It is debatable that a physicist would use those very words, and if they did one would hope their meaning would be the same as for a mathematician, since it means that they are trying to speak the same language.
Having said, and coming from a Physics background, when I first learnt about filtered objects and associated graded objects, I immediately recognised the following examples from Physics. They all have to do with quantisation/classical limit in one way or another.
The Clifford algebra is filtered and its associated graded algebra is the exterior algebra. Under the "classical limit" map which takes the Clifford algebra to the exterior algebra, the first nonzero term in the graded commutator of two elements defines a Poisson structure on the exterior algebra. You can then view the Clifford algebra as the quantisation of this Poisson superalgebra. In Physics the exterior algebra is the "phase space" for free fermions and Clifford modules (=representations of the Clifford algebra) are Hilbert spaces for quantized fermions. Things get more interesting when the underlying vector space is infinite-dimensional, since not all Clifford modules are physically equivalent. (The relevant buzzword is Bogoliubov transformations; although you would not guess this from the wikipedia page.)
The algebra of differential operators on $\mathbb{R}^n$, say, is also filtered and the associated graded algebra is the algebra of functions on $T^*\mathbb{R}^n \cong \mathbb{R}^{2n}$ which are polynomial in the fiber coordinates (=the "momenta"). Again the first nonzero term in the commutator of two differential operators defines the standard Poisson bracket on $T^*\mathbb{R}^n$ and one can view the algebra of differential operators as a quantisation of this Poisson algebra. In Physics, this corresponds to quantising $n$ free bosons.
In both cases there is no unique section to the map taking a filtered algebra to the associated graded algebra, but one has to make a choice. There are number of more or less standard ones: Weyl ordering for the bosons, complete skewsymmetrisation for the fermions,...
By the way, this (and a lot more) is explained in the fantastic paper Symplectic reduction, BRS cohomology, and infinite-dimensional Clifford algebras by Kostant and Sternberg.
Edit (inspired by Mariano's answer)
Kontsevich's deformation quantization is not just of interest to physicists, but has a quantum field theoretical reformulation due to Cattaneo and Felder. It is basically the perturbative computation of the path integral of the Poisson sigma model. (This is analogous to how the perturbative evaluation of the path integral of Chern--Simons theory gives the Vassiliev invariants of (framed) knots.)
The picture that seems to be emerging is that indeed quantisation (be it deformation or path-integral or what have you) of a classical physical system gives rise to a filtered object, filtered by powers of $\hbar$.
Bounding angles in projector-valued Hamiltonians
Suppose that each of the
$h_k$ are projectors, and suppose that we shift the total energy so that
$\lambda_1 = 0$. Further suppose that this eigenspace is known to be non-degenerate for every finite
$N$. Then one proof technique that sometimes works (at least in one dimension, but it can be generalized) to prove an
$N$-independent lower bound is the following, which to the best of my knowledge was first discussed in
M. Fannes, B. Nachtergaele, R. F. Werner, Finitely correlated states on quantum spin chains, Comm. Math. Phys. Vol. 144, Num. 3 (1992), 443-490. MathSciNet:MR1158756
The bound is as follows. Let
$\theta_{j,k}$ be the smallest non-zero angle between the pair of projectors
$h_j$ and
$h_k$. That is,
$\cos^2(\theta_{j,k})$ is the largest eigenvalue not equal to one of
$h_j h_k h_j$. Define
$\theta = \min_{j,k} \theta_{j,k}$. Then FNW show that
$$ \Delta \ge 1-2 \cos(\theta) .$$
As long as
$\cos(\theta)$ is less than 1/2, there is an
$N$-independent lower bound on
$\Delta$, since this involves only local information. The proof proceeds by squaring
$H$ and then bounding how negative the anti-commutators can be by using this angle
$\theta$.
Best Answer
The quantum PCP conjecture (nobody has proved it, so you can't call it a theorem) is possibly (there are a few different ways of generalizing the classical PCP theorem to the quantum regime, and I don't believe any of them deserves the name of the quantum PCP conjecture) given below. Here $k$, $c$, and $d$ are small fixed integer constants, and $\epsilon$ is a constant between 0 and 1.
If you're using qubits, you should probably take $k=4$, by Bravyi's results on Quantum $k$-SAT. Certainly $k\geq 4$, unless Quantum 3-SAT is QMA-complete. (EDITED June 2013: Quantum 3-SAT is indeed QMA complete. See this recent paper.)
This would be an incredibly important development in the theory of quantum computing, but I'm not sure whether a proof has any practical implications for simulation of Hamiltonians. What it would show is that it is QMA-complete to tell whether an $n$-term local Hamiltonian has energy $0$ or at least $\epsilon n$.
Begin(rant)
Asking whether method $X$ is the best way to attack a big open mathematical conjecture is not a question anybody can answer. If you asked in this forum, for example, what is the best way to prove the Riemann hypothesis, and what mathematics you needed to learn in order to do this, your question would probably be promptly closed.
End(rant)
If I had to guess whether a quantum generalization of PCP was even true, I'd probably guess "no." It seems like an incredibly strong statement to me. On the other hand, the classical PCP theorem was also an incredibly strong statement. But just because a miracle happens in the classical regime, should you really be expecting the same miracle in the quantum regime?