As we know that a matrix $H$ is said to be Hermitian if $H=H^\dagger$, and if so, then all the eigenvalues are real.

While a non-hermitian matrix $P$ is said to be pseudo-hermitian if $ \eta P \eta^{-1} =P^\dagger$, ($\eta$ is some constant metric), in this, eigenvalues may also be real.

Could anyone please throw some light that - from where this condition of pseudo-hermiticity $ \eta P \eta ^{-1}=P^\dagger$, comes??

  • $\begingroup$ Check this paper. One of initial theorems. You will easily find the answer. arxiv.org/abs/1512.00447 $\endgroup$ – Chetan Waghela Mar 7 at 23:22
  • $\begingroup$ @ChetanWaghela, Thanks I will take a look. :) $\endgroup$ – math Mar 8 at 3:50

I am not familiar with the term "pseudo-Hermitian" for such matrices (I would call them self-adjoint with respect to $\eta^{-1}$) but there is strong motivation for this condition: it is the natural extension of symmetry to arbitrary non-Euclidean inner products.

That is, given the inner product $\eta$, one might ask that an operator $P$ satisfy $$\langle v, Pw\rangle_{\eta} = \langle Pv, w\rangle_{\eta}$$ for all vectors $v,w$. Unpacking this condition gives $$w^{\dagger}P^{\dagger}\eta v = w^{\dagger}\eta P v$$ or, since equality must hold for all vectors, $$ P^{\dagger} \eta = \eta P$$ or $$\eta P \eta^{-1} = P^{\dagger}.$$ As a special case, you recover ordinary Hermitian matrices when $\eta=I$, and self-adjoint matrices have positive eigenvalues precisely because the spectral theorem works under any inner product.

Note that your condition is self-adjointness with respect to $\eta^{-1}$ rather than $\eta$, for some reason.

  • $\begingroup$ Thank you for your answer! Yes, I have corrected the typo. Now it is self-adjoint with respect to $\eta$ $\endgroup$ – math Sep 5 '18 at 6:26
  • $\begingroup$ One more ques is there: As one can show that if a matrix is Hermitian(Self-adjoint) then its eigenvalues are ALWAYS real. Can it be shown that if it is pseudo-hermitian, then eigenvalues may also be real? $\endgroup$ – math Sep 5 '18 at 6:45
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    $\begingroup$ They are real because a pseudo-hermitian matrix (or linear mapping) is hermitian in just a different inner product. You can see that differently: Setting $A :=\eta^{1/2}$, the equality $\eta P \eta^{-1} = P^\dagger$ implies $APA^{-1} = A^{-1}P^\dagger A = (APA^{.-1})^\dagger$. So, $APA^{-1}$ is hermitian in the usual way. Thus, its eigenvalues are real. But they are the same as those of $P$. $\endgroup$ – amsmath Sep 5 '18 at 7:24
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    $\begingroup$ Since you wrote that "$\eta$ is some constant metric" I assumed that it's hermitian positive definite. That's usually what a "metric" is (defined by). So, $A = \eta^{1/2}$ is also hermitian positive definite by definition of the square root. $\endgroup$ – amsmath Sep 6 '18 at 12:07
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    $\begingroup$ @Sachin typically if you call a matrix a "metric" it will be understood that the matrix is symmetric (or Hermitian) positive-definite. Your matrix is not positive-definite. $\endgroup$ – user7530 Sep 7 '18 at 5:13

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