Deficiency indices

Remarks

Equivalently, $K_\pm$ is the orthogonal complement of the image of $A \pm iI$. Some authors exchange the names of $K_+$ and $K_-$; the pair $(n_+, n_-)$ is then reversed. The operator $A$ is essentially self-adjoint if and only if $n_+ = n_- = 0$. Example: on $L^2([0,1])$, let $A = -i \, d/dx$ on the smooth functions with compact support in $(0,1)$. Then $A^*$ is $-i \, d/dx$ on the Sobolev space $H^1([0,1])$, and $A^* u = \pm i u$ means $u' = \mp u$. So $K_+$ is spanned by $e^{-x}$ and $K_-$ by $e^{x}$, and $(n_+, n_-) = (1, 1)$. Example with unequal indices: on $L^2([0, \infty))$, the same construction gives $K_+$ spanned by $e^{-x}$, while $e^{x}$ is not square integrable, so $(n_+, n_-) = (1, 0)$. Non-example: the deficiency indices are not defined for an operator that is not densely defined, because then the adjoint is not a well-defined operator.