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The Schrodinger equation is of the form [itex] \frac{d^2 \psi}{dx^2}+[\varepsilon-v(x)] \psi=0[/itex].
In a lecture, the lecturer said that if we have in a point [itex] x_0 [/itex] , [itex] \psi(x_0)=\psi'(x_0)=0 [/itex], then [itex] \psi(x)=0 [/itex].(for a smooth [itex]v(x)[/itex]!)
Can anyone give a proof of this?
Is it only for a equation of the form given above?
Thanks
In a lecture, the lecturer said that if we have in a point [itex] x_0 [/itex] , [itex] \psi(x_0)=\psi'(x_0)=0 [/itex], then [itex] \psi(x)=0 [/itex].(for a smooth [itex]v(x)[/itex]!)
Can anyone give a proof of this?
Is it only for a equation of the form given above?
Thanks