Adding versus multiplying identical photons' wavefunctions? I am currently confused with understanding many identical photons' wavefunctions. I think that photon wavefunctions are supposed to be multiplied together to describe the total state of all bosons. However, through some magic I am confused with, the total boson wavefunction must yield a field strength proportional to the sum of all of the boson wavefunctions?
 A: "Multiplying the wavefunctions" is a pretty nebulous term. Let's work with some definite vocabulary here, shall we?
$(1)$ The states of one QM particle are elements of some Hilbert space $\mathcal{H}$. If we care only about position on a line as completely defining the state (which we can for a scalar boson), i.e. demand that the space be spanned by the orthogonal position basis $|x\rangle,x\in \mathbb{R}$, this space is isomorphic to the square integrable functions $\psi : \mathbb{R} \rightarrow \mathbb{C}$ by sending arbitrary states $|\psi\rangle$ to the function $\psi(x) = \langle \psi | x \rangle$. [We have ignored some technicalities above, i.e. normalization, Gelfand tripel, distribution...and we will continue to ignore them, they don't matter for this argument]
$(2)$ The states of $N$ such particles is given by the $N$-fold tensor product space $\mathcal{H} \otimes \dots \otimes \mathcal{H}$. It's basis is formally given by 
$$|x_1,\dots,x_N\rangle :=|x_1\rangle \otimes \dots \otimes |x_N\rangle, x_i \in \mathbb{R} \forall i \in \{1,\dots,N\}$$
We can map any state $|\psi\rangle \in \bigotimes_{i = 1}^N \mathcal{H}$ to a wavefunction again by
$$ \psi : \mathbb{R}^N \rightarrow \mathbb{C} , (x_1,\dots,x_N) \mapsto \langle \psi |x_1,\dots,x_N\rangle$$
This can only be written as $\langle\psi_1|x_1\rangle\dots\langle\psi_N|x_N\rangle = \prod_i \psi_i(x_i)$ if the state $\psi$ is a state constructed from the single states $|\psi_i\rangle$ of the particles as $|\psi\rangle = |\psi_1\rangle \otimes \dots \otimes |\psi_N\rangle$. Not all states in $\bigotimes_{i = 1}^N \mathcal{H}$ can be written that way.
So, if you indeed know all individual states, the total wavefunction is indeed a multiple of the single wavefunctions. However, for so-called entangled states, this is impossible.
So, taking the tensor product is multiplying the wavefunctions, which enlarges the space of all possible states. Adding the wavefunctions is just using the addition on the Hilbert space $\mathcal{H}$. It does not enlarge the space of states, it only superposes two states of the same particle/system.
I have no idea what "field strength" you are talking about in the OP, though.
