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Hamilton; help

  1. Jun 16, 2004 #1
    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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  3. Jun 16, 2004 #2
    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.
     
  4. Jun 16, 2004 #3

    selfAdjoint

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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.
     
  5. Jun 16, 2004 #4
    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?
     
  6. Jun 17, 2004 #5

    selfAdjoint

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    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..
     
  7. Jun 17, 2004 #6
    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.
     
  8. Jun 18, 2004 #7

    turin

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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.
     
  9. Jun 18, 2004 #8

    Dr Transport

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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.
     
  10. Jun 19, 2004 #9
    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
     
  11. Jun 20, 2004 #10
    Another use I have uncovered is that self-adjoint operators are not necessarily Lorentz invariant, wheras Hermitian operators are.
     
  12. Jun 20, 2004 #11

    reilly

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

    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
     
  13. Jun 20, 2004 #12
    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?
     
  14. Jun 20, 2004 #13

    selfAdjoint

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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.
     
  15. Jun 20, 2004 #14

    robphy

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    http://mathworld.wolfram.com/Self-Adjoint.html
    Near the end of this description, there is a comment confirming what turin said above.
     
  16. Jun 21, 2004 #15

    reilly

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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
     
  17. Jun 22, 2004 #16
    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.
     
  18. Jun 22, 2004 #17

    selfAdjoint

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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!
     
  19. Jun 22, 2004 #18
    What is exactly this "LSZ" formalism?
     
  20. Jun 22, 2004 #19
    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.
     
  21. Jun 23, 2004 #20

    robphy

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    I don't have a chance to read these carefully... but these may be useful
    http://www.phys.psu.edu/~collins/525/self-adj.pdf [Broken]
    and
    http://www.cithep.caltech.edu/~fcp/physics/quantumMechanics/ideasOfQM/ideasOfQM.pdf [see pg 16 and 17]
    and
    http://merlin.fic.uni.lodz.pl/LFPPI/konf/WKarwowski.pdf [Broken] [makes a comment on p 3 with reference to an Am J Phy article: G. Bonneau, J. Faraut, and G. Valent, “Self-adjoint extensions of operators and the teaching of quantum mechanics,” Am. J. Phys. 69, 322–331 (2001) ]

    possibly related: http://arxiv.org/abs/quant-ph/9412007

    ...enough googling. (Back to work!)
     
    Last edited by a moderator: May 1, 2017
  22. Jun 24, 2004 #21

    selfAdjoint

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    LSZ stands for Lehmann, Symanzik, and Zimmerman, who described the method in Nuevo Cimento in 1955.

    Basically you Fourier transform the coordinate states of each of the external particles, finding a pole in the transform, and you argue that these poles represent asymptotic one-particle states, "that is states given by the limit of well-separated wavepackets as they become concentrated around definite momenta. Taking the limit in which all external particles go on-shell, we can then interpret the coefficient of the multiple pole as an S-matrix element" (Peskin & Schroeder).
     
    Last edited: Jun 24, 2004
  23. Jun 24, 2004 #22
    Self-Adjoint versus Hermitian

    When the Hilbert space is infinite-dimensional, a difficulty arises in that many linear operators cannot be defined on the entire space. For example, think of the position operator Q for a particle moving in one dimension. Since Q takes the function f(q) to qf(q), there will be many square-integrable functions f(q) such that qf(q) is not square-integrable. This means that Q cannot be defined on the entire Hilbert space. On account of this sort of phenomenon, the mathematician is forced to take into consideration the domain (a linear subspace of the Hilbert space) on which a linear operator is defined.

    So, for example, to say that A = B means two things:

    (i) Domain(A) = Domain(B);

    and

    (ii) |Af> = |Bf>, for all |f> in the common domain.

    In a similar vein, A + B can only be defined on the intersection of Domain(A) and Domain(B).

    Also, AB can only be defined on those elements |f> of Domain(B) such that |Bf> is an element of Domain(A).

    Physicists normally do not worry about such matters.
    _______________________

    Now, what does all of this have to do with "self-adjoint" and "hermitian" operators?

    Well, first of all, let A and B be two linear operators acting in a Hilbert space H. Then, A and B are said to be adjoint to one another iff:

    <Bg|f> = <g|Af> for all |f> in Domain(A) and all |g> in Domain(B).

    Note that B is "an" adjoint of A, and not necessarily "the" adjoint of A. "The" adjoint of A (call it A'), if it exists, is a maximal extension of all possible B which are adjoint to A. A' exists iff Domain(A) is "big enough" (specifically, "big enough" means: the (set theoretic) closure of Domain(A) equals the entire Hilbert space H).

    I'm not going to give any more details than this on A' (... but if anyone would like more, feel free to ask).

    We are now in a position to display the definitions of "hermitian" and "self-adjoint".

