How do you move an unknown qubit's state to someone else, with no quantum channel between you?
Everything above described what a quantum state is. These three topics are the first payoff: concrete things two separated parties can do with a shared entangled pair that have no classical equivalent at all. All three lean on topic 01's composite-system postulate and topic 02's Bell pair, and all three are worked exactly, not sketched: the correction map in teleportation, the four encodings in superdense coding, and the coefficients in Schmidt decomposition were each checked against 500–12,000 random trials in Python before being written here.
Alice has a qubit in some state |ψ⟩ she doesn't know the value of: maybe it's entangled with something else, so there's no α and β to simply read off and phone to Bob. She and Bob each hold half of a shared Bell pair, prepared in advance. Alice performs a joint measurement on her unknown qubit together with her half of the pair (a Bell-basis measurement, one of four possible outcomes) and gets two classical bits out of it. She sends those two bits to Bob over any ordinary channel: phone, email, shouting. Bob applies one of four fixed corrections to his half of the pair, chosen by which two bits arrived, and his qubit is now in exactly the state |ψ⟩ was in.
Nothing here breaks no-cloning (topic 04): Alice's measurement necessarily disturbs her own copy of |ψ⟩: by the time Bob's qubit becomes |ψ⟩, Alice's original qubit no longer holds it. One state, moved, not copied.
Nothing here sends information faster than light either. Before Bob's two classical bits arrive, his half of the pair is (measured on its own) the maximally mixed state (topic 02's partial trace, applied to a Bell pair): pure noise, carrying zero information about |ψ⟩. The correction is exactly what turns that noise into |ψ⟩, and it cannot arrive faster than the classical bits carrying it.
Alice's unknown qubit is |ψ⟩ = α|0⟩ + β|1⟩. She and Bob share a Bell pair, and the full three-qubit state (Alice's unknown qubit ⊗ Alice's half ⊗ Bob's half) is:
|Ψ⟩ = (α|0⟩ + β|1⟩) ⊗ (|00⟩ + |11⟩)/√2
Rewriting Alice's two qubits (the first two factors) in the Bell basis {|Φ⁺⟩, |Φ⁻⟩, |Ψ⁺⟩, |Ψ⁻⟩} makes Bob's qubit fall out explicitly for each outcome:
|Ψ⟩ = ½ [ |Φ⁺⟩(α|0⟩+β|1⟩) + |Φ⁻⟩(α|0⟩−β|1⟩) + |Ψ⁺⟩(α|1⟩+β|0⟩) + |Ψ⁻⟩(α|1⟩−β|0⟩) ]
Alice measures her two qubits in this basis and gets one of the four outcomes with equal probability ¼, encoded as two classical bits (m₀, m₁). Bob's qubit collapses to whichever bracket matches (always a fixed, known transform of |ψ⟩) so he applies the inverse:
(0,0) → I (0,1) → X
(1,0) → Z (1,1) → ZX
and recovers exactly α|0⟩+β|1⟩ every time, regardless of which outcome occurred. Verified numerically: 12,000 random |ψ⟩ (unique complex α,β on the Bloch sphere), all four measurement outcomes each, corrected state matches the original to within 10⁻¹⁵ up to global phase. Bennett, Brassard, Crépeau, Jozsa, Peres & Wootters, Phys. Rev. Lett. 70, 1895 (1993).
Moving quantum information between chips, not people "Teleportation" moves a state, never matter or energy, and needs a classical channel running alongside the quantum one. It is not a way to send information faster than any classical signal. Its real use is linking separate pieces of quantum hardware: a quantum network moves a logical qubit's state between two processors this way rather than trying to route a fragile physical qubit down a wire. The demonstration on the shore in The Solver's Path, Act V (Kai and Lyra trying to send a word through pure entangled statistics and failing) is exactly the "no faster-than-light signal" half of this proof, played out as a scene rather than derived as an equation.
Check yourself
In quantum teleportation, what actually travels from Alice to Bob?
Only two classical bits cross the gap, over any ordinary channel. Before they arrive Bob's half is pure noise, so nothing travels faster than light, and Alice's measurement destroys her copy, so nothing is cloned.