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Euler-Lagrange equations in QFT?

  1. Jun 21, 2007 #1
    Euler-Lagrange equations in QFT??


    I have a problem with a Wikipedia entry::bugeye:

    The equations of motion in your quantized theory (2nd quantization) are d/dtF^=[F^,H^] i.e the quantized version of d/dtF={F,H}. My notation: F^ is the operator for the former classical quantity F.
    There exists no quantized version of the E-L-eqautions!
    I haven't seen the Euler-Lagrange equations of motion in any QFT book yet, and I mean, what should be the derivative with respect to an operator? It's not a functional derivative, don't know if mathematicians have defined a "operator-derivative"...
    Ok the Wikipedia article doesn't speak of operators, but what is presented as "Euler-Lagrange equations in QUANTUM Field theory" are Euler-Lagrange equations of classical field theory or not? They don't play the role of equations of motion in QFT i.e they don't describe the quantum dynamics. If you don't like 2nd quantization ( i.e d/dtF^=[F^,H^] ) you can use the Path integral formalism to come to your equations of motion, but Euler-Lagrange equations don't play a role in the Pathintegral formalism.
    So what are they good for in QUANTUMFT? Where is the point at which they come into play?

    Best regards Martin
  2. jcsd
  3. Jun 21, 2007 #2


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    Hey! Are you the one I started a discussion on renormalization a while ago and did not complete? If yes, we should go back to it!!

    To answer your question: in the path integral approach of QFT, the fields are *not* operators, they are ordinary fields! This is one thing that is great about the path integral approach!
  4. Jun 21, 2007 #3
    The Heisenberg picture field operators in QFT are assumed to satisfy the Euler-Lagrange equations of the classical theory. I've seen those all over the place.

    The equations of type d/dtF^=[F^,H^] are the quantum analogue of the Hamilton's equations which are alternative description but not the same as Euler-Lagrange eqs.

    Derivative of operator function with respect to operator is defined in a natural way for polynomial functions or Taylor expansions just like in ordinary calculus. Example:

    [tex] \frac{d}{dx} x^2 = 2x [/tex]

    where x is operator.
    Last edited: Jun 21, 2007
  5. Jun 22, 2007 #4
    Hi smallphi!

    Could you tell me some book with page?
    Maybe Ryder or Peskin

  6. Jun 22, 2007 #5
  7. Jun 22, 2007 #6
    I know :wink:, but this doesn't answer my question, I mean In the Path-Integral Formalism you don't use the EL-eq. to come to your euqations of motion. You use Feynman-diagrams to compute amplitudes for processes, so where do the EL-eq. play a role here?
  8. Jun 22, 2007 #7

    The Euler-Lagrange equation for the field operator of scalar field is Klein-Gordon equation (2.45) in section 2.4 'The Klein-Gordon field in spacetime', Peskin.
  9. Jun 22, 2007 #8
    (2.45) follows from (2.44) which is d/dtF^=[F^,H^].
    Indeed (2.45) is an equation of motion, but not the quantized version of Euler-Lagrange dµ(dL/d(dµPHI)) - dL/dPHI=0
  10. Jun 22, 2007 #9
    Klein-Gordon is exactly the Euler-Lagrange equation for the scalar field Lagrangian density. Peskin and Schroeder 'derive' it as equation (2.7) without showing you the derivation (as usual).

    Of course Euler-Lagrange equations can be obtained from the Hamilton's equations since they are equivalent systems of differential equations describing the system's evolution. Peskin did exactly that in section 2.4 only with operators.
    Last edited: Jun 22, 2007
  11. Jun 23, 2007 #10
    In Quantum-Field Theory gauge fields usually enter the scenario. The theories can be divided in to be abelian and non-abelian. The Euler-Lagrange equations follow then by building them with respect to this gauge fields. You receive the matter free Euler-Lagrange equations with d F=0 or the matter field equations being d F=j. This is not an exact descpriction and by far not the whole story, but I believe you find this in the book by Warren Siegel in the arxiv: http://www.arxiv.org hep-th:9912205
  12. Jun 28, 2007 #11
    The quantum equivalent of the EL equations are the Schwinger-Dyson equations, which say that the EL equations hold as operator equations in expectations values.
  13. Jul 19, 2007 #12
    Euler-Lagrange equations in QFT??

    I am also quite interested in this question. I have reviewed this thread and have had a number of Usenet discussions at sci.physics.foundations and sci.physics.research these past couple of weeks regarding the applicability of the Euler-Lagrange equation in QFT, and am still mildly perplexed about something. Let me see if I can put this concisely and someone can help me see what I might be missing.

    As I understand it, Maxwell's wave equation with source (&_u is a partial derivative):

    -J^v = &^u &_u A^v (1)

    is physically independent of any action principles, and so can be used in any form and in any context.

    Next, the QED Lagrangian density:

    L = -.5 (&^uA_v) (&_u A^v - &_v A^u) + A^u J_u (2)

    in covariant gauge &_u A^u is also employed throughout QED, most notably, to specify an action ($ is a four-dimensional integral from -oo
    to +oo):

    S = $ d^4x L (3)

    and in the exponentials:

    exp[iS] (4)

    which appear in path integrals. Presumably, there is nothing wrong with applying (2)-(4) broadly throughout QED, which is QFT for

    Now, we come to the Euler-Lagrange equation:

    &_u[&L/&(&_uA_v)-&L/&A_u]=0 (5)

    which is specified so as to minimize an action which, classically, is taken to be continuous. And so, using this equation is "verboten" in QFT, unless, of course, one is considering the classical limit, when it is then be cast in terms of expectation values using Schwinger/Dyson.

    What I find perplexing is that (2) is OK to use in QFT and that (1) is OK to use in QFT and that (5) is just another way of stating (1) when L is given by (2), but, we can't use (5) other than for a classical limit. Is (5) somehow really not the same as (1)? If so, can someone please pinpoint the difference?

    Let me put this slightly differently, speaking from a mathematician's viewpoint: if (5) with (2) is just another way of writing (1) -- mathematically -- then -- again, mathematically, if (1) can be used in a given situation then [(5) with (2)] = (1) can be used in the same situation. And yet, physically, it seems as though we can use (1) if we call it (1) but we can't use (1) if we call it [(5) with (2)].


    Jay. :confused:

    Jay R. Yablon
    Email: jyablon@nycap.rr.com
    co-moderator: sci.physics.foundations
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