# Hamilton; help

1. Jun 16, 2004

### Blackforest

Certainly a simple question but I am lost; can the Hamilton operator of a particle be a matrix [H] ? If yes, must this matrix be self adjoint [H = h(i,j)] = [H = h(j,i)]*? And if yes: why or because of why ? Thanks

2. Jun 16, 2004

### heardie

I can't see why the Hamiltonian can't be a matrix. We could still solve the eigenvalue problem (most people would be more used to solve matricies with eigenvalues)

And I guess it would have to be self-adjoint, to ensure it has real e'values, so what we measure (the e'value) is real.

3. Jun 16, 2004

Staff Emeritus
In general the Hamiltonian is an operator in quantum mechanics. It operates on states (or "wavefunctions") to generate time transitions. In simple cases, the states live in a finite dimensional vector space, and the operators, including the Hamiltonian, then become matrices.

4. Jun 16, 2004

### Blackforest

OK; thanks. Concerning the fact that the Hamiltonian operator under its form of a matrix who should be self adjoint to give us real eigenvalues: does it mean -a contrario- that a non selfadjoint matrix doesn't give real eigenvalues and more generaly doesn't give eigenvalues with physical sense and reality? Or with other words: in developping a theory if I find an Hamiltonian operator whose matrix representation is not self adjoint: did I make somewhere a mistake?

5. Jun 17, 2004

Staff Emeritus

Basically yes. Sometimes quantum physicists use Hermitian instead of self adjoint. The differences are subtle, but the upshot is that self adjointness is better. As you say, for the eigenvalues to be relevant to our world, they have to be real, and the way to guarantee that is to insist on self adjointness..

6. Jun 17, 2004

### rick1138

What is the difference between SA and Hermitian operators? I have heard it mentioned a few times, but have never heard anyone state it expicitly.

7. Jun 18, 2004

### turin

I thought that Hermiticity was the more strict condition: being self-adjoint-ness with the additional requirement on the boundary.

8. Jun 18, 2004

### Dr Transport

Because the eigenvalues of the Hamiltonian are measurable quantities, i.e. energy, the Hamiltonian must be Hermetian, this ensures that the energies are real. Self adjoint does not necessarily ensure that the eigen values are real.

9. Jun 19, 2004

### Blackforest

Exactly Dr Transport, this is the subtle difference suggested by selfAdjoint. [Z] is the (N-N) matrix representation of the operator with complex numbers z(a,b) = x(a,b)
+ i y(a,b) where (a,b) denotes the position inside the matrix. [Z]* is the complex conjugated matrix buit with the z*(a,b) = x(a,b) - i y(a,b) and which is not preserving the diagonal. [Z]~ is built with a transposition of the z(a,b) symmetric to the diagonal. A transposition preserves the diagonal. If one write [Z]*~ = [Z] one is sure that the diagonal is built with real numbers. Thank you for your helps

10. Jun 20, 2004

### rick1138

Another use I have uncovered is that self-adjoint operators are not necessarily Lorentz invariant, wheras Hermitian operators are.

11. Jun 20, 2004

### reilly

*****************

I'm very puzzled. A standard definition of a Hermitian operator/matrix H, is that H is Hermitian if and only if H is self adjoint, that is it is equal to the complex conjugate of it's transpose. This is commonly accepted by physicists and mathematicians, and has been for a century or so -- or so i thought.

I would be most greatful for a reference or discussion of the difference between a self-adjoint and a hermitian operator/matrix.

Thank you,
Reilly Atkinson

12. Jun 20, 2004

### Blackforest

Really, I never thought that I could begin such a discussion in asking this „simple“ question. After my last thread here, I believed I could close my books and work again… but I am curious and I have read some more on the subject. And I got a big surprise: an Hamiltonian must not always be an Hermitian operator to describe a reality; e.g.: the transition for a system with two different states, between a stable and an unstable state. It can be made use of a 2 x 2 matrix built with complex terms on its diagonal (the different energy levels). Each part of these complex numbers has an interpretation… This way of doing is more or less connected with the way of doing concerning the matrix needed to make a description of a particle with a spin 1/2 (real or fictive). So it’s quite more complicated as I thought. I know that this place is not the good one to develop a theory; so I ask just about the principle: do you think that vacuum can be considered as a stable state and random fluctuations of the fields in vacuum as unstable states ? And consequently could we make use of these strange Hamiltonian which are no Hermitian to try a description of the vacuum and of the virtual particles?

13. Jun 20, 2004

Staff Emeritus
In elementary physics the vacuum is a stable state, the lowest energy state. In interacting quantum field theory there is a problem because the vacuum retains a history of previous field interactions. QFT practioners have a formalism ("LSZ") to get around this.

There is a tremendous amount of research on Hamiltonian systems. Look up "symplectic" to see some of it.

14. Jun 20, 2004

### robphy

Near the end of this description, there is a comment confirming what turin said above.

15. Jun 21, 2004

### reilly

robfy & turin and self-adjoint-- I checked out the Wolfram page, and I could not quite figure out what the boundary conditions related to -- the variables are undefined. If you go to the Conjugate Transpose page via the given link, you'll find my definition. So I'll stick with von Neumann's definition.

In fact, von Neumann in his Mathematical Foundations of Quantum Mechanics gives the definition Hermitian if and only if self adjoint, in Hilbert Space. His book really defined the mathematics of QM, and provided a highly rigorous treatment of Hilbert space accessible to the physics community.

For 2nd order DEs, boundary conditions are very important, because adjoint operators are defined through sequences of integrations by parts. I assume that's what the Wolfram page is attempting to note. I recommend Byron and Fuller's, Mathematics of Classical and Quantum Physics for a clear and complete discussion of self-adjointness and Hermitianess (p151 and on) and on Sturm Liouville Eq.s (p263 and on), of which the solutions form the basis for Hilbert spaces.

Regards,
Reilly Atkinson

16. Jun 22, 2004

### rick1138

The definitions and notations of group theory were very inconsistent in the pre-WWII days, as they are in the early days of any field. I am intrigued by the hermitian - self-adjoint difference. I have seen defintions where they are considered the same and others were they aren't. I wouldn't quote Mathworld as an authority, there are a number of mistakes I found on the site, and when I pointed them out they failed to correct them.

17. Jun 22, 2004

Staff Emeritus
Months ago I found a good paper on the arxiv about the self adjoint/hermitian difference. Since the question came up on this thread I have been trying to find it again, without success. My dim memories of its title and guesses at keywords in its abstract have proven vain. Does anyone remember it? I posted about it somewhere on PF, but I can't even find that post again!

18. Jun 22, 2004

### Blackforest

What is exactly this "LSZ" formalism?

19. Jun 22, 2004

### slyboy

Just to complicate the discussion further, an operator does not have to be self-adjoint in order to have real eigenvalues. It need only be normal, i.e. commute with its adjoint.

There is a quantum theory with non-Hermitian Hamiltonians that several researchers have been investigating recently. However, it is still an open question as to whether it is applicable to any real physical systems.

20. Jun 23, 2004

### robphy

I don't have a chance to read these carefully... but these may be useful