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Usefulness of Heisenberg picture?

  1. Oct 9, 2007 #1
    I'm a little confused as to why anyone would want to use the Heisenberg picture of time evolution instead of the Schrodinger picture, beyond showing that the equations of motion are similar to those of classical mechanics. For example, consider a free particle. Using the Heisenberg equations of motion, you can show that the position and momentum operators obey equations similar to their classical analogs (i.e., the position is linear in time and the momentum is constant). But what does this really give you? It doesn't seem to me that it has any concrete meaning outside of the fact that it is mathematically equivalent to the Schrodinger picture. I guess what I'm really asking is this: what physical meaning does a time-varying operator have?
  2. jcsd
  3. Oct 9, 2007 #2


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    The Heisenberg picture is very convenient in some problems, in other problems the Schrödinger picture is easier to use. In quite a few problems the interaction picture is the best choice. Occasionaly you change pictures several times while solving a problem.

    Changing picture is a bit like changing coordinate systems in that you use whatever picture works best for a given problem; asking why someone is using the Heisenberg picture is therefore a bit like like asking why someone is using cylindrical coordinates; the answer is usually that it makes the calculations easier or/and makes the problem easier to understand.
  4. Oct 10, 2007 #3
    And what physical meaning does a non time-varying operator have in the Schrodinger picture?
  5. Oct 10, 2007 #4
    Your post is oversimplification. That case isn’t matter of convenience. In QM we have two layers of convenience:

    1)As in CM, choice of the geometrical (space-time) coordinate system which is “natural” for your physical system sharply simplifies the calculations;
    2)Choice of the Hilbert space geometry coordinate system (for a layman, they are called a special functions and they are similar to sin,cos,sh and ch). In QM the set of mutually commuting hermitian operators (observables) is less than in CM; only if every dynamical variable is observable the Hilbert space is rigid. Again, the “natural” choice sharply simplifies the mathematical complexity of the problem.

    However, the existence of Heisenberg and Schrödinger pictures is connected with the dual nature of the quantum mechanical behavior of the physical system and they are complementary to each other. According to the current formulation of the CM there is no difference between the system state and the dynamical variables which define that state. In QM it is not so. We may say that the dynamical variables represent the particle nature of the physical system (Heisenberg’s quantum mechanics) and the states – wave mechanical (Schrödinger’s wave mechanics). It was shown by E. Schrödinger that these two pictures are equivalent in the non-relativistic limit. If you describe the electron by the Dirac equation, it was shown by P.A.M.Dirac that the equivalence (duality) is broken.

    The serious and organized investigation of that and related questions was initiated by P.A.M.Dirac (see, for example, “Lectures on QM” (1964), “Lectures on QFT” (1967), Yeshiva Univ. Press). For the detailed discussion I would suggest the papers and books written by N.Mukunda. In addition, recently published paper PRL,97,154101(2006)
    http://docto.ipgp.jussieu.fr/IMG/pdf...t_PRL_2006.pdf [Broken] I consider directly relevant to that problem.

    Regards, Dany.
    Last edited by a moderator: May 3, 2017
  6. Oct 10, 2007 #5
    I agree with you that the choice of the "picture" is just a matter of convenience.

    The question is about calculating the time dependence of expectation values of observables (that's what we actually observe in nature). If the operator of observable is [itex] F [/itex] and the state of the system is [itex] | \Psi \rangle [/itex], then quantum mechanics says that the time dependence of the expectation value is

    [tex] \langle F(t) \rangle = \langle \Psi | \exp(\frac{i}{\hbar} Ht) F \exp(-\frac{i}{\hbar} Ht) | \Psi \rangle [/tex]..........(1)

    where [itex] H [/itex] is the Hamiltonian. This formula can be written in two equivalent ways:

    [tex] \langle F(t) \rangle = \langle \Psi |F(t) | \Psi \rangle [/itex]

    where [itex] F(t) = \exp(\frac{i}{\hbar} Ht) F \exp(-\frac{i}{\hbar} Ht) [/itex] is the time-dependent operator in the Heisenberg picture


    [tex] \langle F(t) \rangle = \langle \Psi (t)|F | \Psi (t) \rangle [/itex]

    where [itex] | \Psi(t) \rangle = \exp(-\frac{i}{\hbar} Ht) | \Psi \rangle [/itex] is the time-dependent state vector in the Schroedinger picture.

    No matter how you slice eq. (1) the answer is the same.

  7. Oct 10, 2007 #6
    That is correct if:

    1)the Hamiltonian exist;
    2)the Hamiltonian that describe the time evolution in Schrödinger picture is the same as in Heisenberg picture;
    3)the Hamiltonian is hermitian operator.

    Then since the Hamiltonian is obviously commuted with itself, the energy is the conserved quantity. All that defined by how you treat time in QM.

    Regards, Dany.
  8. Oct 11, 2007 #7
    I don't think there is any reason to doubt these statements. Do you agree?

  9. Oct 11, 2007 #8
    No. See P.A.M. Dirac, “Principles of QM”, 4-th ed., Oxford, 1958, par.70. I would say that in comparison with you, I don’t consider the possibility that:

    1)P.A.M. Dirac don’t know the math;
    2)P.A.M. Dirac don’t know to analyze the math and the physical content of the fundamental eq. of motion;
    3)That I am cleverer than P.A.M. Dirac.

    Regards, Dany.

    P.S. From “Symmetry and conserved probability current for wave functions” session (your post #32):

    Meopemuk:” There is no need to introduce Lagrangian to "prove" these things. I still believe that Lagrangians are out of place in quantum mechanics. Yes, you can formally introduce some action, Lagrangian, etc. in quantum mechanics and "derive" the Schroedinger equation and "conservation of probabilities" from them. But this will not tell you anything new that you didn't know already from the basic laws of QM.”

    I suggest to you stop confusing kids discussing the topics where you have no idea what you are talking about.
    Last edited: Oct 11, 2007
  10. Oct 11, 2007 #9
    Unfortunately, I don't have this book. Could you please remind us what were Dirac's arguments that lead him to the conclusion that there is no well-defined Hamiltonian in QM?

  11. Oct 11, 2007 #10
    Now I understand what a problem is. You know to write but forgot how to read. As I mentioned above, the conclusions are derived by P.A.M. Dirac mathematically.

    Regards, Dany.
  12. Oct 11, 2007 #11
    anonym is right, this is also talked about in Dirac's 1964 QFT book
  13. Oct 13, 2007 #12
    This is off topic, but I would like to add a few words. From today discussions in literature and here at PF it is not clear what should be answers to the above questions, depend how time is treated. It should be clear, however, that nobody among the “fathers” was satisfied with the Dirac solution and P.A.M. Dirac in the first place.

    From “philosophical” POV it is clear (“Ockham razor”) that the spinors must be two component and the “sea” is artifact. From the physical POV the equations should describe also quarks, electroweak and QCD. Mathematically, C3 algebra is bad, it is not a math.

    Regards, Dany.
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