Spin and Spatial Wavefunctions

For identical particles with spin, exchange symmetry applies to the total wavefunction, not to the spatial part alone and not to the spin part alone. If

$$q=(\mathbf x,s)$$

collects spatial and spin variables, the exchange rule acts as

$$(q_1,q_2) \longmapsto (q_2,q_1).$$

When a two-particle state factors as

$$\Psi(q_1,q_2) = \psi(\mathbf x_1,\mathbf x_2)\, \chi(s_1,s_2),$$

the exchange symmetry of the total state is the product of the exchange symmetries of the spatial and spin factors.

Total Exchange Symmetry

Let the spatial factor have exchange parity $\eta_{\mathrm{space}}$:

$$\psi(\mathbf x_2,\mathbf x_1) = \eta_{\mathrm{space}}\, \psi(\mathbf x_1,\mathbf x_2), \qquad \eta_{\mathrm{space}}=\pm1.$$

Let the spin factor have exchange parity $\eta_{\mathrm{spin}}$:

$$\chi(s_2,s_1) = \eta_{\mathrm{spin}}\, \chi(s_1,s_2), \qquad \eta_{\mathrm{spin}}=\pm1.$$

Then the total exchange parity is

$$\eta_{\mathrm{total}} = \eta_{\mathrm{space}}\eta_{\mathrm{spin}}.$$

Identical bosons require

$$\eta_{\mathrm{total}}=+1,$$

while identical fermions require

$$\eta_{\mathrm{total}}=-1.$$

This product rule is the main bookkeeping principle of this page.

Two Spin-One-Half Fermions

For two spin-$1/2$ particles, the spin states decompose into a symmetric triplet and an antisymmetric singlet. The triplet states are

$$\chi_{1,1} = \lvert\uparrow\uparrow\rangle,$$ $$\chi_{1,0} = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle + \lvert\downarrow\uparrow\rangle \bigr),$$

and

$$\chi_{1,-1} = \lvert\downarrow\downarrow\rangle.$$

They satisfy

$$P_{12}\chi_{1,m} = \chi_{1,m}.$$

The singlet is

$$\chi_{0,0} = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \bigr),$$

and satisfies

$$P_{12}\chi_{0,0} = -\chi_{0,0}.$$

The angular-momentum derivation and entanglement properties are developed in Singlet and Triplet States and Two Spin-1/2 Particles. The symmetry-volume bridge is Identical Particles and Exchange Symmetry Preview. Here the point is the exchange-symmetry pairing.

Fermionic Pairing Rule

Electrons are spin-$1/2$ fermions, so their total two-electron state must be antisymmetric:

$$\eta_{\mathrm{space}}\eta_{\mathrm{spin}}=-1.$$

There are two factorized possibilities:

$$\begin{array}{ccl} \text{symmetric spatial} &\times& \text{antisymmetric spin singlet},\\ \text{antisymmetric spatial} &\times& \text{symmetric spin triplet}. \end{array}$$

Thus a spin singlet does not by itself make a valid two-electron state; it must be paired with a symmetric spatial factor. A spin triplet must be paired with an antisymmetric spatial factor.

This rule explains why two electrons can share the same spatial orbital only in a singlet spin state. If both electrons occupy the same spatial orbital $\phi(\mathbf x)$, the spatial factor

$$\psi(\mathbf x_1,\mathbf x_2) = \phi(\mathbf x_1)\phi(\mathbf x_2)$$

is symmetric, so the spin factor must be antisymmetric.

Bosonic Pairing Rule

For identical bosons, the total two-particle state must be symmetric:

$$\eta_{\mathrm{space}}\eta_{\mathrm{spin}}=+1.$$

If the particles are spinless bosons, there is no spin factor to compensate an antisymmetric spatial factor, so the spatial wavefunction must be symmetric.

For bosons with spin, the factorized possibilities are

$$\begin{array}{ccl} \text{symmetric spatial} &\times& \text{symmetric spin},\\ \text{antisymmetric spatial} &\times& \text{antisymmetric spin}, \end{array}$$

when the relevant spin states exist. The total exchange symmetry, not either factor alone, is the physical constraint.

