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

Time Dependence in Pictures of QM

  1. Sep 26, 2012 #1
    I have a question about time dependence in the different pictures of QM. In the Schrodinger Picture, I've read that the time dependence is in the state vector, but one can construct Hamiltonians and other operators that are time dependent in the Schrodinger Picture. For instance, one can construct a time dependent potential in Schrodinger wave mechanics, and one would not be using the Interaction Picture. (But the equation would probably have to be solved numerically.)

    Then could states in the Heisenberg Picture be constructed so that they are time dependent as well without invoking the Interaction Picture? I would think it's possible, but I assume it would usually lead to a difficult problem to solve and rarely done. Are there any references that talk about handling such cases?
  2. jcsd
  3. Sep 26, 2012 #2


    User Avatar
    Science Advisor
    Gold Member

    "Pictures" are a bit like coordinate systems: they have no deep physical meaning and you should usually just work in whatever picture is easiest to handle mathematically.

    Now, the Schroeding and Heisenberg pictures are just the two "extremes", you can move (so to speak) time dependences around more or less as you want to simplify your equations and whenever you do that you are in the "interaction picture".
    A good example would be if you have a hamiltonian with a time-varying electromagnetic field, sometimes you can then simplify things considerably by moving into a picture where the whole hamiltonian is rotating "with" the field.
    (and if you then throw away some higher order terms you get the rotating wave approximation)

    Note that you are always free to go back using another transformation.

    Hence, this is not really very different from moving from e.g. carteesian coordinates to cylindrical coordinate to simply a PDE solved over a circle.
  4. Sep 26, 2012 #3
    I think I follow you. So, for instance, in the rotating wave approx. where E is the electric field operator in the heisenberg picture,

    E = [itex]\sum[/itex]kl(akleklfkl(x)exp(-i[itex]\omega[/itex]t) + a+kle*klf*kl(x)exp(i[itex]\omega[/itex]t))

    with <a|E2|a> as the the energy density expectation value, I could choose a state

    |a(t)> = cos(t)|0> + sin(t)|2>

    and compute <a(t)|E2|a(t)> and still get a "correct" answer. Or do I have to adjust the definition of E as well?
  5. Sep 27, 2012 #4


    User Avatar
    Science Advisor

    I think we should start with the following observation:

    The time-dependent Schrödunger equation

    [tex]i\partial_0\,|\psi,t\rangle = H\,|\psi,t\rangle [/tex]

    can be solved using the time evolution operator

    [tex]U(t) = e^{-iHt}[/tex]
    [tex]|\psi,t\rangle = U(t)\,|\psi,0\rangle [/tex]

    which you see by differentiating

    [tex]i\partial_0\,U(t) = H\,U(t)[/tex]

    Now for any operator A where you are interested in eigenvalues, expectation values or something like that and which you apply to a state |ψ>

    [tex]A\,|\psi,t\rangle [/tex]

    you can insert any unitary operator w/o changing physics:

    [tex]A\,|\psi,t\rangle\;\to\;\Omega A \Omega^\dagger \Omega |\psi,t\rangle[/tex]

    This is like a basis transformation which - as in linear algebra - acts both on the states (vectors) and on the operators (matrices). So you can introduce transformed states and operators as follows

    [tex]|\psi,t\rangle\;\to\;{}^\Omega|\psi,t\rangle = \Omega |\psi,t\rangle[/tex]
    [tex]A\;\to\;{}^\Omega A = \Omega A \Omega^\dagger [/tex]

    In that way you can introduce time-independent coordinate transformations like translations (where Ω is represented as eiap using the momentum operator p) or rotations (where Ω is represented as eiθL using the angular momentum operator L).

    But the unitary operator Ω is arbitrary and can be time dependent as well. So we can use something like

    [tex]\Omega(t) = e^{-iGt}[/tex]

    with a selfadjoint operator G.

    1) If G=H the operator Ω generates the transformation between the Heisenberg and the Schrödinger picture, i.e. it fully shifts the motion generated by U(t) from the states to the operators and vice versa
    2) If G=Hfree where Hfree is the free (and hopefully trivial) part of the full Hamiltonian H, the operator Ω generates the transformation between the interaction and the Schrödinger picture. The shift of the motion generated by U(t) is shifted from the states to the operators only partially!
    3) If you have some Hamiltonian H= H°+εh you may chose G=H° which is similar to the interaction picture but now H° need not be the free Hamiltonian and could be a more complex but still solvable Hamiltonian. Like in the interaction picture the idea is to separate the motion in a trivial (= free or solvable) part and a complicated part such that the motion of the operators A is trivial and that only the states are subject to complicated motion.

    But physics doesn't change at all. You are free to use any Ω, i.e. any selfadjoint G (as long as it simplifies the math).

    Remark: time dependent energy eigenstates do not appear in the Heisenberg picture b/c the Heisenberg is defined such that the time dependence of such states is "rotated away". Using an operator Ω which introduces time dependency of such states should be interpreted as changing from the Heisenberg picture to something else. But there's a loophole in this argument: the starting point is the Schrödinger picture where a time dependent solution |ψ,t> of the Schrödinger equation is constructed using U(t). This solution is then transformed to the Heisenberg picture |ψ,0> where Ω(t) is identical to U(t). Therefore eigenstates of H become time independent. But if you start with an arbitrary state |ζ,t> which does NOT solve the Schrödinger equation, then the transformed state Ω|ζ,t> will NOT be time independent. Therefore the Heisenberg picture does contain time dependent states, but they are not eigenstates of H, i.,e. no solutions to the original problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook