Register to reply 
Questions concerning Bhabha scattering 
Share this thread: 
#1
May1012, 05:03 AM

PF Gold
P: 521

Hi,
I am trying to get a better understanding of what aspects of an electronpositron collision can be physically observed as opposed to mathematically inferred. In part, my confusion is based on a number of different sources, e.g. see links below, which seem to adopt different approaches to describing the same thing, i.e.
Graphically, we appear to have an electron [e] and positron [e+] colliding and annihilating at [t1,x], which results in a photon being created, which then ‘transmutes’ back into an electron and positron at [t2,x]. As such, the photon seems to exists for [t2t1] but travels no distance.
We appears to have a positron [e+] scattering at a point [x1,t], while the electron scatters at a point [x2,t]. Based on a lefttoright timescale, it seems to suggest zero time, but a separation [x2x1].
Thanks 


#2
May1012, 11:08 AM

Sci Advisor
HW Helper
PF Gold
P: 2,603

[tex] E^2  \vec{p}^2 = s,[/tex] which is the centerofmass energy of the initial electronpositron pair. A real photon must satisfy [tex] E^2  \vec{p}^2 = 0,[/tex] because it is massless. This is essentially what distinguishes a real particle from a virtual one. The virtual particle does not satisfy the "mass shell condition" that it's 4momentum satisfy [itex]\mathbf{p}^2 = m^2[/itex]. Energy is conserved at every vertex, but a diagram cannot represent a complete process unless all of the initial and final state particles can satisfy the mass shell condition I described above. [tex] \mathcal{M} = \int d\mathbf{x}_1 d\mathbf{x}_2 M(\mathbf{x}_1 \mathbf{x}_2).[/tex] The momentum space expressions given there are what you'd get by using plane wave solutions for the wavefunctions and using standard formulas for Fourier transforms. Note the need to integrate over all space is an inherently quantum mechanical idea that's tied to the interpretation of the wavefunction. This is not really what's being described by the Bhabha process though. Bhabha scattering already assumes that the particles scatter (we have an electron and positron in the final state) and computes the cross section for the scattering process, which we could call [itex]\sigma(e^+e^\rightarrow e^+e^)[/itex]. This is an example of what's called an exclusive cross section. Since we exclude other possible final states. The inclusive cross section would be [itex]\sigma(e^+e^\rightarrow \mathrm{anything})[/itex], which would include real annihilation processes where we have just photons in the final state. 


#3
May1112, 03:03 AM

PF Gold
P: 521

Fzero,
Really appreciated the knowledgeable answers provided to all my questions. While I need to think about and read around some of the issues raised, I had a couple of immediate points I would like to clarify, if possible: Again, really appreciated the helping hand. Thanks 


#4
May1112, 01:36 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603

Questions concerning Bhabha scattering
Another (related) feature comes up when we study higher order corrections, which inevitably involves socalled loop diagrams. A loop diagram is a diagram, as the name suggests, where there is a closed loop of particle lines, like this one: This diagram describes the process where an electron with 4momentum [itex]p[/itex] emits a photon of momentum [itex]l[/itex] and then reabsorbs it a bit later. The Feynman rules tell us to conserve momentum, so in the intermediate state the electron has momentum [itex]p l[/itex]. In the final state, it has momentum [itex]p[/itex] again. We had to introduce the momentum [itex]l[/itex] because of energy conservation, but energy conservation does not fix [itex]l[/itex] to any specific value. In fact the Feynman rules go on to say that we must integrate over all possible values of [itex]l[/itex]. We then see, even more dramatically, how far from being on mass shell the intermediate particles are. The photon mass shell condition is [itex]l^2=0[/itex], but in this diagram [itex]l^2[/itex] actually runs over all values from [itex]0[/itex] to [itex]\infty[/itex]. Maybe the best way to put it is that, when dealing with virtual particles, energy and momentum are conserved in a strict mathematical sense. But because the mass shell condition is violated, normal physical intuition regarding the conservation laws does not hold. [tex] P_\mathrm{scatt} \sim \frac{s^2+u^2}{t^2},[/tex] while that for annihilation is [tex] P_\mathrm{annih} \sim \frac{t^2+u^2}{s^2}.[/tex] I've dropped numerical factors that are common to both processes. These results were computed in the limit where the energy of the electrons is large compared to their mass, so it turns out that we can approximate [itex] s+t+u\approx 0[/itex]. Then [tex] \frac{P_\mathrm{annih}}{P_\mathrm{scatt}} \approx \frac{t^4+t^2(s+t)^2}{s^4 +s^2(s+t)^2}.[/tex] The convenient limit to consider is the one where the centerofmass energy of the particles, [itex]s[/itex] is large, but the momentum transfer between them, [itex]t[/itex] is held fixed, so the ratio [itex]t/s[/itex] is small. Then [tex] \frac{P_\mathrm{annih}}{P_\mathrm{scatt}} \approx \left(\frac{t}{s}\right)^2 \rightarrow 0, ~~~~\frac{t}{s}\rightarrow 0.[/tex] So yes, at large energies, the scattering diagram is a larger contribution than the annihilation diagram. Note that the probabilities (and even the cross section itself) are not solely just functions of the initial momenta. This is because the amount of momentum transferred is a variable. The final state momenta can take a range of values while still satisfying total energy and momentum conservation. So I would say that we cannot ever detect a virtual particle, but we understand the nature of what a virtual particle is too well to just say that it is some mathematical bookkeeping convenience. 


