bhobba said:
Normal ordering is just a way of handling the issue in a way that makes sense - but skirts the main issue - why do we have to resort to it in the first place. Like I said I am hopeful a better understanding of Effective Field Theory on my part will help - at least me anyway.
At last on PF, I see a question worth answering. It is, indeed, a very good question. I am glad that you brought this up, for I am not aware of any textbook or paper that tackles this issue. So, bellow you will see (probably for the first time) how the vacuum-subtraction (or normal ordering) arises almost naturally in QFT.
All good and bad features in QFT have their origin in the process of “
integrating by parts and ignoring surface terms”. A process which almost all authors use as if it is taken
for granted that surface terms do vanish. So, I will avoid this process by using expressions that can be derived with no reference to the behaviour at the boundary.
Let us consider a field theory whose action integral is invariant under the Poincare’ group. Then, by Noether theorem, we have a conserved translation current given by the energy-momentum tensor T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\varphi_{r})}\partial^{\nu}\varphi_{r} - \eta^{\mu\nu}\mathcal{L} \ , and a conserved Lorentz current given by the moment tensor M^{\rho\mu\nu} = x^{\mu}T^{\rho\nu} -x^{\nu}T^{\rho\mu} + S^{\rho\mu\nu} \ , S^{\rho\mu\nu}(x) = - i \frac{\partial \mathcal{L}}{\partial (\partial_{\rho}\varphi_{r})} \left( \Sigma^{\mu\nu}\right)_{r}{}^{s}\varphi_{s}(x) . And the corresponding Noether charges are given by P^{\mu} = \int d \sigma_{\rho}(x) \ T^{\rho\mu}(x) = \int d^{3}x \ T^{0 \mu}(x) \ , \ \ \ \ \ \ \ \ \ (A)J^{\mu\nu} = \int d \sigma_{\rho}(x) \ M^{\rho\mu\nu}(x) = \int d^{3}x \ M^{0 \mu\nu} (x) . \ \ \ \ \ \ \ (B) In a classical field theory, where we can almost always ignore surface terms, we can show that ( P^{\mu} , J^{\mu\nu}) are time-independent Lorentz vector and Lorentz tensor respectively. So, in a classical field theory, this means that ( P^{\mu} , J^{\mu\nu}) generate the proper Poincare’ algebra (i.e., without central charges). In other words, one can show that the classical ( P^{\mu} , J^{\mu\nu}) are unique Poincare generators. Clearly this cannot be true in a QFT because 1) surface integrals may not vanish, and 2) the uniqueness of ( P^{\mu} , J^{\mu\nu}) does not permit the possibility of vacuum subtractions.
Okay, let us start. People call Noether theorem “the beautiful theorem”. However, one property of Noether charge is
more beautiful that the entire theorem. Indeed we can show, with no reference to 1) symmetry considerations (i.e., conservation law), 2) dynamical consideration (i.e., equation of motion) and/or 3) the
behaviour at the boundary, that the Noether charge generates the correct infinitesimal transformation on local operators. In fact, using
only the canonical equal-time commutation relations and the expressions for ( P^{\mu} , J^{\mu\nu}), it is an easy exercise to show that [iP^{\mu} , \varphi (x)] = \partial^{\mu}\varphi (x) \ , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)[iJ^{\mu\nu}, \varphi (x)] = \left( x^{\mu}\partial^{\nu}- x^{\nu}\partial^{\mu} - i \Sigma^{\mu\nu} \right) \varphi (x) . \ \ \ \ \ (2)
So, these are the equations that we can start working with. Now, we use (1) and (2) to evaluate the RH sides of the following Jacobi identities
\big[ [P^{\mu},P^{\nu}] , \varphi (x) \big] = \big[ P^{\mu}, [ P^{\nu} ,\varphi (x)] \big] - \big[ P^{\nu}, [ P^{\mu} ,\varphi (x) ] \big] ,
\big[ [P^{\sigma} , J^{\mu \nu}] , \varphi (x) \big] = \big[ P^{\sigma} , [ J^{\mu \nu}, \varphi (x)] \big] - \big[ J^{\mu \nu} , [ P^{\sigma}, \varphi (x) ] \big],
\big[ [J^{\mu \nu} , J^{\rho \sigma}] , \varphi (x) \big] = \big[ J^{\mu \nu} , [J^{\rho \sigma}, \varphi(x) ] \big] - \big[ J^{\rho \sigma}, [J^{\mu \nu}, \varphi (x) ] \big] . After couple of pages of easy algebra, we obtain the following relations
\big[ [iP^{\mu} , P^{\nu}] , \varphi (x) \big] = 0 ,
\big[ [iP^{\sigma} , J^{\mu \nu}] - \eta^{\sigma \nu}P^{\mu} + \eta^{\sigma \mu}P^{\nu} , \varphi (x) \big] = 0 ,
\big[ [iJ^{\mu \nu} , J^{\rho \sigma}] - \eta^{\mu \rho} J^{\nu \sigma} - \eta^{\mu \sigma} J^{\rho \nu} - \eta^{\nu \rho} J^{\mu \sigma} - \eta^{\nu \sigma} J^{\rho \mu} , \varphi (x) \big] = 0 .
