# Some confusions about interaction picture, in peskin

In peskin chap 4 on interaction field theory, he first introduced some basics about interaction picture(mostly in pg. 83~87) , it seems he assumed the Hamiltonians H=H_0+H_int in Schrodinger picture are all time-independent, because he used quite a lot of notations like exp(iHt), exp(iH_0t) and exp(iH_int t). So does it mean all the perturbation techniques developed later in the book only apply to time-independent interaction?
The specific example he talked about is K-G field with phi^4 interaction, which is indeed time-independent in schrodinger's picture(or not?). However, what happens if we want to calculate a time-dependent interaction? Maybe in practice all scattering processes can be described by time-independent H? If so, how to justify this?

The techniques developed by Peskin and Schroeder only apply to time-independent interactions. Time-dependent interactions require a more careful treatment of the time evolution operators. This is really the realm of non-equilibrium quantum field theory. The formalism that is being used here is the Schwinger-Keldysh formalism or simply the Keldysh formalism. In this formalism you can also are not restricted to force the final state to be the ground state -- which is a trick frequently used in high energy physics.

The Keldysh formalism is frequently applied in condensed matter systems, for instance in the theory of quantum transport. There are plenty of books and review articles out there that treat this subject.

I see, thanks.

Actually I'm curious, what kind of situation can give rise to a time-dependent interaction?

Actually I'm curious, what kind of situation can give rise to a time-dependent interaction?

Usually when the interaction is being driven by some "classical" source (laser/coherent pumping), and you don't want to include that system in your analysis at the same time. Essentially you can have couplings which change with time.

So if we include all particles involved in the interaction, it'll be time-independent? I guess this is because all fundamental interactions are time-independent?

So if we include all particles involved in the interaction, it'll be time-independent? I guess this is because all fundamental interactions are time-independent?

Exactly.

I'm still learning this stuff, so I could be wrong about this, but I'm pretty sure this formalism can handle time-dependent Hamiltonians just fine, can't it? P&S may have cheated a little bit during the derivation and written $$e^{-iH't}$$, but if you mentally replace all of those with $$e^{-i\int{H' dt}}$$, everything still works. By the time they actually get down to the Dyson series itself, they're using integrals, so it seems like the actual equation they're proving can handle time-dependent Hamiltonians correctly.

I've been slowly working my way through Dr. Coleman's QFT lecture videos, and the first model he studies once this equation is derived is $$H'=g\phi(x)\rho(x)$$, where $$\rho(x)$$ is an arbitrary spacetime function--a classical source. This Hamiltonian is easily solved using this equation, so it seems like it doesn't have any problem handling time-dependent terms in the equation. Is this correct, or did I miss something?

I'm still learning this stuff, so I could be wrong about this, but I'm pretty sure this formalism can handle time-dependent Hamiltonians just fine, can't it? P&S may have cheated a little bit during the derivation and written $$e^{-iH't}$$, but if you mentally replace all of those with $$e^{-i\int{H' dt}}$$, everything still works. By the time they actually get down to the Dyson series itself, they're using integrals, so it seems like the actual equation they're proving can handle time-dependent Hamiltonians correctly.

I've been slowly working my way through Dr. Coleman's QFT lecture videos, and the first model he studies once this equation is derived is $$H'=g\phi(x)\rho(x)$$, where $$\rho(x)$$ is an arbitrary spacetime function--a classical source. This Hamiltonian is easily solved using this equation, so it seems like it doesn't have any problem handling time-dependent terms in the equation. Is this correct, or did I miss something?

What you state about the time evolution operator is indeed correct -- it needs to be replaced by the integral version you mentioned.

But what's more subtle is the type of correlator that is being evalued. P&S only treat the ground state-to-ground state amplitudes. That is, they assume the system starts out in the ground state and also ends up in it. This is basically the assumption of the Gell-Mann Low Theorem (which they prove in chapter 4.2. Specifically, the identity 4.31 for the correlator).

This is a very powerful tool, but there are two problems with this. The first is that you do not always want the state to start out in the ground state. You might want to introduce temperature, for instance, which is very sensible if you're considering condensed matter systems or the early universe.

Second, you cannot justify that the system always ends up in the ground state. Especially when you're dealing with a time-dependent interaction. This can drive the system away from the ground state, and you can (partially) end up in some excited states as well. The problem is that the Gell-Mann Low theorem only applies to ground state-to-ground state correlators. It usese the fact that this state has the lowest energy, which can be used to "project out" the remaining contributions.

