Checkpoint: Exchange Symmetry

This checkpoint pulls the module together. Identical particles are indistinguishable, so their states must be eigenstates of exchange with eigenvalue $\pm 1$ : bosons symmetric, fermions antisymmetric. Antisymmetry forces the Pauli exclusion principle and, for two electrons, locks the spin symmetry to the opposite of the spatial symmetry.

What you should be able to do

State the indistinguishability requirement $|\psi(1,2)|^2 = |\psi(2,1)|^2$ and explain why exchange eigenvalues can only be $\pm 1$ .
Classify a particle as a boson or fermion from its spin (spin–statistics theorem) and from the parity of its fermion count for composites.
Build symmetric and antisymmetric two-particle states and show that the antisymmetric one vanishes when both single-particle states coincide (exclusion).
Pair a spin singlet with a symmetric spatial state and a spin triplet with an antisymmetric spatial state so the total two-electron wavefunction is antisymmetric.

The configuration to build

Two electrons share one spatial orbital — say the ground orbital of a well or atom. Their spatial wavefunction $\psi_1(x_1)\psi_1(x_2)$ is symmetric. To keep the total state antisymmetric, the spin part must be antisymmetric, and the only antisymmetric two-spin state is the singlet:

|S\rangle = \tfrac{1}{\sqrt{2}}\big(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle\big) = \tfrac{1}{\sqrt{2}}\big(|01\rangle - |10\rangle\big),

with total spin $S = 0$ . This is the spin state of, for example, the two electrons in a ground-state helium atom or a covalent bond. Your task is to prepare exactly this state on two qubits.

Try it

Build the spin singlet $\tfrac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ on two qubits, reading $|0\rangle$ as spin up and $|1\rangle$ as spin down. The grader checks the full statevector, so make the amplitudes $+1/\sqrt{2}$ on $|01\rangle$ and $-1/\sqrt{2}$ on $|10\rangle$ , with zero elsewhere.

Run your code to see the quantum state.