I am following a problem in a QFT textbook (Srednicki) which asks us to show that the interaction-picture field

$$\phi_I(\textbf{x},t)=e^{iH_0 t}\phi(\textbf{x},0)e^{-iH_0 t}$$

obeys the Klein-Gordon equation. This is implicit in the rest of the text, but the solutions also say that this can be confirmed directly by writing the time derivatives as commutators, and computing them.

Could someone confirm what the author means by it: is it something related to the Heisenberg equation of motion, or the fact that operators commute with their derivatives?

  • $\begingroup$ Please clarify your question. Provide a clear reference to the book. Provide context. Explain what are $H_0$ , $\phi_I$ and $\phi$. There must also be an $H_I$ I guess. Perhaps the book can be found online? What have you tried so far ? $\endgroup$ – my2cts Sep 9 '18 at 10:08
  • $\begingroup$ What is meant is that you should take the second time derivative of the equation, which on the left gives you, well, the second time derivative, and on the right a double commutator of $H_0$ with $\phi$. Plugging in $H_0$ and simplifying should give you the KG equation. $\endgroup$ – Daniel Sep 9 '18 at 12:42
  • $\begingroup$ I see, I just wasn't sure of the meaning. I've written down an attempt at a solution, perhaps you could correct if it's wrong? $\endgroup$ – Tom Sep 9 '18 at 18:05

We take the derivatives of the interaction-picture field and simplify to obtain the Klein-Gordon equation.

$\phi_I(\textbf{x},t)=e^{iH_0 t}\phi(\textbf{x},0)e^{-iH_0 t}$

$\frac{\partial \phi_{I}}{\partial t} = iH_{0} e^{iH_{0}t}\phi(\textbf{x},0)e^{-iH_{0}t} + e^{iH_{0}t}\phi(\textbf{x},0)(-iH_0)e^{-iH_0 t}$

$\frac{\partial^2 \phi_I}{\partial t^2} = -H_0^2 e^{-H_0 t}\phi e^{-iH_0 t} + iH_0 e^{iH_0 t}\phi(-iH_0)e^{-iH_0 t} + iH_0 e^{iH_0 t}\phi (-iH_0)e^{-iH_0 t} + e^{iH_0 t} \phi H_0^2 e^{-iH_0 t}$

$\partial \phi_{Itt} = 2 [H_0,\phi_I] = [\frac{1}{2}\nabla^2 + \frac{1}{2}m^2,\phi_I] = 2[\frac{1}{2}\nabla^2,\phi_I] + 2[\frac{1}{2}m^2,\phi_I] = \nabla^2 \phi_I - m^2 \phi_I$

$\frac{\partial^2 \phi_I}{\partial t^2} - \nabla^2 \phi_I +m^2 \phi_I = 0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.