According to Higgs Mechanism a particle acquires mass when it couples to the higgs field. Now consider Wave-Particle Duality. Suppose I am doing a Young's Double Slit Experiment using electrons. When they pass through the slits they behave like waves and quantum waves have no mass, there is just momentum given by de-Broglie relation. Now when they "impinge" upon a screen their behaviour is like that of a particle. So they must have mass. When did the electrons interact with the Higgs field to acquire mass when they did not have any mass while "passing" through the slits? Does this imply that Higgs field is decoupling and coupling with the electron field? Also, I could make the distance between the slits and the screen arbitrarily small, doesn't that imply that the coupling "speed" (for lack of a better word on my part) of Higgs field is infinite?

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    $\begingroup$ Why are the de-Brogile relations $E=\hbar\omega$ and $\vec p=\hbar \vec k$ inconsistent with mass? Mass means the dispersion relation is $E=\sqrt{(pc)^2+(mc^2)^2}$. $\endgroup$ – JEB Jul 9 '18 at 13:45
  • $\begingroup$ The de-Broglie relation is $\lambda = \frac{h}{mv}$ with $h$ as Planck's constant, $v$ velocity and $m$ the mass of the particle. If the $m$ were zero, the wavelaength $\lambda$ would be infinite which makes no sense. $\endgroup$ – Frederic Thomas Jul 9 '18 at 15:22

The Higgs induced mass is due to the vacuum expectation value or v.e.v of the Higgs. In the cold dead universe we live in, Higgs has a sort of "crystallized" background configuration (b.g.c) with perturbations around it. This b.g.c is everywhere and all the fields that couple to it get their mass from it. So it doesn't really need to propagate. In theory if the energy level of the system was high enough (hot enough) you could destroy the Higgs b.g.c and decouple everything.


quantum waves have no mass, there is just momentum ...

is a fundamental misconception about QM. The probability amplitude wavepacket desribing the motion of the electron fully accounts for the movement of its mass (and charge, and...) because this wavepacket, moving freely is restricted by the dispersive equation of Schroedinger, here in one dimension w.l.o.g., $$ \left (i\hbar\partial_t + \frac{\hbar^2}{2m} \partial_x^2\right ) \psi =0, $$ which is manifestly "aware" of its mass.

The relevant solution is essentially a spreading wavepacket $\psi (x,t)$, yielding a probability $$ \Large \rho(x,t)=|\psi|^2=\frac{1}{\sqrt{1+4(t\hbar/m)^2}} e^{\frac{-2(x-vt)^2}{1+4(t\hbar/m)^2}} , $$ in natural wavepacket length units, where v is the group velocity of the wavepacket.

The relevant probability flow current to the right can be seen to be ~ $v \rho$, and so the rate of transport of mass to the right is $mv \rho$, as you should confirm.

The takeaway is that the "wave" of your mental picture does very much transport mass, charge, spin, etc... to the screen. Calling it "massless" is meaningless. The Higgs mechanism has done its job of giving the electron a mass and completely decoupled from the problem, conceptually. The wave went through the slits and interfered with no less mass than a corpuscle.


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