In Weinberg's QFT volume 1, chapter 3.3, just below equation 3.3.19, he says $\vec P=\vec P_0$ and $\vec J=\vec J_0$ can be(easily) concluded from the definition of Møller wave operator or equivalently the Lippmann-Schwinger equations. However, I don't see how.

In virtually all known field theories, the effect of interactions is to add an interaction term V to the Hamiltonian, while leaving the momentum and angular momentum unchanged: $$H =H_0+ V, \vec P =\vec P_0, \vec J =\vec J_0. \tag{3.3.18}$$ (The only known exceptions are theories with topologically twisted fields, such as those with magnetic monopoles, where the angular momentum of states depends on the interactions.) Eq. (3.3.18) implies that the commutation relations (3.3.11), (3.3.14), and (3.3.16) are satisfied provided that the interaction commutes with the free-particle momentum and angular-momentum operators $$[V,\vec P_0] = [V,\vec J_0] = 0.\tag{3.3.19}$$ It is easy to see from the Lippmann-Schwinger equation (3.1.16) or equivalently from (3.1.13) that the operators that generate translations and rotations when acting on the 'in' (and 'out') states are indeed simply $\vec P_0 $ and $\vec J_0$.

Related commutators:

$$[J^i,J^j]=i\epsilon_{ijk}J^k\tag{3.3.11}$$ $$[J^i,P^j]=i\epsilon_{ijk}P^k\tag{3.3.14}$$ $$[J^i,H]=[P^i,H]=[P^i,P^j]=0\tag{3.3.16}$$

Definition of the wave operators:

The 'in' and 'out' states can now be defined as eigenstates of $H$, not $H_0$, $$H\Psi^\pm_\alpha=E_\alpha\Psi^\pm_\alpha\tag{3.1.11}$$ which satisfy the condition $$\int d\alpha e^{-iE_\alpha\tau}g(\alpha)\Psi^\pm_\alpha\to \int d\alpha e^{-iE_\alpha\tau}g(\alpha)\Phi_\alpha\tag{3.1.12}$$ for $\tau\to-\infty$ or $\tau\to\infty$, respectively. Eq. (3.1.12) can be rewritten as the requirement that: $$e^{-iH\tau}\int d\alpha g(\alpha)\Psi^\pm_\alpha\to e^{-iH_0\tau}\int d\alpha g(\alpha)\Phi_\alpha$$ for $\tau\to-\infty$ or $\tau\to\infty$, respectively. This is sometimes rewritten as a formula for the 'in' and 'out' states: $$\Psi^\pm_\alpha=\Omega(\pm \infty)\Phi_\alpha\tag{3.1.13}$$ where $$\Omega(\tau)=e^{i\tau H}e^{-i\tau H_0}\tag{3.1.14}$$

The Lippmann-Schwinger equations are given by:

$$\Psi^\pm_\alpha=\Phi_\alpha+(E_\alpha-H_0\pm i\epsilon)^{-1}V\Psi_\alpha^\pm\tag{3.1.16}$$

So, my questions are,

  1. Is (3.3.19) assumed to be true in that statement? If yes, then can we not just follow the previous argument backwards and prove 3.3.18 from 3.3.19? Or is he just trying to see if the definitions of wave operators are consistent with 3.3.18 and 3.3.19? If no, then:

  2. How can one conclude $\vec P =\vec P_0, \vec J =\vec J_0$ from 3.1.16 or 3.1.13 and assuming nothing else? I tried taking infinitesimal Poincaré transformation of the equation 3.1.13. It becomes $$\vec P\Psi^\pm_\alpha=\Omega(\pm \infty)\vec P_0\Phi_\alpha.$$ And applying $\vec P$ on 3.1.13, we get $$\vec P\Psi^\pm_\alpha=\vec P\Omega(\pm \infty)\Phi_\alpha,$$ so it follows $$(\vec P\Omega(\pm \infty)-\Omega(\pm \infty)\vec P_0)\Phi_\alpha=0$$ for all $\Phi_\alpha$. So, $$\vec P\Omega(\pm \infty)=\Omega(\pm \infty)\vec P_0.$$ But that's all I can get. How do I conclude $\vec P=\vec P_0$ and $\vec J=\vec J_0$?

  3. Even if it is possible to conclude $\vec P =\vec P_0, \vec J =\vec J_0$ from 3.1.13 or 3.1.16 alone, then according to the comment in parenthesis following equation 3.3.18, with topologically twisted fields, such as those with magnetic monopoles(which I do not have any idea about), do Lippmann-Schwinger equations not hold?

