eigenvalues of $i\frac{d}{dx}$

Question

\begin{array}{l} B:{\mathcal{D}}(B)\subset L^2\big(]-a,a[\big)\longrightarrow L^2\big(]-a,a[\big)\\\\ \displaystyle{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad~~~~~~\varphi \longrightarrow B\varphi(x)=i\frac{d\varphi}{dx}}\\\\ {\mathcal{D}}(B)=\left\{\varphi\in L^2\big(]-a,a[\big)\hbox{such that} \displaystyle \sum_{n=1}^{\infty}\lambda_n\vert\langle\varphi,\varphi_n\rangle\vert^2<\infty\right\} \end{array} Can we say if the eigenvalues of $- \displaystyle{\frac{d^2 }{dx^2}}$ are $\mu_n$, then the eigenvalues of $B$ are $\lambda_n=\sqrt{\mu_n}$?

Answer 1

If $\mu_n$ are the eigenvalues of B, then $B\varphi = \lambda \varphi $. So $$-\frac{\partial^2\varphi}{\partial x^2}=B^2 \varphi = i\frac{\partial}{\partial x}(i \frac{\partial \varphi}{\partial x}) = i\frac{\partial}{\partial x}(\lambda \varphi)= \lambda^2 \varphi$$

So $\lambda^2 = \mu$ is true. Please note that you have only included the positive square root in your expression.

$\mu_n = \pm \sqrt{\lambda_n}$