1
$\begingroup$

In Ch.19 of the textbook An Introduction to Quantum Field Theory by Peskin and Schroeder, on P.680 the property of a quantity

$$\mathcal{A}^{abc}=\mathrm{tr}\left[t^a\{t^b,t^c\}\right]\tag{19.132}$$ is discussed. As a part of the scattering amplitude $$\langle p,\nu,b;k,\lambda,c|\partial_\mu j^{\mu \alpha}|0\rangle=\frac{g^2}{8\pi^2}\epsilon^{\alpha\nu\beta\lambda}p_\alpha k_\beta \cdot \mathcal{A}^{abc}, \tag{19.131}$$ $\mathcal{A}^{abc}$ is a number and it contains all the indices regarding isospin (flavor). Therefore, it is an invariant with respect to any relevant transformation in isospin space.

The textbook reads

For example, in $SU(2)$ the adjoint representation has spin 1. The symmetric product of two spin-1 multiplets gives spin 0 plus spin 2, with no spin-1 component. Thus, there is no symmetric tensor coupling two spin-1 indices to give a spin 1.

As $t^a$s are generators of transformations, trace and product are linear operations, therefore the quantity $\mathcal{A}$ is related to the linear combination of the basis of the representation in question.

Edit

However, I understand that the above rules for angular momentum are applied when direct-product of physical states is involved, namely, $3 \otimes 3= 1\oplus 3\oplus 5$ as discussed in Mike's answer. For the present case, if the same arguments are employed to derive the properties of the l.h.s. of (19.132), an invariant (relate to the base of a scalar representation), from the l.h.s. of the relation, why the latter only involves "normal" matrix product instead of direct product.

Also, as explained in Mike's answer, invariant tensors are simply Clebsh-Gordan coefficients. Therefore, (19.132) can be viewed as a relation between different C-G coefficients. So does that mean the r.h.s. of (19.132) is vanishing since the correponding C-G coefficient is vanishing, which does not involve what stays on the r.h.s. of (19.132)?

I probably have missed something very basic, thanks a lot for pointing out my misunderstanding.

$\endgroup$
2
$\begingroup$

Invariant tensors are just a form of Clebsh-Gordan coefficients. In partiular, if there was a symmetric invariant tensor ${\mathcal A}_{ijk}$ with indices in adjoint (the vector rep of ${\rm SO}(2)$ then for any two vectors $u_i$, $v_i$ the quantity $w_k={\mathcal A}_{ijk}u_iv_i$ would be a vector. We would therefore have found a new (symmetric) way to make a vector out of two vectors that differs from the usual (antisymmetric) vector product. We know this is impossible because we know that the 3 (i.e the spin $j=1$) in $3 \otimes 3= 1\oplus 3\oplus 5$ is antisymmetric.

$\endgroup$
  • $\begingroup$ My confusing is the following. The rules for angular momentum addition is essentially the Clebsh-Gordan coefficients. But why $\mathcal{A}^{abc}$ defined above are Clebsh-Gordan coefficient? Thanks! $\endgroup$ – gamebm Jun 3 at 18:52
  • 2
    $\begingroup$ @gamebm Invariant tensors are Clebsh-Gordan coefficients. For example in ${\rm SO}(3)$ the Levi-Civita symbol is the Clebsh for coupling two spin-1 vectors to get a third vector. The $\delta^{ij}$ couples two spin-one vectors to get a spin-0 scalar. Similarly the Pauli sigma's couple a spin-1/2 to conjugate spin-12/ to get a spin one vector. $\endgroup$ – mike stone Jun 3 at 19:58
  • $\begingroup$ Invariant tensors are C-G coefficients, the relation (19.132) therefore states a relation between C-G coefficients (on both sides). We deduced that the l,h.s. $\mathcal{A}^{abc}$ as a totally symmetric tensor does not exist, and therefore it must vanish. The argument is not from the r.h.s. of the relation but from the properties of the group SU(3). I think I still am wrong. I cannot establish the relationship between the adjoint representation on the r.h.s. and the rule for spin addition. The index of $\mathcal{A}$ are those represent different generators, not basis... what shall I read? $\endgroup$ – gamebm Jun 3 at 22:02
  • 1
    $\begingroup$ @gamebm You can look at Q6 this homework set: courses.physics.illinois.edu/phys509/sp2019/hw7.pdf which is from the group theory part of my math course. It's aimed at SU(3) but the ideas are general. $\endgroup$ – mike stone Jun 4 at 12:40
  • $\begingroup$ thx! Now it becomes clear that it is definitely something beyond my understanding. I thought it were something straightforward, will study. Again, many thanks for the answer and explanation. $\endgroup$ – gamebm Jun 5 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.