Helium Ground-State Preview

In the simplest helium ground-state approximation, both electrons occupy the same spatial orbital $\phi_{1s}(\mathbf x)$. The spatial part is

$$\psi_{\mathrm{space}}(\mathbf x_1,\mathbf x_2) = \phi_{1s}(\mathbf x_1)\phi_{1s}(\mathbf x_2),$$

which is symmetric under exchange. Therefore the spin part must be the singlet:

$$\chi_{\mathrm{spin}} = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle \bigr).$$

The total approximate state is

$$\Psi(q_1,q_2) = \phi_{1s}(\mathbf x_1)\phi_{1s}(\mathbf x_2) \frac{1}{\sqrt2} \bigl[ \alpha(s_1)\beta(s_2) - \beta(s_1)\alpha(s_2) \bigr].$$

This is the same closed-shell structure written by the two-electron Slater determinant built from spin-orbitals $\phi_{1s}\alpha$ and $\phi_{1s}\beta$.

Excited Two-Electron States

Suppose two electrons occupy different orthonormal spatial orbitals $a$ and $b$. Define symmetric and antisymmetric spatial combinations:

$$\psi_+(\mathbf x_1,\mathbf x_2) = \frac{1}{\sqrt2} \bigl[ a(\mathbf x_1)b(\mathbf x_2) + b(\mathbf x_1)a(\mathbf x_2) \bigr],$$

and

$$\psi_-(\mathbf x_1,\mathbf x_2) = \frac{1}{\sqrt2} \bigl[ a(\mathbf x_1)b(\mathbf x_2) - b(\mathbf x_1)a(\mathbf x_2) \bigr].$$

For two electrons:

$$\psi_+ \chi_{0,0}$$

is allowed, because symmetric spatial times antisymmetric spin gives an antisymmetric total state. Also

$$\psi_- \chi_{1,m}$$

is allowed for $m=1,0,-1$, because antisymmetric spatial times symmetric spin gives an antisymmetric total state.

The antisymmetric spatial factor has an exchange node:

$$\psi_-(\mathbf x,\mathbf x)=0.$$

This often changes electron-electron interaction energies and produces singlet-triplet splittings in two-electron systems. The size and sign of actual level splittings depend on the Hamiltonian and the orbitals, not on exchange symmetry alone.

Molecular Bonding Preview

The same bookkeeping appears in the Heitler–London picture developed quantitatively for the hydrogen molecule. Let $a(\mathbf x)$ and $b(\mathbf x)$ be spatial orbitals localized near the two nuclei. The two-electron spatial combinations are schematically

$$\psi_\pm(\mathbf x_1,\mathbf x_2) \propto a(\mathbf x_1)b(\mathbf x_2) \pm b(\mathbf x_1)a(\mathbf x_2).$$

For electrons, the symmetric spatial combination pairs with the spin singlet, while the antisymmetric spatial combination pairs with the triplet. The symmetric spatial state can enhance amplitude in the internuclear region in simple bonding models, whereas the antisymmetric spatial state has a node structure. A realistic molecular calculation also needs Coulomb interactions, nuclear motion, basis choices, and correlation effects, so this is only a preview; Valence Bond Theory develops the normalized nonorthogonal structures, resonance problem, and many-electron spin coupling.

Slater Determinants and Spin Adaptation

Slater determinants automatically enforce antisymmetry of the total spin-orbital wavefunction. A determinant built from spin-orbitals

$$\varphi_i(q)=\phi_i(\mathbf x)\chi_i(s)$$

does not separately ask whether “the spatial part” and “the spin part” are symmetric. It antisymmetrizes the complete spin-orbitals.

For closed-shell two-electron states, a determinant can factor neatly into a symmetric spatial part and a singlet spin part. For open-shell systems, a single determinant may not be an eigenstate of total spin. Spin-adapted configuration state functions are often built from linear combinations of determinants. That technology belongs to later quantum chemistry and computational pages; the exchange-symmetry rule here is the conceptual foundation.

Nonfactorized States

Not every physical state factors as

$$\psi(\mathbf x_1,\mathbf x_2)\chi(s_1,s_2).$$

Spin-orbit coupling, magnetic-field gradients, relativistic corrections, or correlated many-body calculations can produce states where spin and spatial variables are not separable. In such cases, the rule is still simple: exchange the full labels $q_1$ and $q_2$ and check the symmetry of the total wavefunction.

