What I want to do: $\newcommand{\ket}[1]{\left|#1\right\rangle}$ $\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\braket}[1]{\left\langle#1\right\rangle}$

The Gell-Mann Low Theorem tells us that we can get from non-interacting eigenstates to interacting eigenstates by time-evolving in a system where the interaction is turned off adiabatically at $t=\pm\infty$. $$\ket{\psi^\pm}=\frac{\hat{U}_{I}(0, \pm\infty)\ket{\psi_0}}{\bra{\psi_0}\hat{U}_{I}(0, \pm\infty)\ket{\psi_0}}$$

I'm trying to understand how one goes from that expression to the way it's expressed in QFT in terms of Green's functions (at least in Lancaster and Blundell, Chapter 21):

$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}\mathcal{T}\phi_{1I}(x_1)\cdots\phi_{nI}(x_n)S\ket{0}}{\braket{0|S|0}}$$

(where $I$ indicates interaction picture).

Here's what I follow:

If you start from the definition of the Green's function:

$$G^{(n)}(x_1,\cdots, x_n)=\bra{\Omega}\mathcal{T}\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\ket{\Omega}$$

(where $H$ indicates Heisenberg picture), and plug in the Gell-Mann Low Theorem, you get:

$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}U(-\infty,0)\mathcal{T}\left[\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\right]U(0,\infty)\ket{0}}{\braket{0|U(-\infty,0)|0}\braket{0|U(0,\infty)|0}}$$

Here's where I'm lost:

To get from there to the target expression, Kleinert just equates both the numerator and denominator separately, but I don't understand either. Here I'm referencing equations 10.9 to 10.12 of the Chapter 10.

  1. The numerator:

    He uses $\mathcal{T}$ to pull out all of the evolution operators which relate the Heisenberg picture to the interaction picture reorder them, and cancel them together, then bring the two Moller operators together to form the S-matrix.

    • Can you use $\mathcal{T}$ to switch between the Heisenberg/ interaction pictures like that in general?
    • How did the Moller operators move--they're not even inside $\mathcal{T}$!
  2. The denominator:

    He says that $\braket{0|U(-\infty,0)|0}\braket{0|U(0,\infty)|0}=\braket{0|U(-\infty,0)U(0,\infty)|0}$. This is defended in Eq 9.95 of Chapter 9 as a result of adiabaticity. But adiabatic evolution doesn't give you the state $\ket{0}$ at $t=0$, so why should you be able to project out that state there without losing amplitude?

Does anyone know what's going on in this "trivial" proof, or are there other discussions of this theorem that make the issues more clear? Thanks for any insight!


1 Answer 1


$\newcommand{\ket}[1]{\left|#1\right\rangle}$ $\newcommand{\bra}[1]{\left\langle#1\right|}$ $\newcommand{\braket}[1]{\left\langle#1\right\rangle}$ After reading from Yeh's notes on Advanced Condensed Matter Theory, Section II.9, I've decided that Kleinert was right about the numerator, but the denominator actually comes from normalization.

So, you start from the definition of the Green's function. $$G^{(n)}(x_1,\cdots, x_n)=\bra{\Omega}\mathcal{T}\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\ket{\Omega}$$

Tangent...regarding my question on the use of $\mathcal{T}$ to change pictures:

Note that $\mathcal{T}$ in the expression above is implicitly defined to time-order all of the visible symbols inside of its grasp, that is, it operates symbolically on the $\phi_{iH}(t)$. You can't use it to "cheat" like this sketchy misproof: $$\phi_I(t)=\mathcal{T}\left[\phi_I(t)\right]\\=\mathcal{T}\left[e^{i\int_0^t dt'H_0(t')}\phi_Se^{-i\int_0^t dt'H_0(t')}\right]\\\neq \mathcal{T}\left[e^{i\int_0^t dt'H_0(t')}e^{-i\int_0^t dt'H_0(t')}\phi_S\right]\\=\mathcal{T}[\phi_S]=\phi_S$$ The middle "equality" actually fails because the mistaken proofwriter accidentally changed definitions of $\mathcal{T}$ without noticing. The first line works assuming $\mathcal{T}$ is defined to rearrange the set of explicitly visible operators (in this case, just trivially $\phi_I(t)$). But the middle equality would only be possible if the $\mathcal{T}$ had been intended to order all the embedded $H_0(t)$ as well. Since the $H_0(t)$ in line 2 do not happen to be all ordered already, one cannot make this switch of definitions for $\mathcal{T}$ without changing the value of the expression. One should thus be careful to be consistent on how $\mathcal{T}$ is defined to avoid contradictions. It doesn't operate on the value inside the brackets, it operates in a specific manner on the symbols.