The most general solutions of these equations are given by
\big[ iP^{\mu} , P^{\nu}\big] = C^{\mu , \nu} ,
\big[ iP^{\sigma} , J^{\mu \nu}\big] = \eta^{\sigma \nu} P^{\mu} - \eta^{\sigma \mu} P^{\nu} + C^{\sigma , \mu \nu} ,
\big[ iJ^{\mu \nu} , J^{\rho \sigma} \big] = \eta^{\mu \rho} J^{\nu\sigma} + \cdots + C^{\mu \nu , \rho \sigma} , where all the C’s are constants. If these constants are not all zero, we conclude that the Noether charges \big( P^{\mu} , J^{\mu\nu}) generate a
centrally extended Poincare’ algebra. Next, we may consider various Jacobi identities to establish certain algebraic relations for the central charges. For example, the Jacobi identity
\big[ J^{\mu \nu} , [ P^{\rho}, P^{\sigma}] \big] + \big[ P^{\sigma} , [ J^{\mu \nu}, P^{\rho}] \big] + \big[ P^{\rho} , [ P^{\sigma}, J^{\mu \nu}] \big] = 0 , leads to C^{\mu , \nu} = 0 . Another useful Jacobi identity is
\big[ J^{\lambda \tau} , [ J^{\mu \nu} , P^{\rho}] \big] + \cdots \ = 0 . This allows us to express C^{\mu , \rho \sigma} as
C^{\mu , \lambda \tau} = \eta^{\mu \tau} \left( \frac{1}{3} \eta_{\rho \nu} C^{\rho , \lambda \nu}\right) - \eta^{\mu \lambda} \left( \frac{1}{3} \eta_{\rho \nu} C^{\rho , \tau \nu}\right), which may be used to define the constant C^{\mu} \equiv \frac{1}{3} \eta_{\rho \nu} C^{\rho , \mu \nu} . And the last Jacobi identity is between 3 J’s
\big[ J^{\lambda \tau} , [ J^{\mu \nu}, J^{\rho \sigma}] \big] + \cdots \ = 0 . This allows us to define yet another constant in terms of C^{\mu \rho , \nu \sigma} C^{\mu \nu} \equiv \frac{1}{2} \eta_{\rho \sigma} C^{\mu \rho , \nu \sigma} . Now, it is an easy exercise to show that the shifted Noether charges \bar{P}^{\mu} = P^{\mu} + C^{\mu},\bar{J}^{\mu \nu} = J^{\mu \nu} + C^{\mu \nu} , form an ordinary representation of the Poincare’ algebra, i.e., with
no central charges. The boring algebraic details of all this can be found in Weinberg’s book QFT, Vol.1, P(84-86). However, Weinberg
does not explain the important meaning of (C^{\mu} , C^{\mu \nu}) in QFT. So, we will do better than the
old man by showing that C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle \ ,C^{\mu \nu} = - \langle 0 | J^{\mu \nu} | 0 \rangle \ .
