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

[gfd43tg]

Homework Statement

Homework Equations

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

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

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).

 

[diegzumillo]

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.

 

[gfd43tg]

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.

 

[strangerep]

So if I assume $E < V_{min}$, then $\frac {d^{2} \psi}{dx^{2}}$ will be positive.

That's not true. Read the hint again.

If I have a positive second derivative, how do I know that $\psi$ will also be positive?

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.)

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).

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 $x \to \pm\infty$ ?

 

[gfd43tg]

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.

 

[strangerep]

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.)