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Glorfindel
Glorfindel
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This is a socalledso-called chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.

This is a socalled chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.

This is a so-called chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.

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Carlo Beenakker
Carlo Beenakker
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This is a socalled chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring it$M$ to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast$$ $$\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$$$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.

This is a socalled chiral symmetry. The restriction on the symmetry of the spectrum is the only restriction you need, you can then bring it to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast$$ $$\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

This is a socalled chiral symmetry. The restriction on the symmetry of the spectrum of $M$ is the only restriction you need, you can then bring $M$ to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.

Here $U$ is the unitary matrix of eigenvectors of $M$; the eigenvalues are contained in the diagonal matrix $\Lambda$.

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Carlo Beenakker
Carlo Beenakker
  • 188.2k
  • 18
  • 448
  • 651

This is a socalled chiral symmetry. The restriction on the symmetry of the spectrum is the only restriction you need, you can then bring it to the desired off-diagonal form by a unitary transformation: $$M=U\begin{pmatrix}\Lambda&0\\ 0&-\Lambda\end{pmatrix}U^\ast$$ $$\Rightarrow \Omega^\ast U^\ast MU\Omega =\begin{pmatrix}0&\Lambda\\ \Lambda&0\end{pmatrix},$$ for $\Omega=2^{-1/2}\begin{pmatrix}1&1\\ -1 &1\end{pmatrix}$.