Quantum spin, tensor product: a long time relationship

Question

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.

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 04 '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 04 '17 at 22:12
The decomposition of two spin 1/2 in a spin 0 and spin 1 part is unrelated from the previous observation. The fact that the Hilbert space of a composite system is the tensor product of the individual Hilbert spaces (for distinguishable particles) has to be taken as an axiom of quantum mechanics. Some people are trying to derive this axioms from others but as far as I know there’s no definite result yet. — lcv, Nov 04 '17 at 23:26
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 '19 at 08:31

score 0 · Answer 1 · answered Nov 04 '17 at 23:02

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.

The idea of irreducible representations and decomposing tensor products of representations (for finite groups) was well known from crystallography. But then came Wigner, who considerably expanded the use of representation theory in physics.

Quantum spin, tensor product: a long time relationship

1 Answers1