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Quantum Mechanics

Two Spin-1/2 Particles

Two spin-1/21/2 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

H=C2⊗C2.\mathcal H = \mathbb C^2\otimes\mathbb C^2.

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

Section titled “Total Spin Operators”

Let S1\mathbf S_1 act on the first spin and S2\mathbf S_2 act on the second. The total spin is

S=S1+S2.\mathbf S=\mathbf S_1+\mathbf S_2.

The standard coupled basis diagonalizes S2S^2 and SzS_z:

S2∣s,m⟩=ℏ2s(s+1)∣s,m⟩,S^2\lvert s,m\rangle = \hbar^2s(s+1)\lvert s,m\rangle,

and

Sz∣s,m⟩=ℏm∣s,m⟩.S_z\lvert s,m\rangle = \hbar m\lvert s,m\rangle.

Since each subsystem has s1=s2=1/2s_1=s_2=1/2, the allowed total spins are

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

Thus

12⊗12=1⊕0.\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-11 representation and a one-dimensional spin-00 representation.

Triplet States

Section titled “Triplet States”

The spin-11 states form a triplet:

∣1,1⟩=∣↑↑⟩,∣1,0⟩=12(∣↑↓⟩+∣↓↑⟩),∣1,−1⟩=∣↓↓⟩.\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 mm: 1,0,−11,0,-1. These states transform among themselves under rotations.

Singlet State

Section titled “Singlet State”

The remaining state is the spin-00 singlet:

∣0,0⟩=12(∣↑↓⟩−∣↓↑⟩).\lvert 0,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \right).

It satisfies

S2∣0,0⟩=0,Sz∣0,0⟩=0.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)∣0,0⟩=∣0,0⟩.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

Section titled “Symmetry Under Exchange”

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

∣1,m⟩⟼+∣1,m⟩,∣0,0⟩⟼−∣0,0⟩.\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

Section titled “How the Decomposition Is Found”

Start from the highest-weight state. The state ∣↑↑⟩\lvert\uparrow\uparrow\rangle has m=1m=1, so it must be the top of a spin-11 multiplet:

∣1,1⟩=∣↑↑⟩.\lvert 1,1\rangle = \lvert\uparrow\uparrow\rangle.

Applying the total lowering operator S−=S1−+S2−S_-=S_{1-}+S_{2-} gives the m=0m=0 triplet state:

∣1,0⟩=12(∣↑↓⟩+∣↓↑⟩).\lvert 1,0\rangle = \frac{1}{\sqrt2} \left( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \right).

The other independent m=0m=0 state must be orthogonal to it. Normalization fixes the singlet:

∣0,0⟩=12(∣↑↓⟩−∣↓↑⟩).\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. The relative sign between the two product states is physical: it distinguishes the singlet from the m=0m=0 triplet.

Physical Interpretation

Section titled “Physical Interpretation”

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

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

H=J S1⋅S2H = J\,\mathbf S_1\cdot\mathbf S_2

is diagonal in the singlet/triplet basis because

S1⋅S2=12(S2−S12−S22).\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( S^2-S_1^2-S_2^2 \right).

Thus the same tensor-product space can look simple or complicated depending on which symmetry is being used.

For the general basis dictionary beyond two spin-1/21/2 particles, see Coupled and Uncoupled Bases.

Common Mistakes

Section titled “Common Mistakes”
  • Treating 12⊗12=1⊕0\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=0m=0 state a singlet. The triplet also has an m=0m=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.
Section titled “Cross-Links”

References

Section titled “References”
  • 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.
  • A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
  • D. A. Varshalovich, A. N. Moskalev, and V. K. Khersonskii, Quantum Theory of Angular Momentum, World Scientific, 1988.

Exercises

Section titled “Exercises”
  1. Verify that the singlet is orthogonal to the m=0m=0 triplet.
Solution

Compute the inner product:

⟨1,0∣0,0⟩=12(⟨↑↓∣+⟨↓↑∣)(∣↑↓⟩−∣↓↑⟩)=12(1−1)=0.\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}
  1. Show that S1⋅S2\mathbf S_1\cdot\mathbf S_2 has different eigenvalues on the triplet and singlet subspaces.
Solution

Use

S1⋅S2=12(S2−S12−S22).\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( S^2-S_1^2-S_2^2 \right).

For two spin-1/21/2 particles,

S12=S22=34ℏ2.S_1^2=S_2^2=\frac34\hbar^2.

For the triplet, s=1s=1, so

S1⋅S2=12(2ℏ2−34ℏ2−34ℏ2)=14ℏ2.\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=0s=0, so

S1⋅S2=12(0−34ℏ2−34ℏ2)=−34ℏ2.\mathbf S_1\cdot\mathbf S_2 = \frac12 \left( 0-\frac34\hbar^2-\frac34\hbar^2 \right) = -\frac34\hbar^2.