# Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)

1. Nov 17, 2010

### Living_Dog

I have been struggling with this for a long time. I gave up to review GR and came back to Ryder. I started in Chapter 2 and the material was easier and more intuitive on this second pass. But the same topic, "SU(2) and the rotation group" has trapped me yet again. I am bogged down specifically on pages 34 - 36. So here are my questions:

1) From pg. 35: Under SU(2) $$\xi$$ does not transform like $$\xi^{+}$$, but $$$\left( \begin{array}{c} \xi_1 \\ \xi_2 \end{array} \right)$$$ and $$$\left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)$$$ do.
1. where does he get that (is it magic)??
2. I can choose another spinor and it also transforms "the same way"
3. ...what does he mean by "the same way"???

I tried all four possibilities: $$\xi' \equiv U\xi, \xi^{+}' \equiv \xi^{+}U^{+}, ...$$ and the other two with the second spinor. None of them look the same, and they all transform "the same way"!

2) From pg. 36:

$$\xi \xi^{+} \equiv $\left( \begin{array}{c} \xi_1 \\ \xi_2 \end{array} \right)$ $\left( \begin{array}{cc} -\xi_2 & \xi_1 \end{array} \right)$ \equiv -H$$.

• if $$\xi \equiv (\xi_1 \xi_2)$$ then $$\xi^{+}$$ is not what he uses here - he uses that mysterious other spinor that is NOT $$\xi^{+}$$!!
• what's the point of calling this thing 'H'??
• why does he construct 'h' - as if from nowhere then he says 'h' is 'H'!
• since 'h' is a 2x2 matrix, how can it act on r??

If one thing is clear from my post it is that I have no clue what Ryder is talking about on these two short pages.

Thanks in advance to anyone for anything anywhere at any time!!!

(PS: I can't get that column vector times a row vector to equal the 'H' matrix to display properly. Sorry.)

Last edited: Nov 17, 2010
2. Nov 17, 2010

### Fredrik

Staff Emeritus
I don't have the book, so I'm not sure if I'm even talking about the same thing, but it sounds like this is what you need, and maybe this too (but the latter is for the 4-dimensional case).

If the book can be previewed at Google Books, you should link directly to the relevant pages. If not, consider including more information.

I recommend \begin{pmatrix} for your LaTeX matrices. Click the quote button to see how I did this:

$$\xi \xi^{+} \equiv \begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix}\begin{pmatrix} -\xi_2 & \xi_1 \end{pmatrix} \equiv -H$$

3. Nov 18, 2010

### matonski

That section confuses me as well. I know the big picture is that he is explicitly constructing a homomorphism that maps SU(2) onto SO(3).

To any 3-vector $\mathbf v$, you can associate a traceless hermitian 2 x 2 matrix
$$\mathcal V = \mathbf v \cdot \mathbf \sigma}$$.

An element $U \in SU(2)$ induces a transformation
$$\mathcal V' = U \mathcal V U^\dagger$$

The goal is to find a rotation matrix $R \in SO(3)$ such that
$$\mathcal V' = (R\mathbf v) \cdot \mathbf \sigma[/itex]. I believe those two pages in Ryder are going through the details of how to find an R given a U and vice versa. It turns out that there are two different U's for any given R. 4. Nov 18, 2010 ### Jimmy Snyder I have the second edition in front of me. The text you refer to is on page 33 of this edition. The text below eqn (2.40) says: We we see from (2.39) that $\xi$ and $\xi^\dagger$ transform in different ways. Here is eqn (2.39). Note that the arrow means 'transforms to'. I prefer the following notation which the author also uses in eqns (2.41) and (2.42) [tex]\xi^\prime = U\xi$$
$$(\xi^\dagger)^\prime = \xi^\dagger U^\dagger$$

These are the two different ways. For them to transform the same way would be

$$\xi^\prime = U\xi$$
$$(\xi^\dagger)^\prime = U\xi^\dagger$$

but this is not how they transform. However, if you look carefully at eqns (2.41) and (2.42), you will see that the first one says:

$$\xi^\prime = U\xi$$

and the second one says

$$\chi^\prime = U\chi$$
where
$$\chi = $\left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)$$$
This is what the author means by transform the same way. I will address H and h in my next post.

