Consider the linear constant coefficient differential operator $P$ on the Hilbert space $L^2([0,1]^2;\mathbb C^2)$ $$P= \begin{pmatrix} D_{z}+c & a \\ b & D_{z}+c \end{pmatrix}$$ where $D_z=-i \partial_z =- i(\partial_{x_1} -i \partial_{x_2}).$

Here, $a,b,c$ are just some complex numbers.

I wonder whether one can explicitly compute the spectrum of the self-adjoint operator $P^*P$ then?

Consider the linear constant coefficient differential operator $P$ on the Hilbert space $L^2([0,1]^2;\mathbb C^2)$ $$P= \begin{pmatrix} D_{z}+c & a \\ b & D_{z}+c \end{pmatrix}$$ where $D_z=-i \partial_z =- i(\partial_{x_1} -i \partial_{x_2}).$

Here, $a,b,c$ are just some complex numbers.

I wonder whether one can explicitly compute the spectrum of the self-adjoint operator $P^*P$, with periodic boundary conditions, then?

Consider the linear constant coefficient differential operator $P$ on the Hilbert space $L^2([0,1];\mathbb C^2)$ $$P= \begin{pmatrix} D_{z}+c & a \\ b & D_{z}+c \end{pmatrix}$$ where $D_z=-i \partial_z =- i(\partial_{x_1} -i \partial_{x_2}).$

Here, $a,b,c$ are just some complex numbers.

I wonder whether one can explicitly compute the spectrum of the self-adjoint operator $P^*P$ then?

Consider the linear constant coefficient differential operator $P$ on the Hilbert space $L^2([0,1]^2;\mathbb C^2)$ $$P= \begin{pmatrix} D_{z}+c & a \\ b & D_{z}+c \end{pmatrix}$$ where $D_z=-i \partial_z =- i(\partial_{x_1} -i \partial_{x_2}).$

Here, $a,b,c$ are just some complex numbers.

I wonder whether one can explicitly compute the spectrum of the self-adjoint operator $P^*P$ then?