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Normal ordering

  1. Aug 8, 2005 #1
    Hi,

    I have a question regarding normal ordering in QFT.


    I know that it moves dagger operators to the right and that it gets rid of infinite constants turning up and that 'with observables defined in normal products their vacuum expectations vanish.' (Mandl and Shaw)

    I have a couple of questions.

    1) Is this an ad-hoc intruduction, I mean, if it changes the answer then how do you justify doing it (even if it does get rid of infinity).

    2) If the vacuum expectation vanishes then that means it goes to zero right? Then what is the point if your answer vanishes?

    Please help enlighten me here!

    Richard
     
  2. jcsd
  3. Aug 8, 2005 #2

    Haelfix

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    1) Is this an ad-hoc intruduction, I mean, if it changes the answer then how do you justify doing it (even if it does get rid of infinity).

    Mathematically this is an extremely involved and deep question, and the answer to your question is yes and no. A lot of it comes down to rather conceptual problems with 'quantization' in general. Most of the time, you cannot uniquely quantize a classical phase space, unless you satisfy the Stone Von Neumann theorem. Groenewold and Van Hove then proved that in pretty general circumstances, you are guarenteed to not be able to find a unique unitary lie algebra homomorphism between the Heisenberg algebra and the extended (projective) algebra of the symplectic group (or rather the metaplectic group), read there is no way to quantize a general symplectic manifold completely. This is basically the mathematicians way of talking about 'operator ordering ambiguties'. Hence there is an infinite amount of different ways to quantize the subalgebras. So in that sense, yes it is ad hoc, justified to a physicist b/c it satisfies experiment.

    Now, it turns out that 'normal ordering' can sometimes be promoted to a rather special role (you will need to dl a course on geometric and/or deformation quantization to get the details) under certain axioms. Unfortunately the details of this are complicated.

    2) Yes if your vacuum expectation 'vanishes' that means <0|H|0> = 0. I don't know if there is a 'point' to wanting to make it vanish, other than being a nice whole number that everyone likes to do algebra with =), in general for most field theories it doesn't (it tends to be infinite actually, and you have to subtract an infinite reference point quantity to make it finite or zero.. for instance the harmonic oscillator)
     
  4. Aug 8, 2005 #3
    Hi,

    Thanks for the detailed answers! It never ceases to amaze me how many experts are on this forum!

    So, if I have caught the drift here - it is nothing more than a 'quick n easy' way to get rid of the infinities.

    In general I wouldn't be solving a QFT eqtn and just arbitrarily decide to normal order the operators, it would be in the presence of something that would be infinte otherwise that I can make go to zero.

    Does that sound about right?
     
  5. Aug 18, 2005 #4

    Haelfix

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    Hi Robousy.. The answer is yes. Unless you are working with rather sophisticated highly abstract formalisms, you typically don't see the rather peculiar fundamentals going on in the operator algebra. You can lookup some of Gribov's early papers, he goes into a lot of detail about this.

    In most calculation friendly formalisms, like what most people use, the problem usually appears as infinities. Hence the rather *formal* prescription.

    Also, the ordering ambiguties dissappear when your lagrangian is at most quadratic or lower.
     
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