Cayley transform to von Neumann theorem Self-ajointness of an operator can be found using the Cayley transform of the operator, if its unitary,
$$ U = (A - i I)(A + i I)^{-1} $$
From this we can go about finding the deficiency subspaces and deficiency indices (using $D_U^{\perp}$ and $R_U^{\perp}$).
If we have say obtained these indices to $(m,n)$ say. Now from this fact how to obtain von Neumann's theorem on self-adjoint extensions. 
PS : Although I believe there is a proof of this in standard books, most of them are from very rigorous mathematical background, which of course is necessary but quite difficult to follow, since my background is from physics.
 A: An antilinear map (scalars are factorized "out of the map" as their complex conjugates) $C:\mathscr{H}\to \mathscr{H}$ is called a conjugation if it preserves the norm of $\mathscr{H}$ and $C^2=\mathrm{id}$ (the identity operator).

Theorem [von Neumann]. Let $A$ be a symmetric operator. If there exists a conjugation $C$ such that $C: D(A)\to D(A)$ and $AC=CA$; then $A$ has equal deficiency indices, and thus self-adjoint extensions.

Proof. By hypothesis, $C\, [D(A)]\subset D(A)$ (since $C$ maps $D(A)$ into itself); in addition $C \, [C\, [D(A)] ]=D(A)$, since $C^2=\mathrm{id}$. Therefore $C\, [D(A)]= D(A)$ (the range of $C$ restricted to $D(A)$ is all $D(A)$, i.e. it is surjective).
Now let $\varphi_+\in K_+=\mathrm{Ran}(A+i)^\perp$ (the positive deficiency subspace); $\psi\in D(A)$. Then you can take the following scalar product (the complex conjugate above is only for technical convenience)
$$\overline{\langle \varphi_+ , (A+i)\psi\rangle}\; .$$
On one hand, this scalar product is zero, since $\varphi_+$ is in the orthogonal complement of the range of $A+i$; on the other it is equal to the following since $C^2=\mathrm{id}$ and $C$ is anti-linear (it behaves exactly like the complex conjugation operator):
$$0=\overline{\langle \varphi_+ , (A+i)\psi\rangle}=\overline{\langle \varphi_+,C^2(A+i)\psi\rangle}=\langle C\varphi_+,C(A+i)\psi\rangle=\langle C\varphi_+,(A-i)C\psi\rangle\; .$$
Since $C$ maps $D(A)$ into itself, $C\varphi_+\in K_-$ (the other deficiency subspace). Hence $C:K_+\to K_-$; and an analogous reasoning shows that $C:K_-\to K_+$. Now $C$ also preserves norms, therefore $dim[K_+]=dim[K_-]$, i.e. the deficiency indices are equal. $\square$
