#1
Dec2006, 05:00 AM

P: n/a

Happy holidays to all.
I have just finished a formal course based on Peskin and Schroeder's book on Quantum Field theory. So far the Feynman diagrams have had no explicit reference to space or time in them. Earlier in my career the Feynman diagrams I was shown were drawn a certain way with a set of spacetime axes at least implied. The particles proceeding forward through time antiparticles backward in time, Massless bosons propagating at 45 degree angles to the axes and such. In this way not simply specifying the crosssection of a interaction but it's location in spacetime. What is the distinction I should see between a Feynman diagram drawn with spacetime axes implied and without them?  Feynman diagrams brought Quantum Field Theory to the Masses. Misha Stephanov 11/2006. www.geocities.com/hontasfx 


#2
Dec2406, 05:00 AM

P: n/a

Hontas Farmer wrote:
> I have just finished a formal course based on Peskin and Schroeder's book on > Quantum Field theory. So far the Feynman diagrams have had no explicit > reference to space or time in them. Earlier in my career the Feynman > diagrams I was shown were drawn a certain way with a set of spacetime axes > at least implied. The particles proceeding forward through time > antiparticles backward in time, Massless bosons propagating at 45 degree > angles to the axes and such. In this way not simply specifying the > crosssection of a interaction but it's location in spacetime. > > What is the distinction I should see between a Feynman diagram drawn > with spacetime axes implied and without them? Feynman diagrams are just pictorial representations for terms in a perturbation series. Only the topology of the diagram really matters (i.e: which lines are connected where), as well as the various momenta directions and vertex factors annotating the diagram. There's not much point thinking of the internal lines and vertices as being "located" in spacetime. Only the external lines (i.e: the instates and outstates) have any connection to something detectable with experimental equipment. After all, the whole point of the exercise is to calculate transition amplitudes between the *asymptotic* in and outstates. HTH. 


