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I was reading this review paper by Bender, in particular section VI where they show that, despite $\mathcal{PT}$-symmetric Hamiltonians not being hermitian, they can have a real spectra. They go on to show that one can cook a proper inner product by defining an operator $\mathcal{C}$ such that the evolution is unitary.

My question is very simple:

Is it always possible to find a Hermitian Hamiltonian equivalent to any given $\mathcal{PT}$-symmetric Hamiltonian whose evolution is unitary?

By dual here I mean that it reproduces the same physics, i.e. it can reproduce any observables at any time $t$ identically to the $\mathcal{PT}$-symmetric Hamiltonian.

My motivation is simple: For any unitary matrix $U$ of finite size, $U$ can be written as $U = e^{iH}$, where $H$ is a Hermitian matrix. Now, I don't know if this is sufficient to show the existence of this equivalence.

I mean, if this is true, I can't help the feeling that this $\mathcal{PT}$-symmetric Hamiltonians research business is a bit misleading in making the community believe there is more than Hermitian $H$'s. So what am I missing?

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  • $\begingroup$ Suggestion to the post (v2): Replace the word dual with the word equivalent. $\endgroup$
    – Qmechanic
    Commented May 27 at 11:09
  • $\begingroup$ @Qmechanic happy to change it. Is there a reason why dual is not the correct term here? (And why equivalent is?) $\endgroup$ Commented May 27 at 21:53
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    $\begingroup$ @FriendlyLagrangian dual has a mathematical meaning different from the use we give to it in spoken languages. Dual in a math context refers to (usually) an involutive functor, i.e. duality is a 1-1 correspondence between different mathematical objects. Two operators are both operators, so "dual" is not the right term. An example of duality are vectors and 1-forms $\endgroup$
    – LolloBoldo
    Commented Jun 4 at 11:12
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    $\begingroup$ I don't think I know enough about this subject to answer the question, but I think its a big deal (in normal quantum mechanics) that the inner product which we use when computing Born-rule probabilities is the same inner product that self-adjoint Hamiltonians are self-adjoint with respect to. You can cook up an inner product with respect to which these $\mathcal{PT}$ Hamiltonians are Hermitian but the inner product we use for the Born rule doesn't change so the physics will be noticeably different. $\endgroup$
    – ors
    Commented Jun 6 at 14:54
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    $\begingroup$ @FriendlyLagrangian I think within the realm of "normal quantum mechanics" the answer to that question is pretty clearly no (e.g. by Wigner's theorem). If you go slightly beyond that you might find examples. Once place to look that comes to mind is people who think about objective collapse models (where wavefunction collapse is a physical non-unitary process). I think continuous spontaneous localization models can be described like this, but I've never really been interested in those so I can't comment further. $\endgroup$
    – ors
    Commented Jun 7 at 12:06

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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.

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    $\begingroup$ I'm not sure I agree? They mention several times the evolution is unitary (provided the spectrum is real). For instance in the intro or in section C: "In Sec. VI we show that if the eigenspectrum of a $\mathcal{PT}$-symmetric Hamiltonian $H$ is entirely real, then $H$ defines a unitary theory of quantum mechanics." This brings me back to my original question: Is it always possible to find a Hermitian Hamiltonian equivalent to any given $\mathcal{PT}$-symmetric Hamiltonian whose evolution is unitary? $\endgroup$ Commented Jun 10 at 9:44
  • $\begingroup$ I did not notice this remark, thanks. I have had just a quick look at the paper. If they say this, and if the evolution is strongly continuous in addition to unitary, then there is a selfadjoint generator of the time evolution as a trivial consequence of the Stone theorem. Is there a precise spot where they explicitly describe that unitary evolution "generated" by the non -Hermitian Hamiltonian? (Notice that $e^{-ith}$ is unitary. Maybe they are saying that this is the unitary evolution). $\endgroup$ Commented Jun 10 at 13:43
  • $\begingroup$ @FriendlyLagrangian I believe that the statements there refer to a formal or analogue mathematical system that looks like unitary quantum mechanics in every mathematical sense of the word. However, the formal scalar product or the formal Hermitian operators in the new space $\mathcal{H}'$ do not correspond to true physical observables and measurements that you perform on the system in the real world as they would in true quantum mechanics. $\endgroup$
    – Void
    Commented Jun 10 at 14:39
  • $\begingroup$ @ValterMoretti Indeed, your addendum is exactly my concern: I am not sure what their claim means. To me it feels as if the assumption of a strongly continuous one-parameter group of unitary operators has to hold. Otherwise, they would be considering exclusively "pathological" situations. I say "exclusively" because otherwise, as you say, there exists a Hermitian description of the system and then what would be their point? I think I am missing the point of these theories completely or there is some overselling... $\endgroup$ Commented Jun 10 at 16:53
  • $\begingroup$ I think that the authors' claim is that the evolution is unitary if we change the scalar product. I do not know if they use the scalar product they associate to the transform induced by $Q$ in that part of their paper. Perhaps they use what they call "CPT scalar product" (defined somewhere there around). For me, it is very difficult to read that paper in view of the way it is written. $\endgroup$ Commented Jun 10 at 17:11

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