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Reading different books, I've come upon two apparently different definitions of the $n$-point Green's function.

For simplicity, let's consider a real scalar field $\hat{\phi}(x)$ (in the Heisenberg picture) with an interaction term in the Hamiltonian $$\hat{H}=\hat{H}_0+\hat{H}_{\text{int}}\tag{1}\label{1}.$$ In ref.1 and ref. 4 we have the definition (cf. equation $(4.10)$) $$G^{(n)}(x_1,... x_n):=\langle\Omega\rvert T\hat{\phi}(x_1)\hat{\phi}(x_2)...\hat{\phi}(x_n)\lvert\Omega\rangle.\tag{G}\label{G}$$ where

We introduce the notation $\lvert\Omega\rangle$ to denote the ground state of the interacting theory, which is generally different from $\lvert 0\rangle$, the ground state of the free theory.

After some steps, ref.1 (and similarly ref. 4) finally derives the formula expressing \eqref{G} in terms of the free vacuum state and the interaction picture fields $$\langle\Omega\rvert T\hat{\phi}(x_1)\hat{\phi}(x_2)...\hat{\phi}(x_n)\lvert\Omega\rangle=\lim_{t\to(1-i\varepsilon)\infty}\frac{\langle0\lvert T\hat{\phi}_I(x_1)\hat{\phi}_I(x_2)...\hat{\phi}_I(x_n)\exp\{-i\int_{-t}^t H_I(t')dt'\}\lvert0\rangle}{\langle0\lvert T\exp\{-i\int_{-t}^t H_I(t')dt'\} \lvert0\rangle}.\tag{A}\label{A}$$ Where they have used the equation $$\lvert\Omega\rangle=\lim_{t\to(1-i\varepsilon)\infty}\left(\frac{e^{-i\hat{H}t}\lvert0\rangle}{e^{-iE_0t}\langle\Omega\lvert 0\rangle}\right)$$ along with the relation between the interaction picture and the Heisenberg picture scalar field. On the other hand, ref. 2 and ref. 3 define the $n$-point Green's function as (cf. equation $(5.55)$) $$G^{(n)}(x_1,... x_n):=\langle0\rvert T\hat{\phi}(x_1)\hat{\phi}(x_2)...\hat{\phi}(x_n)\lvert0\rangle.\tag{G'}\label{G'}$$ where we have used the same notation. Note that for $n=2$, both\eqref{G} and \eqref{G'} reduce to the Feynman propagator when $\hat{H}_\text{int}=0$. However, here we do have an interaction and the two definitions are clearly different. Now, ref. 2 and ref.3 derive the formula $$\langle0\rvert T\hat{\phi}(x_1)\hat{\phi}(x_2)...\hat{\phi}(x_n)\lvert0\rangle=\lim_{t\to\infty}\frac{\langle0\lvert T \hat{\phi}_I(x_1)\hat{\phi}_I(x_2)...\hat{\phi}_I(x_n)\exp\{-i\int_{-t}^t H_I(t')dt'\}\lvert0\rangle}{\langle0\lvert T\exp\{-i\int_{-t}^t H_I(t')dt'\} \lvert0\rangle}.\tag{B}\label{B}$$ Aside from the small imaginary part in the limit, the RHS of \eqref{A} and \eqref{B} are the same, althought the LHS, resp. \eqref{G} and \eqref{G'} are different. Again, in this case we've used the relation between the interaction picture and the Heisenberg picture scalar field and the following assumption was made

Physically it is clear that, if the vacuum state is stable, applying to it the evolution operator $\hat{U}(+\infty,-\infty)$ we still find the vacuum. Recall however that in quantum mechanics state vectors that differ by a phase still represent the same physical state. Therefore we will have in general $$\hat{U}(+\infty,-\infty)\lvert 0\rangle=e^{i\alpha}\lvert 0\rangle\tag{5.65}$$ with $\alpha$ a phase.

Why do we get the same result with two different definitions? Is there any reason to prefer either definition or to justify it?


