Okay, I'll take this one. First let me say that the English term is "completely bounded" (or "complete isometry", etc.).

About the term "quantum". The general principle is that analyzing some aspect of a physical system typically involves very different kinds of mathematical structures, depending on whether the system is classical or quantum. Thus if you study a classical system using topological techniques, or measure theory, or Lie groups, or graphs, the analogous analysis of a quantum system would probably involve C*-algebras, or von Neumann algebras, or quantum groups, or operator systems. (For instance, here's a paper of mine that talks about how operator systems arise in quantum error correction in the same way that graphs arise in classical error correction.)

There is no simple rule for translating classical structures into quantum structures, but we have a pretty large dictionary and one can identify broad themes. My take is that operator spaces fit into this scheme as the quantum analog of relations on a set. See Section 2 of the paper linked above. I'm not aware of any really compelling direct connections with real-world quantum mechanics in this instance, but the word "quantum" may be appropriate just because the subject fits into the general scheme of mathematical quantization which does include many direct links.

Now, as to the "beautiful results" you ask for. Well, there's a lot of good stuff so the selection would be idiosyncratic. One general principle is that when you're studying things related to Hilbert space (single operators, C*-algebras, etc.) it is often the case that you get much nicer results if you assume the "complete" (at all matrix levels) version of your hypothesis. For example, if two C*-algebras are linearly isometric, are they isomorphic as C*-algebras? Not in general, but yes if they are completely linearly isometric, i.e., linearly isometric at all matrix levels. (I think this is folklore.)

The result of Pisier that you refer to was actually a counterexample, showing that a polynomially bounded operator need not be similar to a contraction. The positive result is due to Paulsen and says that any *completely* polynomially bounded operator *is* similar to a contraction.

Here's another nice result, this one due to Zhong-Jin Ruan, one of the giants of the subject. B. Johnson proved that a locally compact group $G$ is amenable if and only if the Banach algebra $L^1(G)$ is amenable. Ruan showed that this happens if and only if the Fourier algebra $A(G)$ is *completely* amenable.

A good place to learn more is the book *Operator Spaces* by Effros and Ruan.

References:

G. Pisier, A polynomially bounded operator on Hilbert space which is not similar to a contraction, *J. Amer. Math. Soc.* **10** (1997), 351–369.

V. Paulsen, Every completely polynomially bounded operator is similar to a contraction, *J. Funct. Anal.* **55** (1984), 1–17.

Z-J. Ruan, The operator amenability of A(G), *Amer. J. Math.* **117** (1995), 1449–1474.