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Interpret the high-success formulation as completeness at least $1-\delta (n)$ and soundness at most $\delta (n)$, where $\delta (n)\le 1/p(n)$ for a growing polynomial $p$. This is the standard error-reduction claim. Literally placing $O$ around a success probability approaching one supplies no useful lower bound; it is the failure probability that must be inverse-polynomially small.

Let $M_x$ be the quantum verifier acceptance operator acting on one witness register:Acceptance on $\rho$ is $\mathrm{Tr}(M_x\rho )$. On a NO instance, $\Vert M_x\Vert \le 1/3$; on a YES instance, an honest witness has acceptance at least $2/3$.

Ask for $r$ witness registers, with $r$ odd, run $V_x$ independently on each register with fresh ancillas, and accept a strict majority. The QMA parallel repetition with entangled witnesses acceptance operator isDiagonalize $M_x$. In the tensor-product eigenbasis, an eigenvalue of $M_x^{(r)}$ is the majority probability for independent Bernoulli random variables with parameters equal to the selected eigenvalues of $M_x$. On a NO instance every parameter is at most $1/3$. Coupling these variables using independent uniform random numbers shows that their upper majority tail is bounded by the tail with every parameter $1/3$. This bounds the operator norm and therefore covers every supplied witness, including a witness entangled across the registers. No independence assumption about a dishonest witness is being made.

For a valid Chernoff bound, if $S$ is a sum of $r$ independent Bernoulli variables of parameter at most $1/3$, Markov's inequality applied to $2^S$ yieldsAn honest product witness on a YES instance gives independent trials with rejection probability at most $1/3$, so its failure probability has the same bound. The printed concentration bound is not valid as stated: for five trials at parameter $1/3$, the upper-majority probability is $17/81>e^{-5/3}$. The valid bound above suffices for a rigorous proof.

Choose the next odd integer above $c^{-1}\ln p(n)$. ThenThe witness length and circuit size increase by only $O(\log p(n))$; the majority calculation has polynomial overhead. Conversely, inverse-polynomial failure probability is at most $1/3$ for all sufficiently large input lengths. Finitely many smaller lengths can be handled by hard-wired verifiers, giving the usual constants. Hence the two definitions give the same class QMA. The same QMA error reduction argument also yields exponentially small failure with polynomially many copies.

QMA is the class of promise problems with a polynomial-size quantum witness checked by a uniform polynomial-size quantum circuit. YES instances have an accepting witness with probability at least $2/3$; NO instances have acceptance at most $1/3$ for every witness, including mixed states. The verifier's acceptance is linear in the witness density operator. Inverse-polynomial or exponentially small error can be obtained through QMA error reduction.