Binomial formula for spherical tensors

[francesco75]

We know that the Newton binomial formula is valid for numbers
in elementary algebra.
Is there an equivalent formula for commuting spherical tensors? If so,
how is it?
To be specific let us suppose that A and B are two spherical tensors
of rank 1 and I want to calculate (A + B)⁴ and I want
the result to be a scalar. The two tensors commute, AB=BA.

If I apply naively the binomial formula I have the usual expansion,
with one of term being

$$\frac{4!}{2! (4-2)!} A^2 B^2$$,

where the superscript denotes the number of tensors, i.e. the order.
But according to the rules of the tensors coupling algebra, there are different
ways to couple 4 tensors of rank 1 to a scalar. So my wild guess is that the above
terms should be
$$(\frac{4!}{2! (4-2)!} )(\left[ A^2_0 B^2_0\right]_0+<br /> \left[ A^2_1 B^2_1 \right]_0+<br /> \left[ A^2_2 B^2_2 \right]_0)$$,
where the subscripts are the total rank of the couplings, which are 0,1 and 2 for two tensors of rank 1 coupled togheter.
The other terms should go the same way, that is one is putting all the possible
coupling to the scalar according to the orders of the tensors.
Am I correct?
Tank you very much

 

[mfb]

If the tensors commute, the addition is a group, the distributive property is satisfied and the tensors are associative, you can use the binomial formula (and you can follow its proof step by step to show this).

In your example, (A+B)^4 = (A+B)(A+B)(A+B)(A+B) = (AA+AB+BA+BB)(AA+AB+BA+BB) = AAAA+AAAB+... = AAAA+4AAAB+6AABB+4ABBB+BBBB.

 

[francesco75]

thanks He,
but still my problem remains, how to write down explicitly the term
you denote as 6AABB, taking into account that the tensors have a law
of composition or coupling. Is it correct to write AABB as the sum of all the possible
coupling of the four tensors to a scalar as I did in the last equation of my previous message?

 

[mfb]

I don't understand your couplings -> no idea.

 

[francesco75]

the coupling is the one for spherical tensors, with coefficients given by Clebsch-Gordan coefficients.
For instance (again considering A end B as tensor of rank 1)
$\left[AB\right]_0=<1 0 1 0|1 0> A_0 B_0 +<1 -1 1 1|1 0> A_{-1} B_1 +<1 1 1 -1|1 0> A_{1} B_{-1}$,
where the Clebsch are <j1 m1 j2 m2| J M>
and for instance $A_{i}$ is the component i_th of the tensor A,
that we can express in cartesian representation as
$A_{-1} = \frac{1}{\sqrt{2}}(a_x - i a_y)$,
$A_{0} = a_z$,
$A_{-1} = \frac{1}{\sqrt{2}}(a_x +i a_y)$,
and a_x, a_y and a_z are the cartesian components.

So,
$A_0^2 \equiv \left[AA\right]_0$,
$A_1^2 \equiv \left[AA\right]_1$,
and so on

The other couplings can be constructed by using the tensor of rank one and the opportune Clebsch-Gordan
coefficients, according to the total rank of the tensors coupled together