How to show that $\hat {S}$-operator must be lorentz-invariant operator?

$$ |\Psi (t)\rangle = \hat {S} | \Psi (0) \rangle , \quad \hat {S} = \hat {T}e^{-i\int \hat {H}_{I}d^{4}x}. $$ I have read that this result follows from the unitarity of the Poincare group operator $U_{0}(\Lambda , a)$ and the covariance of S-matrix $S_{out, in} = \langle out| \hat {S}|in \rangle$, but I don't understand how do we conclude that from this follows that $U_{0}(\Lambda , a)\hat {S}U^{-1}_{0}(\Lambda , a) = \hat {S}$.

I'm not interesting in the derivation of lorentz-invariance of S-operator from causality principle at this moment.


Indeed, without assuming it from first principles as in Bogoliubov formulation, the invariance property of $S$ operator you mention holds when the interaction Lagrangian does not include derivatives of fields like in QED. That is a consequence of Dyson's expansion in interaction picture. When,as said above, the interaction Lagrangian does not include derivatives of fields, one has: $${\cal H}_I = -{\cal L}_I$$ so that $$S = \sum_{n=0}^{+\infty} \frac{i^n}{n!} \int\cdots \int T \hat{\cal L}_I(x_1)\cdots \hat{\cal L}_I(x_n) \:d^4x_1\cdots d^4x_n\:.$$ It is worth noticing that $\hat{\cal L}_I(x)$ includes only free field operators as we are dealing with the so-called interaction picture, so everything is explicitely known, commutation relations of field operators in particular. Since the Lagrangian functions are scalars, we have: $$U_\Lambda\hat{\cal L}_I(x) U^\dagger_\Lambda = \hat{\cal L}_I(\Lambda^{-1} x)\qquad (1)$$ Moreover, in view of free fields commutation relations one also has:$$[\hat{\cal L}_I(x), \hat{\cal L}_I(y)]=0 \qquad (2)$$ if $x$ and $y$ are spacelike separated. The fact that $S$ is invariant under the action of orthochronous Lorentz group is quite obvious $$U_\Lambda SU_\Lambda^\dagger = \sum_{n=0}^{+\infty} \frac{i^n}{n!} \int\cdots \int U_\Lambda T[ \hat{\cal L}_I(x_1)\cdots \hat{\cal L}_I(x_n)]U_\Lambda^\dagger \:d^4x_1\cdots d^4x_n$$ $$=\sum_{n=0}^{+\infty} \frac{i^n}{n!} \int\cdots \int T[U_\Lambda \hat{\cal L}_I(x_1)U_\Lambda^\dagger\cdots U_\Lambda \hat{\cal L}_I(x_n)U_\Lambda^\dagger] \:d^4x_1\cdots d^4x_n$$ $$=\sum_{n=0}^{+\infty} \frac{i^n}{n!} \int\cdots \int T[\hat{\cal L}_I(\Lambda^{-1} x_1)\cdots \hat{\cal L}_I(\Lambda^{-1}x_n)] \:d^4x_1\cdots d^4x_n$$ $$=\sum_{n=0}^{+\infty} \frac{i^n}{n!} \int\cdots \int T[\hat{\cal L}_I(x_1)\cdots \hat{\cal L}_I(x_n)] \:d^4x_1\cdots d^4x_n$$ due to the Lorentz invariance of the measure $d^4x$. The identity: $$U_\Lambda T[ \hat{\cal L}_I(x_1)\cdots \hat{\cal L}_I(x_n)]U_\Lambda^\dagger= T[U_\Lambda \hat{\cal L}_I(x_1)U_\Lambda^\dagger\cdots U_\Lambda \hat{\cal L}_I(x_n)U_\Lambda^\dagger]$$ is consequence of the definition of $T$-ordinator, (1), and (2) for spacelike separated arguments, considering all cases concerning the time order of $x_1,\ldots, x_n$ and the fact that $\Lambda$ does not change the temporal order for causally related arguments as it belongs to the orthochronous subgroup.

For more complicated theories the result is not obvious and it could be false in its elementary formulation based on canonical quantization, excluding the case of gauge theories, where it can be proved separately.

