Return to Answer

The paper is quite long and technical. It is written without any care for any mathematical rigor, just in the spirit of theoretical physics (in particular there is no distinction between Hermitian, symmetric, and selfadjoint operators which is fundamental to assure the validity of the spectral theorem). However it is evidently physically interesting!

As far as I understand, the strategy is to define a bijective operator, not unitary, from the original Hilbert space ${\cal H}$ to another Hilbert space ${\cal H}'$, of the form $$S:= e^{-Q/2} : {\cal H} \to {\cal H}'\:.$$ The new Hamiltonian operator (Eq. (6.34)) $$h := S H S^{-1}$$ is argued to be Hermitian in ${\cal H}'$, whereas the original on $H$ in ${\cal H}$ is not (but is ${\cal PT}$-symmetric). The transformation $a \to S aS^{-1}$ is often called an equivalence transformation. I do not know (presumably the paper defines it) what is the scalar product in ${\cal H}'$, but $V$ is not unitary, i.e., it does not preserve the scalar product: $$\langle S\psi|S\phi \rangle_{{\cal H}'}\neq \langle \psi|\phi \rangle_{{\cal H}}\quad in\: general,$$ (otherwise also $h$ would be Hermitian and we do know that it is false!).

Physically speaking, immediate drawbacks (presumably this is what you are missing) of this procedure are that

(a) other Hermitian operators in ${\cal H}$ turns out to be non-Hermitian in ${\cal H}'$ under the equivalence transformation induced by $S$;

(b) a direct interpretation of $|\langle \chi|\rho\rangle_{{\cal H}'}|^2$ as transition probability is disputable.

I think that all that is physically acceptable ant it is related to the fact that the system is considered an open system. In any cases, I strongly expect that the paper examines these issues in details, since they pop out immediately and C. M. Bender is a world authority on the subject of the work.

ADDENDUM. I comment also about the claim (somewhere present in the paper) that the evolution is unitary in spite of $H$ being non-Hermitian.

I do not understand well this claim, however, let us assume that it means that $\mathbb{R}\ni t \mapsto U_t := e^{-itH}$ is a strongly continuous one-parameter group of unitary operators in the Hilbert space ${\cal H}$.

If it is the case, the Stone theorem implies that there is a (unique) selfadjoint operator $H' :D(H) \to {\cal H}$ such that $e^{-itH'} = U_t$ for every $t\in \mathbb{R}$. Here $D(H)$ is exactly the dense subspace of vectors $\psi \in {\cal H}$ such that, the derivative $$\frac{d}{dt}|_{t=0} U_t \psi$$
exists. Also, this derivative just defines $-iH'\psi$.

Putting all together (if $H'$ is defined on domain identical to the one of $H$ or larger) we conclude that $H$ is Hermitian as $H'$ is!

We know that it is not the case a priori. So, I cannot see an evident way to interpret the claim that the evolution generated by $H$ is unitary.

The paper is quite long and technical. It is written without any care for any mathematical rigor, just in the spirit of theoretical physics (in particular there is no distinction between Hermitian, symmetric, and selfadjoint operators which is fundamental to assure the validity of the spectral theorem). However it is evidently physically interesting!

As far as I understand, the strategy is to define a bijective operator, not unitary, from the original Hilbert space ${\cal H}$ to another Hilbert space ${\cal H}'$, of the form $$S:= e^{-Q/2} : {\cal H} \to {\cal H}'\:.$$ The new Hamiltonian operator (Eq. (6.34)) $$h := S H S^{-1}$$ is argued to be Hermitian in ${\cal H}'$, whereas the original on $H$ in ${\cal H}$ is not (but is ${\cal PT}$-symmetric). The transformation $a \to S aS^{-1}$ is often called an equivalence transformation. I do not know (presumably the paper defines it) what is the scalar product in ${\cal H}'$, but $V$ is not unitary, i.e., it does not preserve the scalar product: $$\langle S\psi|S\phi \rangle_{{\cal H}'}\neq \langle \psi|\phi \rangle_{{\cal H}}\quad in\: general,$$ (otherwise also $h$ would be Hermitian and we do know that it is false!).

Physically speaking, immediate drawbacks (presumably this is what you are missing) of this procedure are that

(a) other Hermitian operators in ${\cal H}$ turns out to be non-Hermitian in ${\cal H}'$ under the equivalence transformation induced by $S$;

(b) a direct interpretation of $|\langle \chi|\rho\rangle_{{\cal H}'}|^2$ as transition probability is disputable.

I think that all that is physically acceptable ant it is related to the fact that the system is considered an open system. In any cases, I strongly expect that the paper examines these issues in details, since they pop out immediately and C. M. Bender is a world authority on the subject of the work.

