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Lagrangian and quantum field theory

  1. Dec 5, 2013 #1
    Hello,

    I understand the classical Lagrangian which follows the Principle of Least Action(A)

    A=∫L dt

    But what is Lagrangian density? Is it a new concept?

    A=∫Lagrangian density dx^4

    Here 4 is the four vector? One time-like and 3 space-like co-ordinates?

    QFT uses Lagrangian to field theory?

    Thanks.
     
  2. jcsd
  3. Dec 5, 2013 #2

    atyy

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    The Lagrangian density is a also a classical concept. It's used in classical field theory.

    In quantum mechanics, the fundamental formulation uses commutation operators and Hilbert space. Often there is also a Lagrangian formulation (Feynman's path integral) which is much easier to calculate in, but one must keep in mind that not every Lagrangian you can write down has quantum mechanical meaning.
     
  4. Dec 5, 2013 #3
    Yes, in the equation A=∫Lagrangian density dx4 you're performing a quadruple integral over the three spatial dimensions and time.
     
  5. Dec 5, 2013 #4

    WannabeNewton

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    Hi there! Once you start talking about classical field theories like the classical electromagnetic field there is a surface level difference between the lagrangian for the EM field, which will give you the field equations of EM, and the lagrangian for a charged particle interacting with the field which will give you the equations of motion for that particle.

    For example, making no assumptions about non-relatavistic motion, a charged particle interacting with an electromagnetic field will have the usual free lagrangian ##L_{free} = \frac{p^{\mu}p_{\mu}}{m\gamma}## and we can take the interaction term to be ##L_{int} = \frac{q}{mc\gamma}A_{\mu}p^{\mu}## where ##A_{\mu}## is the electromagnetic 4-potential, ##p^{\mu}## is the 4-momentum of the charged particle, and ##\gamma## is the usual gamma factor from special relativity. Thus our total action would be ##S = \frac{1}{m}\int(p_{\mu} + \frac{q}{c}A_{\mu}) p^{\mu}\frac{dt}{\gamma} = \int(p^{\mu} + \frac{q}{c}A_{\mu})\mathrm{d} x^{\mu}##. Varying ##S## and using the principle of stationary action will give you the usual equations of motion for the charged particle ##\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau} =\frac{q}{c}F^{\mu}_{}{}_{\nu}u^{\nu}## where ##F_{\mu\nu} = 2\partial_{[\mu}A_{\nu]}## is the usual electromagnetic field tensor and ##u^{\mu}## is the 4-velocity of the particle.

    We can do a very similar thing for the electromagnetic field itself in order to get the inhomogenous parts of maxwell's equations (the homogenous parts are immediate from the definition of the electromagnetic field tensor in terms of the 4-potential: ##\partial_{[\gamma}F_{\mu\nu]} = 2\partial_{[[\gamma}\partial_{\mu]}A_{\nu]} = 0##). Here we first define a lagrangian density ##\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_{\mu}J^{\mu}## which is a Lorentz invariant scalar field of space-time points (here ##J^{\mu}## is the 4-current); ##\mathcal{L}## is constructed out of Lorentz covariant fields (which are local objects) in such a way so that ##\mathcal{L}## itself is manifestly Lorentz invariant. The lagrangian is then defined in terms of the lagrangian density as ##L = \int \mathcal{L}d^{3}x## so that the action will be ##S = \int Ldt = \int \mathcal{L}d^{4}x##. If you vary the action and apply the principle of stationary action you will in fact get ##\partial^{\nu}F_{\mu\nu} = J_{\mu}## which are in fact the inhomogenous parts of maxwell's equations.
     
  6. Dec 6, 2013 #5
    Hello WannebeNewton,

    Thank you very much for the wonderful answer. You have mention S=1/4......Where the 1/4 factor comes in? Also if you can tell me in brief the basic concept of Lagrangian density.

    Thanks.
     
  7. Dec 6, 2013 #6
    Also please tell me the formula for this Lfree and Linitial.
     
  8. Dec 6, 2013 #7
    What I understand is that The action of the system is defined to be the quantity S =∫Ldt. While studying field which take on different values at different space points Lagrangian is itself an integral:L = ∫d^3LD, where LD=Lagrangian density. In some place there is d^4x and in some places it is dx^4. Can you please explain?
     
  9. Dec 6, 2013 #8
    You can think of the Lagrangian density this way:

    Probably, the first Lagrangian you wrote down was for a single particle,
    [tex]L=\frac{m\dot{q}^2}{2} - V(q)[/tex].
    This [itex]L[/itex] determines what the particle will do and where it will go for all times. For a system of particles, the Lagrangian is a sum of individual Lagrangians,
    [tex]L=\sum_i \frac{m_i \dot{q}_i^2}{2}-V(q_1,q_2,\dots) [/tex]
    and determines what will happen to the entire system. In the limit of many particles in some volume, you can replace the sum over [itex]q_i[/itex] by an integral and capture continuum mechanics.

