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The Jackson-Feenberg identity for evaluating the expectation of kinetic energy is frequently used in the theory of quantum liquids and is given by:

$$<\hat{T}>=-\frac{\hbar^2}{8m}\int d^3r[(\nabla^2\psi^*)\psi+\psi^*(\nabla^2\psi)-2(\nabla\psi)\cdot\nabla(\psi^*)],$$ where $\psi$ is a normalized eigenstate. How to prove this?

I tried to operate $\hat{T}=-\frac{\hbar^2}{2m}\nabla^2$on $|\psi\rangle $ and integrate after operating from left by $\langle\psi^*$|, and noticing $(fg)''=fg''+gf''+2g'f'$, and the integrand I obtained is like $fg''$. I got $\nabla^2({\psi}{\psi}^*)-\psi\nabla^2\psi^*-2(\nabla\psi\cdot\nabla\psi^*)$ as my integrand, apart from the multiplicative constant -$\frac{\hbar^2}{2m}$ outside the integral. This leads to nowhere and I think I am not on the right track. All papers I came across just applied it. How can we derive this?

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Just integrate by parts a few times to get it in the desired form. For the first term in the identity, $$\int d\mathbf{r} \, (\nabla^2 \psi^*) \psi = - \int d\mathbf{r} \, (\nabla \psi^*) \cdot (\nabla \psi) = \int d\mathbf{r} \, \psi^* \nabla^2 \psi.$$ For the third term in the identity, $$- 2 \int d \mathbf{r} (\nabla \psi^*) \cdot (\nabla \psi) = 2 \int d \mathbf{r} \, \psi^* \nabla^2 \psi.$$ Therefore, putting all the terms together, we get $$- \frac{\hbar^2}{8m} \int d \mathbf{r} \, (1 + 1 + 2) \psi^* \nabla^2 \psi = - \frac{\hbar^2}{2m} \int d \mathbf{r} \, \psi^* \nabla^2 \psi = \langle \hat{T} \rangle$$ as desired.

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