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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?

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  • $\begingroup$ This follows from (in fact, is equivalent to) the fact you quote (spectrum symmetric), if you have a sufficiently strong version of this. More precisely, you need that the spectral measures have the same null sets as their reflections, and the multiplicities agree. I believe that should hold here. $\endgroup$
    – Christian Remling
    Commented Apr 30, 2021 at 19:47
  • $\begingroup$ In fact, it should be straightforward to establish what we need along those lines by taking Fourier transforms. $\endgroup$
    – Christian Remling
    Commented Apr 30, 2021 at 19:50

2 Answers 2

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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.

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It may be of interest to note that the $\pm E$ symmetry of the spectrum of the Dirac operator holds not only for a constant $m$, but also for a spatially dependent $m(x,y)$,
$$H = \begin{pmatrix} m(x,y) & -i\partial_x-\partial_y \\ -i\partial_x+\partial_y & -m(x,y) \end{pmatrix}.$$ This follows from the anti-unitary symmetry $$T\bar{H}T^\ast=-H,\;\;\text{with}\;\;T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$ The overline $\bar{H}$ denotes complex conjugation.

Then if $E$ is a (real) eigenvalue of $H$, we have $$0=\det(E-H)=\det(E-\bar{H})=\det(E+T^\ast HT)=\det(E+H),$$ so $-E$ is also an eigenvalue.

In physics this is called particle-hole symmetry (or charge-conjugation symmetry), the matrix $T$ exchanges an electron state $(1,0)$ into a hole state $(0,1)$.
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  • $\begingroup$ thank you, how do I see it is not of the form $THT^*=-H$? $\endgroup$
    – Landauer
    Commented Apr 30, 2021 at 19:41
  • $\begingroup$ the symmetry you wrote down is a socalled chiral symmetry; it is broken by the mass term $m$. $\endgroup$
    – Carlo Beenakker
    Commented Apr 30, 2021 at 19:42
  • $\begingroup$ yes, but could there not be another unitary $T$ that does it?-I of course understand that the one I wrote down will not work, as you also explain. $\endgroup$
    – Landauer
    Commented Apr 30, 2021 at 19:42
  • $\begingroup$ no, the Dirac operator contains all three Pauli matrices $\sigma_x,\sigma_y,\sigma_z$, there does not exist a $2\times 2$ unitary matrix that anticommutes with all three. $\endgroup$
    – Carlo Beenakker
    Commented Apr 30, 2021 at 19:44
  • $\begingroup$ @CarloBeenakker: I believe by "unitary" the OP means a unitary operator on the Hilbert space $L^2\otimes \mathbb C^2$. $\endgroup$
    – Christian Remling
    Commented Apr 30, 2021 at 19:45

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