quantum mechanics - Tensor Product vs. Direct Product for three spin-1/2 particles

Let us consider three spin-1/2 particles and only focusing on their intrinsic spin $S$. The Hilbert space has then to be $\mathcal H = ℂ^2 ⊗ ℂ^2 ⊗ ℂ^2$. The spin can be described by $V ∈ \text{SU(2)}$ and the fundamental representation $\mathcal D_{1/2}$ with $$\vec{S} = \hbar\vec{M} = \frac{1}{2}\hbar\vec{\sigma}.$$ Let us choose for the base of $ℂ^2$ (1 particle): $$\left\lvert\frac{1}{2},\frac{1}{2}\right\rangle = \left(\begin{array}{cc}(1)\\(0)\end{array}\right)≡\lvert\vec{e}_3\rangle, \quad \left\lvert\frac{1}{2},-\frac{1}{2}\right\rangle = \left(\begin{array}{cc}(0)\\(1)\end{array}\right)≡\lvert-\vec{e}_3\rangle. $$ Furthermore according to the Clebsch-Gordan series one gets: \begin{align} \mathcal D_{1/2}⊗\mathcal D_{1/2}⊗\mathcal D_{1/2} & = \mathcal D_{1/2} ⊗ (\mathcal D_1 ⊕ \mathcal D_0) \\ & = (\mathcal D_{1/2}⊗\mathcal D_1) ⊕ (\mathcal D_{1/2}⊗\mathcal D_0) \\ & = \mathcal D_{3/2} ⊕ \mathcal D_{1/2} ⊕ \mathcal D_{1/2}. \end{align} So we are left with 8 states in the combined 3-particle system.

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Questions:

1. If one would simply consider the direct sum of the three particles, i.e. $ℂ^2 ⊕ ℂ^2 ⊕ ℂ^2$ we would only have 6 states, correct?


2. What is the simplest picture to see the consequences of doing this instead of taking the tensor product?


3. Maybe one could also give me a good (physical) example for the difference of $ℝ^3 ⊗ℝ^2$ versus $ℝ^3 ⊕ℝ^2$ (phase space?).

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I have searched for similar questions and found some stuff. However I am not in particular interested in the look of the outcoming states (i.e. their spin momentum) but more in the differences if one were not considering the tensor product.

Answer

The direct sum of Hilbert spaces is not a "good" notion when talking about spaces of states.

The "actual" states are elements of the projective Hilbert space, where every ray is shrunk to a point to reflect that phases and normalization do not alter the state a vector in the Hilbert space is supposed to represent.

Now, on the projective spaces, the direct sum of the original spaces simply does not produce something that does behave correctly when considering this: Observe that, for states $\phi\in\mathcal{H}_1$ and $\psi\in\mathcal{H}_2$, $(c\phi,\psi) \in\mathcal{H}_1\oplus\mathcal{H}_2$ is not on the same ray as $(\phi,\psi)\in\mathcal{H}_1\oplus\mathcal{H}_2$ for $c \in \mathbb{C} - \{1\}$. So the direct sum does not respect the nature of quantum states, it is simply the wrong notion of product to consider.

On the other hand, the bilinearity of the tensor product means that $(c\phi)\otimes\psi = c \cdot(\phi \otimes \psi)$, so the tensor product does indeed map the product of elements of rays to the same ray in the product space, regardless of which representant we choose.

Therefore, the proper quantum notion of product is the tensor product, not the direct sum. (see also this question, and, as another interesting note, "entangled states" arise simply as the failure of a map from the Cartesian product of the projective spaces to the tensor product to be surjective, since the latter is bigger)

The frequent occurence of direct sums is now because representations of groups have the notion of irreducible representations - representations that do not decompose as the direct sum of other representations.

Therefore, if we wish to get a combined quantum system, we first take the tensor product of their spaces of states (which usually carry representations of the symmetry groups of the theory), and then decompose these tensor products into the direct sum of representations, because the irreducible ones are easier to work with, and, in the case of spin, directly correspond to the total spin of a state in that representation.

(To answer your questions literally and directly:

1. Yes, correct.


2. You have a less dimensional space that doesn't behave properly w.r.t. the projective structure.


3. $\mathbb{R}^3\otimes\mathbb{R}^2 = \mathbb{R}^6$, but $\mathbb{R}^3 \oplus \mathbb{R}^2 = \mathbb{R}^5$ )