(Stinespring dilation) Let $A$ be a unital C*-algebra and $\phi: A \rightarrow B(H)$ be a completely positive map. Then, there exist a Hilbert space $H_{1}$, and a *-representation $\pi: A \rightarrow B(H_{1})$ and operator $V:H \rightarrow H_{1}$ such that $\phi(.)=V^{\ast}\pi(.)V$. In particular, $||\phi||=||V||^{2}=||\phi(1)||$.
In general there could be many different Stinespring dilations, but we may always assume that a dilation $(\pi, H_{1}, V)$ is minimal in the sense that $\pi(A)VH$ is dense in $H_{1}$(which holds for the construction used in the proof). Under this minimality condition, a Stinespring dilation is unique up to unitary equivalence.
How to explain the quotation above? How to explain there are many different Stinespring dilation? And the uniqueness of dilation under this minimality.
Let $\pi,H_1$ be a Strinespring dilation of $\phi$ on $A$. Now let $K$ be some Hilbert space with $\dim K>\dim H_1$ and $T\in B(H,K)$ some operator. Let $H_2=H_1\oplus K$, $V_2:H\to H_2$ be given by $V_2:\xi\mapsto V\xi\oplus T\xi$, $\pi_2:A\to B(H_2)$ by $\pi_2(a)=\pi(a)\oplus 0$. Then $$ V_2^*\pi_2(a)V_2\xi=(V^*\oplus T^*)(\pi(a)\oplus 0)(V\xi\oplus T\xi)=V^*\pi(a)V=\phi(a). $$ So $(H_2,\pi_2,V_2)$ is a Stinespring dilation for $\phi$, such that the Hilbert spaces $H_1$ and $H_2$ are not isomorphic. And of course $\pi_2(A)V_2H$ is not dense in $H_2$, as $$ \overline{\pi_2(A)V_2H}=H_1\oplus 0. $$
Now suppose that $(H_1,\pi_1,V_1)$ and $(H_2,\pi_2,V_2)$ are both Stinespring dilations with $\pi_j(A)V_jH$ dense in $H_j$, $j=1,2$. Then you can define $U:\pi_1(A)V_1H_1\to\pi_2(A)V_2H_2$ by $$ U\pi_1(a)V_1\xi=\pi_2(a)V_2\xi. $$ First we need to check that this is well defined. Note that $$\tag{1} \langle \pi_1(a)V_1\xi,\pi_1(b)V_1\eta\rangle=\langle V_1^*\pi(b^*a)V_1\xi,\eta\rangle=\langle\phi(b^*a)\xi,\eta\rangle=\langle \pi_2(a)V_2\xi,\pi_2(b)V_2\eta\rangle. $$ Then, if $\pi_1(a)V_1\xi=\pi_1(b)V_1\eta$, we get from $(1)$ that $$ \|\pi_2(a)V_2\xi-\pi_2(b)V_2\eta\|^2=\|\pi_1(a)V_1\xi-\pi_1(b)V_1\eta\|^2=0; $$ so $U$ is well-defined. The equality in $(1)$ also shows that $U$ is an isometry. So it is bounded and extends to an isometry $H_1\to H_2$. Being an isometry, the fact that its range is dense implies that its range is all of $H_2$, and so $U$ is a unitary. Finally, $$ U\pi_1(a)V_1=\pi_2(a)V_2. $$ Putting $a=I$ (or using an approximate identity if $A$ is not unital) we get $UV_1=V_2$. And, for any $\pi_1(b)V_1\xi\in H_1$ and $a\in A$, $$ \pi_2(a)U\,(\pi_1(b)V_1\xi)=\pi_2(a)\pi_2(b)V_2\xi=\pi_2(ab)V_2\xi=U\pi(ab)V_1\xi=U\pi_1(a)\,(\pi_1(b)V_1\xi). $$ This shows that $\pi_2(a)U=U\pi_1(a)$ on a dense subset of $H_1$, and so everywhere. Multiplying by $U^*$ on the left, $$ U^*\pi(a)U=\pi_1(a),\ \ a\in A. $$