Normalizing $\Psi$
We'll start by demonstrating some properties shared by solutions, $\{ \Psi(x,t) \}$, of the one-dimensionsional Schrödinger equation.
- Born interpretation: $|\Psi(x,t)|^2$ is the probability density for measuring a particle near $x$ at time $t$.
A property of any probability density:
$$\int_{-\infty}^{+\infty}|\Psi(x,t)|^2\,dx=1$$
- If $\Psi$ is a solution to the Sch equation, then so is $A\Psi$.
- Finding $A$ such that the relation above holds is called normalizing the wave function.
Not all solutions of the Sch equation are normalizable. The ones that can be normalized are called square-integrable.
Time dependence
The wave function depends not only on position, but also on time. If we take the time derivative of both sides of that normalization condition, we find...
$$\begineq \frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi(x,t)|^2\,dx
&=\frac{d}{dt}1\\
\int_{-\infty}^{+\infty}\frac{\del}{\del t}(\Psi^*\Psi)\,dx
&=0 \endeq$$
This is not necessarily true for just any ol' function $f(x,t)$ of position and time.
But you'll show that if $\Psi(x,t)$ is a solution to the Sch equation, it will also behave this way. The outline for you to work through...
- Apply the product rule for derivatives to the expression below:
$\frac{\del}{\del t}(\Psi^*\Psi)=$
- Rearrange the Schrödinger equation
$$i\hbar\frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\del^2\Psi}{\del x^2}+V\Psi,
$$
to get an expression for the time derivative below...
$\frac{\del \Psi}{\del t}=$
- Take the complex conjugate of the equation above to write:
$\frac{\del \Psi^*}{\del t}=\left(\frac{\del \Psi}{\del t}\right)^*= $
- Now substitute those two partial derivatives into your expression in i.
$\frac{\del}{\del t}(\Psi^*\Psi)=$
- Confirm that if you carried out the $\frac{\del}{\del x}[]$ below, that it would match what you came up with above in (iv):
$$\frac{\del}{\del t}(\Psi^*\Psi)=\frac{\del}{\del x}\left[\frac{i\hbar}{2m}\left( \Psi^*\frac{\del \Psi}{\del x} - \Psi\frac{\del \Psi^*}{\del x} \right)\right]$$
[We're going to need this result again when we deal with
We're ready to evaluate the integral.
$$\begineq\int_{-\infty}^{+\infty}\frac{\del}{\del t}(\Psi^*\Psi)&=
\int_{-\infty}^{+\infty}
\frac{\del}{\del x}
\left[
...
\right]\,dx = \left[
...
\right]_{-\infty}^{+\infty}
\\
&=\left[\frac{i\hbar}{2m}\left(
\Psi^*\frac{\del \Psi}{\del x} - \Psi\frac{\del \Psi^*}{\del x}
\right)\right]_{-\infty}^{+\infty}.
\endeq$$
In order to be square-integrable in the first place, both $\Psi$ and $\Psi^*$ had better $\to 0$ as $x\to\pm\infty$, so this expression does indeed go to 0, completing our proof.
$$\frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi(x,t)|^2\,dx=0$$ $\Rightarrow$ Once normalized, the wave function stays normalized.
Problem 1.4 - set up the integrals to find the constant $A$... if (! $homepage){ $stylesheet="/~paulmr/class/comments.css"; if (file_exists("/home/httpd/html/cment/comments.h")){ include "/home/httpd/html/cment/comments.h"; } } ?>