# Correlation function of a Klein-Gordon field

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• Gaussian97
In summary, the conversation discusses the notation and operators in Hamiltonian mechanics. It also mentions the Klein-Gordon field and its Heisenberg picture. The conversation raises questions about the disappearance of the term ##e^{iPx}## and the Lorentz invariance of ##|\Omega\rangle##. The response clarifies that ##\hat{P}^\mu## is the normal-ordered operator and explains that assuming ray representations of the proper orthochronous Poincare group does not yield anything new.
Gaussian97
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TL;DR Summary
Evaluating a correlation function for the Klein-Gordon field I found a term like ##\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle## which should be equal to ##\langle\Omega|\phi(0)|\lambda_{\vec{p}}\rangle e^{-ipx}\left.\right|_{p^0=E_\vec{p}}##
First, let me introduce the notation; given a Hamiltonian ##H## and a momentum operator ##\vec{P}##, and writing ##P=(H,\vec{P})##. Let ##|\Omega\rangle## be the ground state of ##H##, ##|\lambda_\vec{0}\rangle## an eigenstate of ##H## with momentum 0, i.e. ##\vec{P}|\lambda_\vec{0}\rangle=0## and ##|\lambda_\vec{p}\rangle## a ##\vec{p}##-boost of ##|\lambda_\vec{0}\rangle##.

If ##\phi(x)## is the Klein-Gordon field, in the Heisenberg picture ##\phi(x)=e^{iPx}\phi(0)e^{-iPx}## and then $$\langle\Omega|\phi(x)|\lambda_{\vec{p}}\rangle=\langle\Omega|e^{iPx}\phi(0)e^{-iPx}|\lambda_{\vec{p}}\rangle$$. Then I understand that ##e^{-iPx}|\lambda_{\vec{p}}\rangle=e^{-ipx}|\lambda_{\vec{p}}\rangle\left.\right|_{p^0=E_\vec{p}}## because ##|\lambda_{\vec{p}}\rangle## is a ##\vec{p}##-boost of a eigenstate of ##H##. But the term ##e^{iPx}## disapears and I don't understand why. The easy answer would be to suppose that ##H|\Omega\rangle=\vec{P}|\Omega\rangle=0##, the second one I think one can argue that must be satisfied by the ground state, but not the first, in general ##H|\Omega\rangle=E_0|\Omega\rangle\neq 0##.

Also, I for another step I need to use that ##|\Omega\rangle## is Lorentz invariant, what I think makes sense, but also I'm not sure how to prove it, so maybe both problems are related.

Thank you

Note that ##x## are c-number valued four-vector comonents and ##P## is the operator of total momentum. Now you have
$$\exp(-\mathrm{i} \hat{P} \cdot x)|\lambda_{\vec{p}} \rangle=\exp(-\mathrm{i} p \cdot \vec{x}) |\lambda_{\vec{p}} \rangle$$
and
$$\exp(-\mathrm{i} \hat{P} \cdot x)|\Omega \rangle=|\Omega \rangle.$$
Now use this to evaluate the one-point function in question.

vanhees71 said:
$$\exp(-\mathrm{i} \hat{P} \cdot x)|\Omega \rangle=|\Omega \rangle.$$
Now use this to evaluate the one-point function in question.

I don't see why ##e^{-\mathrm{i} \hat{P} \cdot x}|\Omega \rangle=|\Omega \rangle.## It's because even if the energy of ##|\Omega\rangle## is not zero it's simply a constant phase and then its unobservable?

Gaussian97 said:
I don't see why ##e^{-\mathrm{i} \hat{P} \cdot x}|\Omega \rangle=|\Omega \rangle.## It's because even if the energy of ##|\Omega\rangle## is not zero it's simply a constant phase and then its unobservable?
Here ##\hat{P}^{\mu}## is assumed to be the normal-ordered operator, so ##\hat{P}^{\mu}|\Omega \rangle=0##.

vanhees71
Gaussian97 said:
I don't see why ##e^{-\mathrm{i} \hat{P} \cdot x}|\Omega \rangle=|\Omega \rangle.## It's because even if the energy of ##|\Omega\rangle## is not zero it's simply a constant phase and then its unobservable?
Indeed. You don't get anything new when assuming ray representations of the proper orthochronous Poincare group. All are equivalent to the usual unitary representations. That's why it's just convenient to normal order energy, momentum and angular momentum, i.e., to make the vacuum the state where all these additive conserved quantum numbers are 0.

## 1. What is the correlation function of a Klein-Gordon field?

The correlation function of a Klein-Gordon field is a mathematical measure of the relationship between two points in space and time within a quantum field theory. It describes the probability amplitude for finding a particle at one point given that another particle is located at a different point.

## 2. How is the correlation function of a Klein-Gordon field calculated?

The correlation function is calculated using the Klein-Gordon equation, which is a relativistic wave equation that describes the behavior of scalar particles. The equation is solved for the field operator at two different points, and the resulting solution is used to determine the correlation function.

## 3. What is the significance of the correlation function in quantum field theory?

The correlation function plays a crucial role in quantum field theory as it allows us to make predictions about the behavior of particles in a quantum system. It provides information about the probability of particle interactions and can be used to calculate physical observables such as scattering amplitudes and decay rates.

## 4. How does the correlation function change with different values of the Klein-Gordon field?

The correlation function is dependent on the values of the Klein-Gordon field at each point in space and time. As the field changes, the correlation function also changes, reflecting the changing probability of particle interactions at different points in the quantum system.

## 5. Can the correlation function of a Klein-Gordon field be experimentally measured?

Yes, the correlation function can be experimentally measured through techniques such as scattering experiments or particle decay experiments. By comparing the measured correlation function with the theoretical predictions, scientists can test the validity of the Klein-Gordon equation and the underlying quantum field theory.

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