advanced28 min studyLesson 469 in Physics

Molecular spin statistics and identical nuclei

Enforce total nuclear exchange symmetry, count symmetric and antisymmetric spin states, and connect allowed rotations to spin-isomer populations.

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What you’ll understand

  • Apply identical-nucleus exchange symmetry to the full molecular state.
  • Count two-nucleus spin functions in both exchange sectors.
  • Distinguish statistical weights, equilibrium populations, and slow spin conversion.

Exchanging identical nuclei cannot produce a physically different labeled molecule, but the molecular wavefunction must still acquire the correct bosonic or fermionic exchange sign. This requirement correlates nuclear spin symmetry with rotational symmetry. It explains why even and odd rotational ladders can have different statistical weights and why hydrogen has distinct ortho and para spin modifications. We will apply the symmetry to the complete state rather than treating nuclear spins as classical parallel arrows, then separate allowed-state counting from temperature-dependent populations and conversion kinetics.

The full exchange sign is the constraint

Identical nuclei of integer spin are composite bosons and those of half-integer spin are composite fermions, provided their internal nuclear states remain fixed. Exchanging the nuclei requires the full molecular wavefunction to be symmetric or antisymmetric accordingly. Electronic, vibrational, rotational, and nuclear-spin factors all contribute to that sign. For ground-state hydrogen and deuterium in the simple electronic and vibrational state considered here, the electronic-vibrational factor is symmetric, leaving rotation and spin to supply the required product.

Exchanging the two nuclei reverses the diatomic internuclear direction. A spherical harmonic of rotational quantum number J changes by (−1)J(-1)^J, so even J is symmetric and odd J antisymmetric. This is a nuclear permutation statement; it should not be substituted indiscriminately for an electronic gerade/ungerade label. Other electronic states can contribute another exchange sign, and different isotopes such as H and D are distinguishable nuclei, so the same identical-pair restriction does not apply to HD.

P^12Ψ=ηNΨ,ηN=(−1)2I,P^12YJM(R^)=(−1)JYJM(R^)\hat P_{12}\Psi=\eta_N\Psi,\qquad \eta_N=(-1)^{2I},\qquad \hat P_{12}Y_{JM}(\widehat R)=(-1)^JY_{JM}(\widehat R)
I is each identical nucleus’s spin quantum number. Eta N is plus one for integer I and minus one for half-integer I; the product of all molecular exchange factors must match it.↔ Scroll wide equations horizontally.

Count spin states by symmetry

Each nucleus has 2I+12I+1 spin-projection basis states. Two equal projections make symmetric product states. For each pair of unequal projections, the sum and difference of their two orderings make one symmetric and one antisymmetric spin function. This counting gives (2I+1)(I+1)(2I+1)(I+1) symmetric functions and (2I+1)I(2I+1)I antisymmetric functions, adding to the expected square of the one-nucleus state count. It concerns dimensions of exchange sectors rather than classical alignment probabilities.

For two protons, I is one half: the symmetric sector is the three-state triplet and the antisymmetric sector the one-state singlet. In ground electronic-vibrational H two, fermionic exchange requires even J to pair with the antisymmetric singlet, called para, and odd J with the symmetric triplet, called ortho. For two deuterons, I is one and the required total sign is bosonic; the symmetric six-state sector pairs with even J and the antisymmetric three-state sector with odd J under the same spatial assumption.

gsym=(2I+1)(I+1),gantisym=(2I+1)I,NJ∝gJ(2J+1)e−EJ/(kBT)g_{\mathrm{sym}}=(2I+1)(I+1),\qquad g_{\mathrm{antisym}}=(2I+1)I,\qquad N_J\propto g_J(2J+1)e^{-E_J/(k_BT)}
The population formula additionally assumes thermal equilibrium, including interconversion between the relevant spin sectors. The factor g J is selected by total exchange symmetry; the rotational M degeneracy is a separate factor.↔ Scroll wide equations horizontally.