References:

  1. M.Peskin & D. Schroeder, An Introduction to Quantum Field Theory, 1995. Section 4.2.
  2. M. Maggiore, A Modern Introduction to Quantum Field Theory, 2005. Section 5.3.
  3. C. Itzykson& J. Zuber, Quantum Field Theory, 1980. Section 6-1-1.
  4. M. D. Schwartz, Quantum Field Theory and the Standard Model, 2014. Chapters 6-7.
  5. M. Srednicki, Quantum Field Theory, 2007. Problem 9.5.
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    $\begingroup$ Related. $\endgroup$ Commented Apr 19, 2023 at 21:05
  • $\begingroup$ How does that related question does not answer this one? $\endgroup$
    – Mauricio
    Commented Apr 19, 2023 at 22:04

1 Answer 1

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  1. No argument we could make here can ever hold up to rigorous scrutiny. Haag's theorem tells us there can be no unitary equivalence that maps free fields to interacting fields. Hence the core idea of the interaction picture that there is some operator $U$ such that $\phi = U^\dagger \phi_I U$ and $\pi = U^\dagger \pi_I U$ where $\phi_I$ and $\pi_I$ are the canonical operators of a free field theory and $\phi$ and $\pi$ are the canonical operators of an interacting theory, is a lie. Therefore, all arguments that try to establish some relation between $\phi$ and $\phi_I$ on the basis of the interaction picture must be flawed somewhere if we scrutinize them closely enough.

    In both cases the most elementary manifestation of this fact will be that the denominator $\langle 0\vert T\mathrm{e}^{-\mathrm{i}\int H_I}\vert 0\rangle$ - which we treat in both approaches as a clearly finite complex number - will turn out to be divergent, and not a finite number until after renormalization. So both approaches are manipulating ill-defined quantities here and all we can hope for is some sort of physical explanation, but not an actual mathematical derivation.

  2. The difference between $\lvert \Omega\rangle$ and $\lvert 0\rangle$ can be stated easily but dissolves into vagueness when we attempt to get a consistent picture. The reason we even want to compute an expectation value of the form $\langle T\prod_i\phi(x_i)\rangle$ is that we - in one way or another - have convinced ourselves that the LSZ formula holds that looks like this (with a bunch of irrelevant terms omitted): $$ \langle p_1,\dots,p_l;\text{in}\vert p_{l+1},\dots,p_n;\text{out}\rangle = \int \mathrm{e}^{\mathrm{i}(p_k x^k)} \left(\prod_i (\Box_i + m^2)\right)\langle T\prod_i \phi(x^i)\rangle.$$

    The crucial question is: What do we think the $m$ that appears in that formula is?

    The approach in which there is an interacting vacuum $\lvert \Omega\rangle$ constructs the LSZ formula such that this $m$ is the lowest-lying pole of the full interacting propagator, and the expectation value is with respect to a vacuum state from which the asymptotically free "in-fields" and "out-fields" create the free states $\lvert p_i;\text{in}\rangle$ etc. such that these 1-particle states have mass $m$.

    But if you think that this $m$ is the mass $m_0$ that you have written into your Lagrangian, then this defines $\lvert \Omega\rangle$ to be a vacuum for the "in-fields" from which 1-particle states with the mass $m_0$ are created, which is the same as being a free vacuum $\lvert 0\rangle$ (somewhere when discussing what a vacuum even is you should have made an assumption that the vacuum is unique - you can't have two 1-particle states with the same mass whose only difference is that they'd yield different vacuum states when acted upon by the same annihilation operator). So in spirit the two approaches simply disagree about whether it is the free mass or the interacting mass that appears in the LSZ formula.

    Now, you might expect that surely this must matter at some point - but, alas, it does not: Once you realize the denominator of our expression to compute the expectation values is divergent, you must invent renormalization to deal with this, which effectively stops $m_0$ and $m$ from simultaneously being meaningful quantities - you have to make at least one of the two formally divergent, too, and the physical mass $m$ becomes a "renormalization parameter", i.e. an input to the theory rather than a prediction that could depend on a finite $m_0$.

  3. As already mentioned, a rigorously consistent approach to scattering theory in quantum field theory must avoid the fiction of the interaction picture. Such an approach exists, but rarely mentioned in standard texts due to its technicality - it was developed by Haag himself, and is known as Haag-Ruelle scattering theory (the canonical reference for this is Reed and Simon's Methods of modern mathematical physics, vol. III, Scattering Theory).

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