Green's function of the Klein-Gordon operator

In summary, the author argues that the free field propagator of perturbation theory in vacuum (defined as the vacuum-expectation value of free field operators) satisfies the differential equation of the Green's function of the free Klein-Gordon operator. This is done by taking the derivatives of the expectation value of the time-ordered field-operator product, and using the canonical equal-time commutation relations for the field operators.
  • #1
guillefix
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Again, from the Peskin and Schroeder's book, I can't quite see how this computation goes:

See file attached

The thing I don't get is how the term with [itex](\partial^{2}+m^{2})\langle 0| [\phi(x),\phi(y)] | 0 \rangle[/itex] vanishes, and also why they only get a [itex]\langle 0 | [\pi(x),\phi(y)] | 0 \rangle[/itex] from the [itex]\partial_{t}\langle 0 | [\phi(x),\phi(y)] | 0 \rangle[/itex] and not also a [itex]\langle 0 | [\phi(x),\pi(y)] | 0 \rangle[/itex]
 

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  • #2
Me neither. Peskin/Schroeder is pretty unclear concerning propagators and its analytic properties.

First of all, one has to specify which propagator one is talking about, and this depends on what you want to do with it. In the case of perturbation theory in vacuum qft you need the time-ordered propagator, which is defined as the vacuum-expectation value of free field operators (here for an uncharged Klein-Gordon field)
[tex]\mathrm{i} D(x-y)=\langle 0|\mathcal{T}_c \hat{\phi}(x) \hat{\phi}(y)|0\rangle.[/tex]
Now you plug in the expansion of the field operator in terms of creation and annihilation operators
[tex]\hat{\phi}(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega(\vec{p})}} [\hat{a}(\vec{p}) \exp(-p \cdot x) + \hat{a}^{\dagger}(\vec{p}) \exp(+p \cdot x) ]_{p^0=\omega(\vec{p})}.[/tex]
Then you can write the propgator after some algebra with vacuum expectation values of annihilation and creation operator products as
[tex]\mathrm{i} D(x-y)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3 2 \omega(\vec{p})} \left [\Theta(\xi^0) \exp(-\mathrm{i} p \cdot \xi) + \Theta(-\xi) \exp(+\mathrm{i} p \cdot \xi) \right]_{p^0=\omega(\vec{p}),\xi=x-y}.[/tex]
Now you take the Fourier transform of this wrt. [itex]\xi[/itex] with a regulating factor [itex]\exp(-\epsilon |\xi^0|)[/itex], which leads you to
[tex]\tilde{D}(p)=\int_{\mathbb{R}^4} \mathrm{d} \xi D(\xi) \exp(+\mathrm{i} p \cdot \
\xi)=\frac{1}{p^2-m^2+\mathrm{i} \epsilon}.[/tex]
The [itex]\mathrm{i} \epsilon[/itex] has to be understood to be taken in the weak limit [itex]\epsilon \rightarrow 0^+[/itex].

For a more detailed explanation, why one has to use this time-ordered propagator, and also this derivation, see my QFT manuscript,

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Chapter 3.
 
  • #3
guillefix said:
The thing I don't get is how the term with [itex](\partial^{2}+m^{2})\langle 0| [\phi(x),\phi(y)] | 0 \rangle[/itex] vanishes, and also why they only get a [itex]\langle 0 | [\pi(x),\phi(y)] | 0 \rangle[/itex] from the [itex]\partial_{t}\langle 0 | [\phi(x),\phi(y)] | 0 \rangle[/itex] and not also a [itex]\langle 0 | [\phi(x),\pi(y)] | 0 \rangle[/itex]
It is because of the free field eqn. (∂2+m2)[itex]\phi[/itex]=0,the last term vanishes.Note also that ∂μ will act only on operators whose argument is x,not y.Also the first term will involve a by part to get the result.Rest is just simple calculation based on commutator relation and some property of step function.
 
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  • #4
andrien said:
Note also that ∂μ will act only on operators whose argument is x,not y.

That was the key I was missing, probably missed due to writting it so abstractly :P. Thanks!
 
  • #5
That's not writing it "abstractly" but just bad notation. If there are more than one space-time argument involved, you always should write explicitly wrt. which one you take partial derivatives.

Now, I guess what you want to show is that the Feynman propagator of free fields fulfills the differential equation of the Green's function of the free Klein-Gordon operator. For this you just take the derivatives of the expectation value of the time-ordered field-operator product, writing the time-ordering symbol out in terms of Heaviside unitstep functions. Then you use
[tex]\partial_{t_1} \Theta(t_1-t_2)=-\partial_{t_2} \Theta(t_2-t_1)=\delta(t_1-t_2)[/tex]
and the canonical equal-time commutation relations for the field operators.
 

1. What is the Klein-Gordon operator?

The Klein-Gordon operator is a mathematical operator used in quantum field theory to describe the behavior of scalar particles. It is a relativistic generalization of the Schrödinger operator and is often used to study the dynamics of particles with spin 0.

2. What is the Green's function of the Klein-Gordon operator?

The Green's function of the Klein-Gordon operator is a mathematical function that represents the response of the system to a point source. It is a solution to the Klein-Gordon equation with a delta function source term.

3. How is the Green's function of the Klein-Gordon operator used?

The Green's function of the Klein-Gordon operator is used to solve for the wave function of a particle in a potential. It is also used to calculate scattering amplitudes and to study the behavior of particles in a quantum field theory.

4. What is the relationship between the Klein-Gordon operator and the Dirac operator?

The Klein-Gordon operator is related to the Dirac operator through the Dirac equation, which is a relativistic wave equation for particles with spin 1/2. The Klein-Gordon operator is a special case of the Dirac operator when the spin is 0.

5. What are some applications of the Green's function of the Klein-Gordon operator?

The Green's function of the Klein-Gordon operator has a wide range of applications in physics, including in quantum field theory, statistical mechanics, and solid state physics. It is also used in engineering and other fields to study wave propagation and scattering in various systems.

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