Why is $\frac{\hbar}{m}\Im((\nabla \psi)^*\nabla \psi + \psi^*\Delta \psi)=-\frac{2}{\hbar}\Im(\psi^*\frac{\hat{p}^2}{2m}\psi)$ in Quantum Mechanics? Consider a quantum mechanical system with wave function $\psi$. Furthermore, let $\hat{p}$ denote the momentum operator. Show that the identity $\frac{\hbar}{m}\Im((\nabla \psi)^*\nabla \psi + \psi^*\Delta \psi)=-\frac{2}{\hbar}\Im(\psi*\frac{\hat{p}^2}{2m}\psi)$ holds where $\Im(z)$ denotes the imaginary part of $z\in\mathbb{C}$.

My attempt: $-\frac{2}{\hbar}\Im(\psi^* \cdot\frac{\hat{p}^2}{2m}\psi)=-\frac{2}{\hbar}\Im(\psi^*\frac{1}{2m} (-\hbar^2)\Delta \psi)=\frac{\hbar}{m}\Im(\psi^*\Delta\psi)$ so it seems that I am missing the $(\nabla \psi^*)\nabla\psi$ part from above.
Have I made a mistake and if so, where?
 A: First of all, there is really no need to focus on the imaginary part. The kinetic energy operator acting on a wave function is well defined, for both the real and imaginary part that result.
Secondly, the notation in terms of pre-factors on RHS is unusual and rather ugly. The minus sign, factor 2 and Planck's constant are normally placed on the left.
Thirdly, the key question: what is the origin of the first term on the left, with the inner product of two gradient terms? Answer: This term is obtained by partial integration of the second term on the left, with the Laplace operator. Partial integration is allowed because in Quantum Mechanics the wave function is required to be analytic and square integrable.
The idea to retain half of the usual term (with the Laplace operator), and to apply partial integration on the other half (to create the double gradient term) was proposed by German physicist Wigner. The resulting expression has exactly the same expectation value, but better properties when considered as a probability density. For example, there are fewer regions with a negative probability. Hence it makes the physics better understandable.
Note that there is a minus sign missing between the two terms on the left, and also a factor 2.
