# SU(2) ~ O(3) identification

1. Jan 19, 2012

### Matterwave

Hello, I'm reading Ryder's Quantum Field Theory book, and I'm reading the preliminary part where he discusses a little bit of Groups before he introduces the Dirac equation. So, as an example, he is talking about the identification between the SU(2) group and the O(3) group. Before he gets there though, he talks about the transformation properties of the spinors. He makes a statement in equation 2.40 that the outter product of the spinor with it's hermitian conjugate transforms as:

$$\xi\xi^\dagger \rightarrow U\xi\xi^\dagger U^\dagger$$

Later, he introduces another matrix that transforms like this one in equation 2.47. It's the outter product between $(\xi_1, \xi_2)$ (imagine this is a column vector, I forget how to make a column vector on Latex) and $(-\xi_2, \xi_1)$ (this is a row vector). I get that the outter product of these two objects is a matrix which transforms according to the above law, but, I don't see why Ryder has to make this matrix. He later makes the identification with O(3) by identifying this matrix with one he creates from an O(3) vector dotted with the pauli matrices.

I just don't get why he uses this matrix instead of the one in equation 2.40.

2. Jan 19, 2012

### dextercioby

I don't have Ryder right now. From your writings, it looks like he's a little sloppy. O(3) is not <identified> to SU(2). Not even close. From the QFT perspective, Ryder doesn't need O(3), but SU(2), generally SU(n).

3. Jan 19, 2012

### Matterwave

I know that there is a mapping between SU(2) and O(3) right, that SU(2) "double covers" O(3)? I think that's what Ryder is trying to get at here, except I don't get every step in the derivation.

4. Jan 19, 2012

### strangerep

Over on the next page he shows how the elements of that matrix can be identified with (x,y,z) and that an SU(2) transformation on the spinor corresponds to an O(3) transformation on (x,y,z).

I.e., he introduced the matrix because he wants to get to a vector quantity as an outer-product of spinors. (This is analogous to how we build up a 2nd-rank tensor by outer products of vectors, etc.)

5. Jan 19, 2012

### Matterwave

Right, I don't see a problem with doing that step. I was just wondering why he didn't use 2.40 instead? That earlier matrix transforms correctly just as this new matrix. Is there no way to get a correspondence with the first matrix? Is there some theorem in Group theory that proves this or something?

6. Jan 19, 2012

### strangerep

He needs to show that the property of unit determinant in SU(2) corresponds to preservation of vector length in O(3) rotations [eq(2.52)].

7. Jan 19, 2012

### Matterwave

And this cannot be accomplished using the matrix in 2.40? I guess I just see him introduce this new matrix when he had a perfectly good one to start with and got a little side-tracked on why he's doing that.

8. Jan 19, 2012

### strangerep

It could be done with 2.40, and the identification 2.53.

Welcome to the world of physics textbooks (sigh).

Math textbooks usually state things in this order: (1) heuristic motivation for theorem (optional), (2) theorem, (3) proof of theorem.

Ryder's words near 2.53, starting at the black square are the theorem. But he's mixed motivation and proof in the stuff that precedes it. Physics textbooks are often like that. I don't like it much. I prefer a proof to follow the theorem statement. Some folks love Ryder, but I'm not one of them.

9. Jan 19, 2012

### Matterwave

So...did he introduce this new matrix simply to motivate the identifications in 2.53? It seems 2.53 is the important equation here.

10. Jan 19, 2012

### strangerep

It makes it easier to see that the determinant of that matrix is the same thing as the length of the 3-vector. It also makes it easier to see than one can "factorize" an O(3) rotation into that $U$ and $U^\dagger$ acting on either side of the matrix.

But since I have no telepathic ability, I can't be reliably sure why Ryder (or anyone else) does absolutely anything....

(Probably best to move on now, if you understand the math ok.)

11. Jan 20, 2012

### Matterwave

Ok, seems legit. I'll move on then. Thanks.

