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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?

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This is just multivariate integration by parts: \begin{align} \int_V f\, \nabla^2 g \,d^3\mathbf{x} = &\int_{V}\boldsymbol\nabla\cdot\left(f\boldsymbol{\nabla} g\right) d^3\mathbf{x} -\int_V\left(\boldsymbol{\nabla} f\right)\cdot\left(\boldsymbol{\nabla} g\right)d^3\mathbf{x} \\= &\int_{\partial V}f\left(\boldsymbol{\nabla} g\right) \cdot d^3\mathbf{A} -\int_V\left(\boldsymbol{\nabla} f\right)\cdot\left(\boldsymbol{\nabla} g\right)d^3\mathbf{x}, \end{align} Since the wavefunction goes to zero at infinity, the boundary term vanishes, and you're left with $$ \int_V f\, \nabla^2 g \,d^3\mathbf{x} = -\int_V\left(\boldsymbol{\nabla} f\right)\cdot\left(\boldsymbol{\nabla} g\right)d^3\mathbf{x} $$

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