Weights do not fix a universal mixture

The three-to-one proton spin-state count does not mean the ortho-to-para population ratio is always three. Rotational energies and their degeneracies enter the partition sums, and at low temperature the lowest para rotational state is favored in equilibrium. A high-temperature limit can approach the familiar statistical ratio when many even and odd rotational states contribute comparably. Pairwise level ratios and complete spin-isomer fractions are different calculations, so a two-level estimate must not be presented as the full high-temperature partition sum.

Spin conversion can be slow because ordinary electric-dipole or Raman interactions often preserve nuclear spin to high accuracy. A cooled sample may therefore retain a nonequilibrium ortho/para composition until a suitable conversion mechanism acts. Spectral line alternation also depends on thermal populations and transition strengths in addition to spin weights. The symmetry rule defines which states are allowed; kinetics determines whether those allowed sectors equilibrate during an experiment. Neither the equilibrium rule nor a degeneracy factor by itself supplies a conversion timescale.

Compare two hydrogen rotational populations

For ground electronic-vibrational H two, assume spin conversion has equilibrated. Compare the J = 1 population with J = 0 when (E1−E0)/(kBT)=ln⁡3(E_1-E_0)/(k_BT)=\ln3. Include nuclear-spin and rotational degeneracies, but do not infer a complete ortho fraction from only these two levels.

  1. Even J = 0 pairs with the antisymmetric proton singlet, so its nuclear-spin factor is 1 and rotational factor is 2(0)+1=12(0)+1=1.

  2. Odd J = 1 pairs with the symmetric proton triplet, giving nuclear-spin factor 3 and rotational factor 3. Their degeneracy product relative to J = 0 is therefore 9.

  3. The Boltzmann factor is e−ln⁡3=1/3e^{-\ln3}=1/3. Hence N1/N0=9/3=3N_1/N_0=9/3=3. Other J levels can still contribute to the total spin-isomer populations.

The result

The J = 1 level has three times the J = 0 population at the stated temperature ratio. This pairwise result is not a temperature-independent ortho/para mixture rule.

Explore the model

Interactive lab

Interactive lab · quantum

Spin-sector labels require external symmetry rules

Treat the two basis labels as two already allowed complete molecular states, not as individual proton arrows. Reset, set Mixing angle θ = 0° and Relative phase φ = 0°, and predict basis-zero weight 1. For a fresh comparison Reset, set θ = 90° and φ = 0°, and predict equal pure-state weights.

Spin-sector labels require external symmetry rulesA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab is a coherent pure two-level model, not a thermal mixture or two-nucleus spin solver. It does not impose exchange symmetry, generate singlet/triplet multiplicities, or simulate ortho–para conversion.

Check your understanding

For ground electronic-vibrational H two with the stated symmetric spatial factor, which pairing satisfies proton exchange statistics?

Try it yourself

For ground electronic-vibrational D two, each identical deuteron has I = 1. Find symmetric and antisymmetric nuclear-spin counts and the equilibrium ratio N1/N0N_1/N_0 when the J = 1 to J = 0 Boltzmann factor is 1/3.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Deuterons require a symmetric total state; include both the selected nuclear-spin factor and the 2J plus one rotational degeneracy.

Compare with the solution

The symmetric spin count is (3)(2)=6(3)(2)=6 and antisymmetric count (3)(1)=3(3)(1)=3. Even J = 0 has spin factor 6 and rotational factor 1; odd J = 1 has spin factor 3 and rotational factor 3. Thus N1/N0=[3(3)/(6(1))](1/3)=1/2N_1/N_0=[3(3)/(6(1))](1/3)=1/2. The result differs from H two because the required total exchange sign and the spin-sector dimensions differ.

Take it with you

  • Total molecular exchange symmetry correlates nuclear spin with rotation and other spatial factors.
  • Spin-sector dimensions differ for proton and deuteron pairs and are not classical arrow counts.
  • Equilibrium populations require Boltzmann sums and equilibration, while conversion rates are separate dynamical information.

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