When we want to calculate the probability of finding a particle in a small region of space given the time-dependent Schrödinger equation, it should be equal to $\vert\psi(x)\vert^2$ times $dx$ times the normalization constant $A^2$,right?
for example $$ \psi (x,t)=A e^{-x^2} e^{i(kx-wt)} $$
What’s the probability?
On
The usual interpretation of a nonzero solution $u(x,t)$ to the free Schrodinger equation $i\partial_tu - \Delta_x u = 0$ in $\mathbb R^d\times\mathbb R$ is that $|u_t|^2 = |u(\cdot,t)|^2$ represents the probability density function for the "position" of a quantum mechanical particle $P$ at time $t$. For a region $\Omega\subset\mathbb R^d$ of space, the appropriately normalized integral $\int_\Omega|u(x,t)|^2\,dx$ gets the interpretation $$ \int_\Omega|u(x,t)|^2\,dx = \mathrm{Prob}(P\in\Omega\ \text{at time $t$}). $$ A feature of the normalized solution $u$ is that $\int_{\mathbb R^d}|u(x,t)|^2\,dx = 1$ for every $t$.
The probability $\rho$ of finding a particle in a region $D$ in one dimension is:
$$ \rho = \int_D \psi \psi^* dx $$
with $\psi$ normalised and $\psi^*$ the conjugate of $\psi$. Given the wave function:
$$ \psi(x,t)=Ae^{-x^2} e^{i(kx-wt)} $$
We get:
$$ \rho = \int_D Ae^{-x^2} e^{i(kx-wt)} Ae^{-x^2} e^{-i(kx-wt)} dx$$ $$ \rho = A^2 \int_D e^{-2x^2} dx$$
We get the normalisation constant $A$ by solving:
$$ 1 = A^2\int_{- \infty}^{\infty} e^{-2x^2} dx = A^2 \sqrt{\frac{\pi}{2}}$$
for $A$ since the integral of a probability density over its entire region is equal to $1$. And so the probability of finding a particle in some region $D$, i.e. an interval $[a,b]$, is:
$$ \rho = \sqrt{\frac{2}{\pi}} \int_a^b e^{-2x^2}dx $$