Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does the hermiticity of Hamiltonian restrict its Lagrangian?

  1. Jan 28, 2010 #1
    The hermiticity of Hamiltonian comes up as a result of requiring real energy eigenvalues and well-defined inner-product for correlation amplitudes.

    In the corresponding Lagrangian picture (path-integral), I am not clear about the explicit restriction that the above hermiticity of Hamiltonian impose on the Lagrangian. Its the basic path-integral formulation itself, in addition to the integral depending non-trivially on a special path-integral measure, exponential of Lagrangian and so on that makes it unclear. I don't see why Lagrangian should be hermitian.
    Any pointer?
    Last edited: Jan 28, 2010
  2. jcsd
  3. Jan 28, 2010 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    The Lagrangian is just the Legendre transformation of the Hamiltonian:

    [tex]H = \vec q \cdot \vec p - L[/tex]

    So if H, q, and p are Hermitian...then L had better be.
  4. Jan 29, 2010 #3


    User Avatar
    Science Advisor

    I allways thought that the equivalent is the invariance of the action under symplectic transformations.
  5. Jan 29, 2010 #4
    @ Ben Niehoff
    [tex]\vec q \cdot \vec p[/tex] is hermitian only if q and p commute, which they don't always do in quantum theory.

    That sounds plausible. But that takes care of only unitarity, but what about the other conditions (real eigenvalues etc)?
  6. Jan 29, 2010 #5


    User Avatar
    Science Advisor

    Yes, the problem with the statement in #2 is that it refers to the classical Hamiltonian for which the term hermitian is not defined.

    To my post: If time evolution is unitary, then its generator (H) is self-adjoint (Hermitian) and the eigenvalues are real: Theorem of Stone.

    Whether it is sufficient for the Propagator to be unitary that the Action is symplectic, I don't know for sure, but I doubt it. Remember vaguely having read an interesting article by Zhu and Klauder, "Classical symptoms of quantum illnesses", Am J Phys (1993) vol 61, pp. 605
  7. Jan 30, 2010 #6
    @ DrDu

    I am also not sure about your earlier statement now.
    We don't even need a symplectic manifold to define the lagrangian right?
    Also, in hamiltonian mechanics, unitarity is required in symplectic transformations of the flow induced by the hamiltonian vector field. For lagrangian, which flow/transformation to take?

    That article was interesting and satisfies a long-held curiosity about quirks of unbounded operators in physics.
  8. Jan 31, 2010 #7
    I've heard a lot of Mathematical descriptions of what happens to the lagrangian, but they're all so complicated that I can't understand what implications it might have physically

    The hamiltonian is hermitian because time evolution is unitary (these statements are equivalent --- oh, is this true in infinite dimensional vector spaces? it probably isn't).

    Unitarity of Time Development means that if you have 2 distinct quantum states (orthogonal), they stay distinct after any length of time, and probability among all states is conserved. This is equivalent to Liouville's Theorem of classical mechanics, and it's an assumption of quantum mechanics.
    Physically, this means that if I had a decision between 2 measurably distinct states (like |0> and |1>) and one of them was taken, after that point, the 2 states stay distinct.

    I think I can understand that.

    The hamiltonian being hermitian means that all eigenvalues are real and positive. Therefore, any measurable amount of the energy is is positive. I've read that this is good in classical mechanics because energy is always being radiated into the environment, which sends states into lower and lower energy states... if our energy can be negative, we'd get everything collapsing into lower and lower states, which is something we don't observe.
    Therefore, when we generalize to quantum mechanics, we want the same sort of thing. We want our system to be able to stabilize and not decay forever.

    I can understand this too... I think... I think I'm missing something important, but I think I almost get it.

    The lagrangian measures a weighting on the path from one state to the next... Why is it hermitian? What does this mean?


    And I don't think anyone actually knows this very well, so I challenge you all to give a good physical description of what it means that the Lagrangian is hermitian.
    Last edited: Jan 31, 2010
  9. Jan 31, 2010 #8
    Whoops. I walked away and realized that I was completely off. The Hamiltonian being hermitian means that its eigenvalues are real and its eigenvectors are orthogonal.

    I understand that the energy is the angular velocity of the phase, which governs probability. If I make the eigenvalues complex, I seem to get my system to decay immediately.. I don't understand why it's important that the eigenvectors are orthogonal though.

    Perhaps I should make another post on this after I think about it for a while.
  10. Feb 1, 2010 #9
    Orthogonality is naturally satisfied if the Hilbert space is separable (ie. if the vectors generating the Hilbert space are countable). The question of whether to consider an uncountable generating basis (or not) sounds more of a philosophical question.
    As for the theorem, it only says self-adjointness implies reality and orthogonality - not the converse. There may be a weaker requirement than self-adjointness that makes this theorem bidirectional. Supporting this argument is the field of non-self-adjoint, but PT-symmetric, Hamiltonians that have real eigenvalues (quant-ph/0501052).

