Following Peskin & Schroeder's Sec.7's notation, I would like to compute the matrix element

$$ \left<\lambda_\vec{p}| \phi(x)^2 |\Omega\right>\tag{1} $$

where $\langle\lambda_{\vec{p}}|$ is obtained by boosting the state $\langle\lambda_0|$ whose momentum eigenvalue is zero i.e. $\langle\lambda_0|\vec{P} = 0$ while its energy eigenvalue is denoted by $\langle\lambda_0|H = \langle\lambda_0|m_\lambda$. The state $|\Omega\rangle$ is the non-perturbative vacuum of the theory.

Now consider the following OPE $$ \phi(x)\phi(y) = \langle\Omega| T\{\phi(x)\phi(y)\} |\Omega\rangle - :\phi(x)^2:\tag{2} \label{ope} $$ where $:\mathcal{F}(\phi):$ is the usual normal ordering. If we sandwhich the above equation with $\langle\lambda_{\vec{p}}|$ and $|\Omega\rangle$, seemingly we would get $$ \left<\lambda_\vec{p}| \phi(x) \phi(y) |\Omega\right> = \sum_{\lambda'}D_F(x-y,\lambda') \langle\lambda_{\vec{p}}|\Omega\rangle |\left<\lambda_0'| \phi(0) |\Omega\right>|^2\tag{3} $$ where $$ D_F(x-y,\lambda')=\int \frac{d^4 q}{2\pi^4}\frac{e^{-iq\cdot(x-y)}}{q^2 - m_\lambda'^2+i\epsilon}\tag{4} $$ is the Feynman propagator. The normal order vanishes by definition (please let me know if there's a problem). Herein we run into a problem, if we send $x\rightarrow y$ first then the matrix element diverges. But if we set the vacuum decay rate $\langle\lambda_\vec{p}|\Omega\rangle = 0$, the matrix element vanishes.

On the other hand, if we write $\phi(x) = e^{-iP\cdot x} \phi(0) e^{iP\cdot x} $, then the matrix element would be $$\begin{align} & \;\;\;\;\;\left<\lambda_\vec{p}| \phi(x)^2 |\Omega\right> \\ &= \left<\lambda_\vec{p}| e^{-iP\cdot x} \phi(0) e^{iP\cdot x} e^{-iP\cdot x} \phi(0) e^{iP\cdot x} |\Omega\right> \\ &= e^{-ip\cdot x} \left<\lambda_\vec{p}| \phi(0)^2 |\Omega\right>\\ &=e^{-ip\cdot x} \left<\lambda_0| \phi(0)^2 |\Omega\right> \label{eq:derivation} \end{align}\tag{5}$$ which follows by the exact same procedure as 7.4 in P&S QFT textbook.

Evidently the results derived from the two procedures don't agree, I wonder if there's a way to reconcile them.

  • $\begingroup$ I can think of a counter argument for the first approach, the time order is not under control non-perturbatively, we simply cannot apply Wick’s theorem in a non-perturbative approach $\endgroup$
    – Mmmao
    Feb 19 at 19:49
  • $\begingroup$ Comments to the post (v2): There seem to be missing or implicitly written operator orderings in eqs. (1)-(3). E.g. $\phi(x)^2$ in eq. (1) is a mathematically ill-defined singular object. $\endgroup$
    – Qmechanic
    Feb 22 at 11:48
  • $\begingroup$ eqs(1) doesn need time ordering since the two operators are evaluated at the same point in space time, eqs(2) is implicit, eqs(3) should contain time ordering on the LHS within the braket but eventually we're sending $x$ to $y$ thus it shouldn't make a difference. $\endgroup$
    – Mmmao
    Feb 23 at 1:01

1 Answer 1


First, some remarks :

  • Wick's theorem only applies to free (and interaction picture) fields, hence it is fundamentally perturbative.

  • the normal ordered product $:\phi(x)\phi(y):$ contains a term with two creation operators, whose matrix elements between $\langle \lambda_p|$ and the vacuum need not vanish.

  • Lastly, if you find a divergent integral, it is probably best to regularize and renormalize the theory before making any kind of conclusion.

