We know that the Newton binomial formula is valid for numbers(adsbygoogle = window.adsbygoogle || []).push({});

in elementary algebra.

Is there an equivalent formula for commuting spherical tensors? If so,

how is it?

To be specific let us suppose thatAandBare 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

[tex]

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

[/tex],

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

[tex]

(\frac{4!}{2! (4-2)!} )(\left[ A^2_0 B^2_0\right]_0+

\left[ A^2_1 B^2_1 \right]_0+

\left[ A^2_2 B^2_2 \right]_0)

[/tex],

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

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# Binomial formula for spherical tensors

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