Bad Qubits — Learn Quantum by Doing

Multiple choice

Why are magic states needed for fault-tolerant universal quantum computation, given that many codes implement the Clifford group transversally?

Clifford gates cannot be made transversal, so magic states replace them entirely
By the Gottesman-Knill theorem, Clifford-only circuits on stabilizer inputs are efficiently classically simulable, so a non-Clifford resource (e.g. the T gate via a magic state) is required to gain quantum advantage
Magic states are needed only to prepare the stabilizer states |0> and |1>
The Clifford group by itself is already universal, but magic states make it run faster