# Normal ordering

1. May 18, 2009

### Bobhawke

My understanding is this: In the canonical approach to QFT there is an ambiguity in the order in which we write operators when calculating matrix elements. The different choices just correspond to different vacuum energies, which we are free to ignore since we only measure energy differences. So we are free to choose any ordering prescription we like - normal ordering is the special choice where all the creation operators are to the left of annihalation operators.

My questions are:

1. I heard a claim that normal ordering, defined as placing all the creation operators to the left of the annihalation operators, fails to remove the infinity from all diagrams. In particular the diagram in $$\phi^4$$ theory which I will now try to describe:

Draw a straight line between 2 points x and y. Now bring a circle down so it makes contact with the line in only one place, forming a 4-vertex with the line. Now take 2 points on that circle, and draw 2 separate lines connecting those 2 points.

Is this true? And if it is, couldnt we just define a different ordering in which all the divergences go away?

2. Can normal ordering be used to remove the infinities in any theory? I heard a claim that it can fail in interacting theories, and that is when we bring in the heavy artillery of renormalisation.

3. Is normal ordering equivalent to a particular renormalisation scheme?

2. May 18, 2009

### meopemuk

Normal ordering has no physical significance. It is just a bookkeeping convention.

3. May 18, 2009

### genneth

Infinities crop up at various points in quantum field theory, for different reasons. Normal ordering gets rid of a constant offset to the energy. Quite a few others remain. Normal ordering and renormalisation have nothing to do with each other.

4. May 18, 2009

### xepma

What you are talking about is divergencies that pop up due to ultraviolet contributions of loop diagrams. The diagram that you are describing has a loop in it. This means that momentum can flow along this loop. Naively, there is no restriction on "how much momentum" may flow along this loop. As a consequence, there is infinite amount of momentum contributing to this particular Feynamn diagram, and so contribution of the Feynman diagram to the overall process comes out as infinity.

To solve it, you need this thingy called Renormalisation.

But all this still pops up irrespective of what your conventions are for "normal ordering".

5. May 18, 2009

### Bob_for_short

I would like to say that in my opinion the infinite (or big when regularized) perturbative corrections to the fundamental constants are due to too bad initial approximation (free particles). I wrote a couple of popular papers on this subject and I propose another initial approximation where electron "interaction" with quantized electromagnetic field is taken into account exactly. The perturbative corrections in such an approach are different - they are finite and no corrections to the fundamental constants appear (http://arxiv.org/abs/0811.4416, http://arxiv.org/abs/0806.2635). So there is no IR and UV divergences at all.

Beware that I was severely criticized for not completing the relativistic calculations and for not comparing my series with the QED's ones. According to many, my claims are too optimistic without these complete calculations. So I propose to take my findings critically. I personally am sure in my results.

Bob.

Last edited by a moderator: Apr 24, 2017
6. May 20, 2009

### Bobhawke

Ah yes I was confused by a comment one of my professors made. Thanks for the replies everyone