Symmetric and Antisymmetric Wavefunctions

Symmetric and antisymmetric wavefunctions are the explicit coordinate-space form of the symmetrization postulate. This page is the canonical home for constructing them from one-particle wavefunctions and for keeping track of normalization, spin, and same-orbital limits.

The important warning is simple: exchange acts on the complete one-particle labels. If a particle has position and spin, write

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

and exchange $q_1$ with $q_2$, not just $\mathbf x_1$ with $\mathbf x_2$.

Two-Particle Construction

Let $\varphi_a(q)$ and $\varphi_b(q)$ be normalized one-particle wavefunctions. The slot-labeled product

$$\varphi_a(q_1)\varphi_b(q_2)$$

does not by itself have definite exchange symmetry. Exchanging the slots gives

$$\varphi_a(q_2)\varphi_b(q_1) = \varphi_b(q_1)\varphi_a(q_2),$$

where the second equality just reorders scalar factors.

For two orthonormal one-particle states, the symmetric and antisymmetric combinations are

$$\Psi_S(q_1,q_2) = \frac{1}{\sqrt2} \bigl[ \varphi_a(q_1)\varphi_b(q_2) + \varphi_b(q_1)\varphi_a(q_2) \bigr],$$

and

$$\Psi_A(q_1,q_2) = \frac{1}{\sqrt2} \bigl[ \varphi_a(q_1)\varphi_b(q_2) - \varphi_b(q_1)\varphi_a(q_2) \bigr].$$

They satisfy

$$\Psi_S(q_2,q_1)=\Psi_S(q_1,q_2),$$

and

$$\Psi_A(q_2,q_1)=-\Psi_A(q_1,q_2).$$

The symmetric state is the two-boson construction for different occupied one-particle states. The antisymmetric state is the two-fermion construction for different occupied complete one-particle states.

Projector View

The same construction can be expressed with exchange projectors. For two slots,

$$\Pi_S = \frac12(I+P_{12}), \qquad \Pi_A = \frac12(I-P_{12}).$$

Given an unsymmetrized product vector

$$\lvert a\rangle_1\lvert b\rangle_2,$$

the projected vectors are

$$\Pi_S\lvert a\rangle_1\lvert b\rangle_2 = \frac12 \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr),$$

and

$$\Pi_A\lvert a\rangle_1\lvert b\rangle_2 = \frac12 \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

These projected vectors are not automatically normalized. After projection, one must divide by the norm unless the projection gives zero.

Normalization Caveats

The factor $1/\sqrt2$ is correct when the one-particle states are orthonormal and distinct. If the overlap

$$s = \langle a\vert b\rangle = \int dq\,\varphi_a^*(q)\varphi_b(q)$$

is not zero, the correct normalized combinations are

$$\Psi_\pm(q_1,q_2) = \frac{ \varphi_a(q_1)\varphi_b(q_2) \pm \varphi_b(q_1)\varphi_a(q_2) }{ \sqrt{2(1\pm \lvert s\rvert^2)} },$$

provided the denominator is nonzero. Here $+$ denotes the symmetric combination and $-$ denotes the antisymmetric combination.

This formula follows from

$$\left\| \lvert a\rangle_1\lvert b\rangle_2 \pm \lvert b\rangle_1\lvert a\rangle_2 \right\|^2 = 2(1\pm \lvert s\rvert^2).$$

There are two common traps:

• if $\langle a\vert b\rangle\ne0$, the naive $1/\sqrt2$ factor is wrong;
• if $a=b$, the antisymmetric numerator vanishes and there is no normalized antisymmetric state of that form.

