1
$\begingroup$

I study by myself with the QFT, in the page 197 of book of Lewis H. Ryder (2nd edition), The author wrote that he take the functional derivative of equation 6.69:

$$\frac {\delta\widehat{Z}[\phi]}{\delta\phi}$$

where $$\widehat {Z}[\phi]=\frac{{e}^{iS}}{\int{{e}^{iS}}{\cal D}\phi}\tag{6.69}$$

and

$$S=-\int{\left[\frac {1}{2} \phi(\Box+{m }^{ 2 })\phi -{\cal L }_{ int } \right] { d }^{ 4 }x }.\tag{6.71} $$

The result in Eq. 6.72 is:

$$\frac { \delta }{ \delta \phi } \left\{ \exp\left[ -i\int { \left[ \frac { 1 }{ 2 } \phi (\Box +{ m }^{ 2 })\phi-{\cal L}_{ int } \right] } { d }^{ 4 }x \right] \right\} { \left[ \int { \exp\left[ iS \right] } {\cal D}\phi \right] }^{ -1 }\\= \left( \Box +{ m }^{ 2 } \right) \phi \widehat { Z } [\phi ]-\frac { \partial { \cal L }_{ int } }{ \partial \phi }\widehat { Z } [\phi ].\tag{6.72} $$

I don't understand how the calculating procedure taking place. I have known how to calculate the functional derivative to a functional, but I do not know how to take it to a functional integral like $\widehat{Z}[\phi]$. I would be most thankful if anyone help me.

PS: Is there are some detailed textbook or literature about this technique?

|cite|improve this question
$\endgroup$
  • 2
    $\begingroup$ Just like you take any other derivative. But watch out that the denominator (the actual functional integral) does not depend on $\phi$, as it is being integrated over there, so you only have to derive the action in the numerator. $\endgroup$ – David Vercauteren Jul 7 '15 at 8:41
  • $\begingroup$ The denominator is just a normalisation constant independent of $\phi$. So the derivative acts only on the numerator and the result is just $\hat{Z}' = iS' \hat{Z}$ as claimed (with the Factor $i$ missing). $\endgroup$ – Tom Heinzl Jul 7 '15 at 8:45
  • $\begingroup$ Oh, god!thanks! please keeping online, I immediately take a calculation.And I also have some questions about functional derivative involving functional integral $\endgroup$ – alxandernashzhang Jul 7 '15 at 8:48
  • $\begingroup$ I take a calculation just now,I found:$\frac{\delta}{ \delta\phi}\frac{1}{C}{e}^{iS}=\frac{1}{C}\frac{ \partial}{\partial\phi}{e}^{iS}=\frac{1}{C}{e}^{iS }i\frac{\partial}{\partial\phi}S\left[\phi\right]$ with $C$ is denomenator,the derivative $\frac {\partial}{\partial\phi}S\left[\phi \right]$ involving a integral of dx, why the result in Lewish. Ryder's book is lacking this integral? $\endgroup$ – alxandernashzhang Jul 7 '15 at 9:08
  • $\begingroup$ Functional differentiation "kills" space-time integrals: If you write the variation of the action as $\delta S = \int d^4 x \, S'[\phi] \, \delta \phi(x)$, then $S'[\phi] = \delta S/\delta \phi$ is precisely the functional derivative of the action $S$. $\endgroup$ – Tom Heinzl Jul 7 '15 at 9:22
1
$\begingroup$

Hints to the question (v2):

  1. First of all, one should realize the abuse of notation in eq. (6.69) of Ref. 1 where $\phi$ is used in two meanings: both as an external parameter and as internal integration/dummy variable. It is more properly written as $$ \widehat {Z}[\phi]~=~\frac{{e}^{iS[\phi]}}{{\cal N}}, \qquad {\cal N}~:=~\int\!{\cal D}\phi~e^{iS[\phi]} .\tag{6.69'}$$

  2. Similarly, eq. (6.72) is essentially equivalent to $$ \frac { \delta \widehat { Z } [\phi ] }{ \delta \phi(x)} ~=~\frac{1}{\cal N} \frac {\delta {e}^{iS[\phi]} }{ \delta \phi(x)} ~=~\frac{{e}^{iS[\phi]}}{{\cal N}}\frac{\delta S[\phi ]}{\delta\phi(x) } ~=~\widehat { Z } [\phi ]\frac{\delta S[\phi ]}{\delta\phi(x) }.\tag{6.72'} $$ In the first equality we stress that we shouldn't differentiate the denominator ${\cal N}$ wrt. $\phi(x)$ because the denominator ${\cal N}$ doesn't depend on $\phi(x)$, cf. comment by David Vercauteren.

  3. Finally, the functional derivative $\frac{\delta S[\phi ]}{\delta\phi(x) }$ becomes the Euler-Lagrange expression.

References:

  1. L.H. Ryder, QFT; Section 6.4, p. 197.
|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks a lot @Qmechenic,but I also have a puzzle about the notion ":" in front of equality（“=”）in 6.69'， what it is mean? $\endgroup$ – alxandernashzhang Jul 22 '15 at 8:40
  • $\begingroup$ @alxandernashzhang: The notation $A:=B$ means $A$ is defined to be equal to $B$. Similarly, the notation $A=:B$ means $B$ is defined to be equal to $A$. $\endgroup$ – Qmechanic♦ Jul 22 '15 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.