If you look carefully, you'll note that if we multiply the correct relationship from the last problem \(\omega^2=c^2 k^2 + m^2 c^4/\hbar^2\) by \(\hbar^2\) and identify \(E=\hbar \omega, p=\hbar k\) we arrive back at the special relativistic dispersion relation, \(E^2=p^2 c^2 + m^2 c^4\). This is in fact the correct association of energy and momentum in terms of frequency and wavenumber in quantum mechanics. Now, we can see that if free particles are described by such waves they have the correct special relativistic relationship between their energy and momentum. Yay, physics is consistent!

All of this is for particles traveling in empty space. In the real world, of course, there's lots of stuff around, the earth, electromagnetic fields, etc. and various particles and fields can interact. All of our interactions are, however, *local* in that fields and particles must interact at the same point. We model these interactions by *interaction terms* in the wave equation. An interaction term is generated by multiplying our function \(f(t,x)\) that describes the particle by another function \(g(t,x)\) that describes the distribution of some other background or source field. Note that x is the same in both \(f(t,x)\) and \(g(t,x)\) - this is the essence of local interactions. As a result of interactions, our wave equation would get modified to something of the form

\( (\frac{1}{c^2} \frac {\partial^2}{\partial t^2} - \frac {\partial^2}{\partial x^2} + \frac {m^2 c^2}{\hbar^2} + g(t,x)) f(t,x)=0\)

Solving this equation is generally impossible for arbitrary \(g(t,x)\). Consider, however, if there was a field out there, analogous to the electromagnetic field, except that had a constant, non-zero value \(g_0\) *everywhere*. Physicists call such a non-zero value everywhere by a fancy name: *a non-zero vacuum expectation value*. Would a wave solution for \(f(x)\) work in this case? (Remember, we're asking about \(f(t,x)\) when \(g(t,x)=g_0\)).

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