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Heisenberg's uncertainty principle is sometimes invoked in explaining the width of the diffraction pattern as light passes a single slit. In the case of single photons, when a photon is directed through a narrow single slit, its position in the transverse direction is restricted by the slit, leading to an uncertainty in the momentum in the transverse direction.

Now as I reasoned here, there is no change in the color of the photon, because the change in color would constitute a change in energy for which there is no reason to occur because in classical optics the wavelength does not change in Fresnel or Fraunhofer diffraction. Hence for a monochromatic photon, the wavelength $\lambda$ stays constant. (Is this correct?)

However, we know that the spread in the possible momentum direction is changed. If we say the $z$-direction is towards the slit/screen and the $x$-, $y$-directions are parallel to the slit/screen, then before the single slit the photon had momentum $\vec{p} = (h/\lambda) \, \vec{e}_{z}$ but afterwards it is in a superposition of $\vec{p} = (h/\lambda) \, \vec{e}_{n}$ for many possible directions $\vec{e}_{n}$.

(You may notice I'm avoiding any phrasing that implies there are definite trajectories, because it's what I am explicitly not doing, and I hope this is understood.)

Now here is the peculiar part: although it's not surprising that the (certainty in the) $xy$-components of $\vec{p}$ are changed, it also follows that the (certainty in the) $z$-component of $\vec{p}$ is changed (because there is a change in the direction of a vector without a change in its magnitude).

  • In non-relativistic quantum mechanics, we know $[\hat{x}_{i}, \hat{x}_{j}] = 0$, $[\hat{p}_{i}, \hat{p}_{j}] = 0$, and $[\hat{x}_{i}, \hat{p}_{j}] = i\hbar\delta_{ij}$, so while it's expected that a change in certainty of the $xy$-position (due to the slit) changes the certainty of the $xy$-momentum, we have a case here where a change in certainty of the $xy$-position changes the certainty of the $z$-momentum. Perhaps this is where non-relativistic quantum mechanics doesn't apply, and we have to use the formalism of quantum optics. If so, how does quantum optics formalism shed light on this relationship between $xy$-position certainty and $z$-momentum certainty?
  • Suppose a photon diffracts through a single slit and then it is measured on the $x<0$ half of the final screen. By momentum conservation, we surmise that some momentum in the $+x$-direction was given to the slits. However, we also can tell that $\vec{e}_{z}\cdot\vec{p}_{\text{after slit}} < \vec{e}_{z}\cdot\vec{p}_{\text{before slit}}$. It must follow that some momentum in the $+z$-direction is also transferred to the slit. Is there an account that makes this apparent/more obvious? I don't see any obvious interaction that makes this plausible.
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    $\begingroup$ your statement "there is no change in the color of the photon, because the change in color would constitute a change in energy E=ℏω for which there is no reason to occur. " is not completely true at the photon level. The complete quantum mechanical problem is the scattering of the photon throug the slit, which has electromagnetic fields and what you assume as elastic scattering is true only within the accuracies of measurement. $\endgroup$
    – anna v
    Commented Oct 3, 2022 at 13:03
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    $\begingroup$ In a complete solution with the boundary conditions of width and length of slit there will be a small loss in energy due to the interaction of the photon with the slit. This can be seen also in momentum conservation arguments of photon scattering. small because the mass off which the photon scatters is very large. $\endgroup$
    – anna v
    Commented Oct 3, 2022 at 13:03
  • $\begingroup$ @annav I see what you're saying. I am trying to remember why I thought the energy didn't change and I realized that I had a bit deeper reasoning than what I presented here (I edited my post accordingly). In classical optics, Fraunhofer diffraction does not change wavelength whatsoever. By analogy I thought the same would apply here, as far as the wave nature of photons is concerned. However, the particle-like nature can apply and result in scattering. I'm not convinced this plays a substantial role, but I'm happy to be shown wrong if there's a reference or source that analyzed this in detail. $\endgroup$ Commented Oct 3, 2022 at 18:38
  • $\begingroup$ It's extremely important to distinguish between scattering (particle-like nature) and diffraction (wave-like nature). I am strictly talking about diffraction, which as far as I know doesn't change wavelength (even in the quantum case). As far as I can tell, scattering is a red herring here (but again maybe I'm wrong). $\endgroup$ Commented Oct 3, 2022 at 18:49
  • $\begingroup$ Photons are quantum mechanical entities, and at the moment mainstream physics is based on assuming that all theoretical models of the behavior of matter emerges from the quantum mechanical level. At the quantum level there exists only interactions, modeled with Feynman diagrams (which have rigorous rules to calculate those interactions). Diffraction in optics and the wave solutions are within the limits of measurement classically and the wave is in space and time. The wave nature of quantum entities appears only in probability space, in accumulation of events. For photons see $\endgroup$
    – anna v
    Commented Oct 4, 2022 at 3:38

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In quantum mechanics, The expectation value of the energy is preserved whenever the system is freely evolving, but energy itself is not necessarily conserved.

The electron going through the slit does not need to preserve the expectation value of the energy, as it’s not free evolution. Some electrons go through the slits, some are absorbed by the walls. The ones that go through the slit have a slightly higher expectation value the energy compared to when they were emitted. This shift in the expectations value of the the energy comes from the new uncertainty in momentum in the transverse direction, since $\langle E \rangle \propto \langle p^2 \rangle = (\Delta p)^2$

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  • $\begingroup$ I edited my post to reflect a bit stronger reasoning. My understanding is that diffraction (a wave phenomena) by itself cannot change wavelength. It is scattering (a particle phenomena) changes wavelength. I'm talking about photons instead of electrons (as you mentioned electrons), and you are correct to point out these are not free particles, but scattering is absolutely not the explanation for the diffraction pattern, so I don't see how the expectation energy is changed by means of diffraction. Is there a source or calculation that backs up this claim? $\endgroup$ Commented Oct 3, 2022 at 18:45

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