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The Klein-Gordon Equation or the Klein-Fock-Gordon Equation is an equation in quantum field theory which initially was discovered by Schrodinger but discarded by him soon after.
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Normalization of One-Particle States for Klein-Gordon Field Quantization
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 …
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Klein-Gordon Field Quantization and Bose-Einstein Statistics in Peskin & Schroeder
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 …
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answers
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Hamiltonian Field Theory in Peskin & Schroeder
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 …
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answer
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Derivation of Equation 2.27 from Peskin & Schroeder
In Section 2.3, Peskin & Schroeder discusses the quantization of real scalar field in Schrodinger picture. He writes Eq. (2.25) as follows.
$$\phi(\textbf{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sq …
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$m$ in Klein-Gordon Equation
We start with a free scalar field Lagrangian $\mathcal{L}[\phi, \partial_{\mu}\phi]$:
$$
\mathcal{L} = \frac{1}{2} \partial^{\mu}\phi \partial_{\mu}\phi - \frac{1}{2} m^2 \phi^2
$$
where, $m$ is calle …
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$m$ in Klein-Gordon Equation
The Klein-Gordon equation is given by
$$
(\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 …
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votes
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answer
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Problem of Klein-gordon Equation
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 …
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Reason for considering the positive root
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 $\omega$?