Basic QFT Question Driving Me Crazy

[latentcorpse]

I know that the full 2 point Green's function is given by $G_2(p,-p)=\frac{i}{p^2+m^2}$ since this is exactly how we define the physical mass (it is taken to be the pole of $G_2$).

So I know that's what the 2 point green's function should be - I just can't actually SHOW that it's equal to that!

Should I be using momentum space or euclidean space feynman rules to show this? Or should I be showing it using something other than Feynam rules e.g. explicit calculation?

Thanks for your help!

 

[dextercioby]

Use the generating functional. This method is quite spread around the literature. Check out Bailin & Love's text on gauge field theory.

 

[latentcorpse]

Use the generating functional. This method is quite spread around the literature. Check out Bailin & Love's text on gauge field theory.

Ahh. We have the definition $\langle \phi(x_1) \phi(x_2) \rangle = \frac{(-i)^2}{Z[0]} \frac{\delta^2 Z[J]}{\delta J(x_1) \delta J(x_2)} \vline_{J=0}$

where $Z[J]=e^{iS[\phi]+i \int d^dx J(x) \phi(x)}$

So doesn't this just get me back to the usual path integral representation:
$\langle \phi(x_1) \phi(x_2) \rangle = \frac{\int [d \phi(x) \phi(x_1) \phi(x_2) e^{i S[\phi]}}{\int d^dx e^{iS[\phi]}}$?

How does that help?

 

[dextercioby]

Well, Z[J] has a certain form in terms of sources and fields. You've spelled it out already

$$Z[J] = \exp \left\{ i\int d^4 x ... + i\int d^4 x J(x)\phi(x) \right\}$$

Now compute the functional derivatives and set the sources to 0.

 

[latentcorpse]

Well, Z[J] has a certain form in terms of sources and fields. You've spelled it out already

$$Z[J] = \exp \left\{ i\int d^4 x ... + i\int d^4 x J(x)\phi(x) \right\}$$

Now compute the functional derivatives and set the sources to 0.

Given that this is the full 2 point function, I assume there is an interaction so we should take
$Z[J]=e^{-i \int d^dx V(-i \frac{\delta}{\delta J(x)}) e^{i S_0[\phi] + i \int d^dx J(x) \phi(x)}$

Won't the V get in the way?

Ignoring the V though, I find:

$\langle \phi(x_1) \phi(x_2) \rangle = \frac{(-i)^2}{Z[0]} i^2 \phi(x_1) \phi(x_2) Z[0] = \phi(x_1) \phi(x_2)$

How's that?

 

[dextercioby]

The 2-point function is equal to the propagator (that you spelled out in the first post) iff there's no interaction in the Lagrangian density.

You said and I quote: <So I know that's what the 2 point green's function should be - I just can't actually SHOW that it's equal to that!>.

I've just hinted you on how to show that: for a free theory of a scalar field, you only need the definition of the 2 point function and 2 simple functional differentiations. For the interacting theory, you need to say to which order in the coupling constant (assumed tiny, so that the series expansions make sense until the order you need) you want to compute the quantum corrections to the free scalar particle's propagator.

More precisely, you need to know how the generating functional for the unconnected Green functions looks like and then how is the 2-point function related to it, irrespective of the presence of a self-interaction.

 

[latentcorpse]

The 2-point function is equal to the propagator (that you spelled out in the first post) iff there's no interaction in the Lagrangian density.

You said and I quote: <So I know that's what the 2 point green's function should be - I just can't actually SHOW that it's equal to that!>.

I've just hinted you on how to show that: for a free theory of a scalar field, you only need the definition of the 2 point function and 2 simple functional differentiations. For the interacting theory, you need to say to which order in the coupling constant (assumed tiny, so that the series expansions make sense until the order you need) you want to compute the quantum corrections to the free scalar particle's propagator.

More precisely, you need to know how the generating functional for the unconnected Green functions looks like and then how is the 2-point function related to it, irrespective of the presence of a self-interaction.

Sorry, I don't understand. Is my answer in the last post right? How do I show it's equal to the propagator I had in my original post then?

Thanks.

 

[dextercioby]

Sorry, I don't understand. Is my answer in the last post right? How do I show it's equal to the propagator I had in my original post then?

Thanks.

I've attached the formulae for the free scalar field in the coordinate representation. Passing to momentum representation is through a Fourier transformation of the propagator. The text is in Romanian and reads roughly

<For the field theory, the unconnected Green functions are defined through something similar to the finite degrees of freedom case

Formula 1

through analogy

Formula 2

If we integrate the Hamiltonian form of the path integral, we get the Lagrangian version

Formula 3>

You need the case n=2.

 

[latentcorpse]

I've attached the formulae for the free scalar field in the coordinate representation. Passing to momentum representation is through a Fourier transformation of the propagator. The text is in Romanian and reads roughly

<For the field theory, the unconnected Green functions are defined through something similar to the finite degrees of freedom case

Formula 1

through analogy

Formula 2

If we integrate the Hamiltonian form of the path integral, we get the Lagrangian version

Formula 3>

You need the case n=2.

I'm obviously getting a bit confused. Isn't that exactly what I've already done?

Also, why is there not going to be any interaction for the full propagator? Surely the full propagator includes everything (interactions or not interactions)?

 

[latentcorpse]

I think I've successfully shown this is true using the above relation.

I completed the square in the exponent to get $Z_0[J]=Z_0[0]\Delta_F(x-y)$

where we can show $\Delta_F(x-y)=\frac{-1}{p^2+m^2}$ by using the fact that it is a the inverse klein gordon operator i.e. a green's function for the klein gordon operator.

This then means that $\langle \phi(x_1) \phi(x_2) \rangle = (-i)^2 \frac{Z_0[0]}{Z_0[0]} \frac{-1}{p^2+m^2} = \frac{1}{p^2+m^2}$

This seems correct to me. Is it?

One slight problem is that I have had to use $\langle \phi(x_1) \phi(x_2) \rangle = (-i)^n \frac{1}{Z_0[0]} Z_0[J] \vline_{J=0}$ rather than $\langle \phi(x_1) \phi(x_2) \rangle = (-i)^n \frac{1}{Z[0]} Z[J] \vline_{J=0}$. Is it true to assume this is a free theory? As I mentioned in my above post, this seems a bit sketchy to me?

Thanks.