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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.

Solidification
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  • If it is not an eigenstate, then what is the meaning of $\vec{P}$ and $\vec{p}_{1,2}$? – Vladimir Kalitvianski Mar 08 '21 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. – Richard Myers Mar 08 '21 at 19:54

1 Answers1

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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 $\vec k+\vec p$: $$ \vec Pa^\dagger_{\vec p}|\alpha\rangle=([\vec P,a^\dagger_{\vec p}]+a^\dagger_{\vec p}\vec P)|\alpha\rangle\equiv (\vec p+\vec k)a^\dagger_{\vec p}|\alpha\rangle $$

AccidentalFourierTransform
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  • 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. – Solidification Mar 09 '21 at 13:59
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    Just like in the Classical Mechanics, an interaction changes the particle momentum: $\vec{p}=\vec{p}(t)$, so you may label your operators like $a_{\vec{p}(t)}$. – Vladimir Kalitvianski Mar 09 '21 at 14:32