# Powers of an exponential in the pertubation expansion.

+ 2 like - 0 dislike
663 views

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).

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification