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In eqn. (3.11) of Srednicki's QFT book only the positive root is considered; i.e., $ \omega = + \sqrt{(k^2 + m^2 )} $ Why the negative root is not considered? And what is the $\omega$?

Ryder in his QFT book writes in eqn (2.20): Probability density, $\rho = \frac{i\hbar}{2m}(\phi^*\frac{\partial \phi}{\partial t} - \phi \frac{\partial \phi^*}{\partial t})$ Then in the next paragra …

We start with a free scalar field Lagrangian $\mathcal{L}[\phi, \partial_{\mu}\phi]$: $$ \mathcal{L} = \frac{1}{2} \partial^{\mu}\phi \partial_{\mu}\phi - \frac{1}{2} m^2 \phi^2 $$ where, $m$ is calle …

The Klein-Gordon equation is given by $$ (\square + m^2) \phi(x) = 0 $$ where $\square$ is the d'Alembertian operator, $m \in \mathbb{R}$ and $\phi$ is a scalar field. Question: What is $m$ in the K …

In Section 2.3, Peskin & Schroeder discusses the quantization of real scalar field in Schrodinger picture. He writes Eq. (2.25) as follows. $$\phi(\textbf{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sq …

I am trying to understand how Klein-Gordon particles obey Bose-Einstein statistics from Peskin & Schroeder's QFT textbook (page no. 22). The excerpt is given below: From this passage it is clear to m …

In Section 2.2 of their QFT textbook, Peskin & Schroeder introduce the Lagrangian and Hamiltonian field theories of a classical scalar field. While defining the action $S[\phi]$ and deriving the Euler …

Peskin & Schroeder in their QFT textbook discusses how we may normalize one-particle states $|\textbf{p}\rangle$ for Klein-Gordon field quantization in pages 22-23. The excerpts are given below. I u …