Same-Orbital Case

If two identical bosons occupy the same normalized one-particle state $\varphi(q)$, the symmetric two-particle wavefunction is simply

$$\Psi_S(q_1,q_2) = \varphi(q_1)\varphi(q_2).$$

It is already normalized because

$$\int dq_1\,dq_2\, \lvert\varphi(q_1)\rvert^2 \lvert\varphi(q_2)\rvert^2 = 1.$$

For identical fermions, trying to put both particles in the same complete one-particle state gives

$$\Psi_A(q_1,q_2) = \frac{1}{\sqrt2} \bigl[ \varphi(q_1)\varphi(q_2) - \varphi(q_1)\varphi(q_2) \bigr] = 0.$$

The zero vector is not a physical state. This is the wavefunction version of the Pauli exclusion principle.

Spatial Wavefunctions and Spin States

For particles with spin, the total wavefunction may often be written schematically as

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

Exchange acts on both factors at once:

$$(\mathbf x_1,s_1;\mathbf x_2,s_2) \longmapsto (\mathbf x_2,s_2;\mathbf x_1,s_1).$$

For two identical fermions, the total state must be antisymmetric. Therefore the allowed factorized symmetry pairings are

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

For two identical bosons, the total state must be symmetric. In a factorized description, the total symmetry can come from

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

when the relevant spin states exist.

For two spin-$1/2$ fermions, the spin singlet is antisymmetric and the triplet states are symmetric. Thus a singlet spin state must be paired with a symmetric spatial wavefunction, while a triplet spin state must be paired with an antisymmetric spatial wavefunction.

This is the standard bookkeeping behind the simple helium ground-state approximation: both electrons occupy the same spatial orbital, so the spatial part is symmetric and the spin part must be the antisymmetric singlet.

Exchange Nodes for Spinless Fermions

For spinless identical fermions, the wavefunction itself must be antisymmetric in the spatial variables:

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

Setting $\mathbf x_1=\mathbf x_2$ gives

$$\psi(\mathbf x,\mathbf x) = -\psi(\mathbf x,\mathbf x),$$

so

$$\psi(\mathbf x,\mathbf x)=0.$$

This exchange node is an immediate consequence of antisymmetry. It should not be overread as saying that two electrons can never be found at the same position: electrons have spin, and opposite-spin electrons may share a spatial orbital while occupying different complete spin-orbitals.

Multi-Particle Extension

For $N$ identical particles, the formal slot-labeled wavefunction is

$$\Psi(q_1,\ldots,q_N).$$

Given one-particle orbitals $\varphi_1,\ldots,\varphi_N$, a symmetric wavefunction can be formed by summing over all permutations:

$$\Psi_S(q_1,\ldots,q_N) = \mathcal N_S \sum_{\pi\in S_N} \prod_{j=1}^{N} \varphi_{\pi(j)}(q_j).$$

An antisymmetric wavefunction is formed with permutation signs:

$$\Psi_A(q_1,\ldots,q_N) = \mathcal N_A \sum_{\pi\in S_N} \operatorname{sgn}(\pi) \prod_{j=1}^{N} \varphi_{\pi(j)}(q_j).$$

If the $N$ one-particle orbitals are orthonormal and distinct, the fermionic expression has normalization

$$\mathcal N_A = \frac{1}{\sqrt{N!}},$$

and can be written as a determinant:

$$\Psi_A(q_1,\ldots,q_N) = \frac{1}{\sqrt{N!}} \det \bigl[ \varphi_j(q_i) \bigr]_{i,j=1}^{N}.$$

Exchanging two particle slots exchanges two rows of the determinant, so the wavefunction changes sign. If two occupied one-particle orbitals are identical, two columns are identical and the determinant vanishes. The detailed determinant construction is the canonical topic of Slater Determinants.

For bosons, repeated orbitals do not make the state vanish. They change the normalization. Occupation-number notation packages that normalization more cleanly by specifying how many particles occupy each one-particle mode.

What This Page Owns

This page owns the explicit wavefunction construction and its normalization. The broader postulate belongs to Symmetrization Postulate. The operator-level treatment of permutation operators and projectors belongs to Exchange Operators. The spin-spatial pairing rules belong to Spin and Spatial Wavefunctions. The exclusion principle belongs to Pauli Exclusion Principle. Occupation-number and Fock-space notation belong to the Occupation-Number Basis and the later Fock-space pages.

