Why the rank of an irreducible tensor must be an integer?

[wdlang]

why not half-integer?

according to the definition, such as [J_z,T^k_q]=q T^k_q

it is quite possible that k can be a half-integer.

 

[DrDu]

Well, in that context, a tensor has usually two indices, i.e. it transforms as the product of two equal irreducible representations of the rotation group these two representations may be either both integer or both half integer. In any case, the Clebsch Gordon decomposition will yield only integer representations.

 

[Entropia]

However, the rank of an irreducible tensor must be an integer because it represents the maximum angular momentum that a physical system can possess.

In quantum mechanics, angular momentum is quantized and can only take on discrete values. This means that the rank of an irreducible tensor, which is related to the angular momentum, must also be a discrete value. Half-integer values would not be allowed in this context.

Furthermore, the rank of an irreducible tensor is also related to the symmetry of a physical system. In order for a physical system to have a well-defined symmetry, the rank must be an integer. Half-integer values would not satisfy this requirement.

Moreover, the mathematical operations involving irreducible tensors, such as addition and multiplication, only work with integer values. Half-integer values would not be compatible with these operations.

Overall, the rank of an irreducible tensor must be an integer because it is a fundamental property of a physical system and is necessary for the proper functioning of mathematical operations and the representation of symmetry. Half-integer values would not fulfill these requirements and therefore cannot be the rank of an irreducible tensor.