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??
Best Answer
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.