For identical fermions,

$$\Psi(q_2,q_1) = -\Psi(q_1,q_2),$$

and for identical bosons,

$$\Psi(q_2,q_1) = \Psi(q_1,q_2).$$

The factorized pairing table is a useful special case, not the definition of exchange symmetry.

Common Mistakes

• Symmetrizing or antisymmetrizing only the spatial wavefunction while forgetting spin.
• Saying that opposite-spin electrons are distinguishable particles. Their spin states differ, but the electrons are still identical fermions.
• Thinking a triplet spin state is forbidden for electrons. It is allowed when the spatial factor is antisymmetric.
• Thinking a singlet spin state is always the ground state. Energetics depend on the Hamiltonian.
• Treating a Slater determinant as if it always factors into a simple spatial part times a spin part.
• Forgetting that exchange symmetry applies to all one-particle degrees of freedom, not just position.

Cross-Links

• Symmetric and Antisymmetric Wavefunctions
• Pauli Principle in Atoms applies the pairing rule to equivalent-electron terms, shell closure, and atomic periodicity.
• Fermions
• Exchange Operators
• Singlet and Triplet States
• Valence Bond Theory
• Slater Determinants
• Pauli Exclusion Principle
• Symmetrization Postulate
• Identical Particle Exercises
• Occupation-Number Basis
• Fermionic Fock Space
• Two Spin-1/2 Particles
• Identical-Particle Scattering
• Entanglement in Quantum Chemistry
• Helium Atom
• Variational Estimate for the Helium Atom
• Formula Sheet

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• J. C. Slater, “The Theory of Complex Spectra,” Physical Review 34, 1293-1322, 1929.
• 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. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover, 1996.
• C. Cohen-Tannoudji, B. Diu, and F. Laloe, Quantum Mechanics, Wiley, 1977.

Exercises

1. Two identical spin-$1/2$ fermions have an antisymmetric spatial wavefunction. Which spin sector is allowed?

Solution

For fermions,

$$\eta_{\mathrm{space}}\eta_{\mathrm{spin}}=-1.$$

If $\eta_{\mathrm{space}}=-1$, then $\eta_{\mathrm{spin}}=+1$. The spin state must be symmetric, so it lies in the triplet sector.

2. Two electrons occupy the same spatial orbital $\phi(\mathbf x)$. Why is the triplet spin state not allowed?

Solution

The spatial factor

$$\phi(\mathbf x_1)\phi(\mathbf x_2)$$

is symmetric. A triplet spin state is also symmetric. Their product is symmetric, but two electrons are fermions and require an antisymmetric total state. Therefore the triplet pairing is not allowed for two electrons in the same spatial orbital.

3. Construct the allowed singlet and triplet spatial factors for two electrons in different orthonormal orbitals $a$ and $b$.

Solution

The symmetric spatial factor is

$$\psi_+ = \frac{1}{\sqrt2} \bigl[ a(\mathbf x_1)b(\mathbf x_2) + b(\mathbf x_1)a(\mathbf x_2) \bigr],$$

and it pairs with the singlet spin state. The antisymmetric spatial factor is

$$\psi_- = \frac{1}{\sqrt2} \bigl[ a(\mathbf x_1)b(\mathbf x_2) - b(\mathbf x_1)a(\mathbf x_2) \bigr],$$

and it pairs with any triplet spin state.

4. Show that $\psi_-(\mathbf x,\mathbf x)=0$ for the antisymmetric spatial combination.

Solution

Set $\mathbf x_1=\mathbf x_2=\mathbf x$:

$$\psi_-(\mathbf x,\mathbf x) = \frac{1}{\sqrt2} \bigl[ a(\mathbf x)b(\mathbf x) - b(\mathbf x)a(\mathbf x) \bigr] = 0.$$

5. A pair of identical spinless bosons has no spin factor. What exchange symmetry must the spatial wavefunction have?

Solution

The total bosonic state must be symmetric. With no spin factor available to contribute a minus sign, the spatial wavefunction itself must be symmetric:

$$\psi(\mathbf x_2,\mathbf x_1) = \psi(\mathbf x_1,\mathbf x_2).$$