A question in Weinberg's QFT

[ndung200790]

In &5.6 writes:
"An (A,A) field (A is spin) contains terms with only integer spins 2A,2A-1,...,0, and corresponds to a traceless symmetric tensor of rank 2A.(Note that the number of independent components of a symmetric tensor of rank 2A in four(space-time) dimensions is:
{4.5...(4+2A-1)}/(2A)!=(3+2A)!/6(2A)!
and the tracelessness condition reduces this to:
{(3+2A)!/6(2A)!}-{(1+2A)!/6(2A-2)!}=(2A+1)^2
as expected for an (A,A) field"

I can not derive the number of components of a tensor of rank 2A in four dimensions is
4.5...(4+2A-1) and the number of tracelessnes conditions is (1+2A)!/6(2A-2)!

 

[ndung200790]

And I do not understand why a (1,0) or (0,1) field corresponds to antisymmetric tensor
F$^{\mu\nu}$ and satisfies the condition:
F$^{\mu\nu}$=$\pm$$\epsilon$$^{\mu\nu\lambda\rho}$F$_{\lambda\rho}$
for (1,0) and (0,1) fields,respectively.

 

[Bill_K]

And I do not understand why a (1,0) or (0,1) field corresponds to antisymmetric tensor

In a footnote on the previous page, Weinberg mentions an "alternative formalism", in which the representations are written as symmetric SU(2) tensors with two-valued indices of two types - dotted and undotted. Thus, the (1,0) representation is the symmetric tensor φ_AB, while the (0, 1) representation is φ_A'B'. In addition we have available the antisymmetric symbol ε_AB which is used to raise and lower. From these quantities, a Lorentz vector index can be constructed by means of the Pauli matrices σ^μ_AB'

So, given a (1, 0) quantity F_AB, we construct F^μν = σ^μ_AB' σ^ν_CD' F^AC ε^B'D' which is the antisymmetric tensor that corresponds to it.

 

[ndung200790]

In four dimensions a tensor of rank 2A means there are 2A indexes and each index runs on 0,1,2,3.Then there are all 4^2A components.But in Weinberg's there are 4.5...(4+2A-1) components in this tensor.Where is the wrong in my argument?

 

[Bill_K]

In four dimensions a tensor of rank 2A means there are 2A indexes and each index runs on 0,1,2,3.Then there are all 4^2A components.But in Weinberg's there are 4.5...(4+2A-1) components in this tensor.Where is the wrong in my argument?

It's a symmetric tensor. In n dimensions a symmetric tensor of rank r has n^r components, but only some of them are independent. The number of independent components is given by the binomial coefficient, (n + r - 1)!/r!(n - 1)!

 

[ndung200790]

And what is the difference (in tensor forms) between (1,0) field and (0,1) field?

 

[Bill_K]

And what is the difference (in tensor forms) between (1,0) field and (0,1) field?

A (1, 0) field is represented by a rank-two quantity F_AB, which has three independent complex components. So does the F_μν that you can construct from it. See this earlier thread.

 

[andrien]

In &5.6 writes:
"An (A,A) field (A is spin) contains terms with only integer spins 2A,2A-1,...,0, and corresponds to a traceless symmetric tensor of rank 2A.(Note that the number of independent components of a symmetric tensor of rank 2A in four(space-time) dimensions is:
{4.5...(4+2A-1)}/(2A)!=(3+2A)!/6(2A)!
and the tracelessness condition reduces this to:
{(3+2A)!/6(2A)!}-{(1+2A)!/6(2A-2)!}=(2A+1)^2
as expected for an (A,A) field"

I can not derive the number of components of a tensor of rank 2A in four dimensions is
4.5...(4+2A-1) and the number of tracelessnes conditions is (1+2A)!/6(2A-2)!

This is something you will not expect from Weinberg to explain because it is a simple combinatorics.Anyway,when you have a symmetric tensor of order say r into N dimensional space then you can simply specify it by how many indices are zero or 1 and so on.There order does not matter(if you want you can take an ascending order with repetition allowed to find it).So the problem is basically to partition r objects into N sets ignoring the relative positioning.It requires N-1 boundaries among r objects.This boundary and objects together can be arranged in N+r-1! ways and the object itself can be arranged in r! ways.Besides it the boundary can be permuted in N-1! ways.Hence the number of independent components are N+r-1!/(N-1!)(r!).(A simple intution is obtained from second rank tensor in which all the elements above or below the diagonal+diagonal elements are the numbers which is simply 6+4=10).
Also a traceless tensor will have the condition that if you chose any two indices for contraction,then the condition is simply
Tr(aabcd...)=0.so here the elements are now only (r-2) because first two indices are summed over for trace.So just put r-2 in place of r in the above formula and you have N+r-3!/(N-1!)(r-2!).Now this gives your result for N=4.

 

[Bill_K]

Good explanation!