# I Non-Hermitian wavefunctions and their solutions

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1. Jan 18, 2018

### SemM

I was wondering if anyone has worked with non-Hermitian wavefunctions, and know of an approach to derive real and trivial values for their observables using numerical calculations?

Cheers

2. Jan 18, 2018

### Staff: Mentor

What's a non-Hermitian wave function?

3. Jan 18, 2018

### PeroK

Your posts are fascinating. You seem to be ploughing through QM without having learned the basics - and having no intention of learning the basics. Perhaps eventually you will be able to put things together consistently without ever having learned, say, the definition of and difference between a vector and an operator!

Personally, I doubt it. I recommend a first course in linear algebra.

4. Jan 18, 2018

### SemM

Haha, thanks , it is indeed true, but its not true that I don't want to learn the basics. I did an engineering in chemistry, with plenty of math, but zero QM. All QM is by self-learning, so, given the interest in important subjects, coupled with no experience in courses in QM, a bunch of strange questions do indeed arise.

I thought about a solution to this original question, but I am not sure:

What if I expand the function which has no hermitian counter-part in terms of a power-series, then I perform the evaluation of each series component using the form $\langle \psi | d/dx | \psi*\rangle$ and at the end sum up only those that have a hermitian counterpart? Sounds like a lazy form of shortcutting to me, but maybe its the only way to get some estimate. However, a closer thought suggests that each component of a series of say $e^{-kx}$ would be something like $1/x-2/x^2-3/x^4-1/x^8+..$ which have no hermitian counterpart. So back at the original question.

Last edited: Jan 18, 2018
5. Jan 18, 2018

### SemM

6. Jan 18, 2018

### SemM

7. Jan 19, 2018

### Staff: Mentor

I'm not aware of any significant work on the hermicity of wave functions. In QM, what is important is the hermicity of operators, which is what the link you posted above is about. This is also, I think, what has triggered @PeroK's reply. You seem to be confusing operators and wave functions.

To come back to the OP, most often the eigenfunctions of Hamiltonians can be chosen purely real, so they are definitely not Hermitian functions.

This notation doesn't make sense to me. If you are using the Dirac notation, then there is no particular basis, so the $d/dx$ of a ket doesn't make sense. Likewise for the inclusion of a $^*$ in the ket, if it is to mean complex conjugation.

8. Jan 19, 2018

### SemM

It is difficult to use a common language here, because some are physicsts and some are mathematicians. This Dirac notation can also be given as

\int \psi p \psi^*dx

and is the expectation value of the momentum operator.

Thanks

9. Jan 19, 2018

### Orodruin

Staff Emeritus
This ha nothing to do with it. What you wrote simply does not make any sense. It is definitely not the same as you wrote in your last post. I would suggest taking three steps back and learn the fundamentals of Hilbert spaces and Dirac notation to represent elements in the Hilbert space.

10. Jan 22, 2018

### SemM

Anyway, the solution to this problem is to construct wavepackets, for those who are wondering, or should the original Hamiltonian be non-Hermitian, then refer to "Making sense of non-Hermitian Hamiltonians" by Carl Bender.

These conversations must yield an answer and not just critics, otherwise PF becomes a gladiator arena, with no pedagogic or informational role.

11. Jan 22, 2018

### Staff: Mentor

I don't get it. The most common wave packet, a Gaussian wave packet, can be purely real, hence non-Hermitian.

The problem is the questions you ask, along with the replies you give, show a lack of basic knowledge. It is extremely pedagogical for people to reply that you must take steps back and try and get proper basic knowledge first.

Let me translate what these threads of yours sound to me. OP: "What is the colour of a hydrogen atom?" Followed up by question asking what do you mean by color of an atom, and a back and forth about basic principles, to be concluded by you saying "You should simply have answered blue."

12. Jan 22, 2018

### SemM

I didnt say the wavepacket must be non-hermitian. I said that constructing wavepackets is a way to find a solution to a model that does not give real and trivial expectation values for its observables.