Show that E must exceed $V_{min}$

1. Feb 28, 2015

Maylis

1. The problem statement, all variables and given/known data

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. Feb 28, 2015

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.

Last edited: Feb 28, 2015
3. Feb 28, 2015

Maylis

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
4. Feb 28, 2015

strangerep

That's not true. Read the hint again.

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 $x \to \pm\infty$ ?

5. Feb 28, 2015

Maylis

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.

6. Mar 1, 2015

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.)