Okay, let us commute the Noether charge P^{\sigma} with the moment tensor M^{\rho \mu \nu}(x). By the linearity of the bracket, we get
\big[ i P^{\sigma} , M^{\rho\mu\nu}(x) \big] = x^{\mu} \big[iP^{\sigma} , T^{\rho \nu}(x) \big] - x^{\nu} \big[ iP^{\sigma} , T^{\rho \mu}(x) \big] + \big[iP^{\sigma} , S^{\rho \mu \nu}(x) \big] . Using the fact that Eq(1) holds for any local operator, we get \big[ iP^{\sigma} , M^{\rho \mu \nu}(x) \big] = x^{\mu}\partial^{\sigma}T^{\rho \nu} - x^{\nu} \partial^{\sigma}T^{\rho \mu} + \partial^{\sigma}S^{\rho \mu \nu} . We rewrite this as \big[ iP^{\sigma} , M^{\rho \mu \nu}(x) \big] = \partial^{\sigma}M^{\rho \mu \nu} - \eta^{\sigma \mu}T^{\rho \nu} + \eta^{\sigma \nu}T^{\rho \mu} . Integrating this over the hyper-surface \int d \sigma_{\rho}(x), and using the definitions of the Noether charges [Eq’s (A) and (B)], we obtain \big[ iP^{\sigma} , J^{\mu\nu} \big] = \eta^{\sigma \nu}P^{\mu} - \eta^{\sigma \mu}P^{\nu} + \int d \sigma_{\rho}(x) \ \partial^{\sigma}M^{\rho \mu \nu}(x) . Now, we rewrite this in terms of the shifted Noether charges (\bar{P}^{\mu} , \bar{J}^{\mu\nu}). Using P^{\mu} = \bar{P}^{\mu} - C^{\mu} and J^{\mu\nu} = \bar{J}^{\mu\nu} - C^{\mu\nu}, we find \big[ i\bar{P}^{\sigma} , \bar{J}^{\mu\nu} \big] = \eta^{\sigma \nu}\bar{P}^{\mu} - \eta^{\sigma \mu}\bar{P}^{\nu} + \left( \eta^{\sigma \mu}C^{\nu} - \eta^{\sigma \nu}C^{\mu} + \int d \sigma_{\rho}(x) \ \partial^{\sigma}M^{\rho \mu \nu}(x) \right) .
Since the shifted Noether charges satisfy the ordinary Poincare’ algebra (no central charges), we must have
\eta^{\sigma \nu}C^{\mu} - \eta^{\sigma \mu}C^{\nu} = \int d \sigma_{\rho} \ \partial^{\sigma}M^{\rho \mu \nu}(x) . \ \ \ \ \ \ (3)
Of course, if M^{\rho \mu \nu}(x) \to 0 as |\vec{x}| \to \infty, then we can apply the Schwinger identity \int d \sigma^{\rho}(x) \ \partial^{\sigma}F(x) = \int d \sigma^{\sigma}(x) \ \partial^{\rho}F(x) on the RHS of (3) and obtain \eta^{\sigma \nu}C^{\mu} - \eta^{\sigma \mu}C^{\nu} = \int d \sigma^{\sigma}(x) \ \partial_{\rho}M^{\rho \mu \nu}(x) = 0, because of the conservation law \partial_{\rho}M^{\rho \mu \nu} = 0. This then leads to C^{\mu} = 0. However, in QFT it is not always true that the operator M^{\rho \mu \nu} vanishes at infinity. So, we cannot always use the Schwinger identity. Instead, we will stay away form the behaviour at infinity and try to determine the constant C^{\mu} from Eq(3). Contracting Eq(3) with \eta_{\sigma \nu} and doing the differentiation on the moment tensor, leads us to
3C^{\mu} = \eta_{\sigma \nu} \int d \sigma_{\rho} \left( x^{\mu} \partial^{\sigma}T^{\rho \nu} - x^{\nu}\partial^{\sigma}T^{\rho \mu} + \partial^{\sigma}S^{\rho \mu \nu} \right) - 3P^{\mu}. Now, if we take the vacuum expectation value, the integrand vanishes by translation invariance of the vacuum: \langle 0 | \partial \mathcal{O}(x) | 0 \rangle = \partial \langle 0 | \mathcal{O}(0) | 0 \rangle = 0. Thus, we obtain C^{\mu} = - \langle 0 | P^{\mu}| 0 \rangle , and the vacuum subtraction is, therefore, justified \bar{P}^{\mu} = P^{\mu} - \langle 0 | P^{\mu} | 0 \rangle .
Similar, but more complicated calculation, leads to \bar{J}^{\mu\nu} = J^{\mu \nu} - \langle 0 | J^{\mu \nu} | 0 \rangle .
