Comments on: When Feynman Diagrams Fail Thoughts on work and life from particle physicists from around the world. Tue, 06 Oct 2015 08:06:00 +0000 hourly 1 By: Alec Battles Thu, 23 Feb 2012 12:13:07 +0000 awesome blog! i just came across it via a google image search.

one gripe. i find the first feynman diagram *much* uglier than the second.

By: Luboš Motl Wed, 15 Dec 2010 05:09:43 +0000 Dear Flip, just a detail: L_{string}, referring to the typical length of the string moving in 10 dimensions, always appears in the even powers, just like the fine-structure constant.

It’s just the 11-dimensional L_{Planck}, the typical length scale of the 11-dimensional M-theory which contains no strings but only M2-branes, M5-branes etc., where you get powers that are multiples of 3. That’s a part of why M-theory (in d=11 etc.) remains more mysterious than (perturbative) string theory (in d=10).

A recent “membrane minirevolution” describing some exotic 3D theories with a Lagrangian has somewhat reduced this mystery but much of it remains.


By: Flip Tanedo Wed, 15 Dec 2010 02:55:28 +0000 Ah, now I understand why the alpha expansion was so interesting. :-) I didn’t know about the third power of L(string) in string theory, though admittedly my string background is rather limited.

Thanks for the link about the Michigan conference.

By: Luboš Motl Sun, 12 Dec 2010 21:19:18 +0000 By the way, this is the LHC blog, and I am very curious whether we will hear anything nontrivial from Michigan (LHC First Data) on Tuesday.

However, we already know what we will hear from your competitors in the Fermilab:

The bbb MSSM channel is exciting, indeed. 😉

By: Luboš Motl Sun, 12 Dec 2010 19:40:27 +0000 Dear Flip, retroactively, I am actually sure you knew the correct answers before I wrote my comments 😉 so my addition was purely pedagogical. Your scratching is giving too much credit to me. 😉

Reviewing the expansions, note that the cross sections are always expansions in alpha=e^2, not sqrt(alpha) – either because of squaring before summation, or because of pair-appearance of the vertices in sums that may interfere. This is no coincidence – only “alpha” is a real physical parameter. Funnily enough, there is a corresponding statement in M-theory but the exponent is not 2 but 3.

All simple enough physical expressions in M-theory are expansions in L_{Planck}^3. Note that Newton’s constant is L_{Planck}^9 – the exponent is the dimension minus two (just like it is 2 in 3+1 dimensions, area may be divided by Newton’s constant to get dimensionless entropies).

Also, the M2-branes have tension 1/L_{Planck}^3 anfd the M5-branes have tension scaling like 1/L_{Planck}^6 because the exponents are the worldvolume dimensionalities – all the exponents are multiples of 3. This appearance of multiples of 3 may be understood from some kind of U-duality argument, also reflected in del Pezzo surfaces via the mysterious duality.

However, the M-theory case is much more complicated because it’s not just about some perturbative counting of pairs of vertices etc. These are seemingly modest patterns but they do indicate that we don’t understand a certain conceptual argument that will make these things as obvious as the counting of vertices in perturbative expansions.


By: Flip Tanedo Sun, 12 Dec 2010 07:08:51 +0000 Actually, I guess I should mention that comments like these are one of the reasons why I really like blogs as a medium for discourse; we can have nice discussions to clarify subtle points (usually points that I miss originally).

It is also why I try to make corrections with the original text in “strike out” and with attributions to those who contribute constructive comments. Apart from intellectual honesty and issues of citations*, it is a nod to those who contribute polite corrections/further information that adds to the post.

(*–by ‘citation’ I mean hyperlinks, not actual citations from reputable works! This is not a peer/editor-reviewed journal!)

Anyway, thanks for the discussion everyone.

By: Flip Tanedo Sun, 12 Dec 2010 07:00:46 +0000 Hi Lubos — yep, I (and I think ChriSp) completely agree with what you’ve written. I’ve updated the original post to something more innocuous and have attributed you and ChrisSp.

The point that you make about there is indeed a term linear in e from the cross term is what I meant by “modulo interference terms” and the subsequent points about whether one takes an exclusive versus inclusive cross section are what I was nodding to when I mentioned soft photons and such. But you’re right that I didn’t make this explicit and that I certainly didn’t think too carefully about it when I wrote the original post. :-)

Anyway, I think (hope) everyone now agrees and further that what we agree upon is correct. :-) For the sake of the intended outreach audience of the blog, I think I’ll leave the details of this discussion in the comments rather than making extensive revisions to the main post. 😀

By: Luboš Motl Sun, 12 Dec 2010 06:43:12 +0000 Dear Flip, I wanted to correct you about statement B but thought you were just popularly simplifying things.

