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Why we don't see any Gibb's phenomenon in quantum mechanics?

EDIT

At sharp edges (discontinuities), we usually find ringing. This can be observed in many physical phenomenon (eg. shock waves). Naturally, whenever there is a shrap discontinuity in wave functions, I'd expect a ringing in the probability of finding a particle around that edge.

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    $\begingroup$ What do you mean? What does the Gibb's phenomenon have anything to do with quantum mechanics? A little (or a lot) more context would help. $\endgroup$ – Michael Jul 9 '13 at 6:42
  • $\begingroup$ aren't the interference fringes in the electron double slit ? en.wikipedia.org/wiki/… $\endgroup$ – anna v Jul 24 '15 at 15:06
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First of all, the Gibbs phenomenon is a mathematical effect – it is the appearance of narrow but intense oscillations around the right value whenever a function with a discontinuity is approximated by its Fourier expansion that is truncated.

This phenomenon doesn't occur when the relevant function is a wave function $\psi(x,y,z)$ in quantum mechanics for a simple reason: the wave functions don't have such discontinuities. If a wave function had a jump of this sort, then its derivative $\psi'$ or $\nabla x$ would have a $\delta$-function at the point of the jump, and its square would integrate to infinity ($\delta^2$ is infinitely times greater than just $\delta$).

But this integral is proportional to a formula for the expectation value of the kinetic energy $$\int_{-\infty}^\infty |\psi'|^2 dx = -\int_{\infty}^\infty \psi^* \psi'' dx$$ by integration by parts so whenever it diverges, it means that the energy is infinite which is physically impossible. That's why real-world, finite-energy wave functions can't have jump-like discontinuities as a function of spatial coordinates although their first derivatives are already allowed to have such jumps.

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  • $\begingroup$ If there is no such things in wave function, then how come quantum particles are created and annihilated in particle physics? Also how does wave function collapse? $\endgroup$ – Rajesh D Jul 9 '13 at 7:20
  • $\begingroup$ In any real-world process described by QFT, particles are never created at a strictly defined point with $\Delta x =0$ because of the very same reason I have already explained: that would require $\Delta p = \infty$ by the uncertainty principle and that would also require an infinite energy. In reality, particles are always created in wave functions that are non-vanishing in a whole region of the position space. Wave functions that would be localized at points wouldn't even be normalizable. $\endgroup$ – Luboš Motl Jul 9 '13 at 13:35
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    $\begingroup$ Moreover, when you localize to smaller distance than the Compton wavelength, new particle production (and other relativistic effects) becomes extreme. ... There is nothing such a "wave function collapse". What actually happens when misguided people talk about a "collapse" is that we learn the results of a measurement so it's practical for us to replace the original complicated probabilities ready for all options by the conditional probabilities in which the particular measured values have already been taken into account and the remaining options forgotten. Nothing "real" collapses in Nature. $\endgroup$ – Luboš Motl Jul 9 '13 at 13:38
  • $\begingroup$ Instead of defining kinetic energy as square summation of $\psi'$, we can regard it as dot product of $\psi'$ with itself, which amounts to the same for any any sane wave function, and then define dot product of delta function with itself as $1$. And dot product of two deltas different from each other (two different positions) as $0$. It makes sense for kinetic energy of a wave function with a jump discontinuity. $\endgroup$ – Rajesh D Jan 11 '16 at 4:50
  • $\begingroup$ Hi Rajesh, you can't "define" the dot product of $\delta(x)$ with itself to be one, it's just like trying to "define" $5+5$ to be $3$. But $5+5=10$ may be calculated and in the same way, the dot product of $\delta(x)$ with itself is the integral of $\delta(x)^2$ which is infinite because it's $\delta(0)$ if one treats one of the factors as the test function $f(x)$, and $\delta(0)$ is infinite which may be explained in tons of ways. If you "redefined" the inner products, you would destroy the distributive law on the Hilbert space and other things. You just cannot redefine calculable results. $\endgroup$ – Luboš Motl Jan 11 '16 at 11:18
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Remember that

$$p\Psi=-i\hbar\nabla\Psi $$

Also,

$$E\Psi=i\hbar\partial_t\Psi $$

If there is discontinuity in the wavefunction, the 4-momentum would be infinite.

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