Now, if we assume that we have a fully renormalized non perturbative theory, what can we say ? We have a Hilbert space with a representation of the Poincaré group and an invariant vacuum state $|\Omega\rangle$, as well as an operator valued distribution $\phi(x)$.

On one hand, because $\phi(x)$ (even fully renormalized) is an operator-valued distribution (and not a function), the product $\phi(x)\phi(y)$ is well-defined (as a distribution, it needs to be smeared with a Schwartz function $f(x,y)$ to yield an operator). We can try to get a spectral representation for the matrix element $\langle \lambda_p|\phi(x) \phi(y) |\Omega\rangle$ following the same kind of calculations as in P&S section 7.1 : \begin{align} \langle \lambda_p|\phi(x) \phi(y) |\Omega\rangle &= \langle \lambda_p|\phi(x) |\Omega\rangle\langle \Omega|\phi(y) |\Omega\rangle + \sum_{\lambda'}\int\frac{\text d^3 p'}{(2\pi)^3 2E(p')}\langle \lambda_p|\phi(x) |\lambda'_{p'}\rangle\langle \lambda'_{p'}|\phi(y) |\Omega\rangle \\ &= e^{-ip\cdot x}\langle \lambda_0|\phi(0) |\Omega\rangle\langle \Omega|\phi(0) |\Omega\rangle + \sum_{\lambda'}\int\frac{\text d^3 p'}{(2\pi)^3 2E(p')}e^{-i(p-p')x}\langle \lambda_{p}|\phi(0) |\lambda'_{p'}\rangle e^{-iy\cdot p'}\langle \lambda'_{p'}|\phi(0) |\Omega\rangle \end{align} Because we can replace $\phi(x) \to \phi(x) - \langle \Omega|\phi(0)|\Omega\rangle$, we can assume that $\langle \Omega|\phi(0)|\Omega\rangle$ vanishes, so that : \begin{align} \langle \lambda_p|\phi(x) \phi(y) |\Omega\rangle &= e^{-ip\cdot x} \sum_{\lambda'}\int\frac{\text d^3 p'}{(2\pi)^3 2E(p')}e^{-i p'\cdot (y-x)}\langle \lambda_{p}|\phi(0) |\lambda'_{p'}\rangle\langle \lambda'_{0}|\phi(0) |\Omega\rangle \end{align} I don't think that there is much more we can do. (One idea would be to smear $\phi$ with a function whose Fourier transform is supported near the 1-particle mass shell, so that the operator $\phi_f = \int f(x)\phi(x) \text dx$ would only produce 1 particle states when acting on the vacuum). The limit as $y\to x$ has no reason to be well behaved.

On the other hand, we can regularize and renormalize the theory in such a way that there is an operator $\mathcal O_2(x)$ which we can interpret as a renormalized version of the ill-defined $\phi^2(x)$. In a free theory, for example, normal ordering would be enough so $\mathcal O_2(x) = :\phi(x)^2:$ (and we can write a well-defined Hamiltonian operator). Assuming the renormalization process was done properly, this operator should be a Lorenz scalar, and therefore the second calculation from OP works perfectly well : $$\langle \lambda_p|\mathcal O_2(x) |\Omega \rangle = e^{-ip\cdot x}\langle \lambda_0|\mathcal O_2(0) |\Omega \rangle$$

The fact that this does not match with the singular behavior of the matrix element $\langle \lambda_p|\phi(x)\phi(y)|\Omega\rangle$ is just a trace of the fact that the renormalized "$\phi^2$" operator is not defined as the limit of $\phi(x) \phi(y)$ as $y\to x$.

  • $\begingroup$ Thank you so much for the answer, that's exactly what I'm looking for. One minor question, for the final spectral representation, is there any thing we could say about the matrix elements $\langle\lambda_{p'}|\phi(0)|\lambda_{p}\rangle$ and $\langle\lambda_{p'}|\phi(0)|\Omega\rangle$? Since we could freely perform boost on $\phi$ $\endgroup$
    – Mmmao
    Feb 23 at 0:17
  • $\begingroup$ For the second one, you would see that it is equal to $\langle \lambda_0|\phi(0)|\Omega\rangle$. For the first one, you can also perform a boost but in general, you cannot bring both $p'$ and $p$ to $0$. $\endgroup$ Feb 26 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.