Range of values for $\frac{\langle f | \mathbf n \cdot \boldsymbol \sigma |f \rangle}{\langle f | f \rangle}$, with $|\mathbf n| = 1$? I'm trying to find the range of possible values for the expression
$$w := \frac{\langle f | \mathbf n \cdot \boldsymbol \sigma |f \rangle}{\langle f | f \rangle},$$
where $\mathbf n \in \mathbb R^3$, $|\mathbf n| = 1$, $| f \rangle \in \mathbb C^2$, and $\langle f| := | f \rangle^\dagger$ is the hermitian conjugate of $| f \rangle$. The expression $\mathbf n \cdot \boldsymbol \sigma$ is physics shorthand for $\sum_{i=1}^3 n_i \sigma_i$, where the $\sigma_i$ are the Pauli matrices. Note that the Pauli matrices are hermitian, and hence $w \in \mathbb R$.
Alternatively I could settle for just the bounds on $w^2$. There is a bunch of useful identities for the Pauli matrices, but I haven't been able to get much out of them in this case. The most progress I've had is
$$
\begin{split}
\langle f | f \rangle^2 w^2 &= \sum_{i=1}^3 \sum_{j=1}^3 \langle f | n_i \sigma_i | f \rangle \langle f | n_j \sigma_j | f \rangle \\
&= \sum_{i=1}^3 n_i^2 \langle f | \sigma_i | f \rangle^2 + \sum_{i=1}^3 \sum_{\substack{j=1 \\ j \ne i}}^3 n_i n_j \langle f | \sigma_i | f \rangle \langle f | \sigma_j | f \rangle \\
&\le \langle f | f \rangle^2 + \sum_{i=1}^3 \sum_{\substack{j=1 \\ j \ne i}}^3 n_i n_j \langle f | \sigma_i | f \rangle \langle f | \sigma_j | f \rangle,
\end{split}
$$
using $n_i^2 \le 1$ and the easily verifiable fact that $\sum_{i=1}^3 \langle f | \sigma_i | f \rangle^2 = \langle f | f \rangle^2$. But I'm not sure how useful this form is, because I'm not making much headway with that final sum.
Is there a neat way to find the range of possible values for $w$, or at least bounds for $w^2$? I feel like there should be, but I'm not quite getting there. All help is appreciated.
 A: I think the range of values for $w$ is $[-1,1]$.
The problem can be significantly simplified using the adjoint action of elements of $SU(2)$ on Pauli vectors. The adjoint action $g^{-1}(\mathbf n \cdot \boldsymbol \sigma)g$ of an element $g\in SU(2)$ on a Pauli vector $\mathbf n \cdot \boldsymbol \sigma$ corresponds to a rotation of $\mathbf n$ (with each rotation in $O(3)$ corresponding to two elements of $SU(2)$ with opposite signs).
That means that we don’t have to vary $f$ and $\mathbf n$ independently, since a change from $f$ to $gf$ can be replaced by a change from $\mathbf n \cdot \boldsymbol \sigma$ to $g^{-1}(\mathbf n \cdot \boldsymbol \sigma)g$. Thus we can fix $\mathbf n \cdot \boldsymbol \sigma=\sigma_1=\pmatrix{0&1\\1&0}$ and only vary $f$. Moreover, since the expression of interest is normalized with $\langle f|f\rangle$, we only need to consider unit vectors $f$ with $\langle f|f\rangle=1$; also, multiplying $f$ by a phase factor $\mathrm e^{\mathrm i\phi}$ leaves the expression invariant.
With $f=\pmatrix{a\\b}$, we have
$$
w=\pmatrix{a^*&b^*}\pmatrix{0&1\\1&0}\pmatrix{a\\b}=a^*b+b^*a=2\mathfrak{Re}(a^*b)\;.
$$
This can be made $0$ by making $a^*b$ imaginary, and for any $w\gt0$ we can obtain the corresponding value $-w$ by replacing $b$ by $-b$; also, $w$ is continuous in $a$ and $b$, so the range of values of $w$ is of the form $[-x,x]$. To obtain the maximal value $x$, $a$ and $b$ need to be in phase, e.g. both real; then since $f$ is a unit vector we can parametrize it by $\pmatrix{\cos\phi\\\sin\phi}$. The maximum is attained at $\phi=\frac\pi4$, i.e. $f=\frac1{\sqrt2}\pmatrix{1\\1}$, where $w=1$.
