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Show that E must exceed ##V_{min}##

  1. Feb 28, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-2-28_16-29-51.png

    2. Relevant equations
    Equation 2.5
    $$ - \frac {\hbar^{2}}{2m} \frac {d^{2} \psi}{dx^{2}} + V \psi = E \psi $$

    Equation 1.20
    $$ \int_{-\infty}^{\infty} \mid \Psi (x,t) \mid^{2} dx = 1 $$

    3. The attempt at a solution
    So it is easy enough to do the algebra to show that equation 2.5 can be rewritten as the one in the problem statement.

    So if I assume ##E < V_{min}##, then ##\frac {d^{2} \psi}{dx^{2}}## will be positive. If I have a positive second derivative, how do I know that ##\psi## will also be positive?

    For a function to be normalized

    $$ \int_{-\infty}^{\infty} \mid \Psi (x,t) \mid^{2} = 1 $$

    But I have not figured out how to connect the dots to know how the hint will mean that this function cannot be normalized (i.e. the integral cannot be true).
     
    Last edited: Feb 28, 2015
  2. jcsd
  3. Feb 28, 2015 #2
    Just solve the differential equation :) Assume E is smaller than every possible value of V, then you have a differential equation of the form

    $$f''=af$$

    where a is a positive number. While previously a was negative the solution was a complex exponential (periodic) and now the solutions are real exponentials. Try to normalize that.

    edit: I know you are not solving the exact same equation, since V is a function of x, but the behavior should be similar. When the coefficient is positive you have exponential, when it is negative you have complex exponential.
     
    Last edited: Feb 28, 2015
  4. Feb 28, 2015 #3

    Maylis

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    Alrighty then, I will solve the differential equation

    $$ \frac {d^{2} \psi}{dx^{2}} - \frac {2m}{\hbar^{2}} \Big (V(x) - E \Big ) \psi = 0 $$
    $$ r^{2} - \frac {2m}{\hbar^{2}} \Big (V(x) - E \Big ) = 0 $$
    Therefore, ##r_{1} = \sqrt { \frac {2m(V-E)}{\hbar^{2}}}## and ##r_{2} = - \sqrt { \frac {2m(V-E)}{\hbar^{2}}}##
    So the solution to the differential equation is
    $$ \psi = c_{1}e^{\sqrt { \frac {2m(V-E)}{\hbar^{2}}}} + c_{2}e^{- \sqrt { \frac {2m(V-E)}{\hbar^{2}}}} $$

    To normalize, I use equation 1.20
    $$ \int_{-\infty}^{\infty} \mid c_{1}e^{\sqrt { \frac {2m(V-E)}{\hbar^{2}}}} + c_{2}e^{- \sqrt { \frac {2m(V-E)}{\hbar^{2}}}} \mid^{2} dx $$

    Well, I don't exactly know what ##V(x)## is, so I can't really integrate this.
     
    Last edited: Feb 28, 2015
  5. Feb 28, 2015 #4

    strangerep

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    That's not true. Read the hint again. :wink:

    Wait... let's take a little detour...

    Let ##z## be a real variable, and let ##\alpha## be a real constant. If I tell you that ##\alpha z > 0##, and that ##\alpha## is positive, what can you tell me about the sign of ##z##? Similarly, if I told you that ##\alpha## is negative, what could you tell me about the sign of ##z## in that case?

    Then look at your Schrodinger equation again. I think you've already figured out that ##(V-E)>0##, so... can you now see that ##\psi## and its 2nd derivative must have the same sign? (If that's not obvious, just consider the cases ##\psi>0## and ##\psi<0## separately.)

    Consider the case where ##\psi>0## and ##d^2\psi/dx^2 > 0##. What, therefore, do you know about the curve ##\psi(x)## in the neighbourhood of any (arbitrary) point ##x##.

    Hint: remember the stuff about concave-up, and concave-down in relation to 2nd derivatives? If a function ##\psi(x)## is positive for all ##x##, and its graph is concave-up everywhere, what can you deduce about the behaviour of ##\psi(x)## as ##x \to \pm\infty## ?
     
  6. Feb 28, 2015 #5

    Maylis

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    Okay, now I understand the hint, so really since ##V - E## is positive, then the second derivative will of course have the same sign as ##\psi##. And with that said, then ##\psi(x) \rightarrow \pm \infty## as ##x \rightarrow \pm \infty##, assuming ##\psi(x) < 0## or ##\psi(x) > 0##. Then of course ##\mid \psi(x) \mid^{2}## will also approach infinity. Thus it is not normalizable.
     
  7. Mar 1, 2015 #6

    strangerep

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    That's the idea.

    (Strictly speaking, one must also assume that ##\psi(x)## is continuous -- with a well-behaved 1st derivative. But that's a standard assumption when working with the Schrodinger eqn in QM.)
     
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