Common Mistakes

• Using $1/\sqrt2$ for symmetrized combinations of nonorthogonal one-particle states.
• Symmetrizing the spatial wavefunction while forgetting spin or another internal degree of freedom.
• Treating the slot labels $1$ and $2$ as physical particle names after symmetrization.
• Saying two fermions cannot be at the same position without specifying spin and complete one-particle states.
• Thinking an antisymmetrized product is automatically useful entanglement; identical-particle entanglement requires an operational subsystem or mode split.
• Forgetting that repeated bosonic orbitals require different normalization from distinct orthonormal orbitals.

Cross-Links

• Indistinguishability
• Symmetrization Postulate
• Exchange Operators
• Bosons
• Fermions
• Pauli Exclusion Principle
• Slater Determinants
• Permanents
• Identical-Particle Entanglement Cautions
• Position-Space Two-Particle States
• Spin and Spatial Wavefunctions
• Identical Particle Exercises
• Singlet and Triplet States
• Occupation-Number Basis
• Bosonic Fock Space
• Fermionic Fock Space
• Two Spin-1/2 Particles
• Formula Sheet

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• W. Pauli, “The Connection Between Spin and Statistics,” Physical Review 58, 716-722, 1940.
• A. Messiah and O. W. Greenberg, “Symmetrization Postulate and Its Experimental Foundation,” Physical Review 136, B248-B267, 1964.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.

Exercises

1. Verify directly that $\Psi_S(q_1,q_2)$ is symmetric and $\Psi_A(q_1,q_2)$ is antisymmetric.

Solution

Exchange $q_1$ and $q_2$ in the symmetric combination:

$$\Psi_S(q_2,q_1) = \frac{1}{\sqrt2} \bigl[ \varphi_a(q_2)\varphi_b(q_1) + \varphi_b(q_2)\varphi_a(q_1) \bigr].$$

Reordering scalar factors gives the original expression, so $\Psi_S(q_2,q_1)=\Psi_S(q_1,q_2)$. For the antisymmetric combination, the same exchange reverses the order of the two terms, so

$$\Psi_A(q_2,q_1) = -\Psi_A(q_1,q_2).$$

2. Let $\langle a\vert b\rangle=s$. Show that the norm squared of

$$\lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2$$

is $2(1+\lvert s\rvert^2)$.

Solution

Expand the inner product. The two diagonal terms give $1+1$. The cross terms are

$$\langle a\vert b\rangle \langle b\vert a\rangle = s s^* = \lvert s\rvert^2,$$

and the complex conjugate cross term gives another $\lvert s\rvert^2$. Therefore the norm squared is

$$2+2\lvert s\rvert^2 = 2(1+\lvert s\rvert^2).$$

3. What happens to the antisymmetric two-particle wavefunction when $\varphi_a=\varphi_b$?

Solution

The numerator becomes

$$\varphi(q_1)\varphi(q_2) - \varphi(q_1)\varphi(q_2) = 0.$$

The projected state is the zero vector, not a normalizable physical state. This is the two-particle wavefunction form of Pauli exclusion for identical fermions in the same complete one-particle state.

4. Two identical spin-$1/2$ fermions have a symmetric spatial wavefunction. Which spin symmetry is required?

Solution

The total state of two identical fermions must be antisymmetric. If the spatial part is symmetric, the spin part must be antisymmetric. For two spin-$1/2$ particles, that means the spin singlet.

5. For spinless identical fermions, show that $\psi(\mathbf x,\mathbf x)=0$.

Solution

Antisymmetry gives

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

Set $\mathbf x_1=\mathbf x_2=\mathbf x$. Then

$$\psi(\mathbf x,\mathbf x) = -\psi(\mathbf x,\mathbf x),$$

so $2\psi(\mathbf x,\mathbf x)=0$, hence

$$\psi(\mathbf x,\mathbf x)=0.$$