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Here's the methodology: We first write out the equation of motion for $\phi(x,t)$ in the Heisenberg picture. This is identical to classical field equation. Now, we solve the equations perturbatively, pretending that we're perturbatively solving the interacting Euler Lagrange equations from classical field theory. This perturbation series will involve only tree diagrams.

After we have the perturbation series, we can re-interpret $a_p$ and $a^{\dagger}_p$ appearing in the series as operators satisfying $[a, a^{\dagger}]=\delta$. The correspondence between Poisson brackets and commutators guarantees that we can re-interpret the variables as operators.

After we have the full $\phi(x,t)$ solution in the interacting theory, we can plug it into the LSZ formula to get interaction amplitudes.

This method is computationally complicated, which is probably why we use the full interaction picture for QFT. But nevertheless, doesn't it show that QFT involves only tree diagrams identical to CFT, provided that we work in the Heisenberg picture? Or is it possible that this method is wrong?

Ryder Rude
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    "After we have the perturbation series, we can re-interpret and † appearing in the series as operators satisfying [,†]=" --- Are you sure? – Andrew Oct 05 '22 at 14:54
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    @Andrew My reasoning is that : The classical equation of motion is $\frac {d\phi}{dt}=(\phi, H)$. ( ) is the Poisson bracket. The quantum equation of motion is the same thing, but with the commutator. The same perturbation series should solve both equations. – Ryder Rude Oct 05 '22 at 14:58
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    @Andrew It's because $(\phi, H)$ from the classical theory and $[\phi, H]$ from the Quantum theory have the same functional form. So the same perturbation series should solve both differential equations. – Ryder Rude Oct 05 '22 at 15:01
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    Yes but how do you define $a$ for an interacting theory? (Even classically) – Andrew Oct 05 '22 at 15:35
  • @Andrew All the $a$'s and $a$ daggers that appear in the perturbation series of the classical field theory are the free $a$'s and $a$ daggers. This is because the perturbation series is expressible solely in terms of $\phi_0$ where $\phi_0$ is the free field solution. – Ryder Rude Oct 06 '22 at 00:32
  • @Andrew Schwartz's book also mentions time dependent $a$'s and $a$ daggers for interacting theories. To define them, you just take the free field solution and write the $a$'s as $a(t)$. This time dependence makes sure that you can express arbitrary solutions $\phi(x,t)$ in terms of these time dependent creation and annihilation operators. But I don't think these time dependent creation operators are relevant for my post. This is because the perturbation series only has the free solution $\phi_0$ – Ryder Rude Oct 06 '22 at 00:35
  • It might be true that if you define $a$ and $a^\dagger$ as a Fourier transform involving the interacting fields $\phi$ and $\pi$, then $a$ and $a^\dagger$ will obey the usual raising and lower commutation relations with each other, but they won't obey the usual raising/lowering commutation relationship with the full Hamiltonian. So you won't be able to interpret them as creating or annihilating particles. At some point this is going to cause an issue that you need to address in your formalism, I'm not exactly sure where, but I think one goal of the interaction picture is to avoid this problem. – Andrew Oct 06 '22 at 01:23
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    In other words, I think $a^\dagger$, as you define it in the Heisenberg picture, acting on the vacuum state, will probably end up create a superposition of a single particle state and various multiparticle states, and you'll need to account for this in your analogue of the LSZ formula. I don't see exactly how the details will work, but this is an area I'd be very sure I understood carefully if I was pursuing your program. – Andrew Oct 06 '22 at 01:27
  • @Andrew Yeah, but these time dependent creation-annihilation operators don't seem relevant for my method. The perturbation series only involves the free field solution $\phi_0$. So the creation and annihilation operators in the series are just the usual ones from free theory – Ryder Rude Oct 06 '22 at 02:06
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    There is no such thing as "the usual ones from the free theory" in an interacting theory. There are operators $a$ and $a^\dagger$ in the free theory with a list of properties. You can choose to define $a$ and $a^\dagger$ however you like in the interacting theory, but they will only have a subset of the list of properties that they have in the free theory. If you define them the way you said in the comments and in your answer, they won't do what you seem to think they will do, because their commutator with the Hamiltonian is different than in the free theory. – Andrew Oct 06 '22 at 04:36
  • @Andrew Yes, but the method does not rely on the creation operators creating eigenstates of the full Hamiltonian. All we want is the solution to $\frac{d\phi}{dt}=[\phi,H]$. The solution to this is the perturbation series carried over from the classical theory, This gives you $\phi (x,t)$. But when we later plug this into the LSZ formula, we have to act this $\phi (x,t)$ on the vaccuum. This would require the $a^{\dagger}$'s in the perturbations series to have the same vaccuum as the vaccuum in the LSZ formula. This needs to be justified. Is this your concern? – Ryder Rude Oct 06 '22 at 05:02
  • Let us continue this discussion in chat. – Ryder Rude Oct 06 '22 at 05:04

1 Answers1

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  1. In the path integral formalism, the loop-expansion is the same as the $\hbar$-expansion if the action $S$ does not depend explicitly on $\hbar$, cf. e.g. my Phys.SE answer here. In other words, the classical theory ($\hbar=0$) corresponds to tree-level in the path integral.

  2. The operator formalism (and in particular the interaction picture) is formally equivalent to the path integral formalism. The consistent co-existence of the Heisenberg equations of motion and the contact-terms in the Schwinger-Dyson (SD) equations follows from different time-orderings $T$ and $T_{\rm cov}$, cf. e.g. my Phys.SE answers here & here. Therefore the operator formalism contains loop-diagram contributions of the path integral formalism.

  3. Now the notion of tree-level depends on the formalism and context. Examples:

    • (A sum of all possible) connected diagrams is (a sum of all possible) trees of connected propagators and (amputated) 1PI vertices, cf. e.g my Phys.SE answer here.

    • One can argue that the perturbation theory with the Lippmann-Schwinger equation/Born series can be viewed as series of diagrams with the topology of trees. The first term, the Born-approximation, usually represents the classical theory.

    In the above 2 examples, the trees contain the full QFT, cf. OP's title question.

  4. Note that the $\hbar$-dependence in itself is subtle. Example: It makes a difference if in the free Klein-Gordon equation $$ (\pm\Box-\mu^2)\phi(x)~=~0, $$ we treat $\mu\equiv\frac{mc}{\hbar}$ or $m$ as $\hbar$-independent. There are similar issues with the interaction terms.

  5. Related post by OP: "Tree diagrams appear in classical field theory and non-relativistic QM because of the absence of the delta in the Schwinger-Dyson equation"

Qmechanic
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