How does the particle's total energy affect the interference pattern? Individual Particle Double Slit Experiment

Question

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 ($E_{RuleThemAll} = m.c^2 + KE + E_{thermal} + E_{whatever}$)

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

Answer 1

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.

Answer 2

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.