But ChriSp has corrected you and you have revealed that you meant it literally but the statement is not 100% accurate. So let me tell you why you statement is inaccurate in some more detail.

The probability is computed as the probability amplitude squared (absolute value of that). The probability amplitude looks like

(Term0 + e.Term1 + e^2.Term2 + …)

Now, try to square it. What will you get?

Term0^2 + 2e.Term0.Term1 + e^2.(Term1^2 + 2.Term0.Term2) + …

Now, look that the leading contribution goes like Term0^2. But the subleading one, the first one that depends on the diagram Term1 with an extra vertex, is suppressed by one power of “e” only. That’s because the squaring of the amplitude contains the mixed terms that are products of the old term in the amplitude – without the new vertex – and the new one – with the vertex.

The probability formula also contains terms suppressed by e^2 but they come after that.

Just to be sure, we must also realize that the actual pure QED amplitudes for a process with fixed external particles actually contain either “only odd powers of e” or “only even powers of e” because one can only add new cubic QED vertices in pairs.

So in the real world, the expansion of the amplitude is

(Term0 + e^2.Term1 + e^4.Term2 + …)

and its square is

Term0^2 + 2e^2.Term0.Term1 + e^4.(Term1^2 + 2.Term0.Term2) + …

Note that the adjacent terms in the actual probability above differ by a factor scaling like e^2 i.e. alpha. However, the first subleading term, proportional to Term0.Term1, while it’s suppressed by alpha – and there is never sqrt(alpha) relative factor appearing in the formulae for probabilities (in this respect, Flip would be always right) – is actually coming from mixing the original minimal diagram Term0 and another diagram that however has “two more QED vertices” and not one as Flip wrote.

So the simplest fix is to replace

“For example, each photon vertex gives a factor of roughly α=1/137″


For example, each pair of photon vertices (and for diagrammatic reasons, they can only be added in pairs) gives a factor of roughly α=1/137 …

If one computes inclusive cross sections, then the external states don’t have to be identical among the diagrams. And one can add probabilities one by one. By in that case, there is no interference (there can’t be any interference between states with different numbers of particles). The total probability has the form

Term0^2 + e^2.Term1^2 + …

because the diagrams Term0, Term1 – with different numbers of external photons – are squared before they’re added. So in this inclusive context, Flip’s original statement was totally right. 😉

But I am probably just repeating what you said differently now.

Best wishes

Best wishes

By: ChriSp Sun, 12 Dec 2010 06:30:28 +0000 Hi Flip, thank you for correcting the text. I apologize that I added to the confusion by writing amplitude when I meant cross section/probability in my second comment.

By: Flip Tanedo Sun, 12 Dec 2010 00:28:24 +0000 Hi ChriSp, thanks for the correction! I can’t believe I let that slip by me… and I used the phrase incorrectly several times. -_-‘ I’ve now corrected them above, leaving the original text crossed-out as a reminder of my shame. :-)

Regarding your comment (B), I knew someone would bring this up! I was trying to keep things brief (unsuccessfully) so didn’t want to get into things short. I did try to be careful though, and said that each QED vertex gives a factor of alpha to the *probability* (i.e. cross section), not the amplitude. This, I believe, is a correct statement modulo interference terms, which I didn’t want to get into. :-)

And yes, it is true that for a given process one should include *two* additional vertices if you don’t want an additional external photon. (Of course, one should in principle include external soft photon emissions and such… but I suppose soft gluons are already taken care of by the hardonization programs?)

Anyway, the statement about the alpha suppression to the probability was meant to be very hand-wavy just to demonstrate the main idea of the expansion. You are correct that there are subtleties (cross terms, matching precise final states, etc).

Thanks for the comments,

By: ChriSp Sat, 11 Dec 2010 23:13:11 +0000 Actually, there is no interference since the initial or final state is has an extra photon. But the argument would be true if you add two vertices.

By: ChriSp Sat, 11 Dec 2010 23:05:50 +0000 Great post, but two comments:
A) When experimentalists talk about the “underlying event” they mean everything but the hard scattering.
B) More nit-picky: Adding a new diagram with an extra vertex in QED doesn’t always come with a factor of alpha in the amplitude. The new diagram can interfere with the others, and then the new contribution goes like sqrt(alpha).

By: Luboš Motl Sat, 11 Dec 2010 13:58:37 +0000 Well-written, indeed. I suppose you would prefer longer comments, wouldn’t you? :-)

By: Mel Jennings (Meltronx) Sat, 11 Dec 2010 12:58:51 +0000 Nicely done. Thanks!

By: paolo Sat, 11 Dec 2010 08:14:57 +0000 Nice, thanks.