12. Jan 20, 2012

### dextercioby

For the record and the sake of completeness, also for the people who know their stuff in group theory,

there's no direct mapping of SU(2) onto O(3), but only onto SO(3), which is a different group (and also topological space) than O(3). From the group perspective, O(3)={+1,-1} x SO(3) (direct product). Also, there's a surjective mapping from SU(2) to SO(3) taking 2 different matrices from SU(2) into one single matrix from SO(3) and this thing is heavily exploited in the theory of symmetry and angular momentum in quantum mechanics.

As I said, from the perspective of Standard Model, SU(2) and SU(3) are useful, not rotation groups, so I don't know why Ryder uses them in a misleading way.

13. Jan 21, 2012

### IRobot

And I would add that if you really wanted to do an identification, it could be between $SU(2)/Z_2 \;\text{and}\; SO(3)$.

14. Jan 22, 2012

### lugita15

How exactly is the double cover used in QM? I'm familiar with the mathematics of angular momentum, at least at the level of Sakurai. Are you just talking about how you have rotate a spin-1/2 angular momentum state by 720 degrees to get it back to where it started?

15. Jan 22, 2012

### dextercioby

Yes, out of symmetry requirements, SU(2) is to describe general angular momentum, SO(3) is ok only for orbital angular momentum.

16. Jan 22, 2012

### samalkhaiat

There are two group theoretical reasons why the identification in (2.53) cannot be obtained from the matrix (2.40);
(I)(not mentioned in the book) The tensor product $\chi \chi^{\dagger} \in \ ( [2] \otimes [2^{*}])$ is reducible and therefore cannot be identified with the traceless Hermitian matrix $\sigma_{i}x_{i}$ of eq(2.49).
So, if you want to start from the matrix(2.40), you need to decompose the space of tensor product into invariant subspaces;
$$[2] \otimes [2^{*}] = [3] \oplus [1].$$
This is done by subtracting the trace,
$$\chi \chi^{\dagger} = \{ \chi \chi^{\dagger} - \frac{1}{2} Tr(\chi \chi^{\dagger})I_{2 \times 2}\} + \frac{1}{2}Tr(\chi \chi^{\dagger})I_{2 \times 2}.$$
Now that you have a traceless Hermitian matrix, i.e., irreducible tensor;
$$[3] = \chi \chi^{\dagger} - \frac{1}{2}Tr(\chi \chi^{\dagger})I_{2 \times 2},$$
you can identify it with the traceless Hermitian matrix, $\sigma_{i}x_{i}$, in eq(2.49).

(II) But, you are not done yet, in order to arrive at eq(2.53), you need to use the unique and most important fact about the group $SU(2)$, that is the equivalence between the two fundamental representations $[2]$ and $[2^{*}]$, i.e., the identification
$$\chi \sim (-i\sigma_{2}) \chi^{*}.$$
This is the 2nd reason why the matrix $\chi \chi^{\dagger}$ of eq(2.40) was useless for the job; it considers the conjugate representation, $\chi^{\dagger}\in [2^{*}]$, as a different, inequivalent irreducible representation! This is of course not the case. This point is partially explained in Ryder’s book; see the few lines after eq(2.40).

