# Quantum spin, tensor product: a long time relationship [duplicate]

Anyone who has studied quantum mechanics know the following relation: $2 \otimes 2 = 3 \oplus 1$

But how did a man woke up and said "Hell yeah, I'll use tensor product of two spin $1/2$ to simulate the interaction of two particles with spin $1/2$" ? Why didn't he start with the direct sum ?

(And then, group theory made the magic leading to the relation above)

In fact, i'm wondering this because i don't fully understand why we use the tensor product to unite the two Hilbert's space.

## marked as duplicate by Jon Custer, ZeroTheHero, Kyle Kanos, Cosmas Zachos, Aaron StevensAug 28 at 17:45

• Are you referring to Clebsch–Gordan expansion? – Sunyam Nov 4 '17 at 22:07
• If two particles each have $n$ states, the total number of possible states the two particles can be in is reasonably $n^2$, not $2n$. So that favors tensor product over direct sum. – knzhou Nov 4 '17 at 22:12
• Not really, I just wonder how did Pauli (I assume it's him) knew he had to do a tensor product (and then the CG expansion) and not something else (like a direct sum) – user65854 Nov 4 '17 at 22:12
• Total spin of two spin-1/2 particles – Frobenius Nov 4 '17 at 23:03
• Possible duplicate of Should it be obvious that independent quantum states are composed by taking the tensor product? (and several other related questions). – Emilio Pisanty Aug 19 at 8:31

The decomposition $\frac{1}{2}\otimes \frac{1}{2}\to 0\oplus 1$ is useful when the Hamiltonian (or at least its dominant part) is $SU(2)$-invariant. The part that mixes $SU(2)$ irreps can then be treated perturbatively.
Group theory notwithstanding, it’s a pretty natural thing to do from the perspective of finding eigenstates of commuting operators, since the eigenstates of the total angular momentum $\hat J^2$ naturally break into a (degenerate) triplet and a singlet.