Symmetric and Antisymmetric States

The indistinguishability requirement from the previous lesson said the probability density must be unchanged when we swap two particles. Here we turn that into a precise statement about the wavefunction itself, and we construct the two kinds of states that satisfy it.

The exchange operator

Define the exchange operator $\hat{P}_{12}$ that swaps the labels of particles 1 and 2:

Applying it twice returns the original state, so $\hat{P}_{12}^2 = \mathbb{1}$. The eigenvalues of an operator that squares to the identity can only be $\pm 1$. For identical particles the Hamiltonian treats both labels symmetrically, so $\hat{P}_{12}$ commutes with $\hat{H}$ and physical states can be chosen to be eigenstates of $\hat{P}_{12}$. There are exactly two possibilities:

Both choices give $|\psi(x_1,x_2)|^2 = |\psi(x_2,x_1)|^2$, so both respect indistinguishability. The overall sign is unobservable on its own but governs how the particles correlate.

Symmetrizing two product states

Take two distinct single-particle states $\psi_1$ and $\psi_2$. The two product orderings, $\psi_1(x_1)\psi_2(x_2)$ and $\psi_1(x_2)\psi_2(x_1)$, are not eigenstates of exchange. Their normalized symmetric and antisymmetric combinations are:

Acting with $\hat{P}_{12}$ on $\psi_S$ leaves it unchanged; acting on $\psi_A$ flips its sign. These are precisely the $+1$ and $-1$ eigenstates we wanted.

A qubit model

We can mirror this structure with two qubits. Let qubit 0 record whether particle A sits in state "1," and qubit 1 record whether particle B sits in state "1," so the basis state $|01\rangle$ means "the first slot is empty, the second is filled" and $|10\rangle$ means the reverse. The two exchange eigenstates become familiar entangled pairs:

Swapping the two qubits (exchanging the particles) leaves $|\psi_S\rangle$ unchanged and sends $|\psi_A\rangle \to -|\psi_A\rangle$. The antisymmetric combination $|\psi_A\rangle$ is the famous spin singlet; the symmetric one is a member of the triplet.

Try it

Build the antisymmetric state $\tfrac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ on two qubits. The grader checks the full statevector, so you need amplitude $+1/\sqrt{2}$ on $|01\rangle$, $-1/\sqrt{2}$ on $|10\rangle$, and zero on $|00\rangle$ and $|11\rangle$.

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