For a general $SU(n)$ group, the complex conjugate representation $[n^{*}]$, acting on upper spinor with the matrices $U^{*}=(U^{-1})^{T}$, is a different inequivalent irreducible representation. But, in the case of $SU(2)$ the two representations, $[2]$ and $[2^{*}]$, are equivalent. Indeed, in $SU(2)$ the mapping $U \rightarrow U^{*}$ is an inner automorphism, i.e., one obtained by $SU(2)$ similarity transformation. To see this, consider a general group element
$$U = \exp (- \frac{i}{2} \lambda_{i}\sigma_{i}),$$
and the matrix
$$C = -i\sigma_{2}.$$
Since
$$C\sigma_{i}C^{-1} = - \sigma^{*}_{i},$$
thus, conjugating $U$ by $C$ gives
$$CUC^{-1}= U^{*}.$$
This (true only in $SU(2)$) shows that the two representations $[2^{*}]$ and $[2]$ are equivalent, i.e., related by similarity transformation.
OK, let me translate what I have said so far into math and do properly what Ryder is trying to do in his book. Before that though, I need to make the following remark about the relation between the groups $SO(3)$ and $SU(2)$. In group theory, we DO NOT identify different groups; the relation
$$SO(3) \approx SU(2)/Z_{2},$$
means that the homomorphic mapping $SU(2) \rightarrow SO(3)$ has a non-trivial kernel, $K = \{I,-I\}$, that happens to coincide with the centre of $SU(2)$.That is, the two group share the same local structure, i.e., having isomorphic algebras, but differ globally.
For the benefit of those who are studying particle physics, I will do the math using quarks and pions. If we take
$$\chi = \left( \begin{array}{c}u \\ d \end{array} \right) \in [2],$$
$$\chi^{\dagger} = ( u^{*}\ d^{*}) \in [2^{*}],$$
we can form the following tensor product ( in Ryder’s it is eq(2.40))
$$\chi \times \chi^{\dagger} = \left( \begin{array}{cc} uu^{*} & ud^{*} \\ du^{*} & dd^{*} \end{array} \right).$$
Now, according to (I) above, we have, for the triplet (adjoint) representation, the following traceless Hermitian matrix,
$$[3] = \frac{1}{2}\left( \begin{array}{cc} |u|^{2} - |d|^{2} & 2ud^{*} \\ 2du^{*} & |d|^{2} - |u|^{2} \end{array} \right). \ \ \ (1)$$
Now, we take three real numbers $(\pi_{1},\pi_{2},\pi_{3}) \in \mathbb{R}^{3}$ and form the following matrix (eq(2.49 in Ryder’s book)
$$\Pi = \sigma_{i}\pi_{i} = \left( \begin{array}{cc} \pi_{3}& \pi_{1} - i \pi_{2} \\ \pi_{1} + i \pi_{2} & - \pi_{3} \end{array}\right). \ \ \ (2)$$
Since this 2 by 2 matrix is traceless and Hermitian, it must belong to the triplet (adjoint) representation $[3]$ of eq(1), i.e., for all $U \in SU(2), \ \Pi$ transforms as
$$\Pi \rightarrow U \Pi U^{-1}.$$
From this it follows (by taking the determinants of both sides) that
$$\pi_{1}^{2} + \pi_{2}^{2} + \pi_{3}^{2} \ \ \mbox{is invariant}.$$
But this is nothing but the defining action of $SO(3)$ on the real vector $\vec{\pi}\in \mathbb{R}^{3}$.

Now, letting $\Pi = [3]$ we find (the quarks content of the pions)
$$\pi^{+} \equiv \pi_{1} - i \pi_{2} = ud^{*},$$
$$\pi^{-} \equiv \pi_{1} + i \pi_{2} = du^{*},$$
$$\pi^{0} \equiv \pi_{3} = \frac{1}{2}( uu^{*} - dd^{*}).$$
Solving for $\pi_{i}$, we find
$$\pi_{1} = \frac{1}{2}(ud^{*} + du^{*}), \ \ (3a)$$
$$\pi_{2} = \frac{1}{2i}(du^{*} - ud^{*}), \ \ (3b)$$
$$\pi_{3} = \frac{1}{2}(uu^{*} - dd^{*}). \ \ (3c)$$
We said in (II) above that in SU(2) we have the equivalence relation $[2] \sim [2^{*}]$, so for the purpose of transformations, we can put
$$\left( \begin{array}{c} u \\ d \end{array} \right) = \left( \begin{array}{c} - d^{*} \\ u^{*} \end{array}\right). \ \ (4)$$
Putting this in eq(1), we find
$$[3] = \left( \begin{array}{cc} ud & -uu \\ dd & - ud \end{array} \right).$$
This meant to be the same as eq(2.47) in Ryder’s book, but it differs by a minus sign! Where did this minus sign come from? I am too tired now to find out.
Putting eq(4) in (3) we find
$$\pi_{1} = \frac{1}{2}( d^{2} - u^{2} ),$$
$$\pi_{2} = \frac{1}{2i}( u^{2} + d^{2}),$$
$$\pi_{3} = ud.$$
These are exactly eq(2.53) in Ryder’s book. Ok, I think this should be enough. I hope you got the idea.

Regards
Sam

Last edited: Jan 22, 2012