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I am reading the fantastic QED Feynman book. He talks in chapter 3 about a formula he considers too complicated to be written in the book. I would like to know which formula he talks about, although I have a vague idea. Please don't shy away from writing it properly, or giving a non-popular description or reference, I am conversant with basic QM, although not with the variational formulation nor QFT in general.

He talks about a formula for E(A to B), the amplitude for an electron to go from B to A in a space-time diagram), and states that it

(...)can be represented by a giant sum of a lot of different ways an electron could go from point A to point B in space-time. The electron could take a 'one-hop flight' going directly from A to B; it could take a 'two-hop flight', stopping at an intermediate point C; it could take a 'three-hop flight' (...) The amplitude for each 'stop' is represented by $n^2$ (...) The formula for E(A to B) is thus a series of therms (...) for all possible intermediate points

Additionally, I would like to know whether this book is another literal transcription of some lectures video footage, as it happens with, for instance, the 1964 Messenger Lectures at Cornell (published as The Pleasure of Finding Things Out). This is secondary and can be answered in a comment for instance, if you like it so.

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A timely question - it is Feynman's birthday today! What better day to ask a question about his most celebrated work! –  innisfree May 10 '13 at 23:52
    
@innisfree, I didn't quite understand all the Feynmanmania, until I recently discovered that Messenger lectures... He is great. So deep, so pedagogic and, at the same time, so funny (he reminds me of Jack Lemmon in that lectures). Now I am reading this book and really enjoying it. –  Mephisto May 11 '13 at 0:05
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For a deeper (graduate / post-graduate) description of this matter (with Feynman's spark), look Feynman, Hibbs: Quantum Mechanics and Path Integrals and Feynman: Quantum Electrodynamics. The first one will link things to the Schrödinger picture. –  firtree May 11 '13 at 9:22
    
@firtree, thanks for the references! (the second one is the book I refer to in the question). –  Mephisto May 11 '13 at 14:25
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No, it is not. There are two books: QED: The Strange Theory of Light and Matter is the popular lecture cycle, and Quantum Electrodynamics is the graduate course. Please do not confuse them :-) And in a third sense, "Feynman's Quantum Electrodynamics" is one of the names for the Quantum Electrodynamics theory per se. –  firtree May 11 '13 at 15:22
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up vote 2 down vote accepted

I don't have the book with me right now, but from the passage you highlighted I gather that he's talking about his famous Feynman path integral formula. The partition function for a particle going from point A to B is given by \begin{align} Z = \int_{A \to B} [\mathcal{D}x]~ e^{iS[x]}, \end{align} where $\mathcal{D}x$ is the measure that sums up over all paths going from $A$ to $B$ (one hop, two hops, three hops), and $e^{iS[x]}$ is the weight of each path, $S[x]$ is the action functional $S = \int dt L $, where $L$ is the Lagrangian. From this partition function all desirable quantities can be obtained.

I'm not sure about your second question.

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Is it straightforward to write $<B|A>$ in terms of that formula? That would make your answer perfect. I mean $<B|A> = ...$ (some expression involving that Z... sorry, I am also short on Statistical Mechanics for the moment) –  Mephisto May 11 '13 at 0:00
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$Z$ is a generator of Green's functions, which means that, with some work, you can extract amplitudes for interactions from it. $Z \propto <0,t=-\infty | 0, t=+\infty>$ - vacuum to vacuum. $Z$ is related to $<A,t=-\infty | B, t=+\infty>$ by the LSZ reduction formula, which is a little complicated. But the quickest way to calculate these things is with Feynman diagrams, that result from an expansion of the exponential in powers of the coupling constant. I think Zee or Ryder are good pedagogical, introductory QFT textbooks. The wiki articles don't look very nice. –  innisfree May 11 '13 at 1:24
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In general, the path integral is an integral over paths in phase space; i.e. there should be a $\mathcal Dp$ in you measure too (see for example Polchinski String Theory vol. 1 eq. A.1.12). –  joshphysics May 11 '13 at 1:55
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@joshphysics There are two slighly different versions. Lagrangian path integral, first invented, is over $\mathcal{D}x$, and the action is based on $L$. Hamiltonian path integral, most used in QFT, is over $\mathcal{D}x\mathcal{D}p$, and the action is based on $H$ (under Legendre transformation). –  firtree May 11 '13 at 17:13
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@joshphysics There is no $\mathcal{D}p$ measure because I used the Lagrangian formalism, not the Hamiltonian formalism. Equivalently one can think of it as me already having done the $\mathcal{D}p$ integral to get the above path integral expression. –  nervxxx May 12 '13 at 18:56
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