1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Mechanics: Normalization Requirements (E>Vmin)

  1. Oct 11, 2008 #1
    I have just begun using the 1-D Schrodinger Equation in my quantum mechanics course. We are assuming the potential, V, is solely a function of x (V(x)). I have been examining the solution to the differential equation through seperation of variables. Ψ(x,t) = ψ(x)φ(t)

    (Assuming Ψ(x,t) is seperable) I understand the proof to show that the solutions of Ψ(x,t) produce probabilitiy densities that are invariant to time. And that |Ψ(x,t)|^2 = |ψ(x)|^2. I also understand why E must be real.

    When the assumption of seperability is made we have:

    -(h^2)ψ''(x)/2m +V(x)ψ(x) = Eψ(x) (h is hbar)

    Which is equivalent to:

    ψ''(x) = 2m(V(x) - E)(ψ(x))/h^2 (1)

    My textbook (DJ Griffiths Introduction to Quantum Mechanics) claims that if E < Vmin => that Ψ(x,t) cannot be normalized. Because |Ψ(x,t)|^2 = |ψ(x)|^2 I can divert my attention to (1) and ignore φ(t).

    Solving the differential equation for:

    ψ''(x) = c^2(ψ(x), c is real

    Gets me the general solution (I think!):

    ψ(x) = Aexp(cx+b) + Bexp(-cx+d) A,B,b,d are constants

    If we take a piecewise solution like ψ(x) = exp(-|x|) the integral of |ψ(x)|^2 over the whole real line is finite (and equal to 1, I believe). |ψ(x)|^2 is continuous as well.

    The textbook also hints that it has to do with the the fact that the second derivative is always the same sign as the function itself. I can't formulate a proof as to why this should necessarily imply the function cannot be normalized. In fact, to the best of my knowledge I've provided a perfectly valid counterexample...

    I've thought about it for an hour or so and I can't figure out why this answer is unacceptable. Can someone please tell me what I have done incorrectly?

    I'm also curious about if there is something unphysical about having E<Vmin. Thanks for the help!
     
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2
    Your "general solution" is not, in general, a solution to the correct equation, with a position-dependent V(x) [your c is not a function of x].

    Also, as long as |V(x)| is finite, the wavefunction's derivative has to be continuous, too.

    So, two reasons for your counter example to be incorrect.

    Oh, and of course there is something unphysical about E being less than the minimum potential energy when E=T+V!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?