Suppose that $X_i$ - i.i.d. $N(\theta, 1)$. It's easy to find a confidence interval (with level $\gamma$) for $\theta$ with length $\frac{C}{\sqrt{n}}$: $\theta \in (\bar{X} - \epsilon, \overline{X} + \epsilon)$ for $\epsilon = n^{-\frac12}u_{\frac{1+\gamma}2}$ where $u_p$ is a $p$-quantile of $N(0,1)$ and where $\overline{X} = \frac{X_1 + X_2 + \ldots + X_n}{n}$.
It looks like there's no confidence interval in $N(\theta, 1)$ with lenth $\approx \frac{C}{n}$ but how may we prove it? And how can we find the interval with the smallest lenght (maybe it doesn't have the next form: $(\overline{X} - \epsilon, \overline{X} + \epsilon)$? )
And what about other standard distributions - how may we find the length of the shortest confidence interval, when we know the distribution of $X_i$?
Could you please recommend a book with good introduction to this theme?
Addition: by confidence interval with confidence level $\gamma$ I mean a pair of statistics $T_1^{(n)}(X_1, \ldots, X_n)$ and $T_2^{(n)}(X_1, \ldots, X_n)$ such that $\mathbf{P}_{\theta}( T_1^{(n)}< \theta < T_2^{(n)}) \ge \gamma$ for all $\theta$. By the shortest confidence interval I mean such confidence interval with confidence level $\gamma$ that $E_{\theta}(T_2^{(n)} - T_1^{(n)})$ takes the least possible value for all $\theta$.