Differentiation isn't the only operation that can be performed on functions. In this problem, we'll consider the operator \(\hat{S}\) that takes the form

\[\begin{equation*}\hat{S}=-i x \frac{d}{d x} \tag{2.95}\end{equation*}\]

which is proportional to the product of the position and momentum operators.

(a) Determine the eigenvalues and eigenfunctions of this operator. That is, determine the values \(\lambda\) and functions \(f_{\lambda}(x)\) such that\[\begin{equation*}\hat{S}f_{\lambda}(x)=\lambda f_{\lambda}(x), \tag{2.96}\end{equation*}\]

where we assume that the position coordinate \(x \in(-\infty, \infty)\). Are the values of \(\lambda\) real? Can the eigenfunctions be \(L^{2}\)-normalized on \(x \in(-\infty, \infty)\) ?

(b) Consider two eigenfunctions \(f_{\lambda_{1}}(x)\) and \(f_{\lambda_{2}}(x)\) for eigenvalues \(\lambda_{1} eq \lambda_{2}\). Are these two functions orthogonal; that is, does

\[\begin{equation*}\int_{-\infty}^{\infty} d x f_{\lambda_{1}}(x)^{*} f_{\lambda_{2}}(x)=0 ?\tag{2.97}\end{equation*}\]

(c) Now, consider exponentiating this operator, in a similar manner to how we can exponentiate the momentum operator \(\hat{p}\) :

\[\begin{equation*}\hat{U}_{S}(\alpha)=e^{i \alpha \hat{S}}=e^{\alpha x \frac{d}{d x}} \tag{2.98}\end{equation*}\]

for some real-valued parameter \(\alpha\). How does this exponentiated operator act on a general function of \(x, g(x)\) ? Just restrict your consideration to functions \(g(x)\) that have a well-defined Taylor expansion about \(x=0\).

How does this exponentiated operator act on the position to power \(n\), \(x^{n}\) ?