ADDENDUM. I comment also about the claim (somewhere present in the paper) that the evolution is unitary in spite of $H$ being non-Hermitian.

I do not understand well this claim, however, let us assume that it means that $\mathbb{R}\ni t \mapsto U_t := e^{-itH}$ is a strongly continuous one-parameter group of unitary operators in the Hilbert space ${\cal H}$.

If it is the case, the Stone theorem implies that there is a (unique) selfadjoint operator $H' :D(H) \to {\cal H}$ such that $e^{-itH'} = U_t$ for every $t\in \mathbb{R}$. Here $D(H)$ is exactly the dense subspace of vectors $\psi \in {\cal H}$ such that, the derivative $$\frac{d}{dt}|_{t=0} U_t \psi$$
exists. Also, this derivative just defines $-iH'\psi$.

Putting all together (if $H'$ is defined on domain identical to the one of $H$ or larger) we conclude that $H$ is Hermitian as $H'$ is! Indeed, $$H'\psi = i\frac{d}{dt} U_t \psi = i \frac{d}{dt} e^{-itH}\psi = H\psi\:.$$

We know that it is not the case a priori. So, I cannot see an evident way to interpret the claim that the evolution generated by $H$ is unitary.

The paper is quite long and technical. It is written without any care for any mathematical rigor, just in the spirit of theoretical physics (in particular there is no distinction between Hermitian, symmetric, and selfadjoint operators which is fundamental to assure the validity of the spectral theorem). However it is evidently physically interesting!

As far as I understand, the strategy is to define a bijective operator, not unitary, from the original Hilbert space ${\cal H}$ to another Hilbert space ${\cal H}'$, of the form $$S:= e^{-Q/2} : {\cal H} \to {\cal H}'\:.$$ The new Hamiltonian operator (Eq. (6.34)) $$h := S H S^{-1}$$ is argued to be Hermitian in ${\cal H}'$, whereas the original on $H$ in ${\cal H}$ is not (but is ${\cal PT}$-symmetric). The transformation $a \to S aS^{-1}$ is often called an equivalence transformation. I do not know (presumably the paper defines it) what is the scalar product in ${\cal H}'$, but $V$ is not unitary, i.e., it does not preserve the scalar product: $$\langle S\psi|S\phi \rangle_{{\cal H}'}\neq \langle \psi|\phi \rangle_{{\cal H}}\quad in\: general,$$ (otherwise also $h$ would be Hermitian and we do know that it is false!).

Physically speaking, immediate drawbacks (presumably this is what you are missing) of this procedure are that

(a) other Hermitian operators in ${\cal H}$ turns out to be non-Hermitian in ${\cal H}'$ under the equivalence transformation induced by $S$;

(b) a direct interpretation of $|\langle \chi|\rho\rangle_{{\cal H}'}|^2$ as transition probability is disputable.

I think that all that is physically acceptable ant it is related to the fact that the system is considered an open system. In any cases, I strongly expect that the paper examines these issues in details, since they pop out immediately and C. M. Bender is a world authority on the subject of the work.

The paper is quite long and technical. It is written without any care for any mathematical rigor, just in the spirit of theoretical physics (in particular there is no distinction between Hermitian, symmetric, and selfadjoint operators which is fundamental to assure the validity of the spectral theorem). However it is evidently physically interesting!

As far as I understand, the strategy is to define a bijective operator, not unitary, from the original Hilbert space ${\cal H}$ to another Hilbert space ${\cal H}'$, of the form $$S:= e^{-Q/2} : {\cal H} \to {\cal H}'\:.$$ The new Hamiltonian operator (Eq. (6.34)) $$h := S H S^{-1}$$ is argued to be Hermitian in ${\cal H}'$, whereas the original on $H$ in ${\cal H}$ is not (but is ${\cal PT}$-symmetric). The transformation $a \to S aS^{-1}$ is often called an equivalence transformation. I do not know (presumably the paper defines it) what is the scalar product in ${\cal H}'$, but $V$ is not unitary, i.e., it does not preserve the scalar product: $$\langle S\psi|S\phi \rangle_{{\cal H}'}\neq \langle \psi|\phi \rangle_{{\cal H}}\quad in\: general,$$ (otherwise also $h$ would be Hermitian and we do know that it is false!).

Physically speaking, immediate drawbacks (presumably this is what you are missing) of this procedure are that

(a) other Hermitian operators in ${\cal H}$ turns out to be non-Hermitian in ${\cal H}'$ under the equivalence transformation induced by $S$;

(b) a direct interpretation of $|\langle \chi|\rho\rangle_{{\cal H}'}|^2$ as transition probability is disputable.

