 + C^2 + D^2) \end{align}
which are all real.
The unitary transformation $k^\prime = UkU^\dagger$ preserves determinant, meaning $t^2 - x^2 - y^2 - z^2$ is an invariant.
It is readily found that x, y, z transform as before, as in ryder's (2.54), and $t^\prime=t$. So it is a rotation in 3-space that is being performed.
I hope this has been helpful. But if you are to continue with the book, please be warned - I have found the error rate to be high.

7. Sep 1, 2016

Glenn Rowe

Thanks feld, that helps a bit. I have a feeling that I need more background in group theory to really understand what's going on. At the moment, the only experience I've had with groups is what's in Ryder's chapter 2, so I have no experience with Lie algebras or groups, so the more more technical explanations above are beyond me.
Thanks also for the warning about the errors. Sadly there seems to be no errata page for Ryder's book (at least a quick Google didn't show any). My search for a QFT book that is understandable for self-study (something in a style similar to Griffiths's great textbooks on quantum mechanics and electrodynamics would be ideal) continues.....

8. Sep 1, 2016

vanhees71

A very nice book on this topic is

Roman U. Sexl and Helmuth K. Urbandtke. Relativity, Groups, Particles. Springer, Wien, 2001.

9. Sep 1, 2016

feld

I think ryder is trying to justify the use of dirac matrices. Once he 'explains' that the SU(2) matrices describe rotations in 3-space, he then introduces imaginary angles to describe boosts. It is known from special relativity that boosts are rotations through an imaginary angle. However, with SU(2) the transformations are confined to 3-space; time is supposedly not present, so I fail to see how boosts can be described!
Nevertheless, dirac matrices are constructed from the SU(2) matrices, and they work well to describe rotations and boosts, even though ryder's argumentation would suggest otherwise.
So I suspect ryder's argument is simply misleading, and shouldn't be taken too seriously.
Have you looked at weinberg's 'the quantum theory of fields', volume 1? I know it's not very accessible, but it is quite definitive, thoroughly rigorous, and in my opinion a safe introduction to the subject, even if it may take a few reads to fully appreciate his argument.

10. Sep 1, 2016

Demystifier

For a QFT book at a level and style similar to Griffiths, I would suggest
Greiner & Reinhardt, Field Quantization
https://www.amazon.com/Field-Quantization-Walter-Greiner/dp/3540780483

In the same series from the same principal author, you can also find a good book on symmetries, groups and algebras in quantum physics
https://www.amazon.com/Quantum-Mech...990&sr=1-1&keywords=Greiner+Muller+symmetries

Last edited by a moderator: May 8, 2017
11. Sep 1, 2016

vanhees71

The analogue of SU(2) representations of rotations (in fact SU(2) is the covering group of the rotation group SO(3)) for the proper orthochronous Lorentz group is $\mathrm{SL}(2,\mathbb{C})$.

12. Sep 3, 2016

feld

After discussing SU(2) and orthogonal rotations in 3-space, ryder discusses SL(2,ℂ) and lorentz transformations in spacetime.
The rotation operators are stated as $J_i =\sigma_i/2$ and the boost generators $K_i= -i\sigma_i/2$. This is standard theory, and correct, but the only justification he gives for this choice is that they obey the correct lorentz group commutation relations. To be more convincing, he could have demonstrated their transformation properties. Also, he makes errors along the way.
Ryder's argument goes as follows.
Introducing the spinor $\xi=(\xi_1,\xi_2)$, it is clear that under a unitary transformation $\xi^\prime=U\xi$, the quantity
$$K=\xi\xi^\dagger= \left(\begin{array}{cc}\xi_1\xi^*_1 & \xi_1\xi^*_2\\\xi_2\xi^*_1 & \xi_2\xi^*_2\end{array}\right)$$
transforms according to $K^\prime=UKU^\dagger$.
Ryder shows that under a unitary transformation, $\xi$ and $(-\xi^*_2,\xi^*_1)$ transform in the same way. So the quantity
$$H=\left(\begin{array}{cc}\xi_1\xi_2 & -\xi_1^2\\\xi_2^2 & -\xi_1\xi_2\end{array}\right)$$
will transform in the same way as $K$.
He then introduces the quantity
$$h=\boldsymbol{\sigma}\cdot\mathbf{x}=\left(\begin{array}{cc}z & x-iy\\x+iy & -z\end{array}\right)$$
and identifies $h$ with $H$. However, $h$ is hermitian (for real $\mathbf x$), but $H$ cannot be. Furthermore, as pointed out in post #1 by Glenn Rowe, $H$ has zero determinant, which can cause interpretational difficulty.
As mentioned in post #6, I suspect that ryder should instead have worked with $K$, and identified it with
$$k=t+\boldsymbol{\sigma}\cdot\mathbf{x}=\left(\begin{array}{cc}t+z & x-iy\\ x+iy & t-z\end{array}\right)$$
Under the general transformation $M=\exp(i\boldsymbol{J}\cdot\boldsymbol{\theta}+i\boldsymbol{K}\cdot\boldsymbol{\phi})$, involving rotations and boosts, the determinant $\det(k)=t^2-{\mathbf r}^2$ is invariant, essentially because the pauli matrices are traceless. This establishes that $M$ performs lorentz transformations. But I consider it would be worthwhile demonstrating that the transformation $k^\prime=\exp(-i\mathbf{J}_3 \theta) k \exp(i\mathbf{J}_3 \theta)$ say corresponds to the rotation
\begin{align} x^\prime&=x\cos\theta - y\sin\theta \nonumber\\ y^\prime&=x\sin\theta + y\cos\theta \nonumber\\z^\prime&=z,~t^\prime=t\nonumber\end{align}
and the transformation $k^\prime=\exp(-i\mathbf{K}_1 \phi) k \exp(i\mathbf{K}_1 \phi)$ say corresponds to the boost
\begin{align} x^\prime&=x\cosh\phi + t\sinh\phi\nonumber \\ t^\prime&=x\sinh\phi+t\cosh\phi\nonumber\\y^\prime&=y,~z^\prime=z.\nonumber\end{align}
I think that with these additions, ryder's sections on spinors and lorentz transformations become more comprehensible.

13. Sep 3, 2016

samalkhaiat

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

Last edited: Sep 3, 2016