#3
Dec2406, 05:00 AM

P: n/a

Hontas Farmer wrote:
> Happy holidays to all. > > I have just finished a formal course based on Peskin and Schroeder's book on > Quantum Field theory. So far the Feynman diagrams have had no explicit > reference to space or time in them. Earlier in my career the Feynman > diagrams I was shown were drawn a certain way with a set of spacetime axes > at least implied. The particles proceeding forward through time > antiparticles backward in time, Massless bosons propagating at 45 degree > angles to the axes and such. In this way not simply specifying the > crosssection of a interaction but it's location in spacetime. > > What is the distinction I should see between a Feynman diagram drawn with > spacetime axes implied and without them? The short answer is that there is no distinction. Any attempt to endow internal vertices of Feynman diagrams with time ordering is purely for illustratory purposes. The slightly longer answer is that the distinction may have mattered when at some point, long ago. Namely, Feynman's contribution to perturbation theory was to make every step of the calculation relativistically invariant (that would be Poincare invariant). That is why the diagrams we are all familiar with carry his name. I believe that diagrams were used to represent individual terms in perturbative series even before Feynman; unfortunately I don't have a ready reference. The perturbation calculations used to be done in the canonical formalism, which required the use of a Hamiltonian and hence a choice of preferred time direction. In those days, diagrams with all possible time orientations would have to be considered, hence the relative ordering of intermediate vertices was important. The significantly longer answer asks to keep in mind the origin of the perturbative contributions represented by Feynman diagrams, from which the importance (or lack thereof) of various spacetime labels in an individual diagram may be deduced. The first step is to remember that QFT, as any quantum theory, features states and operators representing physical observables. Expectation values of these observables and transition probabilities between different states can be related to experimental measurements. In usual perturbative QFT, states are provided by Fock space, while all observables of interest are built out of the field operators from different spacetime points. At this point I'm neglecting both renormalizability issues as well as the mathematical necessity to treat fields as operator valued distributions. In this setting, the calculation of both the expectation values and the transition amplitudes can be reduced to the evaluation of socalled npoint correlation functions or Green functions: G(x_1,x_2,...,x_n) = < phi(x_1) phi(x_2) ... phi(x_n) >, where <...> denotes the Fockvacuum expectation value and phi(x_i) is the Heisenbergpicture field operator at spacetime point x_i. In short, any physically meaningful quantity must be expressible in terms of these correlation functions. For simplicity, lets restrict ourselves to twopoint functions G(x,y) only. This correlation function is evaluated by expanding it as a series in hbar. G(x,y) = G_0(x,y) + (hbar) G_1(x,y) + (hbar)^2 G_2(x,y) + ... . Each term G_k(x,y) is represented by a finite number of Feynman diagrams. The power of hbar accompanying each term in the expansion happens to coincide with the number of loops in the corresponding diagrams. In QFT books, the above expansion is usually obtained in one of two ways. One way is the path integral approach, where each term in the series corresponds to a Gaussian integral. The other is the canonical approach: write the Hamiltonian as a spatial integral over the Hamiltonian density, write the time evolution operator in terms of Dyson's timeordered exponential, change operators from the Heisenberg to the interaction picture, represent the interactionpicture interacting vacuum state as the Heisenbergpicture