Undergrad Is the uncertainty principle applicable to single slit diffraction?

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SUMMARY

The discussion centers on the application of the uncertainty principle to single slit diffraction, specifically how narrowing the slit affects particle momentum and diffraction patterns. Participants highlight that while the first minimum of the diffraction pattern can be used to derive the uncertainty relation, this approach may seem arbitrary. A more rigorous explanation involves the transformation of the wavefunction as it passes through the slit, acting as an infinite square well, leading to a superposition of momentum eigenstates. This explanation clarifies the relationship between slit width and the resulting diffraction pattern beyond the basic uncertainty principle.

PREREQUISITES
  • Understanding of the uncertainty principle in quantum mechanics
  • Familiarity with single slit diffraction and its mathematical representation
  • Knowledge of wavefunctions and their transformation in quantum systems
  • Basic grasp of momentum eigenstates and their significance in quantum mechanics
NEXT STEPS
  • Study the mathematical derivation of the uncertainty principle in quantum mechanics
  • Explore the concept of wavefunctions in quantum mechanics, particularly in potential wells
  • Investigate the relationship between slit width and diffraction patterns in detail
  • Learn about the role of excited states in quantum systems and their impact on wavefunction behavior
USEFUL FOR

Students and professionals in physics, particularly those focused on quantum mechanics, wave-particle duality, and diffraction phenomena. This discussion is beneficial for anyone seeking a deeper understanding of the implications of the uncertainty principle in practical scenarios.

greypilgrim
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Hi.

I've seen single slit diffraction being brought up as an example of the uncertainty principle: Narrowing the slit restricts the particles more in one dimension, which means the momentum in this dimension is more uncertain, which results in a more spread-out diffraction pattern.

I've even seen "derivations" of the uncertainty relation like the following:
1709223774039.png


They use the 1st minimum of the pattern to define ##\Delta\vec{p}_x##, and then with ##\sin(\alpha_1)\approx\tan(\alpha_1)## and the de Broglie wavelength successfully arrive at ##\Delta x\cdot\Delta p_x\approx h##.

Well just taking the 1st minimum seems arbitrary. But if I'm not mistaken, a correct derivation of ##\Delta p_x## diverges since ##x^2\sinc^2 (x)## isn't integrable. This of course doesn't contradict the uncertainty principle, but is there a more rigorous way to make sense of it in the case of the single slit?

It's kind of weird that this "spreading out" of the pattern while narrowing the slit isn't reflected in ##\Delta p_x## at all which is always infinite.
 
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Hi,

Not a real answer, but:

The colleagues have a thread on your subject.

There is also Sheet 24 here

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greypilgrim said:
This of course doesn't contradict the uncertainty principle, but is there a more rigorous way to make sense of it in the case of the single slit?

It's kind of weird that this "spreading out" of the pattern while narrowing the slit isn't reflected in ##\Delta p_x## at all which is always infinite.
The uncertainty principle is a quick way to justify diffraction, but it doesn't explain the detail. A more detailed explanation is:

When the particle reaches the slit it has effectively the uniform wavefunction of a plain wave. The slit acts like an infinite square well and the wavefunction transforms to a linear combination of momentum eigenstates appropriate to the width of the well. When it emerges from the well, that superposition evolves as a superposition of free particle states. The width of the central band corresponds to the ground state of the well. The narrower the slit, the wider the band. The other bands correspond to the excited states. For a wider slit only the ground state is significant. For a narrow slit, more of the excited energy states become significant.

That's still some way short of a full mathematical treatment. But, it explains more than simply invoking the UP.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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