#5
May1212, 11:00 AM

PF Gold
P: 521

Hi,
Again, thanks for the additional information. I am now trying to work through some of the maths and physics implied in some of the concepts raised. While I guess this might take some time, I just wanted to just table some issues, as I go, in the hope that if I go ‘offshell’ in some of my assumptions, somebody might be able to put me back ‘onshell’. On and Off Shell? As far as I can determine the terminology of ‘onshell’ and ‘offshell’ appears rooted in the relativistic equation [itex]E^2=m^2c^4+p^2c^2[/itex]. Any particle that conforms to this equation is said to be onshell, e.g. [itex]E^2  p^2c^2=m^2c^4[/itex] :real particle [itex]E^2  p^2c^2=0[/itex] :real photon So, by way of a summary of current understanding: a virtual particle is said to be offshell because it does not comply with these equations, i.e. a massless virtual particle, such as a photon, can have (+/) mass and described as off mass shell by having a ‘mass’, which might be better described as ‘borrowed energy’ within the confines of the energy/time variant of the uncertainty principle. So while the conservation of energy and momentum must apply at the vertex of a Feynman diagram, the virtual photon associated with either annihilation or scattering can assume an offshell mass/energy. Mandelstam variables? At the moment, I am just trying to understand how these variables relate to any physical aspect of the collision/scattering to which they are associated, i.e. the schannel appears to be associated with annihilation, while the tchannel seems to be associated with scattering. In the case on annihilation between relativistic particles, i.e. let E=pc, such that I would have thought the total energy/momentum, predicated on the vertex laws of conservation, could be expressed as [itex]E_1+E_2=E_3+E_4[/itex] and [itex]p_1+p_2=p_3+p_4[/itex]. So, from the perspective of energy conservation at the annihilation vertex, would the resultant energy of the photon [itex]E=hf=E_1+E_2[/itex] ignoring any deviation within the uncertainty principle? If so, could the momentum of the photon also be initially quantified as [itex]k=p_1+p_2[/itex]. However, in the case of scattering, would the photon energy at the vertex would be [itex]E=hf=E_1E_3[/itex], while the momentum would be [itex]k=p_1p_3[/itex]? In contrast, the Wikipedia definition of the Mandelstam variables (s) and (t) is: [itex]s=\left(p_1 + p_2 \right)^2 = \left(p_3 + p_4 \right)^2 [/itex] :annihilation [itex]t=\left(p_1 – p_3 \right)^2 = \left(p_2 + p_4 \right)^2 [/itex] :scattering Not sure why the definition of (s) and (t) involves the square of the total energy or momentum other than reflecting its roots in the relativistic equation [itex]E^2=m^2c^4+p^2c^2[/itex]? As yet, I am also not sure I understand the Wikipedia rationalisation of (s) and (t) at the highenergy limit, e.g. [itex]s=\left(p_1 + p_2 \right)^2 = (p_1)^2+(p_2)^2+2p_1.p_2 \approx 2p_1.p_2[/itex] As yet, it is not totally clear to me as to how the Mandelstam variables provide any additional insights into the description of Feynman diagrams over and above the definition of momentum and energy derived from the relativistic equation. However, the answer may well be in the various description of annihilation and scattering in terms of (s,t,u) as described in post#4, which I am still working on. Feynman Rules and other stuff? While I have just got a copy of Mandl and Shaw, plus a few free references, I have found the maths heavy going for a beginner like me. Feynman’s QED book and lectures have helped me gain some initial insights to the basic ideas involved, while an article by Alex Nelson: Notes on Feynman Diagrams has provided a starting point regarding the actual calculations behind the Feynman diagrams in the form of 2 toy models, i.e. tree and loop. 


