General Q.M. question

[emob2p]

I'm trying to prove that E must exceed the minimum value of V(x) for all normalizable solutions to the Schroed. eq. To do this I am going to show that in the case E < V(min), the wave function is not normalizable. Naturally I began with the normalization condition: int(|phi|^2)=1 and started taking derivatives on this. However, I cannot arrive at a contradiction. Any thoughts? Or any other ways to show the same result? Thanks.

[marlon]

I'm trying to prove that E must exceed the minimum value of V(x) for all normalizable solutions to the Schroed. eq. To do this I am going to show that in the case E < V(min), the wave function is not normalizable. Naturally I began with the normalization condition: int(|phi|^2)=1 and started taking derivatives on this. However, I cannot arrive at a contradiction. Any thoughts? Or any other ways to show the same result? Thanks.

this is just of the top of my head here, but you need to solve the schrödinger equation with the fact that E - V < 0. I am gonna assume that you know how to solve second order differential equations so when you solve :

(-h'²/2m * f'' + V * f) = E * f (f is the wavefunction, f" is the second derivative and h' is h devided by 2 * pi)

you get : f" - (2m/h'²)*(V - E)f = 0 and the coëfficiënt (let's call this k²)of f is positive here so the solutions will be a superposition of exp(kx) and exp(-kx). So k² = (2m/h'²)(V-E) and the equation becomes f" = k²f

Now try searching for infinities : when x goes to the positive infinity one of the two exponential will become infinite and the same will occur when x goes to the negative infinite. We are not able to find a solution that is finite everywhere this this corresponds to an unphysical state...

regards
marlon

[emob2p]

Is there a way to prove in general if a function and its second derivative are always the same sign, then the function is not normalizalbe since this is essentially the case with E < V(min)?

[marlon]

Is there a way to prove in general if a function and its second derivative are always the same sign, then the function is not normalizalbe since this is essentially the case with E < V(min)?

Well,
Just plot any function f with these two properties. You will clearly see that |f| will always "grow" without limit when x goes to either the positive or negative infinity. When f and f" > 0 then f will be concave upwards and if they are negative then f will be concave downwards...just check this out...
and suppose that f is zero in some point x then this point is also zero for the second derivative meaning that the function will go from convex (under the x-axis) to concave (above the x-axis)...the switch happens in the point x

marlon