What is the most general version of Sato-Tate ? Like, I know when $f$ is an eigenform (lying in the space of new-forms), without CM, of weight $k$ and level $N,$ then $\frac{a(p)}{p^{(k-1)/2}}'s$ are equidistributed in $[0,1]$ with respect to a mesaure. My question is, when $f$ is arbitary modular form (without CM), or at least if it is an arbitary newform, can we say $\frac{a(p)}{p^{(k-1)/2}}'s$ are equidistributed in (possibly) some larger interval ?



A general question on Sato-Tate
No statement about modular forms is going to be the most general form of Sato-Tate, because we now know Sato-Tate for modular forms on other groups, and conjecture it in much greater generality. 


For the statement you're looking for, there will also be a gap between theorem and conjecture. Anyways, the method to deduce your desired statement from known theorems and believed conjectures will be the same, so let's talk about that.


You will get a conjectural equidistribution statement whenever $f$ is a cusp form, regardless of newness or CM. (For Eisenstein series the coefficients are much bigger and obviously don't equidistribute.)


To see this, write $$f(q) = \sum\_{i=1}^m w\_i f\_i(q^{d\_i})$$ where $f\_i$ is a cuspidal eigenform new of level $N/d\_i e\_i$ for some natural numbers $d\_i$ and $e\_i$. This is the decomposition you get from the eigenbasis of newforms and from writing oldforms in terms of newforms of lower level. Then for all $p$ not dividing $N$, $$a\_p(f) = \sum\_{i=1}^m w\_i' a\_p(f\_i) $$ where we set $w\_i'=w\_i$ if $d\_i=1$ and $0$ otherwise. 


The usual way to state the relevant case of the generalized Sato-Tate conjecture is as a statement about the joint distribution of the $a\_p(f\_i)$s. You could then derive the distribution of a linear combination from this, but it would be more complicated.


For each $f\_i$ appearing, we know $a\_p(f\_i)$ is equidistributed according to the Sato-Tate measure if $f\_i$ is not CM and according to a different measure, proven by Hecke, if $f\_i$ is CM. We conjecture that the distribution of the $f\_i$s are independent from each other, except in the cases when $f\_i$ is a quadratic twist of $f\_j$, or the CM cases when $f\_i$ may be a quartic or sextic twist of $f\_j$, where we predict the simplest joint distribution that follows that. If the level is squarefree and the nebentypus is trivial, you can't get twists, and so the predicted distribution is simply independent in this case. Then your coefficient will look like the weighted sum of independent random variables. Depending on the weights, it may approximate a Gaussian, but it will certainly be bounded as the summands are bounded.


In the cases we can prove this, the method of proof would be the same as the classical case: express the moments of this probability distribution in terms of the asymptotics as $s \to 1$ of $L$-functions, then prove these $L$-functions have meromorphic continuation and calculate their poles. I think the potential automorphy results used to prove Sato-Tate, together with Rankin-Selberg L-functions and cyclic base change, suffice to prove the equidistribution when at most two of the $f\_i$ are non-CM and the rest are CM, but I didn't check.


