Tensor Product vs. Direct Product for Three Spin-1/2 Particles

For three distinguishable spin-1/2 particles, the state space is built by taking a tensor product of the state spaces for the individual particles. Its dimension is eight. When you instead organize those states by total angular momentum, the eight-dimensional space splits into a direct sum of one spin-3/2 space and two spin-1/2 spaces.

The terms can be confusing because “direct product” is sometimes used informally for a tensor product. The distinction below uses the standard linear-algebra meanings.

The composite state space is a tensor product

A single spin-1/2 particle has a two-dimensional Hilbert space with basis states $|\uparrow\rangle$ and $|\downarrow\rangle$. For three distinguishable particles, the composite space is

$$\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2\otimes\mathcal{H}_3.$$

Each factor has dimension two, so

$$\dim\mathcal{H}=2\cdot2\cdot2=8.$$

A product basis consists of the eight kets formed by choosing up or down for each particle, such as $|\uparrow\uparrow\uparrow\rangle$, $|\uparrow\downarrow\uparrow\rangle$, and $|\downarrow\downarrow\downarrow\rangle$. The notation abbreviates a tensor product: $|\uparrow\downarrow\uparrow\rangle=|\uparrow\rangle_1\otimes|\downarrow\rangle_2\otimes|\uparrow\rangle_3$.

The full space also contains superpositions of these basis states. For instance, $(|\uparrow\downarrow\uparrow\rangle+|\downarrow\uparrow\uparrow\rangle)/\sqrt{2}$ is a valid state, and it cannot be described as a choice of one spin state for each particle. In general, a state in a tensor product need not be a product state.

What an ordinary direct product means

For vector spaces, the direct product of a finite number of spaces is equivalent to their direct sum. Thus the ordinary vector-space product $\mathbb{C}^2\times\mathbb{C}^2\times\mathbb{C}^2$ has dimension six, not eight. It describes triples of two-component vectors, but it does not provide the eight-dimensional space of joint quantum states.

There are contexts where “direct product” is used loosely to mean the tensor product of quantum state spaces. If that is what a source means, its intended construction may be the one above. Check the dimension and the definition being used. For independent quantum systems, the standard construction is the tensor product.

Adding the three spins gives a direct sum

Define the total spin operator by $\mathbf{S}=\mathbf{S}_1+\mathbf{S}_2+\mathbf{S}_3$. You can choose a basis that labels states by the total angular momentum quantum numbers $j$ and $m$, along with a multiplicity or coupling label where needed. In that basis, the tensor-product space decomposes as

$$\tfrac12\otimes\tfrac12\otimes\tfrac12 =\tfrac32\oplus\tfrac12\oplus\tfrac12.$$

Here the symbols on the left describe angular-momentum spaces combined by a tensor product; the symbols on the right describe a direct sum of total-spin sectors. A spin-$j$ sector has $2j+1$ states, one for each $m=-j,-j+1,\ldots,j$. Therefore the dimensions are

$$4+2+2=8.$$

The spin-3/2 sector is a quartet. The two spin-1/2 sectors are separate doublets: they have the same total spin but are distinct subspaces, corresponding to two ways of coupling the particles.

Worked example: identify the sectors

The state with all three spins up has total magnetic quantum number $m=3/2$ and is the highest-weight state of the quartet:

$$|\tfrac32,\tfrac32\rangle=|\uparrow\uparrow\uparrow\rangle.$$

The quartet’s state with $m=1/2$ is the symmetric combination

$$|\tfrac32,\tfrac12\rangle=\frac{1}{\sqrt{3}}\bigl(|\uparrow\uparrow\downarrow\rangle+|\uparrow\downarrow\uparrow\rangle+|\downarrow\uparrow\uparrow\rangle\bigr).$$

There are three product-basis states with $m=1/2$, so two independent combinations remain for the two spin-1/2 sectors. One way to construct one of them is to first combine particles 1 and 2 into a singlet, then combine that pair with particle 3:

$$|\tfrac12,\tfrac12\rangle_{(12)\text{ singlet}}=\frac{1}{\sqrt{2}}\bigl(|\uparrow\downarrow\uparrow\rangle-|\downarrow\uparrow\uparrow\rangle\bigr).$$

This state has total spin 1/2. A second spin-1/2 state comes from coupling particles 1 and 2 into a triplet and then combining that triplet with particle 3. The choice of coupling order changes which basis states represent the two doublets, but it does not change the total space or its dimension.

Common mistakes

• Calling the eight-dimensional state space a direct product. Some sources use that phrase informally, but tensor product is the precise standard term for combining quantum systems.
• Confusing a direct sum with a tensor product. Tensor-product dimensions multiply; direct-sum dimensions add. Here the tensor product builds the space, and the direct sum describes its decomposition into total-spin sectors.
• Assuming every joint state is a product state. The eight product-basis kets span the space, but their superpositions include entangled states.
• Thinking the two spin-1/2 sectors are the same subspace. They have the same total-spin value but are distinct copies, giving the decomposition dimensions $4+2+2$ rather than $4+2$.