QFT question, scalar field and so on

Question

$\newcommand{\bbraket}[3]{\langle #1 | #2 | #3 \rangle} \newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1 |}$ I have a problem with a proof. I'm studying the two point correlation function of the interacting theory, $$\bbraket{\Omega}{\phi(x)\phi(y)}{\Omega}$$

where $\Omega$ is the ground state and $\phi$ is the scalar field. Now they tell me that I can insert a completeness relation $1$ between the two fields operator, written in the form

$$1 = \ket{\Omega}\bra{\Omega} + \sum_{\lambda} \int \dfrac{d^3 p}{2\pi}^3 \dfrac{1}{2 E_p(\lambda)} \ket{\lambda_p}\bra{\lambda_p}$$

Now first question: why the states $|\lambda_p\rangle$ should be eigenstates of the interacting Hamiltonian? Then, by inserting this relation and by making the calculation I get a more suitable form. The problem comes no, because I have to rewrite the term

$$\bbraket{\Omega}{\phi(x)}{\lambda_p}$$ as $$\bbraket{\Omega}{\phi(0)}{\lambda_0}$$ by knowing those things: 1) I can make use of: $\phi(x) = e^{iPx}\phi(0)e^{-iPx}$

2) I can insert $U^{-1}U$ to the left and to the right of the above relation where $U$ is a unitary boost operator.

I have to compute $$\bbraket{\Omega}{U^{-1} U e^{iPx}\phi(0) e^{-iPx}U^{-1}U}{\lambda_p}$$

knowing also that $U |\ket{\lambda_p} = \ket{\lambda_0}$

Answer 1

This exact calculation is done on pages 212-213 of Peskin and Schroeder. Now your questions.

The states $|\lambda_p\rangle$ are defined to be eigenstates of the interacting Momentum operator with eigenvalue $\vec{p}$. But from the Poincaré algebra we have $[H,P]=0$, so $|\lambda_p\rangle$ is also an eigenstate of the interacting Hamiltonian.

As for your own answer, there are some things worth considering.

Your point 2 is correct. Under a Lorentz transformation that takes $x\rightarrow x'=\Lambda x$ a scalar field transforms as $\phi(x)\rightarrow \phi'(x')=\phi(x)$. When the field is promoted to an operator, the Lorentz transformations is implemented by a unitary operator $U$.

$\phi(x)\rightarrow \phi'(x')=U^{-1}\phi(x')U=U^{-1}\phi(\Lambda x)U =\phi(x)$. This holds for any Lorentz transformation. Interchange $U$ and $U^{-1}$ to get the transformation in the other direction, i.e. $x\rightarrow \Lambda^{-1}x$.

Setting $x=0$ then gives your result $U^{-1}\phi(0)U =\phi(0)$.

The rest looks fine given that you've defined $U$ as the transformation that take you from $\vec{p}$ to $0$.

Answer 2

I minded as follows:

$$ \begin{array}{rll} \langle\Omega|\phi(x)|\lambda_p\rangle & = \langle\Omega| e^{iPx}\phi(0) e^{-iPx}|\lambda_p\rangle \\ & = \langle\Omega|\phi(0)e^{-iPx}|\lambda_p\rangle \\ & = \langle\Omega|\phi(0)e^{-ipx}|\lambda_p\rangle \\ & = \langle\Omega|U^{-1}U\phi(0)U^{-1}U|\lambda_p\rangle e^{-ipx} \\ & = \langle\Omega|U\phi(0)U^{-1}|\lambda_0\rangle e^{-ipx} \\ & = \langle\Omega|\phi(0)|\lambda_0\rangle e^{-ipx} \end{array} $$

In which I have used those relations:

1) $<\Omega|U^{-1} = <\Omega|$

2) $U^{-1}\phi(0)U = \phi(0)$ (but I don't know if it holds)

3) $U|\lambda_p> = |\lambda_0>$

4) $<\Omega|e^{iPx} = <\Omega|$

5) $e^{-iPx}|\lambda_p> = e^{-ipx}|\lambda_p>$