Archive for January, 2007

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

January 9, 2007

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 I — Dirac Bracket

January 4, 2007

 Dirac bracket is an extension of Poisson bracket, used to deal with the Hamiltonian with constraints. Especially it is useful in quantizing gauge theories. In detail, Dirac bracket is defined as [A,B]_D= [A,B]_P - [A, \Phi_i]_P (\Delta^{-1})^{ij} [\Phi_j,B]_P, where [,]_P is the Poisson bracket, \Phi is some constraints such as a gauge condition, and \Delta is some matrix we will talk about later.

Suppose there exits a Hamiltonian H with a set of constraints \{ \Phi_i \approx 0 \} (Here “\approx ” means equations hold on shell, say after evaluate all the Poisson brackets then turn on the constraints conditions. We call this “weakly equal to”). For example, in E&M theory, L = F^{\mu \nu}F_{\mu \nu}. So the canonical momenta are \Pi^\mu=F^{0\mu}. We immediately get a set of primary constraints that \Pi^0 \approx 0(we call this a set of constraints, because the equation holds on every point of space time). These constraints are needed to hold all the time, so [\Pi^0,H] = 0 is demanded; hence a set of secondary constraints \partial_i \Pi^i \approx 0 follows. There are no more constraints here, because [\partial_i \Pi_i,H] = 0 reduces to 0 = 0(trivial). If it’s a non-trivial equation, we keep on this process until we get a “0 = 0“.

According to Dirac, we decompose those constraints into two classes: the first-class constraints and the second-class constraints. The first-class constraint means its Poison bracket with all the other constraints vanishes [\Phi_i,\Phi_j]_P \approx 0 on the other hand, the second-class does not, which indicates that no linear combination \sum c_i [\Phi_i,\Phi_j]_P vanishes. As the E&M example above, \Pi^0 and \partial_i \Pi^i are first-class constraints.

In quantizing the gauge field, the primary first-class constraints(such as \Pi^0 \approx 0) or maybe all the first-class constraints are harmless for they can be eliminated by a choice of gauge. So we took linear combination of constraints so that as many constraints can be put into first-class as possible then eliminate them by choosing gauge. After that, we left with a set of second-class constraints.

The existence of the second-class constraints indicates some degrees of freedom are not physically important. The naive idea to deal with them is just to “throw them away” and only keeping those degrees of freedom of physical importance. For example, suppose that we have constraints p_1 \approx 0, q_1 \approx 0, we can just take p_1 and q_1 as identically 0 . This is a simplest example. For generalization, Dirac proposed the following scheme:

Since for second-class constraints no linear combination \sum c_i[\Phi_i,\Phi_j]_P vanishes, then Det(\Delta) \neq 0, where \Delta_{ij} \equiv [\Phi_i,\Phi_j]_P. Hence \Delta is reversible. By introducing Dirac bracket [A,B]_D = [A,B]_P - [A,\Phi_i]_P (\Delta^{-1})^{ij}[\Phi_j,B]_P, we can get all commutation relations for canonical variables. And use them to quantize gauge field theory. We can check by using the p_1 \approx 0, q_1 \approx 0, that this scheme is the same as we throw away p_1 and q_1.

Still use the E&M case as an example. First as we mentioned above , \Pi^0 = 0 and \partial_i \Pi^i = 0 are first class constraints. Now we propose a gauge(here we use coulomb gauge) \partial^i A_i = 0 . Together with \partial_i \Pi^i=\partial_i F^{i0} = 0, we can eliminate A^0(In this case A^0= 0). Then we left with two second-class constraints: \partial_i \Pi^i = 0 and the gauge condition \partial^i A_i = 0, since \Delta^{1 \bold{x},2 \bold{y}} = -\Delta^{2 \bold{y},1 \bold{x}} = [\partial^i A_i, \partial_i \Pi^i]_P = \nabla^2 \delta( \bold{x}-\bold{y}), Δ1x,1y=Δ2x,2y = 0, Det(Δ)≠0. Finally, we use the Dirac bracket we get, [Ai, Πj]D= iδijδ(x-y)+i ∂2/∂i∂j(1/4π|x-y|). And extend this to quantum version, to quantize E&M.

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