# Quantum circuit and Witness preserving amplification for QMA

My question is about the witness preserving amplification scheme for QMA proposed by Marriott and Watrous in 2005 (see http://arxiv.org/abs/cs/0506068 and http://www.cims.nyu.edu/~regev/teaching/quantum_fall_2005/ln/qma.pdf).

The original (quantum verifier) circuit $A$ takes as input $|\psi\rangle \otimes |0^k\rangle$ (where $|\psi\rangle$ is the witness and $|0^k\rangle$ the auxiliary qubits) and it ouptuts (on its first row) the answer of the computation.

The amplifying circuit is defined using $A$ as follow: So it repeats several times the following two measurements (see section 3 of http://arxiv.org/abs/cs/0506068, or http://www.cims.nyu.edu/~regev/teaching/quantum_fall_2005/ln/qma.pdf):

1. apply $A$, measure if output qubit of $A$ is 1, apply $A^{\dagger}$
2. measure if auxiliary qubits are $0$

What I don't understand is how the auxiliary qubits could not be equal to 0. The initial state is $|\psi\rangle \otimes |0^k\rangle$ and then $AA^{\dagger} = I$, so we should always have state $|\psi\rangle \otimes |0^k\rangle$ at the second step (and so always obtaining Yes at this step)?

## 1 Answer

Since we measure the output in step 1, we do not apply $AA^\dagger$, but rather $$A\vert0\rangle\langle 0\vert_1 A^\dagger$$ or $$A\vert1\rangle\langle 1\vert_1 A^\dagger\ ,$$ depending on the measurement outcome, which is generally not equal to the identity.