The Schwinger-Keldysh formalism resolves both issues. At the cost of some extra computational effort.

The first is that you do not always want the state to start out in the ground state.

I thought calculating something like $$\langle 0|a_{k1'}a_{k2'}U(\infty,-\infty)a^\dagger_{k1}a^\dagger_{k2}|0\rangle$$ involved starting with something other than the ground state (a 2-particle free-field state), and evolving to something other than the ground state (another 2-particle free-field state). Is this what you were referring to, or are you talking about something more complex? I don't know anything about condensed matter theory.

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I'm still learning this stuff, so I could be wrong about this, but I'm pretty sure this formalism can handle time-dependent Hamiltonians just fine, can't it? P&S may have cheated a little bit during the derivation and written $$e^{-iH't}$$, but if you mentally replace all of those with $$e^{-i\int{H' dt}}$$, everything still works. By the time they actually get down to the Dyson series itself, they're using integrals, so it seems like the actual equation they're proving can handle time-dependent Hamiltonians correctly.

I've been slowly working my way through Dr. Coleman's QFT lecture videos, and the first model he studies once this equation is derived is $$H'=g\phi(x)\rho(x)$$, where $$\rho(x)$$ is an arbitrary spacetime function--a classical source. This Hamiltonian is easily solved using this equation, so it seems like it doesn't have any problem handling time-dependent terms in the equation. Is this correct, or did I miss something?

No. What you've assumed is that the Hamiltonians at different times commute with each other. If they do not commute, then $$e^{-i\int{H' dt}}$$ does not solve Schrodinger's equation; the solution is a time-ordered exponential, which shares some formal similarities to the expression you wrote.

I thought calculating something like $$\langle 0|a_{k1'}a_{k2'}U(\infty,-\infty)a^\dagger_{k1}a^\dagger_{k2}|0\rangle$$ involved starting with something other than the ground state (a 2-particle free-field state), and evolving to something other than the ground state (another 2-particle free-field state). Is this what you were referring to, or are you talking about something more complex? I don't know anything about condensed matter theory.

The issue is that the two-particle state you made there is not the physical two-particle state of the interacting theory. Luckily, there is a formalism which can generically recover the correct scattering (see LSZ reduction theorem).

No. What you've assumed is that the Hamiltonians at different times commute with each other. If they do not commute, then $$e^{-i\int{H' dt}}$$ does not solve Schrodinger's equation; the solution is a time-ordered exponential, which shares some formal similarities to the expression you wrote.

Oh, sorry, that was me being lazy. $$T(e^{-i\int{H' dt}})$$ is what I meant to say. But my point was just that when they finally get to the Dyson series, they're using the correct expression for the time-evolution operator, and that even though they weren't using it at a few points along the way of its derivation, you could mentally substitute the correct expression in, and the logic of their proof would still hold.

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I don't think you can arrive at the equation $$i\frac{{\partial {\Psi _I}}}{{\partial t}} = {H_I}{\Psi _I}$$ in the first place, unless you assume H_0 and H_int are time independent in Schrodinger's picture.

The issue is that the two-particle state you made there is not the physical two-particle state of the interacting theory. Luckily, there is a formalism which can generically recover the correct scattering (see LSZ reduction theorem).

So what kind of physical experiment does the usual method describe? I mean the simplest perturbation method developed at the very beginning of Chap 4, peskin. I know a two-point correlation function can be interpreted as the amplitude of a localized particle traveling to another space-time point, but this doesn't look like a typical scattering experiment in reality, does it?

The issue is that the two-particle state you made there is not the physical two-particle state of the interacting theory. Luckily, there is a formalism which can generically recover the correct scattering (see LSZ reduction theorem).

How about using $$a_p^\dag |\Omega >$$, where $$|\Omega >$$ is the interacting field vacuum with $$|\Omega > = U({t_0}, - \infty )|0 >$$ (omitting some overall coefficient), will this simple substitution work?

How about using $$a_p^\dag |\Omega >$$, where $$|\Omega >$$ is the interacting field vacuum with $$|\Omega > = U({t_0}, - \infty )|0 >$$ (omitting some overall coefficient), will this simple substitution work?

Almost. The trick is to notice that the ground state has, by definition, the lowest energy. So if you propagate the free vacuum with a slight imaginary time component, you will get the full ground state (assuming that they have any overlap... this is not true for theories with symmetry-broken ground states e.g. QCD). Indeed this is roughly how the LSZ reduction formula starts. This discussion is in P&S, if memory serves...