How does a pseudo-Hermitian model differ from a Hermitian?

[SeM]

Hi, I have not been able to learn how a pseudo-Hermitian differs from a Hermitian model. If one has a hermitian model that satisfies all the fundamental prescriptions of quantum mechanics, a non-Hermitian would not, as it yields averages with complex values. How does a pseudo-Hermitian differ from Hermitian and non-Hermitian models?

Thanks

 

[A. Neumaier]

What do you mean by ''a pseudo-hermitian model"?

 

[SeM]

What do you mean by ''a pseudo-hermitian model"?

A pseudo-Hermitian Hamiltonian.

 

[A. Neumaier]

And when is a Hamiltonian pseudo-Hermitian? Do you mean non-Hermitian?

 

[dextercioby]

While "Hermitian" would be mathematically sloppy for the right word "self-adjoint", I believe it was the Russian school of mathematics who treated the subject of "non-self-adjoint operators" in depth:

https://www.encyclopediaofmath.org/index.php/Non-self-adjoint_operator

 

[SeM]

And when is a Hamiltonian pseudo-Hermitian? Do you mean non-Hermitian?

https://arxiv.org/abs/0810.5643
https://link.springer.com/article/10.1007/BF02108301

but this may be more clear:

https://www.google.no/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwj_rI-A8uHXAhWmPZoKHe4eC2oQFggsMAA&url=http://repository.uwyo.edu/cgi/viewcontent.cgi?article=1362&context=ela&usg=AOvVaw0pv87bdit3bB8FTeDI1O3Q

 

[A. Neumaier]

https://arxiv.org/abs/0810.5643

The difference is quite shallow: If the inner product to be used is not the standard inner product in the Hilbert space but $$\langle \phi|\psi\rangle_{new}=\langle \phi|M|\psi\rangle_{standard},$$ with positive definite $M$ then the observables are required to be Hermitian in the new inner product, which is a condition different from Hermiticity in the standard inner product. Thus in the latter, the Hamiltonian need not be Hermitian, but it doesn't matter since it is not the product defining the physical Hilbert space.

 

[SeM]

The difference is quite shallow: If the inner product to be used is not the standard inner product in the Hilbert space but $$\langle \phi|\psi\rangle_{new}=\langle \phi|M|\psi\rangle_{standard},$$ with positive definite $M$ then the observables are required to be Hermitian in the new inner product, which is a condition different from Hermiticity in the standard inner product. Thus in the latter, the Hamiltonian need not be Hermitian, but it doesn't matter since it is not the product defining the physical Hilbert space.

Maybe this is not directly related, but if the Hamiltonian is not Hermitian, but its matrix elements are in a Hilbert space L^2, and satisfy the condition:

\begin{equation}
\langle x,y \rangle = \int_a^bx(t)\overline{y(t)}dt,
\end{equation}

and the corresponding eigenvectors form an open subset M, in H, thus the Hamiltonian operates in Hilbert space, then, can one still use the hermitiian form <Phi|Phi*>=1 to normalize the non-hermitian solution? Or is the pseudo-hermitian form a method to work around the non-hermitian solution which has to be somehow normalized?

 

[A. Neumaier]

Maybe this is not directly related, but if the Hamiltonian is not Hermitian, [...] can one still use the hermitiian form <Phi|Phi*>=1 to normalize the non-hermitian solution? Or is the pseudo-hermitian form a method to work around the non-hermitian solution which has to be somehow normalized?

In the pseudo-Hermitian case, the correct normalization leading to consistent probabilities must always be done using the physical inner product involving the operator $M$.

 

[SeM]

In the pseudo-Hermitian case, the correct normalization leading to consistent probabilities must always be done using the physical inner product involving the operator $M$.

Can you give an example of this integral equality, i. e:

\begin{equation}
N\int_a^b \psi \Omega \psi* = 1
\end{equation}

 

[A. Neumaier]

I don't understand the connection to your question. All I said was that one uses a different inner product to define the physical Hilbert space, and one need to use it whenever one normalizes.

 

[SeM]

I don't understand the connection to your question. All I said was that one uses a different inner product to define the physical Hilbert space, and one need to use it whenever one normalizes.

Thanks.

The normalization condition, which is normally for a Hermitian system:

\begin{equation}
N \int_a^b \psi\psi* = 1
\end{equation}

is an inner product for the Hermitian type of wavefunction. However, you wrote

\begin{equation}
\langle \phi \mid M \mid \phi \rangle
\end{equation}

is this the inner product of the "pseudo-hermitian" wavefunction?

 

[A. Neumaier]

The inner product would typically be $$\int \phi(x)^*M(x,y)\psi(y)dxdy$$ with a positive definite kernel $M(x,y)$. The case $M(x,y)=\delta(x-y)$ gives the standard inner product.

 

[SeM]

Thanks Neumaier, that was clear!

Cheers