How to calculate the functional derivative of the functional integral?

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.

• 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. – David Vercauteren Jul 7 '15 at 8:41
• 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). – Tom Heinzl Jul 7 '15 at 8:45
• Oh, god!thanks! please keeping online, I immediately take a calculation.And I also have some questions about functional derivative involving functional integral – alxandernashzhang Jul 7 '15 at 8:48
• 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? – alxandernashzhang Jul 7 '15 at 9:08
• 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 $S$. – Tom Heinzl Jul 7 '15 at 9:22

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.
• Thanks a lot @Qmechenic,but I also have a puzzle about the notion ":" in front of equality（“=”）in 6.69'， what it is mean? – alxandernashzhang Jul 22 '15 at 8:40
• @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 $A$. – Qmechanic Jul 22 '15 at 9:03