Thinking of the photon itself as a single particle, and imagining the building up diffraction patterns with one photon in the apparatus at a time will give you an idea of the correspondence between classical wavelike behaviour and the probability waves of the wavefunction, at least for bosons. The following, I believe, is what Akrasia's answer means when Akrasia says "He doesn't mention it, but I believe this starts to explain why light exists..." and what Ron Maimon in the opening paragraph to his excellent answer means.
The "wavefunction" for the photon can be taken to be the vectors:
$$\vec{E}=\left(\left.\left<0\right|\right.\hat{E}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$$
$$\vec{H}=\left(\left.\left<0\right|\right.\hat{H}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3\tag{1}$$
where $E^+_j$ and $H^+_j$ are the positive frequency parts of the electric and magnetic field observables and $\psi$ is the one-photon state, expressed as a superposition of one-photon Fock states. This "wavefunction" is not quite the same as for an electron, as there is no position observable, at least in the same sense as for the nonrelativistic electron Schrödinger equation. (Note that there are difficulties with defining a position observable for the relativistic Dirac electron too, so you can think of the nonexistence of the true "wavefunction" as an assertion that there is no nonrelativistic description of light). But this six-component wavefunction uniquely defines the one-photon state, so that the two can be taken as equivalent. Now, witness the following interesting physical interpretations and theoretical observations about this six-component field:
Its normalised "square magnitude" $\frac{1}{2}\,\epsilon\,\|\vec{E}|^2 + \frac{1}{2}\,\mu\,\|\vec{H}|^2$ is the probability density in space and time to photodetect the photon, i.e. destructively detect a photon by absorption with, say, a PMT. So it defines the interference/diffraction pattern that you will build up over time in a one-photon-in-the-apparatus-at-a-time experiment;
As already stated, it uniquely defines the one-photon state $\psi$ and contrariwise, so that the two can be taken to be equivalent information and thus we can think of this vector field as the one-photon state;
Now we suppose that the light field is in a Coherent State, i.e. the field state is of the form:
$$\psi = \prod\limits_j\,\exp\left(\alpha_j\,a_j^\dagger-\alpha_j^*\,a_j\right)\left.\left|0\right>\right.\tag{2}$$
where $a_j^\dagger$ is the creation operator raising the unique quantum field ground state $\left.\left|0\right>\right.$ to the one photon Fock state in the field mode corresponding to the plane wave $\exp\left(\vec{k}_j\cdot\vec{r}-\omega_j\,t\right)$ and the $\alpha_j$ - the "Displacements" - define the strength of the excitation in each mode. Take heed that in a coherent state, there is the superposition of number states in each mode, so that in a coherent state the photon number is uncertain and Poisson-distributed, unlike for the one-photon Fock state superposition. The "displacements" get their name because they behave like "components" of a displacement "vector" displacing the Wigner quasiprobability distribution away from the origin (ground state) in quantum phase space.
Now withness the means of the field observables $\hat{E}$ and $\hat{H}$:
$$\left(\left.\left<\psi\right|\right.\hat{E}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3$$
$$\left(\left.\left<\psi\right|\right.\hat{H}^+_j\left.\left|\psi\right>\right.\right)_{j=1}^3\tag{3}$$
(take heed of the subtle difference between (1) and (3)) as well as their following extremely interesting properties for coherent states:
- They fulfill the same Maxwell's equations as fulfilled by the "wavefunction" in (1), therefore they uniquely define the one photon states that can exist for the prevailing boundary conditions (even though they define a coherent state and NOT a one-photon state);
- If the coherent displacement parameters $\alpha_j$ are big enough (i.e. if the light field is coherently displaced far enough from its ground state), then these means become the same as what we classically measure the electric and magnetic field to be.
So, in (1), Maxwell's equations and their boundary conditions define probability waves that determine how our diffraction pattern will be built up one photon at a time. The objects defined in (3) are what our classical measurements of the Maxwellian electromagnetic field vectors would measure, and they are the same Maxwell equations and boundary conditions!
So, if you get a vector voltmeter and magnetometer and measure classically the field distributions in a microwave setup, you are measuring the objects in (3) and, by our discussion above, you are also exactly measuring the one-photon states of the quantum field that can exist for the same boundary conditions.
