Functional Derivative: Computing the d'Alembert Solution

[parton]

In the literature (Ryder, path-integrals) I have found the following relation for the functional derivative with respect to a real scalar field $$\phi(x)$$:

$$i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) ( \square + m^2 ) \phi(x)} = ( \square + m^2 ) \phi(x) e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) ( \square + m^2 ) \phi(x)}$$

But how do I compute this? I am just confused about this d'Alembert operator $$\square$$ and I never end up with the right solution as above.

Could anybody explain how to obtain this solution, please?

 

[Cyosis]

It would help if you show us your steps until the point where the d'Alembert operator is causing problems.

Using the Minkowski metric (+---) (c=1) the d'Alembert operator is given by $\partial_{tt}-\partial_{xx}-\partial_{yy}-\partial_{zz}$. With this you can evaluate the integral $\int d^4 x \; \phi(x) \square \delta \phi(x)$ using partial integration. What do you know about the volume integrals that appear due to partial integration? Try it for $\int d^4x \; \phi(x) \partial_{xx} \phi(x)$. The other terms work exactly the same.

 

[parton]

Ok, my problem is the following. If I consider

$$i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) m^2 \phi(x)}$$

I could write the exponent as $$-i \int \mathrm{d}^{4} x \frac{1}{2} m \phi^{2}$$ and use some kind of product rule and obtain:

$$i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) m^2 \phi(x)} = m^2 \phi(x) e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) m^2 \phi(x)}$$

Ok, that seems to be correct, but now I if want to compute

$$i \dfrac{\delta}{\delta \phi(x)} e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) \square \phi(x)}$$

I can not rewrite the exponent in such a form like above and otherwise I don't really know how to build the derivative.

I just differentiated the first $$\phi(x)$$ and obtained


$$= \dfrac{1}{2} \square \phi(x) e^{-i \int \mathrm{d}^{4} x \frac{1}{2} \phi(x) \square \phi(x)}$$

where I have a factor 1/2 which is wrong, but I don't know how to derive this correctly.

 

[Cyosis]

You have treated $\square \phi(x)$ as a constant now, but it is not a constant.

Take the variation of the argument of the exponent.

$$\delta \int d^4x \; \phi(x) \square \phi(x) = \int d^4x \; \left[ \delta \phi(x) \square \phi(x) +\phi(x) \square \delta \phi(x) \right]$$

Now integrate the second term by parts.

 

[parton]

Ok, I integrated the second term by parts (2 times) and finally obtained the correct result.

Thanks a lot for your help