notes on quantization of Gauge field II — De Witt–Faddeev

notes on quantization of Gauge field II — De Witt–Faddeev–Popov Approach

By xil41

There is a good reason to use De Witt–Faddeev–Popov approach in practice to quantize the gauge field. It’s easy. Compare with using Dirac brackets, this functional approach is much easier to get the Feynman rules, especially for non-Abelian case.

Consider partition function

$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha exp(iI[A,\psi,\psi^\dagger])$ .

Here I is the gauge invariant action: $I[A,\psi,\psi^\dagger] = I[{}^\Omega A,{}^\Omega \psi,{}^\Omega \psi^\dagger]$ . And also the measure is assumed to be gauge invariant say, $D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha = D {}^\Omega A^\alpha_\mu D {}^\Omega \psi^{\alpha\dagger} D {}^\Omega \psi^\alpha$ .However, in the gauge field case, $Z$ will not be convergent, for all those fields can be different by a gauge, and leave the physical results unchanged. In this way, the integral will count the same value over and over again hence making $Z$ diverge. So intuitively, we should expect that $Z = N \cdot z$ , where $z$ is the value of the integral for one gauge choice, and N is the number of possible gauges which is of course infinity. If we can write $Z$ as $Z = N \cdot z$ , we can just strip off the unimportant overall factor $N$ and only take care of the value of $z$ to calculate physical interesting stuffs.

The way to realize this is to insert some $\delta$ function to fix the gauge. For example if we use the Lorentz gauge, we’d better be able to insert something like $\delta(\partial^\mu A^\alpha_\mu - \lambda)$ and not change the value of $Z$ .

To proceed, we refresh our mind by reviewing a character for $\delta$ -function:

$\delta(f(x)-f(a)) = \delta(x-a)/Det|f'(a)|$ ,

which leads to $Det|f'(a)| \delta(f(x)-f(a))= \delta(x-a)$ .And the determinant $Det|f'(a)|$ is independent of $x$ .

Here we will encounter almost the same case, except for replacing $x$ by some functional. We are about to insert in $Z$ a constant $\int D \Omega Det|\Delta| \delta(f(\Omega, x)-\lambda) = 1$ , where the $f(\Omega,x)$ is some gauge choice. this obviously leaves $Z$ unchanged. The integral was over all gauges $\Omega$ . Here , $Det|\Delta|$ is called Faddeev–Popov determinant and it’s easy to show that it is independent of gauge choice $\Omega$ .
After doing so and noting those gauge independent elements, we get
$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \Omega Det|\Delta| \delta(f(\Omega, x)-\lambda) exp(iI[A,\psi,\psi^\dagger])$ .

$=\int D \Omega \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha Det|\Delta| \delta(f(\Omega_{fix}, x)-\lambda) exp(iI[A,\psi,\psi^\dagger])$

Now we can drop the overall factor $\int D \Omega$ and if by indroducing a fermion like field (Faddeev–Popov ghost) making

$Det|\Delta|= \int D \omega^\dagger D \omega exp(i \omega^\dagger \Delta \omega)$ ,

we get

$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \omega^\dagger D \omega \delta(f(\Omega_{fix}, x)-\lambda) exp(iI[A,\psi,\psi^\dagger]+i \omega^\dagger \Delta \omega)$ .

Finally we multiply $Z$ by some distribution function, usually guassian-type $\int D\lambda exp(- \frac{i}{2\xi} \lambda^2)$ , and use the fact that $Z$ doesn’t depend on $\lambda$ . So we can integrate out the $\delta$ -function and get
$Z = \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \omega^\dagger D \omega exp(iI[A,\psi,\psi^\dagger]-i \frac{i}{2\xi} f^2+ i \omega^\dagger \Delta \omega)$

$= \int D A^\alpha_\mu D \psi^{\alpha\dagger} D \psi^\alpha D \omega^\dagger D \omega exp(iI_{NEW} [A,\psi,\psi^\dagger, \omega^\dagger, \omega]).$
Here $I_{NEW} [A,\psi,\psi^\dagger, \omega^\dagger, \omega] = I[A,\psi,\psi^\dagger]-i \frac{i}{2\xi} f^2+ i \omega^\dagger \Delta \omega$ . The second term is the gauge fixed term, the third the ghost field.

After all, we should note here that though we choose a gauge, this $Z$ really does NOT care about what kind gauge we have chosen. That is to say after you evaluate a physical interesting thing such as cross section, the $\xi$ will be canceled out and shouldn’t show up in your final results(if it does, you make some mistakes).

Last we take the QCD as an example. We will use both Lorentz gauge( $\partial^\mu A^\alpha_\mu = 0$ ) and axial gauge( $A^\alpha_3 = 0$ ). In QCD, $A^\alpha_\mu$ transform as $A^\alpha_\mu \rightarrow A^\alpha_\mu + D_\mu \Omega^\alpha$ .

So for Lorentz gauge $f^\alpha =\partial^\mu A^\alpha_\mu$ , $\Delta = \frac{\partial f^\alpha}{\partial \Omega^\beta} = \delta_{\alpha \beta} \partial^\mu D_\mu$ .

And for axial case, $f^\alpha =A^\alpha_3$ , $\Delta = \frac{\partial f^\alpha}{\partial \Omega^\beta} = \delta_{\alpha \beta} \partial_3$ (remember that $A^\alpha_3 = 0$ ). So in the axial case, we see that the ghosts decouple and by gauge indepence, they should decouple in general.

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notes on quantization of Gauge field II — De Witt–Faddeev–Popov Approach

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