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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?

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    $\begingroup$ What do you mean by $\partial_{x_1}$ and $\partial_{x_2}$ on a space whose functions depend only on one real variable? $\endgroup$ Commented Jan 7, 2021 at 23:12
  • $\begingroup$ @JochenGlueck sorry, should have been $[0,1]^2.$ $\endgroup$
    – Sascha
    Commented Jan 7, 2021 at 23:18
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    $\begingroup$ You'd need to give some boundary conditions for this to be a well-defined question. $\endgroup$ Commented Jan 8, 2021 at 0:46
  • $\begingroup$ @sorry, yes, periodic ones. $\endgroup$
    – Sascha
    Commented Jan 8, 2021 at 0:47
  • $\begingroup$ @MichaelEngelhardt thank you for your efforts $\endgroup$
    – Sascha
    Commented Jan 8, 2021 at 1:50

1 Answer 1

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By noting that $-i\partial_{x_1} $ is diagonalized by $e^{ik_1 x_1} $ and $-i\partial_{x_2} $ by $e^{ik_2 x_2} $, the problem reduces to a $2\times 2$ diagonalization for each $(k_1,k_2)$-block. The resulting eigenvalues are (denoting $k=k_1-ik_2 $) $$ \frac{1}{2} (|a|^2 + |b|^2 ) +|k+c|^2 \pm\frac{1}{2} \sqrt{(|a|^2 -|b|^2 )^2 +4(|a|^2 + |b|^2 )|k+c|^2 +8\,\mbox{Re}\, (a^*b^*(k+c)^2)} $$ where $k_1 , k_2 \in 2\pi \mathbb{Z} $ in view of the boundary conditions.

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  • $\begingroup$ But how do you know this is all of the spectrum? $\endgroup$
    – Sascha
    Commented Jan 8, 2021 at 4:43
  • $\begingroup$ To explain my question. I am a bit sceptic about the conjugation by just the one-dimensional Fourier basis. Perhaps, we are losing something here? $\endgroup$
    – Sascha
    Commented Jan 8, 2021 at 4:50
  • $\begingroup$ I am essentially doing a 2-dimensional Fourier transform, to the basis $e^{i(k_1 x_1 + k_2 x_2 )}$. That's a unitary transformation on $L^2 $. As is the final $2\times 2$ diagonalization. $\endgroup$ Commented Jan 8, 2021 at 4:53

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