    A linear operator A is hermitian iff <g|Af> = <Ag|f> for all f,g in Domain(A).

    A linear operator A is self-adjoint iff: (i) A is hermitian, (ii) A' exists, and (iii) Domain(A) = Domain(A').

    As it turns out, there exist hermitian operators A, such that A' exist but Domain(A) is only a proper subspace of Domain(A').

    The upshot of all of this is that a hermitian operator will not in general guarantee the so called "closure" property with regard to its "eigenkets" and corresponding "bras"; i.e. a relation like:

    Sigma_n { |n><n| } and/or Integral {ds |s><s| }
    = Identity operator on H .

    A self-adjoint operator will always guarantee this sort of relation.

    Mathematicians generalize and rigorously formulate this notion of "closure" with what is called a "spectral family" (or "spectral function"), arriving at what is called "The Spectral Theorem" for self-adjoint operators.

    ********************************************
    * A note on NOTATION: Let C be any linear operator
    * such that the adjoint C' exists.
    *
    * Then, for all |f> in Domain(C) and all |g> in Domain(C'),
    *
    * <g|Cf> = <g|(C|f>) = <g|C|f> = (<g|C)|f> = <C'g|f> .
    ********************************************
    _____________________

    Note that "self-adjoint" is a stricter condition than "hermitian", "self-adjoint" being "hermitian with the two additional properties mentioned above".

    Somehow Turin got it the other way around:

    And Robphy found someone else who, in an example, (just how???) also gets it backwards:

    _______________________

    Reilly brings out another point:

    I haven't checked von Neumann's book ... but you are right, there do exist sources which define "hermitian" as you say. In that case, any time I mentioned the word "hermitian" above, you should replace it with the word "symmetric" to denote the corresponding property.

    But be careful regarding what you found on the Conjugate Transpose page (http://mathworld.wolfram.com/ConjugateTranspose.html):
    On that page, the Hilbert space is finite-dimensional, for which the above mentioned "domain problems" do not exist. At the outset one can talk about linear operators acting on the entire space, and none of the above distinctions need to be considered.
    _______________________

    Slyboy offers us a complication.

    What (I think) he wishes to say is:

    An operator does not have to be "self-adjoint" in order to satisfy "The Spectral Theorem".

    Note that a normal operator can, in general, have complex eigenvalues.

    This is the definition:

    A linear operator N is normal iff: (i) the adjoint N' exists, (ii) Domain(N) = Domain(N'), (iii) [N,N'] = 0.

    This is a generalization of "self-adjoint" and "unitary" operators.
     
    Last edited: Jun 24, 2004
  24. Jun 24, 2004 #23

    reilly

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    Historically most physicists have often used math in a non-rigorous fashion. This has been particularly true in quantum physics -- a very well known example is Dirac's delta function. And, this touches on the issues of Hermitian and self adjoint operators, particularly for operators defined on infinite domains -- like position or momentum operators. You'll note in the papers cited by robphy, the subtle problems of Hermitean and selfadjoint operators typically involve infinite intervals and boundary conditions, and use quite sophisticated mathematics -- from functional analysis.

    When I learned and taught this stuff, hand waving was the norm -- like not to worry, the rigor will catch up with us, concentrate on the physics, not the math. The LSZ approach, in fact, confirmed an approach already in use to S-Matrix or scattering theory based on physical intuition and sort of sloppy math.

    With my old fashioned ways, hermitian and self adjoint are the same thing -- if not, physical intuition will guide you to the appropriate resolution.

    Regards,
    Reilly Atkinson
     
  25. Jun 24, 2004 #24
    Yes. And that is an example of how a physicist's intuition can create something beautiful, useful, and perfectly consistent.
    _______________

    Yes, often a specification of the domain of a linear operator acting in a Hilbert space of functions amounts to a specification what boundary conditions the functions must satisfy.
    _______________

    Yes, I agree. (... On the other hand, if we are talking about technical definitions, then there are two distinct types of mathematical objects, one described as "self-adjoint" and another described as "symmetric". Some sources equate "hermitian" with "symmetric", while others (apparently, the older ones) equate it with "self-adjoint". The main point is, as far as technical definitions go, "self-adjoint" implies "symmetric", and not the other way around ... and, in a finite-dimensional Hilbert space, such a distinction becomes trivial (in the sense that any linear operator whose domain of definition is not all of H has an extension which acts on all of H.))
     
    Last edited: Jun 24, 2004
  26. Jun 24, 2004 #25

    reilly

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    Eye in the Sky -- I'm glad that we agree on much about operators. And, I'm particularly grateful for your clear and succinct description of the subtlties of Hermitian, symmetric, self adjointness and so on. Always somthing new to learn.

    Thanks and regards,
    Reilly Atkinson
     
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