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.