5. Nov 18, 2010

### Jimmy Snyder

Yes, but onto O(3). Given a position vector r = (x, y, z), use eqn (2.49) to construct h. Then use U to transform h:
$$h^\prime = UhU^\dagger$$
Then if U belongs to SU(2), it has determinant 1 and so det h = det h'. Computing these gives
$$x^{\prime 2} + y^{\prime 2} + z^{\prime 2} = x^2 + y^2 + z^2$$
the new position vector r' = (x', y', z') is r rotated. That is to say, given U in
SU(2) the mapping
$$r \rightarrow h \rightarrow h^\prime \rightarrow r^\prime$$
is a rotation, an element of O(3).

6. Nov 19, 2010

### Living_Dog

From your three posts I have made progress. I still have one question remaining. But first let me summarize to see if I have it correct:

1) If one performs a rotation in R3 then one is also simultaneously performing a rotation in the space for SU(2).

2) This is b/c 'h', a 2x2 matrix in SU(2), is also written with the x, y, z coordinates of r, a vector in R3 since $$h \equiv \sigma_i r^i$$.

3) IOW, a transformation in SU(2) , i.e. h' = UhU+ is also a rotation in O(3), namely: (x,y,z) $$\rightarrow$$ (x',y',z') (i.e. |r'| = |r|.)

4) The last point is that the reason Jimmy's $$\chi$$ vector is transformed "the same way" as the original $$\xi$$ vector is because I can get back the same $$\xi'$$ by a) taking the c.c. of each component, and then taking (-1) times $$\chi_2$$'.

So this is my last question: is Jimmy's $$\chi$$ vector (Ryder's (-$$\xi_2^* \xi_1^*$$) spinor) unique?? I tried finding others, but it seems that since there are only two operations (c.c. and multiplication by -1), then this other spinor is unique.

Thanks guys (esp. to matonski for validating my state of confusion over these short pages). I feel I have made more progress with one post than all the time I spent wasting lead and pulp over this small issue.

May God richly bless you all, in Jesus' name, amen.

-joe

Last edited: Nov 19, 2010
7. Nov 19, 2010

### Fredrik

Staff Emeritus
The h in Jimmy's post isn't a member of SU(2). It's a member of the set of complex self-adjoint traceless 2×2 matrices. Since that set is closed under addition and closed under multiplication by a real number, it's a vector space over the real numbers. And it's 3-dimensional, since the set of Pauli spin matrices is a basis. That means that it's isomorphic to $\mathbb R^3$.

You should verify that $\det h=-|\vec r|^2$, and that a transformation of the form $h\mapsto UhU^\dagger$ doesn't change the determinant. This is the main reason why this transformation can be interpreted as a rotation. Each member of SU(2) defines a rotation of $\mathbb R^3$ via the isomorphism between the space mentioned above and $\mathbb R^3$. Note also that -U corresponds to the same member of SO(3) as U.

8. Nov 21, 2010

### Living_Dog

I did verify all that.

But let me ask you this: do I need all that stuff with $$\xi$$? The reason is that I just went over it again and found that I can deduce the exact same results skipping all that voodoo by just starting with the Pauli matrices, which are a basis, and r is a vector and then gleefully say that I can form

h = $$\sigma_ir^i$$.​

Then I used the U derived (note, no where is that $$\xi$$ being mentioned) to transform h into h'. I get the same exact results without appealing to magic statements. I say this b/c after finding the "transforms the same way" spinor, he doesn't use it! Instead he uses the "transforms the same way" adjoint and then forms the matrix "-H"!

Sorry, but since all the witchcraft with $$\xi$$ is skipped, on this one I think I am correct. Anyway, the problem is resolved and thanks to all you guys for helping me out... PF rocks.

-joe