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Questions tagged [klein-gordon-equation]

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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Negative probabilities in quantum physics

Negative probabilities are naturally found in the Wigner function (both the original and its discrete variants), the Klein paradox (where it is an artifact of using a one-particle theory) and the ...
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Why would Klein-Gordon describe spin-0 scalar field while Dirac describe spin-1/2?

The derivation of both Klein-Gordon equation and Dirac equation is due the need of quantum mechanics (or to say more correctly, quantum field theory) to adhere to special relativity. However, excpet ...
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Noether's Theorem and scale invariance

Noether's theorem usually considers coordinate/field transformations which leave the Lagrangian invariant up to a divergence term, i.e. $$\mathcal{L} \rightarrow \mathcal{L} + \partial_{\mu}f^{\mu}$$ ...
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The Lagrangian in Scalar Field Theory

This is perhaps a naive question, but why do we write down the Lagrangian $$\mathcal{L}=\frac{1}{2}\eta^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi - \frac{1}{2}m^2\phi^2$$ as the simplest ...
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Is Maxwell's field the wave function of the photon?

In his ArXiv paper What is Quantum Field Theory, and What Did We Think It Is? Weinberg states on page 2: In fact, it was quite soon after the Born–Heisenberg–Jordan paper of 1926 that the idea ...
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Fourier Transforming the Klein Gordon Equation

Starting with the Klein Gordon in position space, \begin{align*} \left(\frac{\partial^2}{\partial t^2} - \nabla^2+m^2\right)\phi(\mathbf{x},t) = 0 \end{align*} And using the Fourier Transform: $\...
17
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1answer
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Schrodinger equation from Klein-Gordon?

One can view QM as a 1+0 dimensional QFT, fields are only depending on time and so are only called operators, and I know a way to derive Schrodinger's equation from Klein-Gordon's one. Assuming a ...
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3answers
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single-particle wavepackets in QFT and position measurement

Consider a scalar field $\phi$ described by the Klein-Gordon Lagrangian density $L = \frac{1}{2}\partial_\mu \phi^\ast\partial^\mu \phi - \frac{1}{2} m^2 \phi^\ast\phi$. As written in every graduate ...
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4answers
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Is it true that the Schrödinger equation only applies to spin-1/2 particles?

I recently came across a claim that the Schrödinger equation only describes spin-1/2 particles. Is this true? I realize that the question may be ill-posed as some would consider the general ...
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2answers
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How to derive the theory of quantum mechanics from quantum field theory?

I have read the book on quantum field theory for some time, but I still do not get the physics underline those tedious calculations. The thing confused me most is how quantum mechanics relates to ...
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2answers
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How can one find the energy eigenfunctions of a particle in a finite square well via the Klein-Gordon equation?

It is said that Klein-Gordon equation is a relativistic version of the Schrodinger equation. In Schrodinger equation, it is straightforward to include potential energy. But for K-G eqn things seem to ...
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Quantizing a complex Klein-Gordon Field: Why are there two types of excitations?

In most references I've seen (see, for example, Peskin and Schroeder problem 2.2, or section 2.5 here), one constructs the field operator $\hat{\phi}$ for the complex Klein-Gordon field as follows: ...
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1answer
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Klein paradox for bosons and fermions

I am reading this paper about the Klein paradox, i.e. transmission of relativistic particles incident on a potential step of height $V_0 > E + mc^2 > 2mc^2$ with $E$ the energy of the incident ...
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Why do we study the scalar field in QFT when there is no such thing in nature?

The Klein-Gordan equation describing a spinless scalar field is one of the first things one studies in a QFT course, but there are no elementary spin-0 fields in nature. Is the scalar field to QFT ...
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3answers
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Evaluating propagator without the epsilon trick

Consider the Klein–Gordon equation and its propagator: $$G(x,y) = \frac{1}{(2\pi)^4}\int d^4 p \frac{e^{-i p.(x-y)}}{p^2 - m^2} \; .$$ I'd like to see a method of evaluating explicit form of $G$ ...
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1answer
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Localized field quanta?

After the canonical quantization of the Klein-Gordon field (for example), we interpret the quantum of excitations of the fields with definite energy and momentum as particles. But our mental image of ...
12
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5answers
947 views

Why can the Klein-Gordon field be Fourier expanded in terms of ladder operators?

