Why tensor opeator?

[KFC]

In books about angular momentum, they introduce the so call tensor operator to deal with angular momentum, but why's that and what does it look like? In the cover page of J.J.Sakurai's textbook, there is a block matrices, is that any relation to tensor operator?

Thanks in advance.

[tim_lou]

I'm not really sure what tensor operator means in your context, my guess is that take two operators in two hilbert spaces, say
a \textrm{ for } \mathcal{H}
and
b \textrm{ for } \mathcal{H'}

then we define
a\otimes b \textrm{ acts on } \mathcal{H} \otimes \mathcal{H'}
(a\otimes b )\left(\sum \left | i\right>\otimes \left | j\right>\right)=\sum a\left | i\right>\otimes b\left | j\right>

For example, we can take a one particle hilbert space, take it's tensor product so that we have a two particle hilbert space. Then we can get operators that acts on this hilbert space by specifiying it's action on the first particle and the second particle. Of course, more general operator may not be tensor products of operators.

Typically, we naturally identify
L_1 = L_1 \otimes \textrm{id}
and
L_2 = \textrm{id} \otimes L_2

if L1 originally acts on the first hilbert space and L2 on the second.

[olgranpappy]

I think the OP means "tensor" in the same sense as the total angular momentum would be called a (rank one) tensor operator. I.e., the total angular momentum is a vector operator.

In quantum mechanics this means that

[J_i,J_j]=i\epsilon_{ijk}J_k\;,

where the J_i etc are components of the total angular momentum and the epsilon_{ijk} is the Levi-Civita symbol.

Other tensor operators obey other types of commutation relations from which useful results can be derived. This is why we care about tensor operators.

See page 193 of sakuri for more regarding vector operators and block diagonal matrices.

[Naty1]

This may be too basic for you, but offers a reasonable introduction to tensors..

http://www.lerc.nasa.gov/WWW/K-12/Numbers/Math/documents/Tensors_TM2002211716.pdf

Page four introduces tensors via vectors:

" Notice that the effect of multiplying the unit vector by the scalar is to change the magnitude from unity to something else, but to leave the direction unchanged. Suppose we wished to alter both
the magnitude and the direction of a given vector. Multiplication by a scalar is no longer sufficient. Forming the cross product with another vector is also not sufficient, unless we wish to limit the change in direction to right angles. We must find and use another kind of mathematical ‘entity.’ "