# Srednicki P67

1. Feb 22, 2010

### LAHLH

Hi,

Srednicki has $$\langle 0 \mid \phi(x) \mid 0 \rangle =\frac{1}{i} \frac{\delta}{\delta J(x)} W_1 (J) \mid_{J=0}$$

Which is the sum of all diagrams with that started with a single source (before the differentiation) now with the source removed.

Since we want this to be zero for the validity of the LSZ formula, we introduce counterm $$Y \phi$$ in the Lagrangian. We can choose this Y appropriatley at various orders of g now such that the vacuum expectation value is zero as we require.

I'm OK with all so far.

Now Srendicki says that because we have done this the sum of all connected diagrams with a single source is zero. Shouldn't this be the sum all diagrams with a single source with the source removed is zero?

Then he also goes on to say if we replace the single source in each of these diagrams with another subdiagram (any subdiagram), the sum of these is still zero. I do not understand why this is the case?

Could anyone please explain, thanks alot if so.

2. Feb 22, 2010

### Avodyne

Sure, but then it's also zero if you put the source back.

That is, let $f(x)$ be the sume of all connected diagrams with a single source removed (and $x$ is the coordinate label where the source was removed). We adjust $Y$ so that $f(x)=0$.

If we put the source back, we have $\int d^4x\,J(x)f(x)$. But since $f(x)=0$, the integral is also zero!
Same argument. Replace $J(x)$ with $h(x)$, where $h(x)$ is some other subdiagram with a source missing and the endpoint labeled $x$. Since $f(x)=0$, we have $\int d^4x\,h(x)f(x)=0$. So the whole diagram is zero if any part of it is zero.

3. Feb 23, 2010

### LAHLH

Thanks alot. that makes sense.

On a maybe related note, on P97, we only sum over the 1PI diagrams, I'm just wondering why only these diagrams?