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Defeated by quantum field theory yet again 
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#1
Nov2512, 11:28 AM

P: 362

I haven't taken a course in qft yet, just looking ahead to see what's to come, and so far things are not looking good, I read the firet few chapters of qft in a nutshell, and jesus christ what is this stuff, where are the postulates? The equations of motion? How do I even do these crazy path integrals??What is this business about everything being harmonic oscillators??
*crawls back into my nonrelavistic cave where things are nice and cozy 


#2
Nov2512, 11:39 AM

P: 1,062

What is your background in quantum mechanics?



#3
Nov2512, 11:58 AM

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We have axiomatical quantum field theory, don't worry. But one should first start with the basics and have a firm grip on nonrelativistic quantum mechanics, including the path integral formulation by Feynman.



#4
Nov2512, 01:03 PM

P: 647

Defeated by quantum field theory yet again
Check out chapter two in Sakurai's quantum book "modern quantum mechanics." This outlines dynamics, and has a good part on the path integral in nonrelativistic quantum mechanics. Also, DON'T GIVE UP! YOU CAN DO IT! :) Start over in Zee's book! and do it again until it works.



#5
Nov2612, 12:51 AM

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One can do quantum field theory in the canonical i.e. Hamiltonian approach (it's not popular in most text books b/c all scattering and Feynman stuff becomes rather complex, but there are applications where this is advantageous). In this approach one can make QFT look like QM with (countable) infinitly many d.o.f. and one does not need to introduce any path integral.



#6
Nov2612, 01:08 AM

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http://www.mit.edu/~levitov/8514/#lecturenotes (Lecture 3) http://research.physics.illinois.edu...ct18a2ndQ.pdf http://arxiv.org/abs/hepth/0409035 (p1315). 


#7
Nov2612, 01:24 AM

P: 362

Could I get some help tho, because I have no where Zee is getting this stuff from:
1. Why is the klein gordon equation relavistic?? 2.Why do the number of indicies on the stress energy tensor tell us that the spin of the gravitons is 2? 3. How do I interperet particle fields? Are they still probability amplitudes? 


#8
Nov2612, 02:01 AM

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1.: b/c it is formulated in terms of scalars and 4vectors
2.: it's the metric tensor which tell's us that it's spin 2 The metric is symmetric, so it has 4+3+2+1 = 10 indep. components; one can identify a gauge invariance allowing us to gauge away 8 components, so we are left with 2 d.o.f. which indicates that we are talking about a massless spin 1,2,... field. The reason why it's spin 2 (not spin 1) can be seen by looking at its multipole expansion: it starts with the quadrupole instead of a dipole 


#9
Nov2612, 07:30 AM

P: 130

"How do I interperet particle fields? Are they still probability amplitudes?"
That's my favorite question ;) (talked about that a lot in the past few weeks) The problem you have with QFT may be at least in part due to the fact that nobody ever tells you what exactly a QFT field and its state is. In QFT, the fields themselves are (in path integral formulation) classical fields, not prob. amplitudes. Consider an elastic membrane  the field is the displacement at each point. In QFT, you now have a probability amplitude for each possible field configuration  the state of a quantum field is a superposition of all possible field configurations, each with its own probability amplitude. Since it is very difficult to calculate with this kind of object (its a wave functional  a wave function with functions as arguments), people usually don't use it, but conceptionally I think it is important to understand this. If you do a fourier analysis, each fourier component of your field behaves like a harmonic oscillator  so in the ground state the probability of finding a value a_k of the k'th fourier mode is given by a gaussian centered at zero. (And that is why people will tell you that the expectation value of the field is zero for a vacuum state  but there is still a probability of measuring a nonzero state, exactly as for the position of a particle in a QHO). If you have a 1particle state in mode k, this means that the prob. amplitude for the fourier coefficient of a_k looks like the first excited function of the QHO. It still has a zero expectation value, but now it has a different prob. amplitude. (And this is why you can read that a state with a definite particle number has a vanishing expectation value of the corresponding classical field.) Hope this helps. 


#10
Nov2612, 01:34 PM

P: 362

also I have no idea how to calculate these runctional integrals, no where have I ever seen them explained, how is it even defined? What does the infitesimal represent now? 


#11
Nov2612, 01:48 PM

P: 130

"what these particle fields represent physically"
To make it concrete, think of the electromagnetic field (slightly simplified): The classical field is nothing but the fourpotential. This is quantised, so in a quantum state, you have different amplitudes for different values of the four potential. If the functional integral (I assume you mean the path integral?) seems complicated, just discretise it in your mind. Think of a spacetime grid with a discrete number of points. Then the iegral becomes a sum over all possible field configurations, i.e., for each field configuration, you calculate exp(iS/hbar) and then you sum up all these to get the total value of the functional integral. It is completely analoguous to the path integral formulation of quantum mechanics (which is explained by Zee): Instead of summing over all possible paths, you sum over all possible field configurations. As so often in physics, QFT may seem so difficult because you have to learn new physics and new mathematical tricks, and many books don't make it clear whether something is just a mathematical trick or "true" physics... Hope this helps a bit more  if not, it would be nice if you could make your questions more specific. 


#12
Nov2612, 02:01 PM

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#13
Nov2612, 02:09 PM

P: 362

Well I never had a firm grasp on the path integral formalism even in quantum mechanics, and most of the texts I've read are difficult to understand for me, I don't understand how the integral is defined. It clearly is no longer some multiple Reinman sum, so what is it?



#14
Nov2612, 02:38 PM

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P: 6,245

Form "Quantum Field Theory: A Tourist Guide for Mathematicians" by Gerald Folland 


#15
Nov2612, 02:46 PM

P: 362

I really dislike this kind of illdefined procedures...what is rhe best text on path integrals?



#16
Nov2612, 03:15 PM

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#17
Nov2612, 04:47 PM

P: 362

so..
1. what do particle fields represent 2. why don't we just use good old kets and bras for QFT? 3. why can't we just solve for a bunch of eigen fields from the sources and then just superimpose them 


#18
Nov2612, 04:55 PM

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