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The Lagrangian formalism of Quantum Field Theory

  1. Jul 11, 2012 #1
    Recently i am reading A.Zee's Quantum Field Theory in a Nutshell 2nd Edition. there is a equation that i can not derive by myself. I suspect its correctness.

    <k1k2|exp(-iHT)|k3k4>=<k1k2|exp(i∫dxL(x))|k3k4>, where the L(x) on the RHS is an operator function of space-time.

    This equation appears on page 64, chapter I.8, under the section 'Scattering Amplitude'.

    Zee used this equation to calculate the corresponding scattering amplitude in λψ^4 theory and thus justified the Feynman Rules given in previous chapters, so i think this equation is of crucial importance in Zee's book.

    Can someone tell me how to derive it?

    Thanks!
     
  2. jcsd
  3. Jul 11, 2012 #2
    -it's part of the path Integrals of Feynman. propagator= <k1k2| exp[iS] | k3 k4> where S is the action.

    your first expression is indeed the propagator, because it's

    <A' t2 | U(T=t2-t1) | A t1>

    where U is the operator of the time development
     
  4. Jul 11, 2012 #3
    how to deduce that <k1k2| exp[iS] | k3 k4> is the propagator?
     
  5. Jul 11, 2012 #4
    first of all I hope you understand that the first is your propagator:

    [itex] K=<q' t' | q t> = <q'| e^{-\frac{i}{\hbar} H (t'-t)} |q>=<q'| e^{-\frac{i}{\hbar} (\frac{p^{2}}{2m} + V(q)) (t'-t)} |q>[/itex]

    [itex] K(q',q;δt) \approx <q'| [1- \frac{i}{{\hbar}} δt (\frac{p^{2}}{2m}+V(q) ) +...] |q>+... \approx δ(q'-q) - \frac{i}{2m{\hbar}} <q'|p^{2}|q> - \frac{i}{{\hbar}} <q'|V(q)|q>+...[/itex]

    [itex] K(q',q;δt) \approx <q|(1- \frac{i}{{\hbar}} \frac{p^{2}}{2m}+...) ( 1- \frac{i}{{\hbar}} V(\frac{q'+q}{2})+...)|q'> [/itex]

    neglecting the terms of order δt2 and more, the propagator is:

    [itex] K(q',q;δt) \approx {K}_{0} (q',q;δt) e^{-\frac{i}{{\hbar}} V(\frac{q'+q}{2})} [/itex]

    Here I will use a property of the propagator which if you like I can prove you too...


    [itex] K(q',q;t',t) = {\prod}_{j=1}^{N-1} \int ({dq}_{j} (\frac{m}{2πi{\hbar}δt})^{1/2} e^{\frac{im}{2{\hbar}δt} ({q}_{j} -{q}_{j-1})^{2} -\frac{i}{{\hbar}} δt V(\frac{({q}_{j}+{q}_{j-1})}{2})} [/itex]

    where now
    [itex]δt= {t}_{j}-{t}_{j-1}/N = (t'-t)/N\rightarrow0 for N\rightarrow∞[/itex]
    the constant infinitesimal time interval step.

    for it going to zero, you can insert a function q(t) which takes the values q(tj)=qj in the interval [t',t]. The boundary values of this function is q(t')=q' and q(t)=q. Since each of the integrating coordinates qj takes values from [-∞,∞], the function q(t) even for N→∞ is not a priori continuous function. However because of the Gauss form exp[...(qj-qj-1)2/δt .....] of the integrating function, while δt→0, only the neighbouring points qj, qj-1 contribute importantly at the integration. So, at the limit δt→0, the q(t) is practically a continous function. Assuming that at this limit we can define derivative we have:

    [itex]\dot{q}= lim_{δt\rightarrow0} (\frac{q(t+δt)-q(t)}{δt})= lim_{N\rightarrow∞} (\frac{{q}_{j}-{q}_{j-1}}{{t}_{j}-{t}_{j-1}}) [/itex]

    for each j. This derivative will replace in the continuous limit the differences that appear on the exponential of the integrating form of the propagator. Also, the sum can change into integral by the rule:
    [itex]δt {\sum}_{j=1}^{j=N-1} \rightarrow {\int}_{t}^{t'} dτ[/itex]

    at the continuum limit the total exponent takes the form of the Classical Action S for the interval [t',t]:

    [itex]{\sum}_{j=1}^{j=N-1} [\frac{im}{2{\hbar}δt} ({q}_{j} -{q}_{j-1})^{2} -\frac{i}{{\hbar}} δt V(\frac{({q}_{j}+{q}_{j-1})}{2}) ]→ \frac{i}{{\hbar}} {\int}_{t}^{t'} dτ [\frac{m}{2}\dot{q}(τ)^{2} -V(q(τ)) ][/itex]
    Where you see it's the Action...
    Putting then the integral on the expression of the propagator you get:

    [itex] K(q',q;t',t) = {\int}_{q(t)=q}^{q(t')=q'} [dq] e^{\frac{i}{{\hbar}} {S}_{c}[q;t',t]} [/itex]

    where [dq] is:
    [itex] [dq]= {\prod}_{j=1}^{N-1} {dq}_{j} (\frac{m}{2πi{\hbar}δt})^{1/2} [/itex]
     
  6. Jul 11, 2012 #5
    the property I used which i didn't prove, just tells you that if you have several infinitesimal propagators, then the total one is the multiplication of each of them. Then I put in the interval the Π which sees the dq giving [dq] as well as the exponential giving you the exp^{ SUM i used at the 3rd from the end equation}...
     
  7. Jul 15, 2012 #6
    Well, i am an undergraduate at Tsinghua University. Currently A.Zee is giving a short-term lecture on Quantum Field Theory here. I've asked Zee about this equation in his book. He said that this is not exactly right if the Lagrangian is understood as an operator. He just provided a sloppy way to understand how the cross section comes out.

    So we need not be confused about this equation if we have known the more rigorous formulae about cross sections.
     
  8. Jul 15, 2012 #7
    Well, i am an undergraduate at Tsinghua University. Currently A.Zee is giving a short-term lecture on Quantum Field Theory here. I've asked Zee about this equation in his book. He said that this is not exactly right if the Lagrangian is understood as an operator. He just provided a sloppy way to understand how the cross section comes out.

    So we need not be confused about this equation if we have known the more rigorous formulae about cross sections.
     
  9. Jul 15, 2012 #8

    vanhees71

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    2016 Award

    Yes, this book is very sloppy. I don't understand why people like it at all.

    It is the more important to note that the Lagrangian formalism for the S matrix does not always give the right result. You always should start in the Hamiltonian formalism of the path integral and then carefully integrate out the field-momentum components which is quite often possible since they appear only as a quadratic form in the Lagrangian.

    The most simple example, where a naive application of the Lagrangian version of the path integral fails is the ideal gas of charged non-interacting scalar bosons with a chemical potential wrt. the conserved charged, i.e., a quite innocently looking problem. You find the details about this example in

    J. Kapusta, Phys. Rev. D 24, 426–439 (1981)
    http://link.aps.org/doi/10.1103/PhysRevD.24.426
    DOI: 10.1103/PhysRevD.24.426
     
  10. Jul 15, 2012 #9

    tom.stoer

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    This is a very important hint!

    In many cases people write down something like exp(iS) as a defintion of the quantum theory w/o being able to derive this via the ∫Dp integration in the Hamiltonian PI. I agree that the Hamiltonian is the fundamental object and that simply writing down the Lagrangian PI may be too sloppy.
     
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