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The topology for the lowest order $γγh$ interaction in the Standard Model is given by

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where we have an intermediate top quark loop. I am confused with the charge conservation at each vertex. Since both the photon and Higgs boson are neutral, and the top quark has a charge of $(3/2)e$, I think we will need a top quark and a top anti-quark for each vertex. However, how is this possible for this triangle loop?

Also, what's the order of this diagram? If we consider the $γγh$ amplitude at $O(\lambda_te^2)$, does that mean we will have one vertex with a top quark and two with photons (I think I'm very confused with this)?

Qmechanic
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IGY
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  • Why use a quark and not an electron positron pair? – FlatterMann Apr 13 '23 at 13:13
  • @FlatterMann I just saw this example with quark on my lecture note. – IGY Apr 13 '23 at 13:22
  • Interesting. Works just as well, of course... but it's much more messy and a much higher energy term, so it's probably not leading order for low energy photons, is it? Are you trying to estimate an effective photon mass? Does the mechanism create an effective photon mass? – FlatterMann Apr 13 '23 at 13:28
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    the coupling to the top quark from the higgs is indeed much, much higher than the coupling to the electron from the higgs (also the reason the top quark has higher mass). This is the correct leading order diagram for diphoton production from higgs decay - and the photons arent low energy-they carry away the energy of the higgs! As for whether this can be interpreted as a photon mass (considering an effective field theory where the top loop is replaced with an effective vertex)I'm not sure, but I don't think so. I think this vertex rule has dimension -3, and thus is not the same as a mass term. – AXensen Apr 13 '23 at 13:35
  • @AXensen The coupling to the Higgs is stronger, of course, but is the probability of a quark/antiquark pair not much lower than for a less massive electron/positron pair? I am not talking about the high but the low energy limit. Intuitively I don't think this leads to a photon mass, either, but I have never seen a theoretical argument (which is a function of me not reading QFT textbooks, of course). – FlatterMann Apr 13 '23 at 13:43
  • Yeah, I see your point - if we're talking about a process where a low energy photon scatters off a higgs, it may be that lower mass particles going around the loop are leading order because the propogators are small unless $p^2\sim m^2$. But such a process would be extremely uncommon - either suppressed by propogators or by vertices, and in some cases hard to define since the photon frequency might be smaller than the lifetime of the higgs. – AXensen Apr 13 '23 at 13:49

1 Answers1

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Such a great question - an internal fermion line is not distinctly identified as either a particle or an antiparticle. The only necessary condition is that arrows pointing in and arrows pointing out of an an electromagnetic interaction point go in opposite directions.

An incoming fermion, or an outgoing antifermion is associated with a spinor (often denoted $u(p)$). An incoming antifermion, or an outgoing fermion is associated with a dirac adjoint of a spinor (denoted $\bar{u}(p)=u^\dagger(p)\gamma^0$). Then the vertex rule for electromagnetism is $-ie\gamma^\mu$. An outgoing photon is associated with a polarization 4-vector $\epsilon_\mu(p)$. So, for example, if an incoming fermion scatters off a photon, all the matrix operatiosn make sense, the matrix element is proportional to: $$ -ie\bar{u}(p')\gamma^\mu u(p)\epsilon_\mu(p_{\gamma}) $$ An internal fermion line doesn't have a spinor, it has an entire dirac matrix associated with it. The propogator is: $$ \frac{i(p_\mu\gamma^\mu_{ij}+m I_{ij})}{p^2-m^2+i\varepsilon} $$ Here I've taken some care to write the spinor indices of this dirac matrix, and $I$ is an identity matrix. Then the vertex rule+other fermion that the arrow points at is contracted with the $j$ index, and the vertex rule that the arrow points away from is contracted with the $i$ index.

This is the sense in which people say the horrific and poorly defined phrase "an antiparticle is a particle moving backward in time." The vertex on top could be interpreted as making an antifermion that goes down and a fermion that goes to the right. Or it could be phrased as an incoming fermion from the bottom and an outgoing fermion going right. Or maybe it's an incoming fermion from the bottom and an incoming antifermion from the right which annihilate. Who's to say? There's no well defined direction that time goes in this diagram (consider, for example, moving that top vertex of the triangle to the right of the rightmost vertex, and you'll find that both of their time-flow directions have been changed). And besides, for most of the loop integral the quarks propagators are off-shell and aren't really particles anyway. There may be some way to write the propogators when they're near-on-shell as $\bar{u}_i(p)u_j(p)$ and then there might be a way to identify them as a specific linear combination of particle and antiparticle for that particular value of $p$ - I'm not sure.

At the end of the day, all this crap about making sure lorentz indices ($\mu,\nu$) and spinor indices ($i,j$) are all summed over or "contracted" with another vector or matrix like object is the reason that charge is conserved. In other words, if you tried to make a Feynman diagram that didn't conserve charge, you would simply be drawing a diagram that didn't make any sense matrix-wise - like multiplying two vectors together (what would that even mean??)

AXensen
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  • Thanks so much for the answer!! This means internal particles are neutral in charge, right? Even if we have a diagram without loop, the spinor indices of internal fermion are still cancelled with the incoming and outgoing fermions? – IGY Apr 13 '23 at 15:37
  • And what does the 'order' mean? Here we have a loop with 3 vertices, should we say the order is $(_^2)$? – IGY Apr 13 '23 at 15:40
  • Sorry I missed that part of the question. "Order" refers to a back of the envelope way of calculating matrix elements where you only include terms that are very small. You could say its $O(\lambda_te^2)$, but since $\lambda_t\sim 1$, you could probably also say $O(e^2)$. Since $m_t\sim m_H\sim E_\gamma$, the internal lines are not suppressing the diagram by order $(m_h/m_t)^{(3or6)}$ and can be ignored. Keep in mind this is the matrix element though, so the cross section (or decay rate or whatever you're calculating) is $O(e^4=\alpha^2)$ – AXensen Apr 13 '23 at 16:57
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    hmmm... unfortunately "are the internal particles neutral" feels wrong to me. I think, actually, their charge depends on which way time flows. You'll have some integral over the unconstrained momentum in the loop when you calculate the cross section, and for different values of momentum each fermion line will be going either forward or backward in time. So which ones are positive and which ones are negative will change while you integrate over momentum. I'll add a bit to the answer about this actually. – AXensen Apr 13 '23 at 17:00
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    @IGY There are no specific states associated with internal lines and there is no "time direction" for a Feynman diagram (only a labeling of which external lines are "in" or "out") on a formal level, so that question is ill-defined. What would it mean for such an internal line to have a specific charge? (see e.g. https://physics.stackexchange.com/q/230113/50583, https://physics.stackexchange.com/q/297004/50583 and their linked questions for more on the meaning of the diagrammatics) – ACuriousMind Apr 13 '23 at 17:05
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    @ACuriousMind has swayed my opinion. I think my response about "are they neutral" was not good, and I've revised my commentary in the answer a bit. – AXensen Apr 13 '23 at 17:16
  • Thanks so much for clarifying :) – IGY Apr 13 '23 at 19:23
  • @ACuriousMind Thanks you!! – IGY Apr 13 '23 at 19:23