free vacuum state evolved from the infinite past/future, form the expectation value and expand all timeordered exponentials in powers of hbar. In fact, the pathintegral and canonical approaches are proved to be equivalent based on the fact that they give the same answer in these calculations. Either way, each term G_k(x,y) ends up being a multifold spacetime integral. In the pathintegral approach, each spacetime integral comes from a factor of the action, whose exponential is being expanded. In the canonical approach, each spacetime integral comes from a factor of the Hamiltonian (giving the spatial integral) and from the expansion of the time ordered exponential (giving the temporal integral). The presence of the spacetime integrals is what gives these perturbative calculations stepbystep relativistic invariance. Using Wick's theorem, the integrand can be written as a product of freefield twopoint functions, with each of the two arguments being either x, y, or one of the spacetime coordinates being integrated over. Each of the coordinates (either x, y, or one being integrated over) can be represented by a dot on a piece of paper. Each freefield twopoint function can be represented by a line connecting the two dots whose coordinate labels constitute its arguments. This is how Feynman diagrams are born. Hontas, if you're still with me, you can now see an answer to your question. Given a Feynman diagram, there should be a spacetime label on each internal vertex (these are the coordinates being integrated over) and at the end of each external leg (that would be x or y). I hope it is by now obvious that the relative position of internal vertices of a Feynman diagram is quite arbitrary, as each one will assume all possible spacetime coordinates in the course of integrations necessary to evaluate the diagram. The spacetime labels on the external legs are fixed, and you may, if you wish, position them on the page keeping some kind of chronological order. But that is by no means necessary. The above discussion should also illustrate that a single Feynman diagram may not mean much on its own. After all, it is only one of the (sometimes many) terms constituting the expansion coefficient G_k(x,y), while the latter is one term of the expansion of G(x,y), which ultimately the quantity of physical interest. However, this sober point of view in no way detracts from the more poetic illustrative power of Feynman diagrams. One last piece to complete the puzzle is the transition from positionspace to momenumspace evaluation of Feynman diagrams. Since each freefield twopoint function, say D(x,y), is translation invariant, it may be written as D(xy), that is, as a function of the single argument xy. Note that this simplification is impossible in the presence of some kind of background field or inhomogeneous medium. When the simplification is possible, the twopoint function may also be expressed in terms of its Fourier transform: D(xy) = int D(k) exp[i k.(xy)] dk . Once the above identity is substituted into the Feynman diagram evaluation, the order of momentum and spacetime integrals can be exchanged and the latter evaluated to a bunch of deltafunctions. Now, each line in the diagram is attributed a momentum label that is integrated over, while the deltafunctions enforce "momentum conservation" at each vertex. The external legs keep some position dependent polarization factors, since the external coordinate labels were not integrated over. The change to momentumspace evaluation often simplifies the calculation since it reduces the number of independent integrations. I'm sure that an introductory QFT class has not spent much time dwelling on a large fraction of the steps described above. However, if you are intent on attributing physical meaning to Feynman diagrams, you should be familiar with each of the above steps, since their understanding is necessary to decide when such attribution is justified. Hope this helps. Igor 


