This "research vignette" (whatever that is) claims that the polynomial hierarchy
classifies problems according to a natural notion of logical complexity, and is defined with an infinite number of levels: problems at the zeroth level are the “easiest”, and for every integer $k$, problems at the $(k+1)$-st level have logical complexity “one notch higher” than those at level $k$.
Obviously this is just a heuristic explanation rather than any kind of rigorous claim, but is it true that problems higher in the hierarchy should be thought of as harder than problems lower in the hierarchy?
For example, consider the complexity class BQP. It is not believed to contain NP; in fact, many experts consider (Fig. 3) NP-complete problems to be much harder than the hardest problems in BQP. But the same experts also believe that BQP is not contained within the polynomial hierarchy PH (with Recursive Fourier Sampling as a concrete example of a problem believed to be in BQP \ PH). It therefore seems reasonable to say that for any $n$, there exist problems in $\Sigma_n \setminus \Sigma_{n-1}$ that are easier than RFS.
But putting these two results (that any NP-complete problem is harder than RFS, and that RFS is harder than some problem in $\Sigma_n \setminus \Sigma_{n-1}$ for any $n$) together, we find that there are problems in $\Sigma_1$ (namely, any NP-complete problem) that are harder (in fact, much, much harder) than certain problems in $\Sigma_n \setminus \Sigma_{n-1}$ for all $n$. Therefore, I don't see any way in which the polynomial hierarchy actually ranks problems by complexity.
Is there is mistake in my reasoning? Is there any useful sense in which problems higher up in the hierarchy are always harder than problems lower down? (I know this is a soft question, since there are different ways of formalizing the notion that one problem is "harder" than another. Maybe the answer is just that the PH is one way of ranking certain problems by difficulty, but not generally the best way.)