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General Q.M. question

  1. Sep 24, 2004 #1
    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.
     
  2. jcsd
  3. Sep 24, 2004 #2
    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
     
    Last edited: Sep 24, 2004
  4. Sep 24, 2004 #3
    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)?
     
  5. Sep 25, 2004 #4
    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
     
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