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

Interaction picture

  1. Jul 3, 2011 #1
    So this concept of H = H_o + H_int has been extremely confusing to me. Wikipedia offers the best explanation, but there a couple things that still confuses me
    http://en.wikipedia.org/wiki/Interaction_picture

    Why is the state vector in the Interacting picture defined as
    |[itex]\psi[/itex][itex]_{I}[/itex](t)> = e[itex]^{i H_{O,S}t/h}[/itex]|[itex]\psi[/itex][itex]_{S}[/itex](t)>

    instead of

    |[itex]\psi[/itex][itex]_{I}[/itex](t)> = e[itex]^{i H_{O,S}t/h+ i H_{1,S}t/h}[/itex]|[itex]\psi[/itex][itex]_{S}[/itex](t)>?

    why isn't the schrodinger picture of the perturbation included?

    Similary, why isn't the schrodinger picture of the perturbation included for the equation of the Operators in the Interaction picture?

    Finally, why does the exponential factor that determines the perturbation Hamiltonian include only a H[itex]_{O,S}[/itex] and not a H[itex]_{1,S}[/itex]
     
    Last edited: Jul 3, 2011
  2. jcsd
  3. Jul 4, 2011 #2
    The key idea here is to bundle up the "usual" time dependence into the state vector, so we can focus on what's different.

    If the Hamiltonian were just [itex]H_{0,S}[/itex], we know the state would evolve with the time-dependent phase factor [itex]e^{iH_{0,S}t/\hbar}[/itex]. The point of the interaction picture is to see how adding [itex]H_{I,S}[/itex] to the Hamiltonian changes the situation. So instead of using [itex]|\psi_S(t)\rangle[/itex] as our point of comparison, we compare to what [itex]|\psi_S(t)\rangle[/itex] would have been if it had just evolved according to [itex]H_{0,S}[/itex] only.
     
  4. Jul 4, 2011 #3
    1. When do we compare what |[itex]\psi[/itex](t)>[itex]_{S}[/itex] would have been if it had just evolved according to H[itex]_{O,S}[/itex] only?

    I will use the equations in David Tong's notes to make things easier:
    http://www.damtp.cam.ac.uk/user/tong/qft/three.pdf

    So if, in eq 3.9, we use H = H_O + H_int, then the difference between |[itex]\psi[/itex]>[itex]_{H}[/itex] and |[itex]\psi[/itex](t)>[itex]_{I}[/itex] is what we're really interested in? If so, why do none of the textbooks I've looked at ever talk about that? They just proceed with Dyson's formula with |[itex]\psi[/itex](t)>[itex]_{I}[/itex], but not the difference between it and |[itex]\psi[/itex](t)>[itex]_{H}[/itex]


    2. If we only care what |[itex]\psi[/itex](t)>[itex]_{S}[/itex] would have been if it had just evolved according to H[itex]_{O,S}[/itex] only, then why is it that in eq. 3.13, he includes the H[itex]_{int}[/itex] part in H[itex]_{S}[/itex]? and does NOT include it in the
    |[itex]\psi[/itex]>[itex]_{S}[/itex]?

    3. So even though it is called the 'INTERACTION' picture, we don't include H[itex]_{int}[/itex] for H? We only look at H[itex]_{O}[/itex]?


    Regards,

    creepypasta13
     
    Last edited: Jul 4, 2011
  5. Jul 4, 2011 #4

    SpectraCat

    User Avatar
    Science Advisor

    You seem to be missing the point that the equations (3.11 in Tong's notes) that you are focusing on are just used to transform the states and operators into the proper format for the interaction picture. The actual calculations are then done using the techniques described further on in those notes.

    Another way of thinking about it is that the interaction picture is devoted to focusing on the effects of the interaction Hamiltonian, so you just roll the effects due to the zero-order Hamiltonian into the states and operators using the formalism described in 3.11.
     
    Last edited: Jul 4, 2011
  6. Jul 4, 2011 #5
    I know that eqs 3.11 are being transformed to the format in the interaction picture (IP). I still don't understand. If the IP is focusing on the effects of the H[itex]_{int}[/itex], why is there the factor e[itex]^{iH_{O}t/h}[/itex] instead of, e[itex]^{iH_{Int}t/h}[/itex] ? where H[itex]_{int}[/itex] is the interacting Hamiltonian in the Schrodinger picture (the H[itex]_{1,S}[/itex] in my 1st post)
     
  7. Jul 4, 2011 #6

    SpectraCat

    User Avatar
    Science Advisor

    Think about what you are doing in 3.11 .. you are simply generating the appropriate form of the state vector (and operator) in the interaction picture. Now, think about the Schrodinger picture ... do the state vectors there take the Hamiltonian into account? No ... they are what is acted upon by the Hamiltonian. It is no different in the interaction picture, once you have "formatted" the state vectors and operators according to 3.11.

    That's what I said in my last post .. you are "getting rid of" the zero-order Hamiltonian by rolling it into the state vector. That process isn't *supposed* to take into account the interaction Hamiltonian .. it is just setting up the problem so that only the interaction Hamiltonian need be considered in later steps.

    I can't really explain it any more clearly than that. You are really just getting hung up on a detail .. I think it will become clear to you if you press on through the notes.
     
    Last edited: Jul 4, 2011
  8. Jul 4, 2011 #7
    I think I am starting to understand this a little better. For instance, if I substitute
    |[itex]\psi[/itex]>[itex]_{S}[/itex]
    from eq 3.9 to 3.11, we get
    |[itex]\psi[/itex](t)>[itex]_{I}[/itex] = e[itex]^{-i(H-H_{O})t}[/itex] | [itex]\psi[/itex](t)>[itex]_{H}[/itex]

    which makes sense considering that we want |[itex]\psi[/itex](t)>[itex]_{I}[/itex] to exclude the H[itex]_{O}[/itex] from the total hamiltonian in the Heisenberg picture H[itex]_{O}[/itex]+H[itex]_{I}[/itex] , right?

    Similarly, I obtained (H[itex]_{int}[/itex])[itex]_{I}[/itex] = e[itex]^{-it(H-H_{O})}[/itex]H[itex]_{int, H}[/itex]e[itex]^{it(H-H_{O})}[/itex]. This is also time dependent, right? I thought (H[itex]_{int}[/itex])[itex]_{I}[/itex] is supposed to be time dependent in the Interaction picture, so then why are the minus and plus signs in the exponential factors flipped? (In general, when making H time dependent by converting it from the Schrodinger to the Heisenberg representation, we should have e[itex]^{+H}He^{-H}[/itex])

    So we say that the Interaction picture is a "HYBRID" of the free and perturbed hamiltonians when in fact we only care about how the state in the "Interaction picture" only gets affected by the perturbed hamiltonian, and not the total hamiltonian. It is called "HYBRID" because the time dependence of states is determined by H_O but the time dependence of operators is determined by H_int, right?
     
    Last edited: Jul 4, 2011
  9. Jul 5, 2011 #8
    In the Schrodinger picture, all operators are still, and the state vector moves.

    In the Heisenberg picture, the operators move, and the state vector stays still.

    The interaction picture is conceptually closer to the Heisenberg picture --- if it wasn't for the perturbation, then the state vector would be stationary. We can get this by applying backwards [tex]H_0[/tex] to the Schrodinger state.
     
  10. Jul 5, 2011 #9
    you mean eq 3.11? Or do you mean applying the exp(-H_o) factor to both sides of eq 3.11 to get the |[itex]\psi[/itex](t)>[itex]_{S}[/itex] by itself on the right hand side?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook