Timeline for Will difference of creation operators in free and interacting theories affect the construction of $\phi_I$ similar to the free-field theory?
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| Dec 29, 2021 at 6:50 | comment | added | Matas Mackevicius | Right, I see the confusion. Formally, the $e^{iH_0t}a_{\bf{k}}e^{-iH_0t}$ is called the interaction picture operator because when the coupling constant in $H_{\text{int}}$ is small (and usually it is assumed to be), most of the time dependence in $a_{\bf{k}}(t)$ will come from $H_0$, hence my previous arguments. If you want to describe the entire field, you then impose unitary operators that have $H$ and $H_0$ both involved. At that point I believe that closed form commutation relations seize to exist. | |
| Dec 29, 2021 at 4:53 | comment | added | AlphaF20 | That's my concern. I can derive $e^{iH_0t}a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{\bf{k}}t}a_{\bf{k}}$, in a way similar to P&S (2.46). By using $e^-B A e^B = A + [A, B] + \cdots$, noticing $[A,B] = [a_{\mathbf{k}}, -iHt] $, and P&S (2.32) $[H, a_{\mathbf{k}} ] = -\omega_{\mathbf{k}} a_{\mathbf{k}}$. But, that's all for free fields. My question is about if the above derivation is also valid in interacting theories, for specifically, $[a^{free}_{\mathbf{k}}, a^{int, \dagger}_{\mathbf{k}'}] = ?$. I'm not sure if I overlooked your point. | |
| Dec 29, 2021 at 3:00 | comment | added | Matas Mackevicius | You can show that $e^{iH_0t}a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{{\bf{k}}}t}a_{{\bf{k}}}$. In the answer you referenced, it is shown that the time derivative of $a_{{\bf{k}}}$ is non-zero only when the Hamiltonian has an interacting component. The $a_{\bf{k}}$ (free) is constant in time when your theory is described only by a free field $H_0$. Whilst $a_{\bf{k}}(t)$ will in general depend on the interacting Hamiltonian and will not necessarily be the same for all time. | |
| Dec 28, 2021 at 21:16 | comment | added | AlphaF20 | $a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{\bf{k}}t}a_{\bf{k}}$ is derived from free field, I think. | |
| Dec 28, 2021 at 20:59 | comment | added | AlphaF20 | Thanks. My concern is, the difference betwen $a^{int}_p$ and $a^{free}_p$, based on Eq. (2) in the answer in physics.stackexchange.com/questions/310353/…, is the non-zero part of $(\partial^2 + m^2)\phi(x) = H_{int} \phi(x)$. The term triggerd by $H_{int}$ can possibly be complicated. I am not sure if we can have $a_k(t) = e^{- i \omega_k t} a_k$ for $a_k^{int}$ | |
| Dec 27, 2021 at 20:49 | history | answered | Matas Mackevicius | CC BY-SA 4.0 |