For the AWGN Channel with input power $P$ and noise power $\sigma^{2}$, and information capacity $C(P)=\frac{1}{2}\log_{2}\left(1+ \frac{P}{\sigma^{2}}\right)$, its operational capacity satisfies: $$C_{op}(P)=C(P)=\frac{1}{2}\log_{2}\left(1+ \frac{P}{\sigma^{2}}\right)$$In other words: Forward Part (Achievability) For any $0<\epsilon<1$, there exist $0<\gamma<2\epsilon$ and a sequence of $(n,M_{n})$ block codes $\mathcal{C}_{n}$ for the channel with $$\frac{1}{n}\log_{2}M_{n}>C(P)-\gamma$$and each codeword $c_{m}=(c_{m1},\ldots,c_{mn})$ in $\mathcal{C}_{n}$ satisfying $$\tag{$*$}\frac{1}{n}\sum\limits_{i=1}^{n}c_{mi}^{2}\le P$$such that $$P_{e}(\mathcal{C}_{n})<\epsilon$$ for $n$ sufficiently large. Converse For any sequence of $(n,M_{n})$ block codes $\mathcal{C}_{n}$ for the channel whose codewords satisfy $*$, we have that if $$\liminf_{n\to\infty} \frac{1}{n}\log_{2}M_{n}>C(P)$$then $$\liminf_{n\to\infty}P_{e}(\mathcal{C}_{n})>0$$i.e. for any rate exceeding channel capacity, our probability decoding error is always non-zero.