# Powers of an exponential in the pertubation expansion.

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I am reading the book called: "Gauge theory of elementary particle physics" by Ta-Pei Cheng and Ling-Fong Li.

On pages 19-20 they write: $(1.78) W[J] = \bigg[\exp(\int d^4x (\mathcal{L}_1 (\frac{\delta}{\delta J}))\bigg] W_0[J]$, where: $W_0[J]=\int [d\phi] \exp\bigg[ \int d^4x (\mathcal{L}_0+J\phi)\bigg]$.

Now, on page 20 they write:" The perturbative expansion in powers of $\mathcal{L}_1$ of the exponential in (1.78) gives:

$$(1.85) W[J] = W_0[J]\bigg\{ 1+\lambda\omega_1[J]+\lambda^2 \omega_2[J]+\ldots \bigg \},$$

where $$(1.86) \omega_1[J] = -\frac{1}{4!}W_0^{-1}[J]\bigg\{ \int d^4x \bigg[\frac{\delta}{\delta J(x)} \bigg]^4 \bigg\} W_0[J]$$

$$\omega_2[J]=-\frac{1}{2(4!)^2} W_0^{-1}[J]\bigg\{ \int d^4x \bigg[ \frac{\delta}{\delta J(x)}\bigg]^4\bigg\}^2 W_0[J] =$$

$$= -\frac{1}{2(4!)} W_0^{-1}[J]\bigg\{ \int d^4x \bigg[\frac{\delta}{\delta J(x)}\bigg]^4\bigg\} \omega_1[J]$$

Now, for my question, after I plug $\omega_1[J]$ into the above last equation I get:

$$\frac{1}{2(4!)^2} W_0^{-1}[J]\{ \int d^4 x \bigg[ \frac{\delta}{\delta J(x)} \bigg]^4 \} W_0^{-1}[J] \{ \int d^4 x \bigg[ \frac{\delta}{\delta J(x)} \bigg]^4 \} W_0[J]$$

The last expression is not the same as the above expression, i.e. of $-\frac{1}{2(4!)^2} W_0^{-1}[J]\bigg\{ \int d^4x \bigg[ \frac{\delta}{\delta J(x)}\bigg]^4\bigg\}^2 W_0[J]$.

Perhaps instead of $\omega_1[J]$ it should be $-W_0[J] \omega_1[J]$ in equation (1.86)?

I am puzzled, what do you think?

asked Oct 3, 2017
edited Oct 3, 2017

I think the opening curly bracket in (1.86) and below should be moved to the left to include the $W_0^{-1}$ term.

@Arnold Neumaier I see your point, but I don't think it's correct.

The $W_0^{-1}[J]$ should be before the integral in order to cancel the $W_0[J]$ in the expansion for $W[J]$ as written in (1.85).

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