    The difference in field theory is that we aren't trying to describe how the coordinates [itex]q[/itex] evolve, rather we want to see how the field evolves. However, the field exists and has a value at all coordinates in space (at least in some region of interest), so to describe the entire field and not just the field at a given point, we need to integrate over space to get the full Lagrangian for the field. The Lagrangian density is thus analogous to the single particle Lagrangian in the particle system above, and represents the "Lagrangian" for the field at one point. But it is not enough to describe the entire field.
     
  10. Dec 6, 2013 #9

    WannabeNewton

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    I gave the formula for ##L_{\text{free}}## above. ##L_{\text{int}}## is the interaction term of the lagrangian (not "initial") and describes in essence how the particle interacts with whatever field you're working with. The example above deals with the electromagnetic field and I wrote down the interaction term for that. Notice that the lagrangians are Lorentz invariant quantities. The ##\frac{1}{4}## term in the action for the electromagnetic field is there so that you get the correct field equation after variation.

    As for your latest question, where exactly have you seen the notation ##dx^4## being used? In general, ##dx^4## refers to the differential of the ##x^4## coordinate. ##d^4 x## on the other hand represents a four-dimensional volume element.
     
  11. Dec 6, 2013 #10
    Let me put it in this way. Can we say the following?:

    A Lagrangian L, in a system (classical) which contains finite degrees of freedom, when generalized to QFT have infinite degrees of freedom and hence need re-normalization.

    In QFT, we consider Lagrangian density (Ld), integrating over all spacetime. As you have mentioned above:

    'to describe the entire field and not just the field at a given point, we need to integrate over space to get the full Lagrangian for the field'.

    Can we state that Action A=integral Ld dx^4 and as 'dauto' has mentioned above x^4 is the 1 time like and 3 space like dimension?

    In studying field can we mention L=integral d^3x Ld where d^3 are the 3 dimensional volume element?

    With the path integral formulation, the action principle is generalized and takes into account the infinite trajectories, considering +infinity and -infinity.

    Once we get the Lagrangian we can use it to get in QFT, Dirac eqn.,QED Lagrangian, QCD Lagrangian.......

    Please correct me where I am wrong.

    -- Thanks
     
  12. Dec 6, 2013 #11
    Well, as has been pointed out earlier in the thread, it could be a classical field theory (such as classical electrodynamics) as well. The same ideas of an infinite number of degrees of freedom and a Lagrangian density apply, because it is still about a field. When you quantize the classical field theory, you get a QFT. Anyway, you should probably try to play around with the Lagrangian WannabeNewton gave you for the classical electrodynamics and see if you can derive Maxwell's equations from it (it is a good practice, and will make some of the field concepts seem less new).

    Surely you mean [itex]d^4 x[/itex], not [itex]dx^4[/itex]? Then yes, the action is the time integral of the Lagrangian, and the Lagrangian is the space integral of the Lagrangian density.

    Yes (assuming three space dimensions). You could do the same with a Hamiltonian and a Hamiltonian density, but that formulation turns out to be less useful for quantum field theories.

    If you start with a Lagrangian, and apply the canonical quantization method you will get to some corresponding QFT, yes. Depending on which Lagrangian you start with, you can end up with these examples or something very, very different.

    The point of the path integral formulation is that all conceivable trajectories contribute (including some going to infinity), not just the one that minimizes the action. How this works is not as obvious for fields as for isolated particles, when you can just draw lines on a paper with the same start and end points. But if you understand how to get the equations of motion for the classical electrodynamics example, it should become less abstract.
     
  13. Dec 7, 2013 #12
    Thank you Hypersphere for the wonderful answer. It's clear.
     
  14. Dec 7, 2013 #13
    Now, just a small question before we conclude.

    As QFT takes all the possible trajectories including the infinite path possibilities but does it follow the Principle of Least Action?
     
  15. Dec 7, 2013 #14

    atyy

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    In the path integral picture, the least action path is the classical path. Quantum theory differs from classical theory by including all paths. The classical path is important in semiclassical approximations, and is a way to see why classical mechanics emerges as some limit of quantum mechanics. http://www.ifi.unicamp.br/~aguiar/Cursos/UFPE-2010/Path-Int-2.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  16. Dec 7, 2013 #15
    I expect that the less classical paths have a tendency to cancel out like in the path integral formulation, but I haven't studied QFT to be sure.
     
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