
#1
Jul311, 06:56 PM

P: 375

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] 



#2
Jul411, 07:51 AM

P: 117

If the Hamiltonian were just [itex]H_{0,S}[/itex], we know the state would evolve with the timedependent 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. 



#3
Jul411, 10:56 AM

P: 375

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 



#4
Jul411, 12:38 PM

Sci Advisor
P: 1,395

Interaction picture
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 zeroorder Hamiltonian into the states and operators using the formalism described in 3.11. 



#5
Jul411, 01:46 PM

P: 375





#6
Jul411, 02:52 PM

Sci Advisor
P: 1,395

That's what I said in my last post .. you are "getting rid of" the zeroorder 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. 



#7
Jul411, 04:41 PM

P: 375

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(HH_{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(HH_{O})}[/itex]H[itex]_{int, H}[/itex]e[itex]^{it(HH_{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? 



#8
Jul511, 09:35 AM

P: 981

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. 



#9
Jul511, 10:30 AM

P: 375




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