#6
May1212, 11:21 AM

Sci Advisor
Thanks
P: 4,160




#7
May1212, 02:06 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603

[tex]s=\left(p_1 + p_2 \right)^2 = (E_1+E_2)^2   \vec{p_1}  \vec{p_2}^2, [/tex] with similar expressions for the other combinations. [tex](p_1)^2=m_1^2,~~~(p_2)^2 =m_2^2,[/tex] because the external particles are on shell. Then [tex] s = 2p_1\cdot p_2 + m_1^2 + m_2^2 = 2p_1\cdot p_2 \left( 1+\frac{m_1^2 + m_2^2}{2p_1\cdot p_2 }\right) .[/tex] In the high energy limit all energies and 3momenta are much larger than the masses, [itex]E,\vec{p}\gg m[/itex], so we ignore the ratio on the RHS. 


#8
May1212, 02:30 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603

My original reply to mysearch's question started with a discussion of the measurement problem in QM and it just degenerated from there. I chose not to go down that path because my own understanding is not sufficient to avoid the esoteric in trying to remain rigourous. I don't really get the impression that the people that study QM foundations agree on everything either. While strictly accurate, your reply really doesn't shed any light on why measured particles are on shell. If you have any insight on why it's a better answer to give someone whose just trying to learn QFT, I'd be happy to listen. There seems to be a some happy medium between giving practical (incomplete, but not strictly incorrect) explanations and giving rigorous (correct, but nevertheless incomplete or opaque) explanations that not everyone agrees on. I tend to think it's better to err on the side of explaining what we do understand than emphasizing the esoteric details that we do not really understand. 


#9
May1212, 03:45 PM

Sci Advisor
Thanks
P: 4,160




#10
May1212, 04:12 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603




#11
May1312, 05:29 AM

PF Gold
P: 521

"Our mission is to provide a place for people (whether students, professional scientists, or others interested in science) to learn and discuss science as it is currently generally understood and practiced by the professional scientific community."As such, I rely heavily on the PF forum to help out when I get stuck on a subject, although by its nature, the PF can be a ‘hit or miss’ method that often depends on member interest and qualifications to answer specific issues. While many appear to like the brevity of a soundbite response, the value of such responses may only be appreciated by those who already know the subject. I mention this, not as a criticism, as each member is entitled to their own approach, but by way of explanation as to why I have appreciate the balanced level of detail provided in response to my questions and why I use this thread, and others, to detail my own understanding for future reference. Many thanks 


#12
May1312, 05:33 AM

PF Gold
P: 521




#13
May1412, 05:44 AM

PF Gold
P: 521

It is accepted that the issue of 4momenta is possibly not the focus of this subforum, but the issue was originally raised in connection to understanding the Mandelstam variables in post #5, to which Fzero responded in post #7 as follows:
[itex]k^2= \left(p_1 \right)^2 + \left(p_2 \right)^2 – 2 \left(p_1 p_2 \right) cos \theta [/itex] [itex]k^2= \left(p_1 \right)^2 + \left(p_2 \right)^2[/itex] when θ=90 While the idea of an analogous billiard ball collision might be inappropriate, it is only intended as a reference model for some of the ideas that still seem to apply, if the conservation of energy and momentum are applicable at the collision vertex within the Feynman annihilation diagram. For example, if we view the Feynman annihilation diagram referenced as a spacetime diagram, we appear to end up with the suggestion that the virtual photon only exists in time, albeit briefly, as there is no implied propagation in space. Just for reference, the scattering diagram appears to suggest the opposite, i.e. the virtual photon propagates in space in zero time. As such, it is not really clear as to how momentum is transferred to the virtual particle, especially if the mechanism actually involves some form of quantum wave superposition. [itex]4vector=[scalar, \left(vector \right)][/itex] [itex]4position~~[\vec{x}] =[ct, \left(x, y, z \right)][/itex] Given the reference above to special relativity, we might describe 4position in terms of a Minkowski spacetime diagram: [itex]s^2= \left(x_0 \right)^2 + \left(x_1 \right)^2 = \left(x_0’ \right)^2 + \left(x_1’ \right)^2[/itex] Here (x0,x0’) are the generalisation of time (ct), while (x1,x1’) represents the 1D separation in space in 2 different frames of reference, such that the spacetime interval or proper time in invariant in all frames of reference. We might simply highlight that the invariance of a spacetime interval [s] is linked to the Lorentz factor [itex]( \gamma )[/itex]. Extending the 4vector concept to velocity and momentum: [itex]4velocity ~~[\vec{v}] = \gamma[c, \left(v_x, v_y,v_z \right)][/itex] [itex]4momentum ~~[\vec{p}] = \gamma m_0[c, \left(v_x, v_y, v_z \right)][/itex] In the case of 4momentum, we might rearrange the definition of (p0), such that it is linked to energy (E): [itex] p_0 = \gamma m_0 c[/itex] [itex] c p_0 = \gamma m_0 c^2=E[/itex] [itex] p_0 =E/c[/itex] At this point we can defined 4moment as: [itex] \vec{p} = [E/c, \gamma m_0 \left(v_x, v_y, v_z \right)][/itex] [itex] \vec{p} = [E/c, \gamma \left(p_x, p_y, p_z \right)][/itex] We might now interpret this equation in two ways. First, by analogy to a Minkowski spacetime diagram, which was shown to lead to an invariant quantity called the spacetime interval. As such, if we plotted energy against momentum, the implication is that we might end up with another invariant quantity, although this might be best presented in terms of 4momentum constrained to 1D: [itex] \vec{p}.\vec{p} = \left(p_0 \right)^2 + \left(p_1 \right)^2 = \left( \gamma m_0 c \right)^2 + \left(\gamma m_0 v \right)^2 = \left(m_0 c \right)^2[/itex] [itex] \left(p_0 \right)^2 + \left(p_1 \right)^2 = \left( E/c \right)^2 + \left( p \right)^2 = \left(m_0 c \right)^2[/itex] [itex] E^2 = \left( m_0 c^2 \right)^2 + \left(pc \right)^2 [/itex] In essence, we see the encapsulation of the relativistic energy equation within the definition of 4momentum, where rest mass energy is the invariant quantity in all frames of reference. If this understanding of 4momentun is correct, it should be possible to reconcile it with the following statement from post#2: [itex] \mathbf{p}^2 = \vec{p}.\vec{p} = \left( E/c \right)^2 + \left( p \right)^2 = \left(m_0 c \right)^2[/itex] Again, if [c=1], we have the relationship suggested between 4momentum and the rest mass? However, making reference back to original descriptive model above, it still appears that the annihilation collision between an electron and positron requires the conservation of energy at the vertex, i.e. E=E1+E2. While a ‘real’ photon would defined this energy in terms of E=hf, a virtual photon has some notional concept of rest mass energy, as defined by [s]? As such, there appears to be some ‘ambiguity’ as to the definition of momentum in terms of a virtual photon as opposed to a real photon, i.e. [itex](p= \gamma m_0 v)[/itex] versus [itex](p=E/c)[/itex]? While I realise that people might not have the time or inclination to wade through all this, any clarifications or corrections would be welcomed. Thanks 


