Given a weighted graph with $n$ vertices and weights $w_{ij}\geq 0$, the max-cut problem is equivalent to
$$ \max_{x \in \mathbb{R}^n} \sum_{i,j} w_{ij} (1-x_i x_j) \quad \mbox{s.t.} \quad x_i \in \{-1,1\}, \ i=1,\ldots,n. $$
It is well-known that this problem is equivalent to either
(A) $\max_{x \in \mathbb{R}^n} x^T L x$ with $x_i \in \{-1,1\}$, $i=1,\ldots,n$, with $L$ being the graph Laplacian (see Max Cut: Form of Graph Laplacian?)
OR
(B) $\min_{X \in \mathbb{R}^{n\times n}} W \cdot X$ s.t. $X_{ii} = 1$ for all $i$, $X \succeq 0$, and ${\rm rank}(X)=1$, with $W = [w_{ij}]$.
From these, we have two relaxed problems:
(A') $\max_{x \in \mathbb{R}^n} x^T L x/ x^T x$ (i.e. dropping the +1/-1 constraint in (A))
(B') $\min_{X \in \mathbb{R}^{n\times n}} W \cdot X$ s.t. $X_{ii} = 1$, for all $i$ and $X \succeq 0$ (i.e. dropping the rank 1 constraint in (B))
If I am not mistaken, (B') is the famous Goemans-William SDP of which a suitably rounded solution achieves an approximation ratio of 0.878.
Question: If I take a dominant eigenvector of the graph Laplacian of $L$ and round the entries to +1 or -1 according to the signs of the entries (assuming no entry is $0$ -- not sure what to do if there is), do we get an approximate solution with a certain approximation ratio?