Hamilton; help


by Blackforest
Tags: hamilton
Blackforest
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#1
Jun16-04, 12:45 PM
P: 445
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
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heardie
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#2
Jun16-04, 07:34 PM
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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.
selfAdjoint
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#3
Jun16-04, 09:06 PM
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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.

Blackforest
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#4
Jun16-04, 10:29 PM
P: 445

Hamilton; help


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?
selfAdjoint
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#5
Jun17-04, 11:08 AM
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Quote Quote by 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?

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..
rick1138
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#6
Jun17-04, 06:06 PM
P: 199
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.
turin
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#7
Jun18-04, 01:40 PM
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selfadjoint,
I thought that Hermiticity was the more strict condition: being self-adjoint-ness with the additional requirement on the boundary.
Dr Transport
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#8
Jun18-04, 06:33 PM
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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.
Blackforest
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#9
Jun19-04, 04:25 AM
P: 445
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
rick1138
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#10
Jun20-04, 12:04 AM
P: 199
Another use I have uncovered is that self-adjoint operators are not necessarily Lorentz invariant, wheras Hermitian operators are.
reilly
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#11
Jun20-04, 12:43 AM
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Quote Quote by selfAdjoint
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..
*****************

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
Blackforest
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#12
Jun20-04, 09:46 AM
P: 445
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?
selfAdjoint
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#13
Jun20-04, 11:15 AM
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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.
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#14
Jun20-04, 12:15 PM
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Quote Quote by 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

http://mathworld.wolfram.com/Self-Adjoint.html
Near the end of this description, there is a comment confirming what turin said above.
reilly
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#15
Jun21-04, 12:22 AM
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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
rick1138
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#16
Jun22-04, 01:10 AM
P: 199
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.
selfAdjoint
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#17
Jun22-04, 09:12 AM
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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!
Blackforest
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#18
Jun22-04, 09:28 AM
P: 445
What is exactly this "LSZ" formalism?


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