Matterwave said:
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?
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,
<br />
\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}.<br />
Now that you have a traceless Hermitian matrix, i.e., irreducible tensor;
<br />
[3] = \chi \chi^{\dagger} - \frac{1}{2}Tr(\chi \chi^{\dagger})I_{2 \times 2},<br />
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))
<br />
\chi \times \chi^{\dagger} = \left( \begin{array}{cc} uu^{*} & ud^{*} \\ du^{*} & dd^{*} \end{array} \right).<br />
Now, according to (I) above, we have, for the triplet (adjoint) representation, the following traceless Hermitian matrix,
<br />
[3] = \frac{1}{2}\left( \begin{array}{cc} |u|^{2} - |d|^{2} & 2ud^{*} \\ 2du^{*} & |d|^{2} - |u|^{2} \end{array} \right). \ \ \ (1)<br />
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)
<br />
\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)<br />
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
