Two Spin-1/2 Particles

Two spin-$1/2$ particles give the simplest nontrivial example of angular momentum addition. Each particle has a two-dimensional spin Hilbert space, so the composite spin space is

$$\mathcal H = \mathbb C^2\otimes\mathbb C^2.$$

Treat the factors as two specified spin degrees of freedom. For identical particles, the constraints on the complete spatial-plus-spin state will be imposed separately. The uncoupled basis labels each spin separately:

$$\lvert\uparrow\uparrow\rangle,\qquad \lvert\uparrow\downarrow\rangle,\qquad \lvert\downarrow\uparrow\rangle,\qquad \lvert\downarrow\downarrow\rangle.$$

Angular momentum addition asks for a different basis: one that diagonalizes the total spin operators.

Total Spin Operators

Let $\mathbf S_1$ act on the first spin and $\mathbf S_2$ act on the second. The total spin is

$$\mathbf S=\mathbf S_1+\mathbf S_2.$$

The standard coupled basis diagonalizes $S^2$ and $S_z$:

$$S^2\lvert s,m\rangle = \hbar^2s(s+1)\lvert s,m\rangle,$$

and

$$S_z\lvert s,m\rangle = \hbar m\lvert s,m\rangle.$$

Since each subsystem has $s_1=s_2=1/2$, the allowed total spins are

$$s=1 \qquad\text{or}\qquad s=0.$$

Thus

$$\frac12\otimes\frac12 = 1\oplus 0.$$

This statement is not ordinary multiplication of numbers. It says that a four-dimensional tensor-product representation decomposes into a three-dimensional spin-$1$ representation and a one-dimensional spin-$0$ representation.

Triplet States

The spin-$1$ states form a triplet:

$$\begin{aligned} \lvert 1,1\rangle &= \lvert\uparrow\uparrow\rangle,\\ \lvert 1,0\rangle &= \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \right),\\ \lvert 1,-1\rangle &= \lvert\downarrow\downarrow\rangle. \end{aligned}$$

They are called triplet states because there are three values of $m$: $1,0,-1$. These states transform among themselves under rotations.

Singlet State

The remaining state is the spin-$0$ singlet:

$$\lvert 0,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right).$$

It satisfies

$$S^2\lvert 0,0\rangle=0, \qquad S_z\lvert 0,0\rangle=0.$$

The singlet is rotationally invariant up to phase. In fact, for ordinary spin rotations it is invariant:

$$U(R)\lvert 0,0\rangle = \lvert 0,0\rangle.$$

This makes the singlet a central object in spin correlations and entanglement. Its rotational meaning is developed in Singlet and Triplet States, while its composite-systems role is developed in Composite-Systems Singlet and Triplet States. Here it is the first place where angular momentum addition becomes visible.

Symmetry Under Exchange

If the two spin labels are exchanged, the triplet states are symmetric while the singlet is antisymmetric:

$$\lvert 1,m\rangle \longmapsto +\lvert 1,m\rangle, \qquad \lvert 0,0\rangle \longmapsto -\lvert 0,0\rangle.$$

This exchange symmetry is not yet the full identical-particle symmetrization postulate. It is a property of the spin part of the two-particle state. The angular-momentum preview is Identical Particles and Exchange Symmetry Preview. For identical particles, the total state also includes spatial and possible internal factors, as developed in Spin and Spatial Wavefunctions.

How the Decomposition Is Found

Start from the highest-weight state. The state $\lvert\uparrow\uparrow\rangle$ has $m=1$, so it must be the top of a spin-$1$ multiplet:

$$\lvert 1,1\rangle = \lvert\uparrow\uparrow\rangle.$$

Applying the total lowering operator $S_-=S_{1-}+S_{2-}$ gives the $m=0$ triplet state:

$$\lvert 1,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \right).$$

The other independent $m=0$ state must be orthogonal to it. Normalization fixes its magnitude, and the standard phase convention chooses:

$$\lvert 0,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right).$$

The minus sign is a convention-dependent phase choice only in the sense that an entire state may be multiplied by an overall phase. Once the one-spin basis phases have been fixed, the relative sign between the two product states is physical: it distinguishes the singlet from the $m=0$ triplet.

Physical Interpretation

The uncoupled basis answers: “What is each spin’s $z$ component?” The coupled basis answers: “What is the total spin and its $z$ component?”

Both bases span the same four-dimensional space. The physics decides which basis is natural. A Hamiltonian such as

$$H = K\,\mathbf S_1\cdot\mathbf S_2/\hbar^2$$

is diagonal in the singlet/triplet basis because

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( S^2-S_1^2-S_2^2 \right).$$

Here $K$ is an energy. The triplet energy is $K/4$, the singlet energy is $-3K/4$, and the gap is $|K|$. This convention keeps the coupling distinct from an angular-momentum quantum number.

For the general basis dictionary beyond two spin-$1/2$ particles, see Coupled and Uncoupled Bases.

Common Mistakes

• Treating $\frac12\otimes\frac12=1\oplus0$ as arithmetic instead of representation decomposition.
• Forgetting that $\lvert\uparrow\downarrow\rangle$ and $\lvert\downarrow\uparrow\rangle$ are not themselves states of definite total spin.
• Calling every $m=0$ state a singlet. The triplet also has an $m=0$ state.
• Confusing exchange symmetry of the spin state with the full identical-particle symmetrization rule.
• Missing the relative sign that separates the singlet from the triplet.

Exercises

1. Verify that the singlet is orthogonal to the $m=0$ triplet.

Solution

Compute the inner product:

$$\begin{aligned} \langle 1,0|0,0\rangle &= \frac12 \left( \langle\uparrow\downarrow| + \langle\downarrow\uparrow| \right) \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right)\\ &= \frac12(1-1)\\ &=0. \end{aligned}$$

2. Show that $\mathbf S_1\cdot\mathbf S_2$ has different eigenvalues on the triplet and singlet subspaces.

Solution

Use

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( S^2-S_1^2-S_2^2 \right).$$

For two spin-$1/2$ particles,

$$S_1^2=S_2^2=\frac34\hbar^2.$$

For the triplet, $s=1$, so

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( 2\hbar^2-\frac34\hbar^2-\frac34\hbar^2 \right) = \frac14\hbar^2.$$

For the singlet, $s=0$, so

$$\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( 0-\frac34\hbar^2-\frac34\hbar^2 \right) = -\frac34\hbar^2.$$

3. In the product state $|\uparrow\downarrow\rangle$, what are the probabilities of measuring total spin zero and one? Does a basis change itself prepare an entangled state?

Solution

Invert the two $m=0$ combinations:

$$|\uparrow\downarrow\rangle =\frac{|1,0\rangle+|0,0\rangle}{\sqrt2}.$$

The two total-spin outcomes each have probability $1/2$. This rewriting leaves the original product state unchanged. A subsequent selective total-spin measurement can prepare a singlet or the entangled $m=0$ triplet, but changing coordinates alone does not change a state.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.