# QFT question, scalar field and so on


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 $U |\ket{\lambda_p} = \ket{\lambda_0}$

• you can choose any orthonormal basis you want, but if you have one which is made of eigenstates of the interacting hamiltonian makes calculations easier. as for the last question, the vacuum the only invariant state, i.e. $U|\Omega\rangle = |\Omega\rangle$ for any symmetry transformation $U$ (not just Lorentz). – Phoenix87 Jan 10 '15 at 15:54
• Huh, funny, at Caltech there's a homework problem that looks almost exactly like this for Mark Wise's class...you wouldn't happen to be in it, @Henry, would you? – Alex Nelson Jan 10 '15 at 16:17
• @AlexNelson lol! Well I think that surely it is a very useful exercise. However I'm from Italy and I don't know any Mark Wise! I tried to solve it by myself but a little help was needed.. – Henry Jan 10 '15 at 16:33
• @Phoenix87 thanks! Anyway I made a misprint! Check now the edit (in the end of the answer!!) – Henry Jan 10 '15 at 16:34
• just use the fact that $|\lambda_p\rangle$ is a $p$-eigenstate of the momentum operator to replace $e^{-iPx}$ by $e^{-ipx}$ (a c-number). $e^{iPx}$, again, leaves the vacuum invariant. – Phoenix87 Jan 10 '15 at 16:39

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.

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 $0$.

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

• This is correct. Relation 2) holds because $\phi$ is a scalar field and therefore invariant under Lorentz transformations, and because $x=0$ is invariant too. – Javier May 31 '16 at 23:21