Is the closure of a symmetric operator unique?

Let $T:D(T)\rightarrow H$ be a densely defined symmetric operator in a Hilbert space H.

The closure $\overline T$ of $T$ is defined as the operator whose graph $G(T)$ is the closure of the graph of $T$, $G(\overline T)=\overline{G(T)}$. For a symmetric operator, a closure always exists.

I am a bit confused as to why the closure is unique (e.g. see http://bit.ly/tFreVy, Lemma 1).

Let's take the momentum operator $p=-id/dx$ on the open interval $(0,1)$ with dense domain $D(p)=C^\infty_0(a,b)$ in $H=L^2((0,1))$. Its closure $\overline p$ exists because $p$ is symmetric. It is given by $p=-id/dx$ with domain $D\left(\overline p\right)=\left\{f\in AC[0,1]:f(0)=f(1)=0\right\}$.

Now there also exists a one-parameter family of self-adjoint extensions $p_\lambda$ of $p$ defined by $p_\lambda=-id/dx$ with domain $D(p_\lambda)=\left\{f\in AC[0,1]:f(0)=e^{i\lambda}f(1)\right\}$. Any self-adjoint operator is closed. Can these $p_\lambda$ not also be considered ''closures'' of p, as in, they are extensions of $p$ that are closed?

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You gave the definition of $\overline{T}$. Closed extensions are closed extensions. They are not the same thing as closures. What's the problem? – Robert Israel Aug 16 '12 at 23:26
Oh thanks. It's just a mix-up with words then. I didn't know the word "closed extension". I thought any closed extension is a closure. – matimo2 Aug 16 '12 at 23:35