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I'm a bit confused about this following issue concerning representations of $SU(2)$.

Denote by 1 the 1-dimensional representation of the group $SU(2)$ (=the spin 0). Similarly, denote by 2 and 3 the 2-dimensional (spin 1/2) and 3-dimensional (spin 1) representations, respectively. Also, denote by $(j,j')$ the representation of $SU(2)\times SU(2)$ given by the tensor product of the $j$ representation with the $j'$ representation. By addition of angular momentum, we know that (2,2) = 1+3, where the + sign denotes the direct product of two representations. But since 1 and 3 both represent the same group $SU(2)$, so does their direct sum (a reducible representation). It follows that 1+3 is a representation of both $SU(2)$ and $SU(2)\times SU(2)$. Am I correct?

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    $\begingroup$ Physically you are rotating both spins simultaneously and thus you have a representation of SU(2) and not of SU(2)$\times$SU(2)! $\endgroup$ – Fabian Nov 21 '14 at 11:10
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Your group theory text probably betrayed you if it did not spend much time contrasting the two cases. A possibly related question is 254461. People use loose language and symbols that aggravate the confusion. Talking abstractly without explicit hands-on formulas clinches it (the confusion)!

Let me stick to your 4-dimensional matrices and vectors, all tensor products of 2x2 matrices and 2-vectors in both cases.

  • A tensor-product (Cartesian product) group such as $SU(2)\times SU(2)$ has group elements of the type $\exp (i\theta^a \sigma_a) \times \exp (i\phi^b \tau_b)$, where I am using σ s and τ s for Pauli matrices acting on the left and right spaces, respectively---they are apples and oranges, think of them as spin rotations and isospin rotations. They are side by side. Importantly, their rotation angles are different and hopelessly unrelated, θ for space/spin rotations and φ for isospin rotations. You might combine them into a 4 dim space acted upon by 4×4 matrices, without absolutely any meaning. Crucially, you chose one of many possibilities for your group. The isospin group might have been supplanted by a different flavor group, say SU(3), so, then $SU(2)\times SU(3)$ and you'd have Gell-Mann λ s instead of τ s and your φ s would now be 8.

  • This is very different from a Kronecker product of two representations of SU(2), so the same group, here chosen to be of two doublet representations: adding two spin 1/2s. The group element may be acting on two different spaces, left and right, but with the same angles, like synchronized swimming, $\exp (i\theta^a \sigma_a) \otimes \exp (i\theta^a \tau_a)$. You could have chosen different sized spaces for left and right, but the generators in the exponent should always be representation matrices of SU(2) in the representation of your choice, with the same angle. To see this, consider the coproduct in mathematese, namely $$ \Delta(J^a) = \sigma^a \otimes 1\!\!1 +1\!\!1\otimes \tau^a .$$ This coproduct is easy to see obeys the same Lia algebra as either representation, left or right, here chosen to be both Pauli matrices, "accidentally", by you. So then $ \Delta(J^a) $ is a fine representation of SU(2). Let's saturate it with three angles and exponentiate it to get the group element, $\exp (i \theta^a \Delta(J^a) )= \exp (i \theta^a( \sigma^a \otimes 1\!\!1 +1\!\!1\otimes \tau^a))= \exp (i \theta^a \sigma^a)\otimes \exp (i \theta^a \tau^a) $, where the left and right terms in the exponent commute among themselves, and so the exponential factors to two exponentials acting on the left and right spaces, respectively, in tandem. Now it turns out the 4-dim rep of the $\Delta(J^a)$s is reducible. That is, if we wrote it down as a 4×4 matrix, it would be reducible: that is, a suitable similarity transformation (Clebsch-Gordan) would transform it into a block matrix; here, kind of silly, since it would have a 3×3 block and a 1×1 block with zero entries, as the spin matrices for the singlet rep are 0s. The 3×3 block would be just the spin-one spin matrices, $j^a$. The group action on this rearranged 3+1 space would be $\exp(i\theta^a j^a)\oplus 1$, since exp(i θ 0)=1, quite exceptionally. Such would happen, of course, for all Kronecker products of any reps, not just your chosen example Pauli matrices. So here it does make sense to rearrange the tensor product space, as it facilitates the reduction of the reps.

  • Takeaway: look at the angles---the transformation parameters: when you are combining groups, each group will have different ones, even if the two groups coincide. Now you are ready to face the Lorentz-group reps: different groups, many angles! If, instead, you are combining reps, the angles are the same, as many as the dimensionality of the group/Lie-algebra, but never more.

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I guess what you are missing is the following:

given a representation $\rho(g)$ of $g\in$SU(2) acting on some vector space $V$. We define the representation $\rho_\otimes$ of SU(2) (not of SU(2)$\times$SU(2)) on $V\otimes V$ as $$\rho_\otimes(g) (v_1 \otimes v_2) = \rho(g) v_1 \otimes \rho(g) v_2.$$ So in fact we are defining the tensor product of two representations as a representation of SU(2).

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    $\begingroup$ @ But, I may also look at a rep. of SU(2)*SU(2), defined as follows: Let p be a rep. of SU(2) on a vector space V, and let l belong to SU(2)*SU(2). So l=(g,h) for some g and h in SU(2). Define the rep. p' of SU(2)*SU(2), on the vector space V*V (vector space tensor product), by p'(l)=p(g)*p(h) (matrix tensor product). When we add two spin 1/2 particles, we treat the two spin operators as unrelated, and we don't require that they both operate simultaneously. Why then isn't my rep. the right one? $\endgroup$ – Lior Nov 21 '14 at 14:26
  • $\begingroup$ @Lior: if the two spins are unrelated the representation does not factor into 1+3 but it is irreducible itself. $\endgroup$ – Fabian Nov 21 '14 at 14:45
  • $\begingroup$ But I don't understand where in the process of adding two spin 1/2's we demanded that the two spin together. We started with two 2-dimensional representation of SU(2), which each have as generators the 2x2 matrices Six,Siy,Siz ( i=1,2). Since we were interested in the system of two spins, we took the tensor product of these two vector spaces. This means that we need to extend the generators to Six*I, I*Six ( i=1,2 and I is the unit 2x2 matrix). So we got the 6 generators of SU(2)*SU(2)! $\endgroup$ – Lior Nov 21 '14 at 16:11
  • $\begingroup$ In the previous comment I meant to write that we need to extend the generators to S1x*I, S1y*I, S1z*I, I*S2x, I*S2y, I*S2z (I is the unit 2x2 matrix). $\endgroup$ – Lior Nov 21 '14 at 16:26
  • $\begingroup$ Also, in the traditional treatment of addition of two spins, we specify the state of two spins by the quantum numbers |j1,m1,j2,m2>. m1 and m2 are two different quantum numbers, which means that they are eigenvalues of two different operators. These operators are precisely S1zI and IS2z mentioned in the previous comment. If we used instead S1z*S2z, we would have just one quantum number to use. $\endgroup$ – Lior Nov 21 '14 at 16:28

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