Consider the Dirac operator


$$ H = \begin{pmatrix} m & -i\partial\_z \\ -i\partial\_{\bar z} & -m \end{pmatrix},$$
where $\partial\_{\bar z}$ is the Cauchy-Riemann operator and $m \ge 0.$


It is not hard to see that the spectrum of this operator is symmetric with respect to zero.


However, does there exist a simple unitary $T$ such that


$$THT^\*=-H?$$


If $m$ was zero, then already $$T = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ would do, what about this general case?



Massive dirac operator symmetric spectrum
With $z=x+iy$, we use the Fourier transformation in $(x,y)$ to see that $H$ is unitarily equivalent to
$$
\frac12\begin{pmatrix}2m&\xi-i\eta\\
\xi+i\eta&-2m\end{pmatrix},
\text{whose eigenvalues are } \lambda\_\pm=\pm\sqrt{m^2+\frac{\vert\zeta\vert^2}{4}}, \ \zeta=\xi+i\eta.
$$
With $\mu=\sqrt{m^2+\frac{\vert\zeta\vert^2}{4}},$ we have
$$
\begin{pmatrix}0&1\\
-1&0\end{pmatrix}\begin{pmatrix}\mu&0\\
0&-\mu\end{pmatrix}\begin{pmatrix}0&-1\\
1&0\end{pmatrix}= \begin{pmatrix}-\mu&0\\
0&\mu\end{pmatrix},
$$
so that $H$ is indeed unitarily equivalent to $-H$. All calculations can be made explicitly.


