Let $\psi_1(r), \psi_2(r)$ be two wave-functions for an electron, which are plotted in the graph below. It may be observed of $\psi_1(r)$, that it possesses greater curvature than $\psi_2(r)$. If, by the Schrodinger Equation, $-\frac{h^2}{2m}\frac{d^2\psi}{dx^2} = K\psi$, may it be concluded that the kinetic energy of the first particle (i.e., that corresponding to $\psi_1(r)$) exceeds the second ($\psi_2(r)$)?

enter image description here

I am reserved on this answer, for the kinetic energy does not strictly depend upon the curvature of the graph, but also on the height of the graph (the value of $\psi(x)$), in such a way that if the height should be larger then the kinetic energy is smaller.

Edit: The potential energy graph applies for both particles.

  • $\begingroup$ $\psi$'s are positive real functions for small $r$ according to the graphics. The curvatures are also positive. Is the energy $K$ negative to fulfill the SE? $\endgroup$ – Claudio Saspinski Aug 28 '20 at 22:25

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