Symmetric and Antisymmetric Wavefunctions

Symmetric and antisymmetric wavefunctions express exchange symmetry directly in coordinates. Starting from one-particle orbitals, we construct normalized two-particle states and see how orbital overlap changes the normalization.

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 $\mathbf x_2$.

Helpful background. Symmetrization Postulate introduces exchange parity. The calculation below uses normalized one-particle inner products.

Two-Particle Construction

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

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

generally does not have definite exchange symmetry unless the two one-particle states are proportional. 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.

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

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

where $\int dq$ includes a sum over discrete spin labels. The denominator must be 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.

Spin and spatial symmetry

Exchange acts on the full labels $(\mathbf x_1,s_1)$ and $(\mathbf x_2,s_2)$. If the state factors as a spatial wavefunction times a spin state, their exchange parities multiply. For electrons, a spatially symmetric factor pairs with a spin singlet, and a spatially antisymmetric factor pairs with a spin triplet. Bosons require the product of the two parities to be positive.

Such factorization is a special case: spin–orbit coupling or spin-dependent preparation can entangle spin and position. Spin and Spatial Wavefunctions gives explicit examples and explains when a determinant is spin-adapted.

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

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

For wavefunctions with well-defined coincident-point values, this exchange node means the same-spin coincidence density vanishes. Exact position outcomes in a continuum already have zero probability individually; the density is the meaningful comparison. 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.

More than two particles

Summing a product of $N$ orbitals over all assignments to the particle slots produces a symmetric state. Weighting each assignment by its permutation sign produces an antisymmetric state. The compact formulas and their normalization are developed in Permanents and Slater Determinants.

These constructions give occupation basis states or symmetrized products. A general correlated bosonic or fermionic state can require a superposition of many such states.

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

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

References

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