Back to the flow of the proof

We want to substitute in our Gell-Mann and Low expression into the definition of the Green's function, but notice that $\ket{\Omega}$ is not just $\ket{\Psi^{\pm}}$... because $\ket{\Psi^{\pm}}$ is not normalized. So really

$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{\Psi^+}\mathcal{T}\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\ket{\Psi^-}}{\braket{\Psi^+|\Psi^-}}$$

(where I'm using the fact that actually $\ket{\Psi^+}=\ket{\Psi^-}$ and are just different expressions for the same ket.) Expanding, we find

$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}U_I(\infty,0)\mathcal{T}\left[\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\right]U_I(0,-\infty)\ket{0}}{\braket{0|U_I(\infty,0)|0}\braket{0|U_I(0,-\infty)|0}}\\\quad \times 1/\frac{\braket{0|U_I(\infty,0)U_I(0,-\infty)|0}}{\braket{0|U_I(\infty,0)|0}\braket{0|U_I(-0,\infty)|0}}$$ The unpaired Moller operators in the denominator cancel: $$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}U_I(\infty,0)\mathcal{T}\left[\phi_{1H}(x_1)\cdots\phi_{nH}(x_n)\right]U_I(0,-\infty)\ket{0}}{\braket{0|U_I(\infty,-\infty)|0}}$$

What we CAN do with $\mathcal{T}$

Now we just have to clean up the numerator. Assume WLOG that, for a given $\{x_i\}$, those Heisenberg operators are time ordered when written from 1 to $n$ as above. Then write them out in terms of the interaction operators using

$$O_H(t)=U^\dagger(t,0)U_0(t,0)O_I(t)U_0^\dagger(t,0)U(t,0)$$ and $$U_I(t, t')=U_0^{\dagger}(t, 0)U(t, t')U_0(t', 0)$$

(where subscripts indicate full, free, and interaction evolution operators) to reach

$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}\mathcal{T}\left[U(\infty,t_1)\phi_{1H}(x_1)U_I(t_1,t_2)\phi_{2H}(x_2)\cdots U_I(t_{n-1},t_n)\phi_{nH}(x_n)U(t_n,-\infty)\right]\ket{0}}{\braket{0|U(\infty,-\infty)|0}}$$

I played some tricks here; let me explain. Formally, the $\mathcal{T}$ was only assumed at the start to order the $\phi_{iH}(t)$ around as unbreakable chunks (as discussed above). But since each of the individual evolution operators contains only the already time-ordered $H_I(t)$ from $t_i$ to $t_{i+1}$, then, if the $\phi_{iH}(t)$ are in order, that means all of the various implicit $H_I(t)$ in the entire expression are actually in time-order for us. So now we can actually treat the $\mathcal{T}$ as ordering the $\phi_{iH}(t)$ and the $H_I(t)$ without changing the value of our expression. Then we can also bring the outermost evolution operators inside the $\mathcal{T}$ since they are already at earlier/later times than all other operators.

Now, having played all those games with the definition of the time-ordering operator to show that we can have it apply to all of the various symbols inside the brakets for this expression...we are free to rearrange all those symbols ourselves. Let's put all the evolution operators together to the right, and voila

$$G^{(n)}(x_1,\cdots, x_n)=\frac{\bra{0}\mathcal{T}\left[\phi_{1H}(x_1)\cdots \phi_{nH}(x_n)S\right]\ket{0}}{\braket{0|S|0}}$$

where the $\mathcal{T}$ is understood to order together all of the $\phi_I(t)$ and the $H_I(t)$ inside the Dyson series for the $S$-matrix (even though the $H_I(t)$ inside the $S$-matrix are, of course, already time-ordered within themselves by the Dyson series!) Phew!

  • $\begingroup$ May 2022: Link is now re-directed to front page. $\endgroup$
    – Qmechanic
    Commented May 22, 2022 at 8:21

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