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Double slit experiment with individual particles has been performed with a variety of particles from photons to molecules of up to 2000 atoms.

For photons, as the frequency increases (i.e. wavelength decreases), fringe width decreases at the same ratio. For instance, if the frequency is doubled (wavelength halved), fringe width is halved. Since $E_{photon} = h.f$, this can be interpreted as 'the fringe width decreases as the energy of the photon increases'.

I wonder if the same principle applies to objects with mass, too. How is the fringe width affected by:

  1. the kinetic energy $(KE)$ of the particle only
  2. the mass $(m)$ of the particle only
  3. the combination of mass energy and kinetic energy ($E_{t} = m.c^2 + KE)$
  4. the total energy of the incoming particle ($E_{RuleThemAll} = m.c^2 + KE + E_{thermal} + E_{whatever}$)

P.S. I am not sure if $m.c^2$ can be applied to molecules and if $eV$s would be more meaningful.

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  • $\begingroup$ What do you mean by the P.S. note ? $\endgroup$
    – SPHerical
    Sep 25, 2021 at 23:42
  • $\begingroup$ @SPHerical As $m.c^2$ is derived by relativity, I didn't know if it would make sense for molecules, which might be classified as quantum objects. But it looks like the mass of a particle is also expressed as $eV/c^2$, which can be seen in the link. $\endgroup$
    – Xfce4
    Sep 26, 2021 at 0:10
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    $\begingroup$ Special relativity is true no matter whether you study a quantum object or a classical object. Afterall the main motivation for QFT has been the search for a relativistic theory of quantum mechanics. The fact that you can express the mass of a molecule in $eV/c^2$ is simply due to the fact that $eV/c^2$ is a unit for mass just like Kg or pounds. $\endgroup$
    – SPHerical
    Sep 26, 2021 at 8:14

2 Answers 2

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Short answer: Look up "De Broglie wave length". You can associate a wavelength $\lambda = h/p$ to particles and then use your intuition from wave optics to think about how the fringe pattern should change with changing $p$ (i.e. $p$ goes up means that the wave length decreases which means that ...). Especially this means that the fringe width is only affected by momentum.

Some Details: The reasoning via the De Borglie wavelength applies to particles that are in momentum eigenstates (states with definite momentum) i.e. satisfying:

$ - i \hbar \partial_x \psi = p \psi$

for some momentum $p$. The equation implies that $\psi$ is of the form

$\psi(x,t) = \exp(i \frac{p}{\hbar} x) \phi(t)$

for some function $\phi(t)$ that is to be determined from the Schrödinger equation i.e. which depends on the total energy. But no matter what $\phi(t)$ is, it will only contribute to a global phase factor that has no meaning for the interference pattern. So only the momentum is relevant and this analysis confirms that the wave length is $h/p$ since the wave vector is $p/\hbar$.

Comments about your $E_{RuleThemAll}$ term:

A. I does not make much sense to speak of thermal energy of a single particle.

B. In general you need to be careful when you try to decompose the energy of a quantum mechanical system into different components. For example while it does make sense to speak of the total energy of the electron in a Hydrogen atom, you cannot decompose this energy into kinetic and potential energy, as these correspond to non commuting operators i.e. quantities that cannot be measured simultaneously.

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  • $\begingroup$ Thank you. (1) Then there is a connection between the interference pattern and both $m.c^2$ & $KE$ , although not in a linear manner as in $m.c^2 + KE$. (2) I thought thermal energy would make a difference for molecules with 2000 atoms. But I don't know. Maybe these molecules come out with the same amount of average vibration. $\endgroup$
    – Xfce4
    Sep 26, 2021 at 0:35
  • $\begingroup$ (1) No this is not correct. The mc^2 term can be added to the Hamilton, but it is only a constant. Therfore it has no impact on the interference pattern. Only the momentum matters! $\endgroup$
    – SPHerical
    Sep 26, 2021 at 8:05
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    $\begingroup$ (2) The problem is that "thermal energy" implies thermal averaging over some ensemble, while what you are really interested in here are the propagation of individual objects in Quantum Mechanics. It would be better to call the thing you presumably mean "excitations of internal degrees of freedom". $\endgroup$
    – SPHerical
    Sep 26, 2021 at 8:08
  • $\begingroup$ (1) But $p = \sqrt{2\times KE \times (m.c^2)/(c^2) }$. We cannot directly convert KE to p. $\endgroup$
    – Xfce4
    Sep 26, 2021 at 17:39
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Interference in the two slit experiment is a spatial phenomenon: different path lengths ($\Delta x$) lead to phase differences at detection:

$$ \Delta\phi = \hbar k\Delta x $$

Clearly, $\Delta x$ is fixed, so the only thing that affects $\Delta \phi$ is the wave vector $k$. That is it.

Of course:

$$ p = \hbar k$$ $$ \omega = \sqrt{(ck)^2 + (mc^2/\hbar)^2}$$

$$ E = \hbar\omega$$

and so on, so anything that changes $k$ changes $\Delta\phi$, and it becomes a matter of semantics as to what affects interference.

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