  4. Can someone refer me to any book/paper/lecture notes where Lorentz invariance of S operator is proven without considering Lagrangian theory or the field operators(this is what Weinberg wants to do in this chapter)? Or, can it be done even in principle?

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    $\begingroup$ Please type out material you want to quote instead of adding pictures, since pictures are not searchable. $\endgroup$
    – ACuriousMind
    Commented Mar 22, 2017 at 12:18
  • $\begingroup$ @ACuriousMind I am new to Stack exchange so I know a little about website-specific typesetting. For example I don't know how to number equations and align them at the center. I may do it, but it will take time. Thanks. $\endgroup$ Commented Mar 22, 2017 at 12:40
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    $\begingroup$ A tutorial for MathJax is here, most of it works like ordinary $\LaTeX$. You can center equations by enclosing them with $$ instead of $ and you add numbers by \tag{equation-number-here}. $\endgroup$
    – ACuriousMind
    Commented Mar 22, 2017 at 12:44

2 Answers 2


Here's an argument to backup Weinberg's "assumption." In particular, I will show that the energy $P_0$ is changed by the presence of a potential, while the linear momentum $P_i$ and angular momentum $L_i$ are not, for a scalar field theory with a potential. I'll also consider a case below where $P_i$ is affected by the presence of a peculiar kind of interaction.

We can use the following definition to compute stress-energy tensor: \begin{equation} T_{\mu\nu} = -\frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu}}\Big|_{g_{\mu\nu}=\eta_{\mu\nu}} \end{equation}

Using the stress energy tensor, we can compute

  • the energy, $P=T_{00}$,
  • the linear momentum, $P_i = T_{0i}$,
  • and the angular momentum, $L_i = \frac{1}{2} \epsilon_{ijk} (x^j T^{0k} - x^k T^{0j})$

Let's consider a scalar field with a potential. Let's write the action for this. I'm intentionally writing it in a way where we can take functional derivatives with respect to $g^{\mu\nu}$ easily, but in the end we are interested in Minkowski space, $g_{\mu\nu}=\eta_{\mu\nu}$. I'm using a mostly plus metric convention, $\eta_{\mu\nu}={\rm diag}(-1,1,1,1)$. \begin{equation} S = \int {\rm d}^4 x \sqrt{-g} \left(-\frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - V(\phi)\right) \end{equation} Taking a variation $g_{\mu\nu}\rightarrow g_{\mu\nu}+\delta g_{\mu\nu}$, we obtain \begin{equation} \delta S = \int {\rm d}^4 x \sqrt{-g} \left( -\frac{1}{2} \partial_\mu \phi \partial_\nu \phi + \frac{1}{4} g_{\mu\nu} (\partial \phi)^2+\frac{1}{2} g_{\mu\nu} V(\phi)\right) \delta g^{\mu\nu} \end{equation} so \begin{equation} T_{\mu\nu} = -\frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu}}\Big|_{g_{\mu\nu}=\eta_{\mu\nu}} = \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} \eta_{\mu\nu}(\partial_\alpha \phi)^2 - \eta_{\mu\nu} V(\phi) \end{equation}

Now let's consider different cases to verify Weinberg's "assumption".

  • The energy is $P=T_{00}=\frac{1}{2} \dot{\phi}^2 + \frac{1}{2} (\partial_i \phi)^2 + V(\phi)$, and depends on the potential, where a dot refers to a time derivative and $\partial_i$ is a spatial derivative.
  • The linear momentum is $P_i = T_{0i} = \frac{1}{2} \dot{\phi}\partial_i \phi$, and does not depend on the potential.
  • The angular momentum is $L_i = \frac{1}{2} \epsilon_{ijk} (x^j P^k - x^k P^j)$ also does not depend on the potential, since the linear momentum does not.

Finally, in the interest of seeing a situation where this assumption can break down, we can consider a case where interactions will affect the linear momentum and angular momentum operators:

\begin{equation} S = \int {\rm d}^4 x \sqrt{-g} \left(-\frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi + \frac{c}{\Lambda^3} (\partial \phi)^2 \square \phi \right) \end{equation} where $c$ is a dimensionless coupling parameter and $\Lambda$ is a scale with units of energy, and this interaction is a so-called "cubic Galileon" interaction.