#14
May1412, 04:27 PM

Sci Advisor
HW Helper
PF Gold
P: 2,603

The whole waveparticle duality is a good means of explaining the need for quantum mechanics. However, once you learn QM, the distinction should really be blurred. Elementary particles behave like point particles, the wave nature is a consequence of the probability amplitudes that quantum mechanics uses to describe them. There's no distinction between particle and wave, the distinction is really between the quantum description and the classical limit. The bare mass corresponds to the parameter that we put into the Lagrangian. Because of the way the renormalization procedure deals with divergences, it turns out that the bare parameters themselves must have an infinite part. Even the finite part of the bare parameters are not themselves physical, since there is still some freedom to choose what is called a "renormalization scheme." While the details are exceedingly technical, the lesson to take away is that when the renormalization scheme is chosen, finite observables like scattering amplitudes emerge. By comparing enough of these to experiments, we use physical data to fix the unknown parameters of the theory (like the finite part of the bare parameters). The "suggestion that the virtual photon only exists in time" is an artifact of both the diagram being drawn in momentum space and it's being drawn symmetrically. The diagram drawn with the photon at an angle is perfectly equivalent. In fact, the only difference between the "annihilation" and "scattering" diagrams is encoded in the way that the external particles are assigned to be "in" or "out" states. Therefore we get different expressions depending on how the 4momentum of the photon is written in terms of the external momenta. This is more apparent in the nomenclature also used on the wikipedia page of [itex]s,t,u[/itex] "channels," which makes this distinction precise in terms of the Mandelstam variables. There is no particularly good reason to correlate the deviation from the mass shell condition with a mass for the virtual particle. This is why I emphasized the "borrowed energy" concept as being more appropriate. The mass shell condition is simply not obeyed by a virtual particle, we can't use it to assign a fictitious mass. There are good physical reasons like gauge invariance that prevent us from concluding that a virtual photon has a nonzero mass. 


#15
May1512, 04:12 AM

PF Gold
P: 521

Fzero,
I have appreciated your insights, and patience. You have raised a number of issues throughout this thread that I now recognise I need to understand better, possibly a lot better. I have found a couple of video lectures, e.g. David Tong and Prof.V.Balakrishnan, which may help. However, the biggest problem I am having is finding a good mathsphysics textbook that will help me with the step function between QM and QFT. Any suggestions welcomed. Thanks 


Register to reply 
Related Discussions  
Bhabha scattering  Advanced Physics Homework  0  
Bhabha Scattering  High Energy, Nuclear, Particle Physics  4  
Bhabha scattering  High Energy, Nuclear, Particle Physics  6  
Bhabha scattering  Advanced Physics Homework  1  
Doubt about Bhabha scattering.  High Energy, Nuclear, Particle Physics  1 