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We know that every progressive wave depends on its position X and time variable T. y= Asin (kx- wt) My doubt is "Even an election is a wave (de broglie's hypothesis), but we cannot write its equation as a function of position because of heisenberg's uncertainty principle. So is there a mistake in our equation or is there some other equation for subatomic particles?

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  • $\begingroup$ The wavefunction of any particel basically gives you the probability of finding a praticle at a given position. There is no contradiction to the uncertainty principle. $\endgroup$ – user_na Jul 9 at 7:05
  • $\begingroup$ The uncertainty principle states that we cannot determine the exact position and velocity of a particle simultaneously. But to define the particle's wave nature, we need to no its exact position right? There should be no uncertainty in that. Then how is it possible to call an electrons path as a wave? $\endgroup$ – Aaradhanb Jul 9 at 7:35
  • $\begingroup$ No, this is a misconception. The wavefunction gives you a tool to calculate the probability to measure a particle at a given position. Without a measurement it doesn't even make sense to ask where it "really" is. When you are doing a measurement you can get the position of the particle with a given uncertaintay. This uncertainty actually gives you the limit for the accuaracy of an additional measurement of its momentum - fully in line with the uncertainty principle. $\endgroup$ – user_na Jul 9 at 7:46
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There is an equation for the de broglie's wave and its depend on the coordinate, for example for the particle which moving with the given energy $E = \frac{\textbf{p}^2}{2m}$ and momentum P along X-axe it will be as $$\psi(x,y,z) = exp\left(\frac{i}{\hbar }(Et - px)\right)$$ But x is not the coordinate of the particle, it is the coordinate of the point of space in which you want to calculate the probability of particle detection.

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The wave function is just a function that tells you what are the probability amplitudes associated with the particle being found in a certain state. The $x$ in $\psi(x,t)$ is NOT the position of the particle; it is a point in space(time) at which $|\psi(x,t)|^2$ tells you the probability density of finding the particle. The fact that at the same $t$, a different point $x'$ has a nonzero $|\psi(x',t)|^2$ reflects that it is not certain that the particle WILL be at $x$(or $x'$), rather there is a certain probability that it will be in any given position.

These positions can now thought to be a random variable, and the standard deviation of that distribution would tell you the uncertainty in position. Moreover, the schrodinger equation tells you only the undisturbed, causal time evolution of the wavefunction- a measurement will collapse the wave function to a value the Schrodinger equation cannot causally determine(it will be in one of the eigenstates).

The specific example of a plane wave associated with a free particle(which, by the way, is not the 'right' solution, since it is not normalizable-the correct wavefunction would be given by a fourier integral of the plane waves) has the wavefunction $$\psi(x,t)=e^{i(kx-\omega t)}$$ tells you that the probability amplitude is $$|\psi|^2=\psi\psi^*=1$$, which is a constant; i.e. the particle is equally likely to be found at all points. This means that there is infinite uncertainty in position-you have no information about where the particle is. This is consistent with the uncertainty principle-the wavefunction above assumes that we know exactly what $k$ (and hence $p$ ) is. The more general result is to introduce a 'variance' in $k$-i.e. consider a superposition of various plane waves, and do a Fourier transform. The true wavefunction would be $$\psi(x,t)=(2\pi)^{-1/2}\int{\phi(k)e^{i(kx-\omega(k) t)}}dk$$ , and now this will show a spread in both $x$ and $k$(i.e.$p$), consistent with the uncertainty principle.

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