Recent content by Plantation

Thanks. Question 1 : Why the final equality is true? ; i.e., why ##\frac{1}{\sqrt{2 \omega_{\vec{p}}}} \frac{1}{\sqrt{2 \omega_{\vec{p}'}}} 2 \omega_{\vec{p}} =1 ## ? Perhaps, $$\langle \vec{p},s |\vec{p}',s' \rangle=(2 \pi)^3 2 \sqrt{\omega_{\vec{p}}} \sqrt{\omega_{\vec{p'}}}...

I am reading the Schwartz's quantum field theory, p.207 and stuck at some calculation. In the page, he states that for identical particles, $$ | \cdots s_1 \vec{p_1}n \cdots s_2 \vec{p_2} n \rangle = \alpha | \cdots s_2 \vec{p_2}n \cdots s_1...

O.K. Again.. How can we perform this integral : ##\int d\bar{\vec{\theta}}d \vec{\theta} e^{-(\bar{\vec{\theta}} - \bar{\vec{\eta}} A^{-1})A( \vec{\theta}-A^{-1}\vec{\eta})} = \operatorname{det}(A)## ? An issue that makes me annoying is the involved objects ##\bar{\vec{\eta}}## (and...

I am reading the Schwartz's Quantum field theory, p.269~p.272 ( 14.6 Fermionic path integral ) and some question arises. In section 14.6, Fermionic path integral, p.272, (14.100), he states that $$ \int d\bar{\theta}_1d\theta_1 \cdots d\bar{\theta}_n d\theta_n e^{-\bar{\theta}_i A_{ij}...

I am reading the Lancaster & Blundell, Quantum field theory for gifted amateur, p.225 and stuck at understanding some derivations. We will calculate a generating functional for the free scalar field. The free Lagrangian is given by $$ \mathcal{L}_0 = \frac{1}{2}(\partial _\mu \phi)^2 -...

O.K. Anyway, thank you :)

O.K. Looking closerly, in (4.77) (Peskin's book), ##\frac{1}{|\frac{\bar{k}_A^{z}}{\bar{E}_A}- \frac{\bar{k}_B^{z}}{\bar{E}_B}|} = \frac{1}{|\frac{d}{d \bar{k}_A^{z}} f(\bar{k}_A^z)|}## seems a 'function' which depends on variable ##\bar{k}_A^{z}##. On the other hand...

By taking modulus, what does you exactly means? Shall we find the roots of $$f(\bar{k}_A^{z}) := (\sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A} $$ directly, and then brutally force them into the formula $$ \int \delta[f(x)] dx =...

O.K. Thank you. And.. I think that I also reached to such a step, as I wrote. I still don't know why $$ \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$ $$\stackrel{?}{=}...

For the final integration ; i.e., $$ \int d \bar{k}_A^z \delta ( \sqrt{\bar{k}_A^2 + m_A^2}+\sqrt{\bar{k}_B^2 + m_B^2} - \Sigma E_f ) |_{\bar{k}^z_B = \Sigma p_f^z - \bar{k}^z_A}$$ $$\stackrel{?}{=} \frac{1}{|\frac{\bar{k}^z_A}{\bar{E}_A}...

Thanks for kind explanation, although there seems to be few typos. And I am somewhat confused since In your derivation of the integral equation ; i.e., Peskin's book (4.77) , it seems that you have used symbols A and B interchangeably. Anyway, I still don't understand why such next integral...

Perhaps, next formula holds true? $$ \int d \bar{k}_A^{x} \int d \bar{k}_A^{y} \int d \bar{k}_B^x \int d \bar{k}_B^{y} \delta ( \bar{k}_A^{x} + \bar{k}_B^{x} - \Sigma p_f^{x}) \delta ( \bar{k}_A^{y} + \bar{k}_B^{y}-\Sigma p_f^{y} ) \delta^{(2)}(k_B^{\perp} - \bar{k}_B^{\perp}) = 1...

Sorry I am late. Thank you. I mean steps marked by the question symbol!. I'm still trying to integrate. I also don't understand the strange notation ##(k_i^{\perp} = \bar{k}_i^{\perp})##.. And what is the definition of ##k_i^{\perp}## ( and ##\bar{k}_i^{\perp})## ? I think that understanding...

I am reading the Horatiu Nastase's Introduction to quantum field theory (https://professores.ift.unesp.br/ricardo.matheus/files/courses/2014tqc1/QFT1notes.pdf ) ( Attached file ) or Peskin, Schroeder's quantum field theory book, p.105, (4.77). Through p.176 ~ p. 177 in the Nastase's Note, he...