|q⟩ Bad Qubits

intermediate · Physics · Addition of Angular Momenta

Checkpoint: Add Two Spins

What this checkpoint tests

You have now seen the complete picture for adding two spin-12\tfrac12 particles: the uncoupled and coupled bases, the total-spin operator S2\mathbf{S}^2, the singlet–triplet split, the Clebsch–Gordan coefficients, the triangle rule, and the exchange symmetry. This checkpoint asks you to build a coupled state explicitly and confirm it on the simulator.

The target state

The middle member of the triplet is the symmetric, maximally entangled combination

∣1,0⟩=12(∣ ⁣↑↓⟩+∣ ⁣↓↑⟩)=12(∣01⟩+∣10⟩).|1, 0\rangle = \tfrac{1}{\sqrt2}\big(|\!\uparrow\downarrow\rangle + |\!\downarrow\uparrow\rangle\big) = \tfrac{1}{\sqrt2}\big(|01\rangle + |10\rangle\big).

It has total spin s=1s = 1 and projection m=0m = 0. Contrast it with the singlet ∣0,0⟩=12(∣01⟩−∣10⟩)|0,0\rangle = \tfrac{1}{\sqrt2}(|01\rangle - |10\rangle), which differs only by the sign of the second term — symmetric versus antisymmetric exchange.

A recipe to recall

A clean way to reach ∣1,0⟩|1,0\rangle from ∣00⟩|00\rangle:

  1. HH on qubit 0 to make 12(∣00⟩+∣10⟩)\tfrac{1}{\sqrt2}(|00\rangle + |10\rangle).
  2. XX on qubit 1 to reach 12(∣01⟩+∣11⟩)\tfrac{1}{\sqrt2}(|01\rangle + |11\rangle).
  3. CX(0,1)\mathrm{CX}(0,1) to convert the ∣11⟩|11\rangle term into ∣10⟩|10\rangle, giving the triplet.

If you instead wanted the singlet, you would insert a ZZ on qubit 0 after step 1 to flip the relative sign — a one-gate difference that encodes the whole symmetry distinction.

Try it

Build the triplet state ∣1,0⟩=12(∣01⟩+∣10⟩)|1,0\rangle = \tfrac{1}{\sqrt2}(|01\rangle + |10\rangle) as a two-qubit statevector and press Check.

Run your code to see the quantum state.

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