Using the plane wave ansatz $$\phi(x) = e^{ik_\mu x^\mu}$$ the solution to the Klein-Gordon equation $(\Box + m^2) \phi(x) =0$ can be written as a sum of solutions, since the equation is linear and ...
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1answer
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Klein-Gordon inner product

Studying the scalar field and Klein-Gordon equation in quantum field theory I came across this definition for the inner product in the space of the solutions of the K.G. equation: $$\langle \Phi_1 | \...
11
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1answer
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Interpreting the Klein-Gordon Annihilation Operator Expression

I can derive $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ for a free real scalar Klein-Gordon field in three ways mathematically: the usual Fourier transform way in ...
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1answer
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Is the Dirac equation equivalent to the Klein-Gordon equation for its left handed component?

The Dirac equation $$(i\gamma^a\partial_a - m)\psi=0\tag{0}$$ is given by a first order operator acting on a Dirac spinor, which is the direct sum of a left handed spinor and a right handed spinor. ...
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Why use Fourier expansion in Quantum Field Theory?

I have just begun studying quantum field theory and am following the book by Peskin and Schroeder for that. So while quantising the Klein Gordon field, we Fourier expand the field and then work only ...
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1answer
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Lorentz transformation of classical Klein–Gordon field

I'm trying to see that the invariance of the Klein–Gordon field implies that the Fourier coefficients $a(\mathbf{k})$ transform like scalars: $a'(\Lambda\mathbf{k})=a(\mathbf{k}).$ Starting from the ...
9
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1answer
763 views

Can't the Negative Probabilities of Klein-Gordon Equation be Avoided?

I came across these notes of Dyson on Relativistic Quantum Mechanics. There on p. 3, he mentions that the issue with the Klein-Gordon equation is that the only way to relate $\psi$ with a probability ...
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Why aren't four-vectors used in the definition of a Klein-Gordon quantum field?

I am a beginner who is learning QFT. When I was going through the quantisation of Klein-Gordon real-field. I got confused about something: The solution to Klein-Gordon equations are of the form $ \...
9
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1answer
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Green's function for the inhomogenous Klein-Gordon equation

I'm trying to solve the massive Klein-Gordon equation in good old Minkowski space-time: $$(\square + m^2) \phi = \rho(t,\mathbf{x})$$ where $\square = \partial_{\mu} \partial^{\mu} = \partial_{t}^2 - \...
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1answer
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Is Bose-Einstein condensate a good example of a classical massive boson field?

Physically, we know that a BEC has formed if a macroscopic number of bosons occupy a single quantum state. The wave-function $\Psi(x)$ of the latter, normalized to the total number of condensed atoms ...
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The idea of analytical continuation method to solve the Klein-Gordon equation, how and why?

For simplicity, let's consider a two dimensional version of Klein-Gorden equation: $$ (\partial_t^2-\partial_x^2-\partial_y^2+m^2) G(\vec{x},t) = -\delta(\vec{x})\delta(t) $$ From the previous posts: ...
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Why isn't the wave equation $\nabla^2 \psi - 1/c^2 \partial_{tt} \psi = (\frac{mc}{\hbar})^2\psi$

Special relativity was well established by the time the schrodinger equation came out. Using the correspondence of classical energy with frequency and momentum with wave number, this is the equation ...
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3answers
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Why is the Klein Gordon equation of second order in time?

I was wondering if there is any way to interpret the fact that the Klein Gordon equation is a 2nd order PDE in time. I mean, normally you would expect that as soon as you fix the initial wavefunction, ...
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2answers
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Why is there a minus sign in this wave equation derivation?

My book on quantum mechanics suggests a derivation of the wave equation $$\left(\Delta - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \right) \psi(\bar{r},t) = 0$$ from the photon energy-impulse ...
8
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1answer
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Difference between field and wavefunction

Can someone give me a clear explanation of what is the difference between a classical field, a wave function of a particle and a quantum field? I haven't find a clear explanation. For example for ...
8
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2answers
486 views

Does relativistic quantum mechanics really violate causality?

The Hamiltonian $H=\sqrt{p^2+m^2}$ defines a one-particle quantum mechanics in the usual way. Let us call this theory RQM for short. Peskin and Schroeder claim that RQM violates causality because ...
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1answer
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Derivation of total momentum operator QFT

The expansion of the Klein Gordon field and conjugate momentum field are $\hat{\phi}(x) = \int \frac{d^3k}{(2 \pi)^3} \, \frac{1}{ \sqrt{2 E_{k}}} \left( \hat{a}_{k} + \hat{a}^{\dagger}_{-k} \right) ...
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How to obtain the explicit form of Green's function of the Klein-Gordon equation?

