2
$\begingroup$

Suppose we want to calculate vacuum expectation $$ \tag 1 D_{lm}(x - y) = \langle \Omega | \hat {T}\left( \hat {\Psi}_{l}(x)\hat {\Psi}_{m}^{\dagger}(y)\right)| \Omega\rangle = \langle \Omega| \hat {T}\left( \hat {\Psi}_{l}(x)\hat {\Psi}_{m}^{\dagger}(y)\right)| \Omega\rangle , $$ where $\hat {\Psi}_{a}(x)$ refers to the irreducible representation of the Poincare group, and we have lagrangian containing interaction $\hat {\Psi}(x)$ with some other fields (including interaction "with itself"). We may insert full basis $1 = \sum_{\mathbf p , \sigma , i}| (\mathbf p , \sigma )_{i}\rangle \langle (\mathbf p , \sigma )_{i}|$ ($i$ numerates the number of particles) between fields operator in $(1)$: $$ \tag 2 D_{lm}(x - y) = \sum_{i, \mathbf p , \sigma}\hat {T}\langle \Omega| \hat {\Psi}_{l}(x) | (\mathbf p , \sigma )_{i}\rangle \langle (\mathbf p , \sigma )_{i}| \hat {\Psi}_{m}^{\dagger}(y)| \Omega\rangle $$ Let's then separate one-particle states from the full basis in $(2)$: $$ \tag 3 D_{lm}(x - y) = \sum_{\mathbf p , \sigma}\hat {T}\langle \Omega| \hat {\Psi}_{l}(x) | (\mathbf p , \sigma )\rangle \langle (\mathbf p , \sigma )| \hat {\Psi}_{m}^{\dagger}(y)| \Omega\rangle + ..., $$ where $...$ marks the contribution of the other (multiparticle) states. Here is one remark: in the theory with interaction we have also the bound states, where mass $m$ refers to the bound states of two and more particles. Even for states of one field $\Psi $ we have masses between $(m, 2m)$, where $m$ refers to the one-particle state.

So, the question: are $...$ summands in $(3)$ not equal to zero in theory with interaction? I.e., the propagator $(1)$ contains not just the $[D_{lm}(x - y)]_{free}$ line, but also the lines of the bound states, doesn't it?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.