Regarding Weinberg's statement about Lorentz covariance of the $S$ matrix and Lorentz invariance of $S$ operator, if I understood well the definition, I think that it works like this.

Let us start from the full (interacting) theory. There are vectors $\Psi^\pm_{\{p_i\}}$ describing states which, at late time (respectively $t\to +\infty$ and $t\to -\infty$) evolve like free particle states with momenta $\{p_i\}$. The correspondingly associated free states, always evolving in accordance witht he free theory, are indicated by $\Phi_{\{p_i\}}$. The $S$-matrix is the matrix of elements: $$\langle \Psi^+_{\{q_i\}}|\Psi^-_{\{p_i\}}\rangle = \langle \Phi_{\{q_i\}}|S\Phi_{\{p_i\}}\rangle\:.\qquad(3)$$ In the RHS the $S$ operator takes place. In view of it, the scattering process is completely described in terms of free states.

To say that the $S$ matrix is Lorentz covariant should mean (as far as I understand): $$\langle \Psi^+_{\{\Lambda q_i\}}|\Psi^-_{\{\Lambda p_i\}}\rangle = \langle \Psi^+_{\{q_i\}}|\Psi^-_{\{p_i\}}\rangle\quad \forall \Lambda \in O(3,1)\uparrow\:, \forall \{\Lambda q_i\}\:, \{\Lambda p_i\}\:.$$ From (3), it immediately entails: $$ \langle \Phi_{\{\Lambda q_i\}}|S\Phi_{\{\Lambda p_i\}}\rangle = \langle \Phi_{\{q_i\}}|S\Phi_{\{p_i\}}\rangle\:.\qquad(4)$$ If $U_\Lambda$ is the unitary representation of $O(3,1)\uparrow$ on free states, so that $\Phi_{\{\Lambda p_i\}}= U_\Lambda \Phi_{\{p_i\}}$, we therefore have: $$\langle U_\Lambda \Phi_{\{ q_i\}}|S U_\Lambda\Phi_{\{p_i\}}\rangle = \langle \Phi_{\{q_i\}}|S\Phi_{\{p_i\}}\rangle\:,$$ that is $$\langle \Phi_{\{ q_i\}}|U^\dagger_\Lambda S U_\Lambda\Phi_{\{p_i\}}\rangle = \langle \Phi_{\{q_i\}}|S\Phi_{\{p_i\}}\rangle\:,$$ and so: $$\langle \Phi_{\{ q_i\}}| \left( U^\dagger_\Lambda S U_\Lambda - S\right) \Phi_{\{p_i\}}\rangle = 0\:,$$ Since the set of vectors $\Phi_{\{p_i\}}$ forms a basis of the Hilbert space (of the free theory) in view of the asymptotic completeness hypotheses, we conclude that: $$U^\dagger_\Lambda S U_\Lambda - S=0$$ i.e. $$S=U_\Lambda S U^\dagger_\Lambda\:, \quad \forall \Lambda \in O(3,1)\uparrow\:.$$ In other words if the $S$ matrix is Lorentz covariant, then the $S$ operator is Lorentz invariant.

  • $\begingroup$ For me it is not obvious since the operation of chronological ordering of two points is poincare invariant only if there points are divided by the timelike or light-like interval (we talk about orthochronous subgroup). $\endgroup$ – Andrew McAddams Apr 20 '14 at 20:24
  • $\begingroup$ In the remaining cases the lagrangians commute so that $T$ can be omitted... $\endgroup$ – Valter Moretti Apr 20 '14 at 20:26
  • $\begingroup$ yes, they must commute if S-operator must be lorentz invariant operator. But one part of my question was about how do we conclude that it is lorentz invariant from the fact of the poincare-covariance of s-matrix. Can you explain it, if you please? $\endgroup$ – Andrew McAddams Apr 20 '14 at 20:30
  • $\begingroup$ No there is no logical loop, since in the interaction picture they commute just in view of the free field commutation relations, and the fact that ${\cal L}_I$ is a scalar. $\endgroup$ – Valter Moretti Apr 20 '14 at 20:33
  • $\begingroup$ Sorry I did not read well that part of your question. I do not understand what 'covariance' of S matrix should mean actually here... $\endgroup$ – Valter Moretti Apr 20 '14 at 20:37

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