# Is $|i\rangle=\sqrt{2\omega_1}\sqrt{2\omega_2}a^\dagger_{p_1}(-\infty)a^\dagger_{p_2}(-\infty)|\Omega\rangle$ a momentum eigenstate?

Define an asymptotic state in the far past as $$|i\rangle=\sqrt{2\omega_1}\sqrt{2\omega_2}a^\dagger_{{\vec p}_1}(-\infty)a^\dagger_{{\vec p}_2}(-\infty)|\Omega\rangle$$ where $$|\Omega\rangle$$ is the ground state of interacting theory and $$a_{\vec p}(t),a^\dagger_{\vec p}(t)$$ are time-dependent creation and destruction operators in the expansion $$\phi(\vec{x},t)=\int\frac{d^3{\vec p}}{(2\pi)^3\sqrt{2\omega_{\vec p}}}[a_{\vec p}(t)e^{-ipx}+a^\dagger_{\vec p}(t)e^{ipx}],~~px=\vec{p}\cdot{\vec x}-\omega_{\vec p}t.$$

The state $$|i\rangle$$ is a time-independent state. It is a momentum eigenstate or not? How do we check that? The problem is that unlike free theory, the commutator $$[{\vec P},a_{\vec p}^\dagger]$$ is not known.

• If it is not an eigenstate, then what is the meaning of $\vec{P}$ and $\vec{p}_{1,2}$? Mar 8 at 17:31
• The answer to this question may depend on the source you use. For example I believe in Weinberg's QFT volume 1, these operators are taken to produce asymptotic particle states (which have definite momentum) as a matter of definition. Mar 8 at 19:54

The commutator $$[\vec P,a^\dagger_{\vec p}]$$ is known. Recall that (cf. this PSE post) $$a_{\vec p}=\int e^{ipx}(\omega_{\vec p}\phi(x)+i\pi(x))\mathrm d\vec x$$ and therefore \begin{aligned} {}[\vec P,a_{\vec p}]&=\int e^{ipx}(\omega_{\vec p}[\vec P,\phi(x)]+i[\vec P,\pi(x)])\mathrm d\vec x\\ &\overset{\mathrm A}=i\int e^{ipx}\partial_x(\omega_{\vec p}\phi(x)+i\pi(x))\mathrm d\vec x\\ &\overset{\mathrm B}=-i(-i\vec p)\int e^{ipx}(\omega_{\vec p}\phi(x)+i\pi(x)\mathrm )d\vec x\\ &=-\vec pa_{\vec p} \end{aligned} where I have used $$[\vec P,O(x)]=i\partial_x O(x)$$ in $$\mathrm A$$, and I have integrated by parts in $$\mathrm B$$.
From this it follows that if a state $$|\alpha\rangle$$ has momentum $$\vec k$$, then the state $$a^\dagger_{\vec p}|\alpha\rangle$$ has momentum $$\vec k+\vec p$$: $$\vec Pa^\dagger_{\vec p}|\alpha\rangle=([\vec P,a^\dagger_{\vec p}]+a^\dagger_{\vec p}\vec P)|\alpha\rangle\equiv (\vec p+\vec k)a^\dagger_{\vec p}|\alpha\rangle$$
• Thanks a lot. So the asymptotic two-particle state in the question is a momentum eigenstate with momentum ${\vec p}_1+{\vec p_2}$. But I think, the momentum eigenstates at time $t_1=-\infty$ are different from the momentum eigenstates at time $t_2=+\infty$ because $a_p(t_1)\neq a_p(t_2)$. Please tell me if I am getting it right. Mar 9 at 13:59
• Just like in the Classical Mechanics, an interaction changes the particle momentum: $\vec{p}=\vec{p}(t)$, so you may label your operators like $a_{\vec{p}(t)}$. Mar 9 at 14:32