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...

  1. Apply the product rule for derivatives to the expression below:

    $\frac{\del}{\del t}(\Psi^*\Psi)=$


  2. 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...

    $\frac{\del \Psi}{\del t}=$


  3. Take the complex conjugate of the equation above to write:

    $\frac{\del \Psi^*}{\del t}=\left(\frac{\del \Psi}{\del t}\right)^*= $


  4. Now substitute those two partial derivatives into your expression in i.

    $\frac{\del}{\del t}(\Psi^*\Psi)=$


  5. 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 momentum.]

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 $A$...