Why is the polarization identity important, intuitively? The polarization identity states, roughly, that a norm satisfying the parallelogram law induces a vector space inner product (and vice versa). This has many nice applications, such as a simple characterization of unitary operators  (which is useful in quantum mechanics), recovering operators from their associated quadratic forms, proving things about expectations of stochastic variables, or showing that the limit of a sequence of quadratic forms is a quadratic form.
What I don't see fully is why, when facing some problem, it would occur to someone to attempt to polarize something. What does polarization mean, philosophically, and why and when would I expect it to make an appearance?
Any other illustrative consequences of the polarization identity, or useful references are also welcome.
 A: The polarization identity tells you that the geometry of a (complex) Hibert space is captured in its norm through the parallelogram law. This was precisely John von Neumann's goal in formulating his result that an inner product could be defined from a norm, assuming that the norm satisfies the parallelogram law. And that allowed simpler characterizations of various operators by considering only their quadratic forms, instead of their full forms. By the way, it was von Neumann who first defined an abstract inner product, even though Hilbert often gets that credit. Hilbert was von Neumann's advisor. von Neumman did many interesting things in his career, including proving the Spectral Theorem for selfadjoint operators, and inventing the CPU, just to name a couple of things.
Another interesting application is to selfadjoint operators. Suppose $X$ is a complex inner product space, and further suppose that $A : X \rightarrow X$ is such that $\langle Ax,x\rangle$ is real for all $x\in X$. Then, $\langle Ax,x\rangle=\overline{\langle Ax,x\rangle}=\langle x,Ax\rangle$, which also gives
\begin{align}
     \langle Ax,y\rangle &= \frac{1}{4}\sum_{n=0}^{3}i^n\langle A(x+i^n y),x+i^ny\rangle \\
     &= \frac{1}{4}\sum_{n=0}^{3}i^n\langle x+i^ny,A(x+i^ny)\rangle \\
     &= \langle x,Ay\rangle.
\end{align}
From this, it also follows that
\begin{align}
  \langle A(\alpha x+\beta y),z\rangle 
   &=\langle \alpha x+\beta y,Az\rangle \\
   &=\alpha \langle x,Az\rangle+\beta\langle y,Az\rangle \\
   &=\alpha \langle Ax,z\rangle+\beta\langle Ay,z\rangle \\
   &=\langle \alpha Ax+\beta Ay,z\rangle.
\end{align}
Hence $A$ is linear because the above holds for all $z$. So real Quantum measurements give rise to linear selfadjoint operators.
