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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!

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  • $\begingroup$ Mohammad I've converted the math in your post to use the MathJax rendering engine that we have running on the site. Please take a look at my work to insure that I haven't messed things up for you. $\endgroup$ – dmckee Dec 17 '16 at 21:29
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  1. $\lvert \psi \rvert^2 (x,t)$ 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.

  2. 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.

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  • $\begingroup$ Thanks for your good anwer. For the plane wave, isn't that $\psi(x,t)^2=A^2 $? If yes, how could your integral value over infinity $\int_S \lvert \psi(x,t)\rvert^2\mathrm{d}x$ be normalized to 1? $\endgroup$ – Mohammad Mostafavi Dec 18 '16 at 3:57
  • $\begingroup$ @MohammadMostafavi It can't - that's why I say that this is only an allowed state if $S$ is a finite region. If it's infinite, you can't normalize the wavefunction. $\endgroup$ – ACuriousMind Dec 18 '16 at 11:21
  • $\begingroup$ @ACuriousMind How can the probability to find a particle at a point be zero? The particle must be physically present somewhere, if I integrate the wavefunction square at point where the particle is actually present, the probability should be one? Does this mean that our laws of integration must be changed or I am severely wrong? $\endgroup$ – Ajinkya Naik Nov 3 at 8:19
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    $\begingroup$ @AjinkyaNaik There are no points where the particle is "actually present", there is just this probability density, that's the whole point of quantum mechanics. $\endgroup$ – ACuriousMind Nov 3 at 9:16
  • $\begingroup$ Some people describe the "probability thing" as a result of our own limitations in extracting a quantum particle's physical parameters without disturbing the others, whereas some describe the probabilistic nature as being intrinsic to the quantum particles. There is a lot of confusion around!! $\endgroup$ – Ajinkya Naik Nov 3 at 9:19
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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.

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