# After normalizing a wavefunction I don't know how to calculate probability on an interval (-0.1 + 0.1)

This is quite large homework where I 1st had to normalize the wavefunction $\psi = Axe^{-x^2/2a}$ and I got a constant $A=\sqrt{2/(a\sqrt{\pi a})}$.

How do I calculate the probability now for the interval (-0.1nm , + 0.1nm) if $a=1nm^2$?

I first tried to calculate the normalisation factor but the units are weird... is this possible? Is most likely my normalisation factor wrong?

$$A=\sqrt{\frac{2}{1nm^2 \cdot \sqrt{\pi\cdot1nm^2}}}=1.06\frac{1}{nm\sqrt{nm}}$$

EDIT: After thinking again I decided to include the derivation of the normalisation factor $A$:

\begin{align} &\phantom{=}\int\limits_{-\infty}^\infty A^2x^2 \exp\left[\tfrac{-x^2}{a}\right]dx = 1 = A^2\int\limits_{-\infty}^\infty \underbrace{x}_f\cdot \underbrace{x\exp\left[\tfrac{-x^2}{a}\right]}_{dg/dx} dx= \\ &=A^2\left[\smash{\underbrace{\left.x\left(-\tfrac{a}{2}\right)\exp\left[\tfrac{-x^2}{a}\right]\right|_{-\infty}^{\infty}}_0}-\int\limits_{-\infty}^\infty\left(-\tfrac{a}{2}\right)\exp\left[\tfrac{-x^2}{a}\right]dx\right]=\\ &=\tfrac{A^2a}{2}\int\limits_{-\infty}^{\infty} \exp\left[\tfrac{-x^2}{a}\right]dx = \tfrac{A^2a}{2}\int\limits_{-\infty}^{\infty} e^{-t^2}\sqrt{a}\,dt = \tfrac{A^2a\sqrt{a}}{2}\int\limits_{-\infty}^{\infty} e^{-t^2}dt = \frac{A^2a\sqrt{a}}{2}\sqrt{\pi} \end{align}

Which means that:

\begin{align} \frac{A^2a\sqrt{a}}{2}\sqrt{\pi} &= 1\\ A&=\sqrt{\frac{2}{a\sqrt{a\pi}}} \end{align}

Your units are perfectly fine. In QM, the units of your wavefunction should be $1/\sqrt{length}$, which they are. The reason for these units is that probability is unitless and to get the probability, you integrate the square modulus of the wavefunction over x. Doing the unit analysis on that easily gives unitless.
Since $a$ is essentially one, and since it seems like you've correctly found $A$, you could drop the units and integrate $x$ from $-0.1$ to $0.1$.
• But I have units $1/(nm \sqrt{nm})$. Take a closer look. This means i have a wrong $A$? – 71GA Aug 1 '13 at 14:56
• You normalization factor has units of $1/(nm\sqrt{nm})$ but it is multiplied by x, which would have units of nm, thus your wavefunction has the right units – Jim Aug 1 '13 at 15:12
• Ok so I now calculated the normalisation factor $A=1.06/(nm\sqrt{nm})$. What follows? What I know is that i have to redo the integral with limits $\int\limits_{-0.1nm}^{+0.1nm}A^2x^2\exp\left[-\frac{x^2}{a}\right]$ but if i do this i cannot evaluate the integral in the end to equal $\sqrt{\pi}$ and even the first part isn't $0$ (see my EDIT). This is where i am lost... What is the easiest way to do this? Do i really have to calculate allover again? I have tried and in the end i end up with an integral $\int_{-0.1nm}^{0.1nm} e^{-x^2/a}$ which I cannot solve... – 71GA Aug 1 '13 at 15:27
• 0.1993 is what you get before you multiply by the $A^2$ outside the integral – Jim Aug 1 '13 at 15:42