Let $X$ be a regular scheme and consider Grothendieck's $\gamma$-filtration $F^nK(X)$ on $K(X)$. For the graded pieces, one has $Gr^0K(X) = CH^0(X)$ and $Gr^1K(X) = \mathrm{Pic}(X) = CH^1(X)$. Does this continue to hold, i.e., do we have $Gr^pK(X) = CH^p(X)$?

I found that for $X/k$ smooth quasi-projective, $CH^q(X,p) \otimes \mathbf{Q} = K_p(X)^{(q)} \otimes \mathbf{Q}$, so this holds after rationalising.