#4
Dec2506, 05:00 AM

P: n/a

Feynman diagrams and spacetime
Thus spake Hontas Farmer <hfarme2@uic.edu>
>Happy holidays to all. > >I have just finished a formal course based on Peskin and Schroeder's book on >Quantum Field theory. So far the Feynman diagrams have had no explicit >reference to space or time in them. Earlier in my career the Feynman >diagrams I was shown were drawn a certain way with a set of spacetime axes >at least implied. The particles proceeding forward through time >antiparticles backward in time, Massless bosons propagating at 45 degree >angles to the axes and such. In this way not simply specifying the >crosssection of a interaction but it's location in spacetime. > >What is the distinction I should see between a Feynman diagram drawn with >spacetime axes implied and without them? The internal ordering of points in a Feynman diagram is irrelevant. Only the topology matters. There is certainly no reason to draw lines at 45deg angles, unless it is one of graphical clarity. However, when used to describe a particular physical interaction, there is a defined in state and a defined out state, and the external lines will be assigned either to the in or the out state. Clearly there is a time ordering in that. Regards  Charles Francis substitute charles for NotI to email 


#5
Dec2706, 05:00 AM

P: n/a

Igor Khavkine wrote:
> Hontas Farmer wrote: >> Happy holidays to all. >> >> I have just finished a formal course based on Peskin and Schroeder's book >> on >> Quantum Field theory. So far the Feynman diagrams have had no explicit >> reference to space or time in them. Earlier in my career the Feynman >> diagrams I was shown were drawn a certain way with a set of spacetime >> axes >> at least implied. The particles proceeding forward through time >> antiparticles backward in time, Massless bosons propagating at 45 degree >> angles to the axes and such. In this way not simply specifying the >> crosssection of a interaction but it's location in spacetime. >> >> What is the distinction I should see between a Feynman diagram drawn with >> spacetime axes implied and without them? > > The short answer is that there is no distinction. Any attempt to endow > internal vertices of Feynman diagrams with time ordering is purely for > illustratory purposes. > > The slightly longer answer is that the distinction may have mattered > when at some point, long ago. Namely, Feynman's contribution to > perturbation theory was to make every step of the calculation > relativistically invariant (that would be Poincare invariant). That is > why the diagrams we are all familiar with carry his name. I believe > that diagrams were used to represent individual terms in perturbative > series even before Feynman; unfortunately I don't have a ready > reference. The perturbation calculations used to be done in the > canonical formalism, which required the use of a Hamiltonian and hence > a choice of preferred time direction. In those days, diagrams with all > possible time orientations would have to be considered, hence the > relative ordering of intermediate vertices was important. Ah so I am not crazy. The diagrams you may be thinking of are diagrams that are used in classical statistical physics. > > The significantly longer answer asks to keep in mind the origin of the > perturbative contributions represented by Feynman diagrams, from which > the importance (or lack thereof) of various spacetime labels in an > individual diagram may be deduced. > > The first step is to remember that QFT, as any quantum theory, features > states and operators representing physical observables. Expectation > values of these observables and transition probabilities between > different states can be related to experimental measurements. In usual > perturbative QFT, states are provided by Fock space, while all > observables of interest are built out of the field operators from > different spacetime points. At this point I'm neglecting both > renormalizability issues as well as the mathematical necessity to treat > fields as operator valued distributions. In this setting, the > calculation of both the expectation values and the transition > amplitudes can be reduced to the evaluation of socalled npoint > correlation functions or Green functions: > > G(x_1,x_2,...,x_n) = < phi(x_1) phi(x_2) ... phi(x_n) >, > > where <...> denotes the Fockvacuum expectation value and phi(x_i) is > the Heisenbergpicture field operator at spacetime point x_i. In > short, any physically meaningful quantity must be expressible in terms > of these correlation functions. > > For simplicity, lets restrict ourselves to twopoint functions G(x,y) > only. This correlation function is evaluated by expanding it as a > series in hbar. > > G(x,y) = G_0(x,y) + (hbar) G_1(x,y) + (hbar)^2 G_2(x,y) + ... . > > Each term G_k(x,y) is represented by a finite number of Feynman > diagrams. The power of hbar accompanying each term in the expansion > happens to coincide with the number of loops in the corresponding > diagrams. > > In QFT books, the above expansion is usually obtained in one of two > ways. One way is the path integral approach, where each term in the > series corresponds to a Gaussian integral. The other is the canonical > approach: write the Hamiltonian as a spatial integral over the > Hamiltonian density, write the time evolution operator in terms of > Dyson's timeordered exponential, change operators from the Heisenberg > to the interaction picture, represent the interactionpicture > interacting vacuum state as the Heisenbergpicture free vacuum state > evolved from the infinite past/future, form the expectation value and > expand all timeordered exponentials in powers of hbar. In fact, the > pathintegral and canonical approaches are proved to be equivalent > based on the fact that they give the