Green's function with same time and spatial arguments

[taishizhiqiu]

Concerning green's function with the same time and spatial argument(i.e. $G_0(x,t;x,t)$, mostly in QFT), I have the following question

1. Is green's function well defined at this point?

2. if green's function is well defined at this point, is it continuous here?

3. In quantum many body theory, I am instructed to view $G_0(x,t;x,t)$ as $G_0(x,t;x,t+0^+)$ in feynman diagrams. Why is this so?

 

[vanhees71]

The Green's functions in vacuum QFT are defined as the time-ordered products of field operators and as such a priori singular when their space-time arguments become equal. That becomes clear from the equal-time commutator relations. For some field $\hat{\phi}$ and its canonical field momentum $\hat{\Pi}$ it reads
$$[\hat{\Phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
In non-relativistic many-body theory you have a Schrödinger field. The free field consists only of a annihilation piece, and thus the special rule given in your OP means that a closed loop connecting the same space-time point in a Feynman diagram should be interpreted as the expectation value of a normal ordered piece in the Hamiltonian, i.e., a density. This automatically subtracts the always diverging vacuum pieces of such tadpole diagrams.

 

[taishizhiqiu]

The Green's functions in vacuum QFT are defined as the time-ordered products of field operators and as such a priori singular when their space-time arguments become equal. That becomes clear from the equal-time commutator relations. For some field $\hat{\phi}$ and its canonical field momentum $\hat{\Pi}$ it reads
$$[\hat{\Phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
In non-relativistic many-body theory you have a Schrödinger field. The free field consists only of a annihilation piece, and thus the special rule given in your OP means that a closed loop connecting the same space-time point in a Feynman diagram should be interpreted as the expectation value of a normal ordered piece in the Hamiltonian, i.e., a density. This automatically subtracts the always diverging vacuum pieces of such tadpole diagrams.

Thanks for your post, still:
1. Why in QFT I am never instructed to view $G_0(x,t;x,t)$ as $G_0(x,t;x,t+0^+)$ in feynman diagrams?
2. Tadpole diagrams is divergent in QFT. Is there any relationship between $G_0(x,t;x,t)$ and the divergence?

 

[vanhees71]

In vacuum QFT a tadpole loop is a connection between field operators within $\mathcal{H}_{\text{int}}$. Usually you can assume normal-ordering of the Hamiltonian and just skipt the diagram. There's one caveat: Usually (naive) normal ordering is not gauge invariant, and you cannot naively leave the diagram out. So it's more convenient to keep the diagrams and renormalize them along with the other divergences (most simple examples: tadpole diagram for the photon polarization in scalar QED; tadpole diagram in gluon polarization in QCD).

 

[taishizhiqiu]

In vacuum QFT a tadpole loop is a connection between field operators within $\mthcal{H}_{\text{int}}$. Usually you can assume normal-ordering of the Hamiltonian and just skipt the diagram. There's one caveat: Usually (naive) normal ordering is not gauge invariant, and you cannot naively leave the diagram out. So it's more convenient to keep the diagrams and renormalize them along with the other divergences (most simple examples: tadpole diagram for the photon polarization in scalar QED; tadpole diagram in gluon polarization in QCD).

Renormalization deals with diagrams that's divergent because of large momentum($k$). Is tadpole diagram divergent because of green's function of the same argument or because of large momentum? And what's the relationship between the two things?

 

[kau]

Greens functions are of two types... one is retarded and other one is advanced.. retarded green's functions is causal meaning it respect Lorentz symmetry.. And then you can define Feynman propagator accordingly. Anyway the question of green's function evaluated at same time at same space physically mean I think what is the probability that a state at that particular spacetime point would collapse to the state defined at other spacetime point. So I think if Space and time both are same then It should be just 1...

 

[vanhees71]

No, the propagator is singular at equal space-time points. This is clear from the equal-time commutation relations for field operators. This is the origin of UV divergences in the Feynman rules.

 

[taishizhiqiu]

No, the propagator is singular at equal space-time points. This is clear from the equal-time commutation relations for field operators. This is the origin of UV divergences in the Feynman rules.

Can you explain more explicitly why equal space-time green's function corresponds to UV divergences? You can refer to peskin's book if necessary.

 

[vanhees71]

Well take a free uncharged Klein-Gordon field for Simplicity. The Lagrangian is
$$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi)(\partial^{\mu} \phi)-\frac{m}{2} \phi^2$$
and the canonical field momentum thus
$$\Pi(x)=\frac{\partial \mathcal{L}}{\dot{\phi}}=\dot{\phi}(x).$$
Thus you have
$$[\phi(t,\vec{x}),\Pi(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
The time-ordered propagator (which is for vacuum QFT the Feynman propagator)
$$\mathrm{i} G(x-y)=\langle \mathcal{T}_c \phi(x) \phi(y)$$
thus has the correct Green's function property
$$(\Box_x+m^2) \mathrm{i} G(x-y) =\delta^{(4)}(x-y).$$
For an explicit expression for the Green's function, which you can evaluate via the Fourier transform from the momentum-space representation
$$G(x)=\int \frac{\mathrm{d}^4 k}{(2 \pi)^4} \frac{1}{k^2-m^2 + \mathrm{i} 0^{+}} \exp(-\mathrm{i} k \cdot x).$$
It's very clear that this integral for $x=0$ does not exist, because it diverges for large four momenta (UV divergence). You find a discussion of this integral in some detail in the beginning sections of Peskin and Schroeder.