    Coming back to the main question of this thread:
    What condition is to be satisfied by the Lagrangian, if the Hamiltonian is to have a real spectrum (or at least, hermitian)?
  11. Feb 1, 2010 #10


    User Avatar
    Science Advisor

    Why? The commonly used position (same for momentum) basis [tex]\{\left|x\right>\}_{x\in\mathbb{R}}[/tex] is uncountable, so it seems like a very practical question?
  12. Feb 1, 2010 #11
    Oh yes, you are right. But these are orthogonal as a distribution because their measures can be defined well (an extension of the basis being countable). I should then say that the question of not being countable (even as a distribution) is not very practical.
    Maybe LukeD can start a new thread on this.
  13. Feb 2, 2010 #12


    User Avatar
    Science Advisor

    I just had a look at my von Neumann, "Mathematische Grundlagen der Quantenmechanik":
    If the Hamiltonian (no explicit time dependence) is self-adjoint then it has a spectral decomposition [tex] H=\int_{-\infty}^{+\infty} \lambda dE(\lambda) [/tex], then the propagator can be written as [tex] U(t-t_0)=\int_\lambda exp(i\lambda (t-t0)) dE(\lambda) [/tex]. Here [tex][E(\lambda),E(\lambda')]=0, \; E(-\infty)=0; \; E(+\infty)=1; E(\lambda) E(\lambda')=E(\mathrm{min}(\lambda, \lambda')) [/tex]. That is the E form a family of projectors. E.g. if the Hamiltonian has a pure point spectrum with eigenvalues lambda, then [tex]E(\lambda)[/tex] projects on the subspace spanned by all eigenvectors with [tex]\lambda'<\lambda[/tex]. Hence E(lambda) is constant most of the time and increases at an eigenvalue. If H has also a continuous spectrum, then E increases continuously. Thus there is no need to bother about a non-countable basis of vectors |x>.
  14. Feb 2, 2010 #13
    @ DrDu
    Ya, thats the spectral theorem.
    The countability in finite-dim cases is required so that we can do the Gram-Schmidt process explicitly. I think, in the infinite-dim case, even when the measure (in this case, dE) exists it might exist abstractly as a Lebesgue measure which does not necessarily lend itself to doing explicit processes. That is, even if we can abstractly define the spectrum as those 'x' where (A - xI)⁻¹ exists, it doesn't say anything about explicitly mapping those 'x'. Or may be I am wrong. This is too much abstractness for me.

    Coming back to the main question of this thread:
    What condition is to be satisfied by the Lagrangian, if the Hamiltonian is to have a real spectrum (or at least, hermitian)?
  15. Feb 2, 2010 #14


    User Avatar
    Science Advisor

    I think the next step would be to define precisely which mathematically well defined path integral representation to use. I don't know much about this. I think, analytic continuation to Euclidean path integrals with imaginary time is an approach which can be applied rather generally. The analytic continuation of the propagator is no longer hermitian but decays exponentially. Any idea what we could learn from this?
  16. Feb 3, 2010 #15
    @ DrDu

    I don't get why you are rotating the time axis. Lets just stick to a real time.
    Now, in deriving the path-integral from time evolution operator (involving a self-adjoint Hamiltonian), in what way is the Lagrangian restricted?
    I believe, this restriction can be seen only on quantizing the path integral.

    I get a feeling that the individual formulations of Lagrangian and Hamiltonian mechanics are rigorous, but the their connection is a little hand-waving.
    Last edited: Feb 3, 2010
  17. Feb 3, 2010 #16


    User Avatar
    Science Advisor

    As far as I know, the continuum version of the path integral (the one that really contains L) is ill-defined for real times. On the other hand, if you stick to a finite number of time slices, then there will be no Lagranian, only p's and q's and H's.
  18. Feb 3, 2010 #17
    Time rotation is more of a calculation trick that simplifies things than anything. As long as we are not doing any explicit calculation with, lets leave time as real.
  19. Feb 4, 2010 #18


    User Avatar
    Science Advisor

    Ok, then please write down the concise formula for the path integral you want to talk about.
  20. Feb 4, 2010 #19
    I was thinking we could just stick to the following original formulation without rotating time:
    \langle \phi_f | e^{-iHt} | \phi_i \rangle = \int_{\phi_i}^{\phi_f} [\mathcal{D} \phi]\ e^{i\int_0^t L\ dx^4 }

    Has the equivalence of hamiltonian and path-integral formalism been established beyond doubt, or just established in few situations (quadratic etc) and just assumed to extend to rest?
    Last edited: Feb 4, 2010
  21. Feb 4, 2010 #20


    User Avatar
    Science Advisor

    It took 20 years to figure out how to solve the Hydrogen atom using path integrals. That was the main reason why Feynman himself did not use it as an alternative introduction to QM in his classes, if I remember correctly. I must say that I know little about the path integral, but I think the problem is that the measure [tex]D \phi [/tex] is not defined.
    No one else out there who knows more?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook