1
$\begingroup$

Consider the field equation $(\partial^2-m_1^2)(\partial^2-m_2^2)\phi =0 $

Now let me make the field redefinition $\psi = (\partial^2-m_2^2) \phi$

The question is, will the S-matrix be invariant under this field redefinition?

The criterion for a valid field redefinition, as usually stated, is that it $\langle \Omega| \psi(x)|p\rangle$ is not zero, where $|p\rangle$ is the one particle state. This follows from the LSZ formula (see QMechanic's answer on equivalence theorem here): $$\left[ \prod_{i=1}^n \int \! d^4 x_i e^{ip_i\cdot x_i} \right] \left[ \prod_{j=1}^m \int \! d^4 y_j e^{-ik_j\cdot y_i} \right] \langle \Omega | T\left\{ \phi(x_1)\ldots \phi(x_n)\phi(y_1)\ldots \phi(y_m )\right\}|\Omega \rangle$$

$$~\sim~\left[ \prod_{i=1}^n \frac{i\langle \Omega |\phi(0)|\vec{\bf p}_i\rangle }{p_i^2-m_1^2+i\epsilon}\right] \left[ \prod_{j=1}^m \frac{i\langle \vec{\bf k}_j |\phi(0)|\Omega\rangle }{k_j^2-m_1^2+i\epsilon}\right] \langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$$

Redefining the fields on both sides leaves the $S$-matrix $\langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$ unchanged.

Now clearly the criterion $\langle \Omega| \psi(x)|p\rangle \neq 0$ holds.

On the other hand the original field theory has two particles of masses $m_1$ and $m_2$. This has a different particle spectrum and it would appear that the two theories are physically inequivalent.

Indeed, if one had started with the original equation one would have obtained a different set of one particle states.

So my questions are, 1. Are the $S$-matrices of the two theories still equivalent? 2. If not, what is the criterion for equivalence under field redefinition?

Is it that the Hilbert space of asymptotic states should remain unchanged?

$\endgroup$
2
  • 1
    $\begingroup$ Why do you believe that the original theory had masses $m_1$ and $m_2$? $\endgroup$
    – Prahar
    Jul 19 '17 at 17:11
  • $\begingroup$ @Prahar: From the poles of the propagator. $\endgroup$ Jul 19 '17 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.