same answer in these calculations. > > Either way, each term G_k(x,y) ends up being a multifold spacetime > integral. In the pathintegral approach, each spacetime integral comes > from a factor of the action, whose exponential is being expanded. In > the canonical approach, each spacetime integral comes from a factor of > the Hamiltonian (giving the spatial integral) and from the expansion of > the time ordered exponential (giving the temporal integral). The > presence of the spacetime integrals is what gives these perturbative > calculations stepbystep relativistic invariance. Using Wick's > theorem, the integrand can be written as a product of freefield > twopoint functions, with each of the two arguments being either x, y, > or one of the spacetime coordinates being integrated over. Each of the > coordinates (either x, y, or one being integrated over) can be > represented by a dot on a piece of paper. Each freefield twopoint > function can be represented by a line connecting the two dots whose > coordinate labels constitute its arguments. This is how Feynman > diagrams are born. > > Hontas, if you're still with me, you can now see an answer to your > question. Given a Feynman diagram, there should be a spacetime label > on each internal vertex (these are the coordinates being integrated > over) and at the end of each external leg (that would be x or y). I > hope it is by now obvious that the relative position of internal > vertices of a Feynman diagram is quite arbitrary, as each one will > assume all possible spacetime coordinates in the course of > integrations necessary to evaluate the diagram. The spacetime labels > on the external legs are fixed, and you may, if you wish, position them > on the page keeping some kind of chronological order. But that is by no > means necessary. In a short over simplified way, you are saying that any spacetime dependence is being integrated out of the calculation at every vertex. > > The above discussion should also illustrate that a single Feynman > diagram may not mean much on its own. After all, it is only one of the > (sometimes many) terms constituting the expansion coefficient G_k(x,y), > while the latter is one term of the expansion of G(x,y), which > ultimately the quantity of physical interest. However, this sober point > of view in no way detracts from the more poetic illustrative power of > Feynman diagrams. > > One last piece to complete the puzzle is the transition from > positionspace to momenumspace evaluation of Feynman diagrams. Since > each freefield twopoint function, say D(x,y), is translation > invariant, it may be written as D(xy), that is, as a function of the > single argument xy. Note that this simplification is impossible in the > presence of some kind of background field or inhomogeneous medium. When > the simplification is possible, the twopoint function may also be > expressed in terms of its Fourier transform: > > D(xy) = int D(k) exp[i k.(xy)] dk . > > Once the above identity is substituted into the Feynman diagram > evaluation, the order of momentum and spacetime integrals can be > exchanged and the latter evaluated to a bunch of deltafunctions. Now, > each line in the diagram is attributed a momentum label that is > integrated over, while the deltafunctions enforce "momentum > conservation" at each vertex. The external legs keep some position > dependent polarization factors, since the external coordinate labels > were not integrated over. The change to momentumspace evaluation often > simplifies the calculation since it reduces the number of independent > integrations. > > I'm sure that an introductory QFT class has not spent much time > dwelling on a large fraction of the steps described above. However, if > you are intent on attributing physical meaning to Feynman diagrams, you > should be familiar with each of the above steps, since their > understanding is necessary to decide when such attribution is > justified. > > Hope this helps. > > Igor Yes that helps very much. As the professor put it it could take many years to fully understand this topic. To be honest I am going to have to reread that reply several times. Oh No has written: >The internal ordering of points in a Feynman diagram is irrelevant. Only >the topology matters. There is certainly no reason to draw lines at >45deg angles, unless it is one of graphical clarity. However, when used >to describe a particular physical interaction, there is a defined in >state and a defined out state, and the external lines will be assigned >either to the in or the out state. Clearly there is a time ordering in >that. It could be that the diagrams that I was shown before taking this course in QFT were based on standard Feynman diagram formalism but not 100% standard. The rationale for drawing the lines certain ways and making reference to spacetime was as follows. A massless Boson (or fermion for that matter) would travel at the speed of light. If space and time units are equal...as in special relativity..then such particles could only propagate with the speed of light. All other material particles would travel at some other sublight speed. Antiparticles would be drawn such that they would approach from a latter time to an earlier time. Particles would seem to travel forwards in time. Basically the spacetime axes and these concepts I have described were used by this one particular professor to denote just what type of particle a given particle was. For the record he NEVER said that any crossections depended on spacetime coordinates. We just used then to discuss particle interactions. Questions like..."Can a photon of sufficient energy decay into an electron and a positron? Why or why not? and such" Thanks for all your help.  Feynman diagrams brought Quantum Field Theory to the Masses. Misha Stephanov 11/2006 


