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- - **Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)**
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Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)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) [tex]\xi[/tex] does not transform like [tex]\xi^{+}[/tex], but [tex]\[ \left( \begin{array}{c} \xi_1 \\ \xi_2 \end{array} \right)\][/tex] and [tex]\[ \left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)\][/tex] do. - where does he get that (is it magic)??
- I can choose another spinor and it also transforms "the same way"
- ...what does he mean by "the same way"???
I tried all four possibilities: [tex]\xi' \equiv U\xi, \xi^{+}' \equiv \xi^{+}U^{+}, ...[/tex] 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: [tex]\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[/tex]. - if [tex]\xi \equiv (\xi_1 \xi_2)[/tex] then [tex]\xi^{+}[/tex] is not what he uses here - he uses that mysterious other spinor that is
**NOT**[tex]\xi^{+}[/tex]!! - 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.) |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)Quote:
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: [tex]\xi \xi^{+} \equiv \begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix}\begin{pmatrix} -\xi_2 & \xi_1 \end{pmatrix} \equiv -H[/tex] |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)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 [itex]\mathbf v[/itex], you can associate a traceless hermitian 2 x 2 matrix [tex]\mathcal V = \mathbf v \cdot \mathbf \sigma}[/tex]. An element [itex]U \in SU(2)[/itex] induces a transformation [tex]\mathcal V' = U \mathcal V U^\dagger[/tex] The goal is to find a rotation matrix [itex]R \in SO(3)[/itex] such that [tex] \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. |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)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 [itex]\xi[/itex] and [itex]\xi^\dagger[/itex] 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[/tex] [tex](\xi^\dagger)^\prime = \xi^\dagger U^\dagger[/tex] These are the two different ways. For them to transform the same way would be [tex]\xi^\prime = U\xi[/tex] [tex](\xi^\dagger)^\prime = U\xi^\dagger[/tex] 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: [tex]\xi^\prime = U\xi[/tex] and the second one says [tex]\chi^\prime = U\chi[/tex] where [tex]\chi = \[ \left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)\][/tex] This is what the author means by transform the same way. I will address H and h in my next post. |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)Quote:
[tex]h^\prime = UhU^\dagger[/tex] Then if U belongs to SU(2), it has determinant 1 and so det h = det h'. Computing these gives [tex]x^{\prime 2} + y^{\prime 2} + z^{\prime 2} = x^2 + y^2 + z^2[/tex] the new position vector r' = (x', y', z') is r rotated. That is to say, given U in SU(2) the mapping [tex]r \rightarrow h \rightarrow h^\prime \rightarrow r^\prime[/tex] is a rotation, an element of O(3). |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)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 R^{3} 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 R^{3} since [tex]h \equiv \sigma_i r^i[/tex].3) IOW, a transformation in SU(2) , i.e. h' = UhU ^{+} is also a rotation in O(3), namely: (x,y,z) [tex]\rightarrow[/tex] (x',y',z') (i.e. |r'| = |r|.)4) The last point is that the reason Jimmy's [tex]\chi[/tex] vector is transformed "the same way" as the original [tex]\xi[/tex] vector is because I can get back the same [tex]\xi'[/tex] by a) taking the c.c. of each component, and then taking (-1) times [tex]\chi_2[/tex]'. So this is my last question: is Jimmy's [tex]\chi[/tex] vector (Ryder's (-[tex]\xi_2^* \xi_1^*[/tex]) 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 |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)Quote:
You should verify that [itex]\det h=-|\vec r|^2[/itex], and that a transformation of the form [itex]h\mapsto UhU^\dagger[/itex] 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 [itex]\mathbb R^3[/itex] via the isomorphism between the space mentioned above and [itex]\mathbb R^3 [/itex]. Note also that -U corresponds to the same member of SO(3) as U. |

Re: Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)Quote:
But let me ask you this: do I need all that stuff with [tex]\xi[/tex]? 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 formh = [tex]\sigma_ir^i[/tex]. Then I used the U derived (note, no where is that [tex]\xi[/tex] 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 [tex]\xi[/tex] 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... rocks.PF-joe |

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