Why can I, in the 2nd quantisation representation of a kinetic energy Hamiltonian
$$ H=\frac { -\hbar ^ { 2 } } { 2 m } \nabla^2 $$ write the Laplace (=Nabla$^2$) operator out like this? $$ \hat { T } = \sum _ { i j } t _ { i j } \hat { a } _ { i } ^ { \dagger } \hat { a } _ { j } = \sum _ { i j } \hat { a } _ { i } ^ { \dagger } \hat { a } _ { j } \int d \mathbf { r } \phi _ { i } ^ { * } ( \mathbf { r } ) \left[ - \frac { \hbar ^ { 2 } \nabla ^ { 2 } } { 2 m } \right] \phi _ { j } ( \mathbf { r } ) = \underline{\frac { \hbar ^ { 2 } } { 2 m } \int d \mathbf { r } \nabla \hat { \psi } ^ { \dagger } ( \mathbf { r } ) \nabla \hat { \psi } ( \mathbf { r } )} $$
Is $\nabla^2$ not only acting on the right wave function?