# Integration on the way to Generating Functional for the free Dirac Field

1. Mar 3, 2009

### Phileas.Fogg

Hi,
if I want to calculate the generating functional for the free Dirac Field, I have to evaluate a general Gaussian Grassmann integral. The Matrix in the argument of the exponential function is (according to a book) given by:

I don't understand the comment with the minus-sign and the integration by parts. I tried to do this integration, but didn't get the same result.

Could anyone tell me, how this integration works explicitly?

Or is there a mistake in the book and in the last line it should be +m (not -m) ?

Regards,
Mr. Fogg

2. Mar 3, 2009

### Avodyne

The minus sign comes from integration by parts:

$$\int d^4x\,[\partial^\mu f(x)]g(x) = -\int d^4x\,f(x)\partial^\mu g(x)$$

where we assume $f(x)g(x)$ vanishes on the surface at infinity. In your case, $f(x)=\delta^4(x'-x)$ and $g(x)=\psi(x)$.

3. Mar 4, 2009

### Phileas.Fogg

Hi,

Let me show you my calculation so far:

$$\int d^4 x' \overline{\psi}(x') \int d^4 x (-i \hbar \gamma \partial^{\mu} \delta^4(x' - x) - m \delta^4(x'-x)) \psi(x)$$

$$= \int d^4 x' \; \overline{\psi}(x') \int d^4 x (-i \hbar \gamma \partial^{\mu} \delta^4(x' - x) \psi(x) - m \delta^4(x'-x) \psi(x))$$

As you suggested

$$f(x) = \delta(x' - x) , g(x) = \psi(x)$$

$$= \int d^4 x' \; \overline{\psi}(x') \left( -i \hbar \gamma( \left[ \delta^4(x'-x) \psi(x) \right]_{-\infty}^{+\infty} - \int d^4 x \delta^4(x'-x) \partial \psi(x)) - m( \psi(x)^2 )_{-\infty}^{+\infty} + m \int d^4 x \psi(x) \delta^4(x'-x) \right)$$

Somewhere must be a mistake. Can you help me to find it?

Regards,
Mr. Fogg

4. Mar 4, 2009

### jensa

Hi,

I am not sure what you did with the mass-term or why you did it. This term can be trivially integrated over $x'$ due to the delta function:

$$-\int d^4x\int d^4x'\bar{\psi}(x)m\delta^4(x-x')\psi(x')=-\int d^4x \bar{\psi}(x) m \psi(x)$$

So you only need to make the partial integration on the derivative term. This you do precisely like Avodyne said, and what your first two terms in your last equation say. So take your last equation, discard the mass-terms, and then use the fact that $\bar{\psi}(x)\psi(x)\rightarrow 0, \ x\rightarrow \pm \infty$ to drop the very first term. Finally the integration over the second term is again trivial and gives you

$$-\int d^4 x'\int d^4 x\bar{\psi}(x')\delta(x-x')(-i\hbar)\gamma\cdot \partial\psi(x)=\int d^4x \bar{\psi}(x) i\hbar\gamma\cdot \partial \psi(x)$$