If we work out the stress energy tensor, we find there is a new term (please double check the signs and coefficients if you actually use this somewhere): \begin{equation} T_{\mu\nu} = T_{\mu\nu}^{(c=0)} -\frac{2c}{\Lambda^3} \left[ \left(\partial_\mu \phi \partial_\nu \phi - \frac{1}{2} \eta_{\mu\nu}(\partial_\alpha \phi)^2\right) \square \phi + (\partial \phi)^2 \partial_\mu \partial_\nu \phi \right] \end{equation} where $T^{\mu\nu}_{(c=0)}$ is the stress energy tensor with $c=0$.

The last term will lead to a correction to the linear momentum \begin{equation} P_i = T_{0i} = P_i^{(c=0)} - \frac{2c}{\Lambda^3} \left(-\dot{\phi}^2 + (\partial_i \phi)^2\right) \partial_i \dot{\phi} \end{equation} Similarly, the angular momentum will also receive a correction.

  • $\begingroup$ So it seems, that the independence of the momentum and angular momentum is valid actually for non-derivative interaction terms. $\endgroup$ Commented Nov 6, 2020 at 6:29
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    $\begingroup$ @spiridon_the_sun_rotator Yes, though it's also worth pointing out that boundary terms, such as the QCD theta term, or instanton contributions more generally, may also contribute by the same arguments. I believe this is the source of Weinberg's assertion about monopoles and twisted topologies. $\endgroup$ Commented Nov 6, 2020 at 7:49
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    $\begingroup$ @spiridon_the_sun_rotator It's also not the case that all derivative interactions will lead to a change in the momentum operator (even neglecting boundary terms). Originally I was going to use as an example example $(\partial \phi)^4$ but it turns out for this $T_{0i}$ is just the free theory version. The key thing is the index structure, you need an interaction term that shows up in $T_{0i}$ which is not just $\eta_{0i}$. I suspect that with vector fields you could create non-YM interactions that modify $T_{0i}$ by having the vector carry one or both of the indices. $\endgroup$
    – Andrew
    Commented Nov 6, 2020 at 9:06

I spent a while thinking about these specific paragraphs when I first went through Weinberg and was able to work out a couple things. Equation (3.3.18), $$ H=H_0+V,\ \ \ \ \boldsymbol J=\boldsymbol J_0,\ \ \ \ \boldsymbol P=\boldsymbol P_0 $$ is, in fact, an assumption. Indeed, he claims just above this equation that this assumption is almost always valid (I have not tried to check the literature on this myself).

From this assumption, however, (3.3.19) can be derived from the commutator algebras. Namely, in order to demand $$ [J^i,H]=[P^i,H]=0, $$ it is both necessary and sufficient to demand $$ [J_0^i,V]=[P^i_0,V]=0 $$ which can be seen just by plugging in $H=H_0+V$ , replacing $J^i$ and $P^i$ with their unperturbed versions, and using $[J^i,H_0]=[P^i_0,H_0]=0$ from the Poincare algebra.

The following statement Weinberg makes about the Lipmann-Schwinger equation and the in/out states is actually essentially the statement $\boldsymbol P\Omega=\Omega \boldsymbol P_0$ you found in the question, but now using $[\boldsymbol P_0,V]=0$ and hence $[\boldsymbol P_0,H]=[\boldsymbol P_0,H_0]$, it's clear that both the momentum operators commute with the S matrix. It follows from this statement that the states (both perturbed and unperturbed) are eigenstates of both $\boldsymbol P$ and $\boldsymbol P_0$ with the same eigenvalues (and hence the unperturbed operators generate what they are supposed to generate on the unperturbed states, which was not assumed a priori). All of this applies equally to $\boldsymbol J$ and $\boldsymbol J_0$.

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    $\begingroup$ If the equation $(3.3.18)$ and the commutation relation $(3.3.19)$ are $\textbf{assumptions}$,then I think we could directly get to the point that the $P$ and $P_0$ all commute with the S matrix and $P \Omega = \Omega P_0$ rather than "following from the Lipmann-Schwinger equation..." right? $\endgroup$
    – Jiahao Fan
    Commented Nov 6, 2020 at 3:27
  • $\begingroup$ It is my understanding that this is exactly what he meant by the "or equivalently..." part of his statement. Though I will note specifically that only one of the two, (3.3.18) and (3.3.19), needs to be assumed. The assumption of one always implies the other. $\endgroup$ Commented Nov 6, 2020 at 7:44

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