7
$\begingroup$

I am trying to study the canonical formulation of Yang-Mills theories so that I have direct access to the $n$-particle of the theory (i.e. the Hilbert Space). To that end, I am following Kugo and Ojima's (1978) 3-part paper.

At the outset, I am confused by their Lagrangian, and their two differences from the conventional one (I write the Lagrangian 2.3 of their paper):

$$\mathcal{L}=-\frac{1}{4}F^a_{\mu\nu}F^{a\,\mu\nu}-i\partial^\mu\bar{c}D_\mu^{ab}c^b-\partial^\mu B^a A_\mu^a+\alpha_0 B^a B^a/2$$

  1. They have rescaled the Ghost field so that its kinetic term has a factor of $i$ in front.
  2. They have integrated by parts on $B^a\partial_\mu A^{a\,\mu}$, effectively making $B$ dynamical.

The authors chose these two differences are so that (1) BRS variations (eq 2.15 in their paper) preserve Hermiticity of the Ghost fields, and (2) to make the Lagrangian BRS invariant.

I am totally confused by their second point. I thought the standard BRS Lagrangian appearing in standard texts, for example, in Peskin and Schroeder was already BRS invariant. Why the $\partial B.A$ term?

$\endgroup$
1

2 Answers 2

5
$\begingroup$

Kugo and Ojima's work was one of the major breakthroughs in understanding the role of BRST in the quantization of gauge theories. Historically BRST was discovered in the path integral formalism. The understanding of this theory as a cohomology theory started from the Kugo and Ojima's work.

Now, the action is BRST invariant with and without the Gaussian integration over the auxiliary field $B^a$ (called the Lautrup-Nakanishi multipliers). They are introduced in order to not have explicit dependence on the gauge parameter in the Ward identities (please see a recent review by Becchi). The BRST invariance Ward identities is a crucial step in the unitarity proof.

Kugo and Ojima actually solved the BRST cohomology problem of the Yang-Mills theory. They actually identified the physical and unphysical states of the theory (in terms of the BRST operator) as follows: The physical states correspond to the states annihilated by the BRST operator and in addition of positive norm.

The unphysical states are arranged in degenerate quartets. This is called the Kugo - Ojima quartet mechanism. One quartet corresponds to a ghost, anti-ghost, longitudinal and temporal gluons. In their formalism these states can be generated from the vacuum by the action of the ghost antighost operators as well as by the field $B^a$ and its conjugate momentum. They also conjectured that since colored quark operators and transversal gluon operators belong to the quartet sector, then these states must be confined.

$\endgroup$
1
  • $\begingroup$ Although this reply does not directly address my question, it is incredibly informative. I always welcome historical anecdotes regarding achievements and breakthroughs made in theoretical physics! $\endgroup$
    – QuantumDot
    Commented Nov 4, 2012 at 2:14
1
$\begingroup$

Adding a total divergence term $-\partial^{\mu}(B^aA_{\mu}^a)$ (and other$^{1}$ total divergence terms) to the Lagrangian density does not change the classical equations of motion. In particular, this change doesn't make the Lautrup-Nakanishi field $B^a$ a propagating field.

One may prove that both the old and the new actions are BRST invariant.

As to why Kugo and Ojima choose to add the total divergence terms in the path integral formalism, if they are not being cavalier about boundary terms, I'm guessing it is linked to their choice of boundary conditions, and to have the most straightforward correspondence to the operator formalism, where all the field variables of the model should have well-defined Hermitian properties wrt. the inner product structure of the Hilbert space, and where the quartet mechanism takes place.

--

$^{1}$ Kugo and Ojima also integrate the Faddeev-Popov determinantal term by part (as compared to Peskin et al.).

$\endgroup$
3
  • $\begingroup$ I don't understand how performing an integration by parts makes "all the field variables of the model have well-defined Hermitian properties wrt. the inner product structure of the Hilbert space". Would you elaborate on this? Thanks! $\endgroup$
    – QuantumDot
    Commented Oct 30, 2012 at 15:16
  • $\begingroup$ Is it straightforward to show that certain representations of the Lagrangians would make all field variables have well-defined Hermitian properties wrt inner product structure of the Hilbert space? What does that mean, anyway? $\endgroup$
    – QuantumDot
    Commented Nov 1, 2012 at 23:39
  • $\begingroup$ Please, please, please; can you tell me what you mean by "all the field variables of the model have well-defined Hermitian properties wrt. the inner product structure of the Hilbert space"? Thanks! $\endgroup$
    – QuantumDot
    Commented Nov 4, 2012 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.