Perfect quantum code 

A non-degenerate code constructed out of \(q\)-dimensional qudits and having parameters \(((n,K,2t+1))\) is perfect if \(n\), \(K\), \(t\), and \(q\) are such that the quantum Hamming bound \begin{align} \sum_{j=0}^{t}(q^2-1)^{j}{n \choose j}\leq q^{n}/K \tag*{(1)}\end{align} becomes an equality. For example, for a qubit \(q=2\) code with one logical qubit (\(K=2\)) and \(t=1\), the bound becomes \(3n+1 \leq 2^{n-1}\). The bound can be saturated only at certain \(n\).

For qubit codes with \(K=2^k\), one can work out an asymptotic Hamming bound in the large-\(n,k,t\) limit, \begin{align} \frac{k}{n}\leq 1-\frac{t}{n}\log_{2}3-h(t/n), \tag*{(2)}\end{align} where \(h\) is the binary entropy function.

A quantum Hamming-like bound exists for degenerate qubit stabilizer codes [1].