What is the proof for this property of solutions?

[ShayanJ]

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

 

[CubicFlunky77]

I'm new to this but I'm assuming you're referring to the time-independent non-relativistic equation. Let [itex]\psi(x_0)[/itex] represent a potential value where kinetic is 0. Let [itex]\psi'(x_0)[/itex] represent kinetic value where potential is 0. Iff [itex]\psi(x_0)[/itex] and [itex]\psi'(x_0)[/itex] are two separable wave functions that subscribe to all applicable axioms of [itex]\psi(x)[/itex], [itex]\psi[/itex] will include all potential and kinetic components. A rough analogy could be drawn from binary code. Anyway I hope this helped steer you in the right direction if it did anything at all. Everything I wrote came from a purely logical standpoint.

 

[ShayanJ]

I'm new to this but I'm assuming you're referring to the time-independent non-relativistic equation. Let [itex]\psi(x_0)[/itex] represent a potential value where kinetic is 0. Let [itex]\psi'(x_0)[/itex] represent kinetic value where potential is 0. Iff [itex]\psi(x_0)[/itex] and [itex]\psi'(x_0)[/itex] are two separable wave functions that subscribe to all applicable axioms of [itex]\psi(x)[/itex], [itex]\psi[/itex] will include all potential and kinetic components. A rough analogy could be drawn from binary code. Anyway I hope this helped steer you in the right direction if it did anything at all. Everything I wrote came from a purely logical standpoint.

I don't understand what you mean by "potential value" and "kinetic value"!
And...[itex] \psi(x_0) [/itex] and [itex] \psi'(x_0) [/itex] are two complex numbers. Not wave functions!
What do you mean by binary code?

 

[HallsofIvy]

This can be proved by an application of the basic "existence and uniqueness" theorem for differential equations: "If [itex]\phi(x,t)[/itex] is continuous in both x and t and differentiable with respect to x in some neighborhood of [itex](t_0, x_0)[/itex] then there exist a unique function x(t) satisfying [itex]dx/dt= \phi(x, t)[/itex] and [itex]x(t_0)= x_0[/itex]."

That can be extended to second (and higher) order equation by writing [itex]d^2\phi/dt^2= F(t,\phi, \phi')[/itex] as [itex]dx/dt= F(x, t)[/itex] by taking x to be the vector function [itex]x= (x_1, x_2)= (\phi, d\phi/dt)[/itex]. Then [itex]d^2\phi/dt^2= dx_2/dt[/itex] and of course, [itex]dx_1/dt= d\phi/dt= x_2[/itex] so that the differential equation becomes the vector equation [itex](dx_1/dt, dx_2/dt)= (x_2, F(t, x_1, x_2))[/itex].

The "initial value conditions", [itex]\phi(x_0)= 0[/itex] and [itex]\phi'(x_0)= 0[/itex] becomes the single vector condition [itex]x(0)= (x_1(0), x_2(0))= 0[/itex].

Obviously [itex]\phi(t)= 0[/itex] is a solution to the given differential equation and obviously [itex]\phi(0)= \phi'(0)= 0[/itex]. The "existence and uniqueness theorem" tells us that this is the only solution.

(This theorem does NOT apply to "boundary value conditions" since we cannot create the single vector condition. The differential equation [itex]d^2\phi/dt^2= -\phi[/itex] has NO solution that satisfies [itex]\phi(0)= 0[/itex], [itex]\phi(\pi/2)= 0[/itex] but has an infinite number of solutions that satisfy [itex]\phi(0)= 0[/itex], [itex]\phi(\pi)= 0[/itex].)

 

[ShayanJ]

This can be proved by an application of the basic "existence and uniqueness" theorem for differential equations: "If [itex]\phi(x,t)[/itex] is continuous in both x and t and differentiable with respect to x in some neighborhood of [itex](t_0, x_0)[/itex] then there exist a unique function x(t) satisfying [itex]dx/dt= \phi(x, t)[/itex] and [itex]x(t_0)= x_0[/itex]."

That can be extended to second (and higher) order equation by writing [itex]d^2\phi/dt^2= F(t,\phi, \phi')[/itex] as [itex]dx/dt= F(x, t)[/itex] by taking x to be the vector function [itex]x= (x_1, x_2)= (\phi, d\phi/dt)[/itex]. Then [itex]d^2\phi/dt^2= dx_2/dt[/itex] and of course, [itex]dx_1/dt= d\phi/dt= x_2[/itex] so that the differential equation becomes the vector equation [itex](dx_1/dt, dx_2/dt)= (x_2, F(t, x_1, x_2))[/itex].

The "initial value conditions", [itex]\phi(x_0)= 0[/itex] and [itex]\phi'(x_0)= 0[/itex] becomes the single vector condition [itex]x(0)= (x_1(0), x_2(0))= 0[/itex].

Obviously [itex]\phi(t)= 0[/itex] is a solution to the given differential equation and obviously [itex]\phi(0)= \phi'(0)= 0[/itex]. The "existence and uniqueness theorem" tells us that this is the only solution.

(This theorem does NOT apply to "boundary value conditions" since we cannot create the single vector condition. The differential equation [itex]d^2\phi/dt^2= -\phi[/itex] has NO solution that satisfies [itex]\phi(0)= 0[/itex], [itex]\phi(\pi/2)= 0[/itex] but has an infinite number of solutions that satisfy [itex]\phi(0)= 0[/itex], [itex]\phi(\pi)= 0[/itex].)

I don't understand.
The vector ODE you mentioned can also be written as [itex] \frac{d^2 x_1}{dt^2}=F(t,x_1,\frac{dx_1}{dt}) [/itex]. This is again a second order ODE and instead of helping us, just gets us back to the main question. I don't see a proof!
Another problem is, the function [itex] y=x^3 [/itex] is the answer to the 2nd order ODE [itex] \frac{d^2 y}{dx^2}-6 \frac{y}{x^2}=0 [/itex] and also satisfies [itex] y(0)=y'(0)=0 [/itex]. Yeah, you may say F has a singularity but you may instead consider the trivial ODE [itex]y''=6x [/itex]!