# Conflict between Lippmann–Schwinger equation and Gell-Mann and Low theorem about energy

+ 2 like - 0 dislike
562 views

Lippmann–Schwinger equation states that scattering state will have the same energy as free state, while Gell-Mann Low theorem says that they have different enery.

Lippmann–Schwinger equation says: That is, if $|f\rangle$ is the eigenstate of interacting picture's free Hamiltonian $\hat{H}_0^{(I)}$ with eigenvalue $E$, then $$|F \rangle := \hat{U}(0,-\infty)|f\rangle$$ is the eigenstale of the full Hamiltonian $\hat{H}^{(I)}(0)$ at $t=0$ with the same eigenvalue $E$, where $\hat{U}(t,t_0)$ is the evolution operator in interacting picture.

On the other hand, Gell-Mann and Low theorem states that :

Let $|f\rangle$ be an eigenstate of $H_0$ with energy $E_0$ and let the 'interacting' Hamiltonian be $H=H_0 + gV$, where $g$ is a coupling constant and $V$ the interaction term. We define a Hamiltonian $H_\epsilon=H_0 + e^{-\epsilon |t|}gV$ which effectively interpolates between $H$ and $H_0$ in the limit $\epsilon \rightarrow 0^+$ and $|t|\rightarrow\infty$. Let $U_{\epsilon I}$ denote the evolution operator in the interaction picture. The Gell-Mann and Low theorem asserts that if the limit as $\epsilon\rightarrow 0^+$ of

$|F_\epsilon \rangle = \frac{ U_{\epsilon I} (0,-\infty) |f\rangle}{\langle f | U_{\epsilon I}(0,-\infty)|f\rangle}$ exists, then $|F_\epsilon \rangle$ are eigenstates of $H$. But the energy $E$ will not be same as $E_0$ and will have a shift.

Here is a proof of my first state that energy will be same

This post imported from StackExchange Physics at 2014-04-16 05:28 (UCT), posted by SE-user user34669
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.