In P&S's QFT book, page 213, the book considered Heisenberg operator's translation under General interacting field: $$ \begin{aligned} \left\langle\Omega|\phi(x)| \lambda_{\mathbf{p}}\right\rangle &=\left\langle\Omega\left|e^{i P \cdot x} \phi(0) e^{-i P \cdot x}\right| \lambda_{\mathbf{p}}\right\rangle \\ &=\left.\left\langle\Omega|\phi(0)| \lambda_{\mathbf{p}}\right\rangle e^{-i p \cdot x}\right|_{p^0=E_{\mathbf{p}}} \\ &=\left.\left\langle\Omega|\phi(0)| \lambda_0\right\rangle e^{-i p \cdot x}\right|_{p^0=E_{\mathbf{p}}} . \end{aligned} \tag{7.4}$$

Also, in Weinberg's book page 458, when he introducing The Källen-Lehmann Representation, he gives the similar formula

$$\begin{aligned} &\langle 0|\Phi(x)| n\rangle=\exp \left(i p_n \cdot x\right)\langle 0|\Phi(0)| n\rangle \\ &\left\langle n\left|\Phi^{\dagger}(y)\right| 0\right\rangle=\exp \left(-i p_n \cdot y\right)\left\langle n\left|\Phi^{\dagger}(0)\right| 0\right\rangle \end{aligned} \tag{10.7.2}$$

My question is why in general interacting picture, we can transform the $\Phi(x)$ in above way?

In free field, P&S proved this transformation in page 26: $$ e^{-i \mathbf{P} \cdot \mathbf{x}} a_{\mathbf{p}} e^{i \mathbf{P} \cdot \mathbf{x}}=a_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x}}, \quad e^{-i \mathbf{P} \cdot \mathbf{x}} a_{\mathbf{p}}^{\dagger} e^{i \mathbf{P} \cdot \mathbf{x}}=a_{\mathbf{p}}^{\dagger} e^{-i \mathbf{p} \cdot \mathbf{x}} \tag{2.48}$$ and therefore $$ \begin{aligned} \phi(x) &=e^{i(H t-\mathbf{P} \cdot \mathbf{x})} \phi(0) e^{-i(H t-\mathbf{P} \cdot \mathbf{x})} \\ &=e^{i P \cdot x} \phi(0) e^{-i P \cdot x} \end{aligned} \tag{2.49}$$

While in interacting field, we may cannot write field as creation and annihilation operators.

By the way, it seems that the field operator transformation law in Weinberg's book have a sign difference with P&S, ($\phi(x)=e^{-i P \cdot x} \phi(0) e^{i P \cdot x}$) is this right?

  • $\begingroup$ Weinberg uses a different sign convention for the Minkowski metric ($\mathrm{d}{s}^2=-\text{d}{t}^2+\text{d}{x}^2+\text{d}{y}^2+\text{d}{z}^2$), which is conventional in general relativity and string theory. P&S uses the standard particle physics/QFT sign convention, which differs from the above by a factor of $-1$. $\endgroup$ Oct 18, 2022 at 0:43

1 Answer 1


I consider the spacetime translation covariance property ($\phi(x) = \mathrm{e}^{\mathrm{i} x\cdot P} \phi(0) \mathrm{e}^{-\mathrm{i} x\cdot P}$) as fundamental in QFTs on Minkowski spacetime (as e.g. in the Wightman axioms of QFT). This definitely makes sense because translations are symmetries on Minkowski spacetime and symmetry transformations should not change predictions of the theory. Thus, we assume that there are some unitary operators $U(x)$, $x\in\mathbb{R}^4$, that implement translations on quantum fields: $$U(x) \phi(y) U(x)^* = \phi(y+x), \ y \in\mathbb{R}^4.$$ By Stone's theorem, the unitaries $U(x)$ have self-adjoint generators that we call $P = (H,\mathbf{P})$, i.e. $U(x) = \mathrm{e}^{\mathrm{i}x\cdot P} = \mathrm{e}^{\mathrm{i}x_0H - \mathrm{i} \mathbf{x} \cdot \mathbf{P}}$. Using some other properties of QFTs (positivity of energy, causality, etc.) it turns out that the choice of $P$ is essentially unique.

In classical mechanics, Noether's theorem relates symmetries to conservation laws, e.g. spacetime translation symmetry is related to conservation of energy-momentum. Therefore, it is more than reasonable to call the generators $P$ the energy-momentum operators. Basically you can say that the energy-momentum operator is defined via the spacetime translation covariance property.


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