I just found this blog post, which gives an interpretation of virtual particles I haven't seen before.
Consider a 1D system of springs and masses, where the springs are slightly nonlinear. A "real particle" is a regular $\cos(kx-\omega t)$ wavepacket moving through the line, where $\omega$ satisfies the dispersion relation $\omega = \omega(k)$. When two real particles collide, the region where they collide temporarily looks really weird, as they interact nonlinearly, pulling and pushing on each other.
Formally, we can write this weird region as the sum of a bunch of $\cos(kx-\omega t)$ waves, but there's no guarantee they'll have the right dispersion relation. Thus, each term in the resulting expansion is a off-shell "virtual particle". If you add up all the virtual particles, you get the actual intermediate field state, which is just a weird ripple in the field where the two particles are interacting.
As another example, consider the statement "a static EM field is made of virtual particles". Under this interpretation, what that really means is, "a static field (e.g. $1/r^2$) is not equal to $\cos(kx-\omega t)$, but may be expanded in terms of such sinusoids", which is much less mysterious sounding. In fact, this is exactly what we do when we consider scattering off a potential in normal QM, e.g. in the Born approximation.
The above gives some intuition for what a 'virtual particle' means in classical field theory. They are the Fourier components of the field with $(\omega, k)$ not satisfying the dispersion relation, and they are useful in classical perturbation theory. However, they are not propagating degrees of freedom, so they only appear during interactions.
The picture above is entirely classical. Does this picture generalize to quantum field theory, giving a physical intuition for virtual particles there?
Best Answer
I will use the expressions "virtual particles" and "internal lines in a Feynman diagram" interchangably in this answer.
This interpretation fails because you can draw Feynman diagrams in both position and momentum space. When you draw them in momentum space and squint really hard, you might be able to convince yourself they have something to do with "Fourier components of the field", but the Fourier transform simply appears because you obtained the momentum space expressions from the position space expressions by a Fourier transform - it's not a characteristic of virtual particles.
In particular, virtual particles do not appear because we expand the field. They appear because we expand the interaction part of the time evolution operator - the n-point function we want to compute is something like $\langle \mathcal{T}\prod_{i = 1}^n \phi_i(x_i)\exp(-\mathrm{i}\lambda\int V(\phi(x))\mathrm{d}^d x)\rangle$ and the virtual particles appear when we expand the exponential as its power series and represent the resulting expressions graphically.
That we expand the time evolution operator suggests another "intuition" however: Virtual particle represent (fictitious) intermediate states over which we must sum. The time evolution operator there essentially means that when we compute the n-point function in an interacting theory, we must not only compute the "naive" n-point function, but also that in the presence of 1, 2, 3, 4, (i.e. finitely many) interactions (the vertices). At weak coupling $\lambda$, each interaction term suppresses the contribution of the summand it appear in, meaning the diagrams with few vertices contribute the most.
Heuristically: If you compare this to usual time-dependent perturbation theory in ordinary quantum mechanics (semi-randomly picked reference from a Google search [pdf]), it should be familiar: Expand the time evolution operator into its Dyson series, insert identies $1 = \sum_m \lvert m \rangle\langle m \rvert$, compute up to desired/feasible order. If we interpret $V(\phi)$ as the vertices, then the internal lines correspond to the inserted identities, meaning you may view virtual particles simply as generic "intermediate states" over which we must sum. You may even draw the same types of diagrams to organize such perturbation series in general.
What I think the blog post by Matt Strassler you're citing is getting at is that there is, regardless of all "virtual particles", of course an actual intermediate state during a QFT scattering, however complicated to describe it may be. He's saying that "virtual particles" are what physicists call that intermediate state which is...close enough, but more precisely we need to keep in mind that the intermediate states of perturbation theory aren't the same as the actual state of the system, they are computationally convenient fictions.
Finally, let me reiterate what I've said many times: Statements like "A static EM field is made out of virtual particles" are nonsense, since virtual particles are a tool in perturbation theory, not a fundamental entity in the theory. If you can manage to compute the relevant quantities without perturbation theory, then you'll never use the notion of virtual particle. Since physics does not depend on the specific mathematical method chosen to evaluate the relevant quantities, such statements have no basis in general QFT, and fail completely at strong coupling where no perturbative expansion is accessible. The only rigorous sense in which a static EM field is "made out of virtual particles" is that the Coulomb potential may be recovered from a diagram with a virtual particle in it. (Note that also in that case the Fourier transform is because we're computing stuff in momentum space, not because of the virtual particle as such.)