What Are the Implications of Srednicki's \langle 0 \mid \phi(x) \mid 0 \rangle?

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SUMMARY

The discussion centers on the implications of Srednicki's expression \langle 0 \mid \phi(x) \mid 0 \rangle =\frac{1}{i} \frac{\delta}{\delta J(x)} W_1 (J) \mid_{J=0}, which relates to the sum of all diagrams initiated by a single source. The introduction of the counterterm Y \phi in the Lagrangian is crucial for ensuring the vacuum expectation value is zero, which is necessary for the LSZ formula's validity. The participants clarify that the sum of connected diagrams with a single source is indeed zero, and this holds true even when substituting the source with another subdiagram, as the integral remains zero if f(x) is zero.

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LAHLH
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Hi,

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

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 [tex]Y \phi[/tex] 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 a lot if so.
 
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LAHLH said:
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?

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

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

If we put the source back, we have [itex]\int d^4x\,J(x)f(x)[/itex]. But since [itex]f(x)=0[/itex], the integral is also zero!
LAHLH said:
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?

Same argument. Replace [itex]J(x)[/itex] with [itex]h(x)[/itex], where [itex]h(x)[/itex] is some other subdiagram with a source missing and the endpoint labeled [itex]x[/itex]. Since [itex]f(x)=0[/itex], we have [itex]\int d^4x\,h(x)f(x)=0[/itex]. So the whole diagram is zero if any part of it is zero.
 
Thanks a lot. 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?
 

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