#6
Dec2806, 05:00 AM

P: n/a

Hontas Farmer wrote:
> Igor Khavkine wrote: [...] > In a short over simplified way, you are saying that any spacetime > dependence is being integrated out of the calculation at every vertex. That's a good way to put it. [...] > > Hope this helps. > Yes that helps very much. As the professor put it it could take many years > to fully understand this topic. To be honest I am going to have to reread > that reply several times. That's a good idea. I certainly didn't grasp all that I talkd about over night. If you're serious about it, I can give textbook references for most of the steps I described (sorry, to my knowledge, no one textbook would be sufficient). If you search for my past posts about Feynman diagrams, you'll find that I've already cited some of them before. Igor 


#7
Dec2906, 05:00 AM

P: n/a

Thus spake Hontas Farmer <hfarme2@uic.edu>
>Igor Khavkine wrote: > >> >> The slightly longer answer is that the distinction may have mattered >> when at some point, long ago. Namely, Feynman's contribution to >> perturbation theory was to make every step of the calculation >> relativistically invariant (that would be Poincare invariant). That is >> why the diagrams we are all familiar with carry his name. I believe >> that diagrams were used to represent individual terms in perturbative >> series even before Feynman; unfortunately I don't have a ready >> reference. The perturbation calculations used to be done in the >> canonical formalism, which required the use of a Hamiltonian and hence >> a choice of preferred time direction. In those days, diagrams with all >> possible time orientations would have to be considered, hence the >> relative ordering of intermediate vertices was important. > >Ah so I am not crazy. The diagrams you may be thinking of are diagrams that >are used in classical statistical physics. That sounds probable, though I am not sufficiently familiar with statistical physics to say. I think it was almost exclusively Feynman who thought that playing with these diagrams could lead to an advance in understanding, but the calculation which removes time ordering from the diagrams and shows them to be equivalent to the canonical formulation was actually done by Dyson, on his famous bus journey (famous for this calculation). > >In a short over simplified way, you are saying that any spacetime >dependence is being integrated out of the calculation at every vertex. Precisely. That is the basis of the calculation. One thing to watch out for  swapping the order of integration isn't strictly legitimate in rigorous analysis. This is called "the abuse of Wick's theorem" by Scharf in Finite Quantum electrodynamics and can be regarded as the origin of the ultraviolet divergence. I think once this is understood, Feynman diagrams can also be understood as having much more physical meaning than is often said. >Oh No has written: >>The internal ordering of points in a Feynman diagram is irrelevant. Only >>the topology matters. There is certainly no reason to draw lines at >>45deg angles, unless it is one of graphical clarity. However, when used >>to describe a particular physical interaction, there is a defined in >>state and a defined out state, and the external lines will be assigned >>either to the in or the out state. Clearly there is a time ordering in >>that. > >It could be that the diagrams that I was shown before taking this course in >QFT were based on standard Feynman diagram formalism but not 100% standard. >The rationale for drawing the lines certain ways and making reference to >spacetime was as follows. A massless Boson (or fermion for that matter) >would travel at the speed of light. If space and time units are equal...as >in special relativity..then such particles could only propagate with the >speed of light. All other material particles would travel at some other >sublight speed. Antiparticles would be drawn such that they would >approach from a latter time to an earlier time. Particles would seem to >travel forwards in time. This isn't actually strictly true. There is a nonzero amplitude for the creation of a particle and its annihilation outside the light cone, even for a massive particle. This is an unobservable process, because it is always matched by the corresponding antiparticle going the other way, and the amplitude for both processes taken together cancels to zero. Each line in a Feynman diagram incorporates the amplitudes for both the particle and the antiparticle. > Regards  Charles Francis substitute charles for NotI to email 


#8
Jan707, 05:00 AM

P: n/a

Happy Gregorian New year to you all!
I have been thinking over what was said about the presence of spacetime on the Feynman diagrams that I saw earlier... In condensed form, spacetime coordinates are summed over and the final amplitude does not depend on them. The same goes for these diagrams when drawn in momentum space. (Personally I find momentum space easier to work with. There is no reason to insert what amounts to a Fourier transform into the process. Among other things.) I get where you are all coming from on that. So to get a better idea of what was going on I decided to reread "QED: The Strange Theory of Light and Matter" by Richard P. Feynman. In this book which is based heavily on a series of lectures on QED he gave he works his way up to drawing Feynman diagrams as a way of calculating the possible outcomes of various particles interacting. For a good portion of the book he explicitly draws in SpaceTime axes and does not trivialize the importance of spacetime. In particular the fact that this space time is presumed to be a nonreactive spectator to what is going on. I have scanned some figures I would like you all to see. <a href="http://photobucket.com/" target="_blank"><img src="http://i57.photobucket.com/albums/g204/Zahara_TS/fyenmanhadST.jpg" border="0" alt="Photobucket  Video and Image Hosting"></a> In the whole thing he alludes in as unmathematical a way as possible to the fact that these diagrams are used in a perturbation series. But that does not change the fact that spacetime is present for all of this. OK. Basically...this is what I get from reading this book right through. He is saying that to compute the full coruscation of an interaction one needs to draw diagrams for every way the interaction can occur. Every way that is allowed to connect external lines, and vertices's with the given set of particle interactions. Then add up the results. That as we all know is how one finds the quantity that most modern particle physicist are interested in. The cross section (units of area) that expresses the probability of a given interaction. THERE IS HOWEVER ONE OTHER QUANTITY WHICH DR. FEYNMAN MENTIONED Often...the "Amplitude". Each arrow on the diagram has a length called it's amplitude (not unlike in classical physics, in fact quite analogous). He describes the process as that of calculating the "final arrow" for the event. It's probability amplitude. Does this difference in what was being calculated (the crosssection of a vertex vs the amplitude of an arrow) effect the answers I got before? It seems to and I am racking my brain to figure out how. Any help will be appreciated and if I am able I will help you someday. Thank you  Feynman diagrams brought Quantum Field Theory to the Masses. Misha Stephanov 11/2006 


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