The definition of the green's function for the Klein-Gordon equation reads: $$ (\partial_t^2-\nabla^2+m^2)G(\vec{x},t)=-\delta(t)\delta(\vec{x}) $$ According to these resources: Green's function ...
8
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1answer
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Quantization of a free field: Klein-Gordon case

I am a beginner and reading this course text on QFT. The author first introduces the KG equation: $$\partial_\mu\partial^{\mu}\phi+m^2\phi=0$$ [with Minkowski signature $(+,-,-,-)$]. Then the ...
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Why is $V=(1/2) m^2 \phi^2$ for a free relativistic scalar field of mass $m$?

Bit of a basic question here but how come for a free relativistic scalar field of mass $m$ such as Klein Gordon theory, we take the potential to be $$V=\frac{1}{2} m^2 \phi^2$$ Is the mass term ...
8
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1answer
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Delocalization in the square root version of Klein-Gordon equation

In this Wikipedia article a relativistic wave equation is derived using the Hamiltonian $$H=\sqrt{\textbf{p}^2 c^2 + m^2 c^4}$$ Substituting this into the Schrödinger equation gives the square root ...
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Noether current for a local gauge transformation for the Klein-Gordon Lagrangian

The Noether current corresponding to the transformation $\phi \to e^{i\alpha} \phi$ for the Klein-Gordon Lagrangian density $$\mathcal{L}~=~|\partial_{\mu}\phi|^2 -m^2 |\phi|^2$$ by finding $\...
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2answers
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Is there a reason why a relativistic quantum theory of a single fermion exists, but of a single scalar not?

When we try to construct the relativistic generalization of non-relativistic time dependent Schroedinger equation, there are at least two possible completions - Klein-Gordon equation and Dirac ...
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1answer
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What is a superfluid in field theoretic terms?

I'm wondering how one precisely defines a superfluid in terms of the effective field theory description. In Nicolis's paper http://arxiv.org/abs/1108.2513 there seems to be an extremely simple ...
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Is there supersymmetry between Dirac and Klein Gordon solutions?

Usually supersymmetry is explained at the level of the action of a quantum field theory, and there are two ways to go down from QFT to relativistic quantum mechanics: either a non-covariant way where ...
7
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1answer
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Free Field theory to Interacting Field theory

Free field theory: Why is it said that different Fourier modes in case of a free field (say, real Klein-Gordon field) are independent of each other? Interacting field theory: How exactly does the ...
7
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2answers
536 views

Contradictory result for scalar-field propagator from Feynman rules and LSZ formula

I am trying to learn how to calculate scattering amplitudes in a Klein-Gordon theory. I am getting stuck with the simplest of the examples: $\phi\to\phi$ in a free scalar-field theory. If I calculate ...
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5answers
645 views

Why Use the Non-Relativistic Momentum Operator in Relativistic Quantum Mechanics?

In deriving the Klein Gordon equation one starts out with the relativistic energy relation $E^2 = p^2 + m^2$ and substitutes the quantum momentum operator that corresponds to non-relativistic QM, $\...
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1answer
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What is $\phi(x)|0\rangle$?

Suppose for instance that $\phi$ is the real Klein-Gordon field. As I understand it, $a^\dagger(k)|0\rangle=|k\rangle$ represents the state of a particle with momentum $k\,.$ I also learned that $\phi^...
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Using a time-like boundary as a computer?

Question and Summary Using classical calculations and the Robin boundary condition I show that one calculate the anti-derivative of a function within time $2X$ (I can compute an integral below) $$\...
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2answers
943 views

What's wrong with the square root version of the Klein-Gordon equation?

The Wikipedia article has a derivation of the Klein-Gordon equation. It gets to this step: $$\sqrt{\textbf{p}^2 c^2 + m^2 c^4} = E$$ and inserts the QM operators to get $$\left( \sqrt{ (-i \hbar \...
6
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2answers
365 views

fitting free QFTs into the Haag-Kastler algebraic formulation

Has the free Klein-Gordon quantum field theory been fitted into the Haag-Kastler algebraic framework? (Actually, John Baez told me "yes", and he should know.) If so, can you describe the basic ...
6
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1answer
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Bessel function representation of spacelike KG propagator

Preliminaries: In their QFT text, Peskin and Schroeder give the KG propagator (eq. 2.50) $$ D(x-y)\equiv<0|\phi(x)\phi(y)|0> = \int\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_\vec{p}}e^{-ip\cdot(x-y)...
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3answers
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Massless limit of the Klein-Gordon propagator

I am working with the propagator associated to the Klein-Gordon equation, as derived in "Quantum Physics a functional integral point of view", James Glimm, Arthur Jaffe or as derived here: http://www....