Confusion about probability of finding a particle The wave representation of a particle is said to be $\psi(x,t)=A\exp\left[i(kx−\omega t)\right]$.
The probability of the particle to be found at position x at time t is calculated to be $\left|\psi\right|^2=\psi \psi^*$ which is $\sqrt{A^2(\cos^2+\sin^2)}$. And since $\cos^2+\sin^2=1$ regardless of position and time, does that means the probability is always $A$? I think I am doing something wrong but I know what!
 A: *

*$\lvert \psi \rvert^2 (x,t)$ is not a probability, it is a probability density which you have to integrate over smoe region of space to get a probability. The probability to find the particle in an interval $[a,b]$ is $\int_a^b \lvert \psi(x)\rvert^2\mathrm{d}x$, which is zero for $a=b$, i.e. the probability to find a particle at a point is always zero.

*The plane wave $\psi(x,t) = A\mathrm{e}^{\mathrm{i}(kx-\omega t)}$ is only an admissible quantum state if the particle is confined to a region of space $S$ of finite volume, and then $A = 1/\sqrt{\mathrm{vol}(S)}$ because we want the probability density to be normalized as $\int_S \lvert \psi(x,t)\rvert^2\mathrm{d}x = 1$. If the particle is not confined, the function is not normalized and does not represent an actual quantum state.
A: It is not true that the probability of finding the particle at $x$ is $|\psi|^2$ (think of it as if you have a continuum of possible values, what is the probability of obtaining an specific value?). As it has been pointed out, $|\psi|^2$ can be interpreted as a probability density, so the probability of finding a particle between $x$ and $x + \mathrm dx$ is $|\psi(x)|^2\mathrm dx$. Integrating you can compute that probability for a specific interval.
