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}
Answer
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 $0.1$.
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