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Is it possible to just factor out the derivative like this:

$$ x \left[ \frac{d^2 \Psi}{d x^2} \cdot \Psi^* - \frac{d^2 \Psi^*}{d x^2} \cdot \Psi \right] = x \, \frac{d}{dx} \left( \frac{d \Psi}{d x} \cdot \Psi^* - \frac{d \Psi^*}{d x} \cdot \Psi \right) $$

$\Psi$ is just some function of $x$ while $\Psi^*$ is its complex conjugate. I am in doubts as we have products inside square brackets...

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HINT. Try to go from the other side: compute the derivative
$$\frac{d}{dx} \left( \frac{d \Psi}{d x} \cdot \Psi^* - \frac{d \Psi^*}{d x} \cdot \Psi \right)$$ using the product rule and see what happens.

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  • $\begingroup$ Thank you for the hint! Was a good one! This equality is TRUE! $\endgroup$
    – 71GA
    Mar 20, 2013 at 12:16

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