I think that all that is physically acceptable ant it is related to the fact that the system is considered an open system. In any cases, I strongly expect that the paper examines these issues in details, since they pop out immediately and C. M. Bender is a world authority on the subject of the work.

ADDENDUM. I comment also about the claim (somewhere present in the paper) that the evolution is unitary in spite of $H$ being non-Hermitian.

I do not understand well this claim, however, let us assume that it means that $\mathbb{R}\ni t \mapsto U_t := e^{-itH}$ is a strongly continuous one-parameter group of unitary operators in the Hilbert space ${\cal H}$.

If it is the case, the Stone theorem implies that there is a (unique) selfadjoint operator $H' :D(H) \to {\cal H}$ such that $e^{-itH'} = U_t$ for every $t\in \mathbb{R}$. Here $D(H)$ is exactly the dense subspace of vectors $\psi \in {\cal H}$ such that, the derivative $$\frac{d}{dt}|_{t=0} U_t \psi$$
exists. Also, this derivative just defines $-iH'\psi$.

Putting all together (if $H'$ is defined on domain identical to the one of $H$ or larger) we conclude that $H$ is Hermitian as $H'$ is!

We know that it is not the case a priori. So, I cannot see an evident way to interpret the claim that the evolution generated by $H$ is unitary.

The paper is quite long and technical. It is written without any care for any mathematical rigor, just in the spirit of theoretical physics (in particular there is no distinction between Hermitian, symmetric, and selfadjoint operators which is fundamental to assure the validity of the spectral theorem). However it is evidently physically interesting!

As far as I understand, the strategy is to define a bijective operator, not unitary, from the original Hilbert space ${\cal H}$ to another Hilbert space ${\cal H}'$, of the form $$S:= e^{-Q/2} : {\cal H} \to {\cal H}'\:.$$ The new Hamiltonian operator (Eq. (6.34)) $$h := S H S^{-1}$$ is argued to be Hermitian in ${\cal H}'$, whereas the original on $H$ in ${\cal H}$ is not (but is ${\cal PT}$-symmetric). I do not know (presumably the paper defines it) what is the scalar product in ${\cal H}'$, but $V$ is not unitary, i.e., it does not preserve the scalar product: $$\langle S\psi|S\phi \rangle_{{\cal H}'}\neq \langle \psi|\phi \rangle_{{\cal H}}\quad in\: general,$$ (otherwise also $h$ would be Hermitian and we do know that it is false!).

Physically speaking, immediate drawbacks (presumably this is what you are missing) of this procedure are that

(a) other Hermitian operators in ${\cal H}$ turns out to be non-Hermitian in ${\cal H}'$ under the equivalence transformation induced by $S$;

(b) a direct interpretation of $|\langle \chi|\rho\rangle_{{\cal H}'}|^2$ as transition probability is disputable.

I think that all that is physically acceptable ant it is related to the fact that the system is considered an open system. In any cases, I strongly expect that the paper examines these issues in details, since they pop out immediately and C. M. Bender is a world authority on the subject of the work.

The paper is quite long and technical. It is written without any care for any mathematical rigor, just in the spirit of theoretical physics (in particular there is no distinction between Hermitian, symmetric, and selfadjoint operators which is fundamental to assure the validity of the spectral theorem). However it is evidently physically interesting!

As far as I understand, the strategy is to define a bijective operator, not unitary, from the original Hilbert space ${\cal H}$ to another Hilbert space ${\cal H}'$, of the form $$S:= e^{-Q/2} : {\cal H} \to {\cal H}'\:.$$ The new Hamiltonian operator (Eq. (6.34)) $$h := S H S^{-1}$$ is argued to be Hermitian in ${\cal H}'$, whereas the original on $H$ in ${\cal H}$ is not (but is ${\cal PT}$-symmetric). The transformation $a \to S aS^{-1}$ is often called an equivalence transformation. I do not know (presumably the paper defines it) what is the scalar product in ${\cal H}'$, but $V$ is not unitary, i.e., it does not preserve the scalar product: $$\langle S\psi|S\phi \rangle_{{\cal H}'}\neq \langle \psi|\phi \rangle_{{\cal H}}\quad in\: general,$$ (otherwise also $h$ would be Hermitian and we do know that it is false!).

Physically speaking, immediate drawbacks (presumably this is what you are missing) of this procedure are that

(a) other Hermitian operators in ${\cal H}$ turns out to be non-Hermitian in ${\cal H}'$ under the equivalence transformation induced by $S$;

(b) a direct interpretation of $|\langle \chi|\rho\rangle_{{\cal H}'}|^2$ as transition probability is disputable.

I think that all that is physically acceptable ant it is related to the fact that the system is considered an open system. In any cases, I strongly expect that the paper examines these issues in details, since they pop out immediately and C. M. Bender is a world authority on the subject of the work.