Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is wrong with this lagrangian

  1. Jul 9, 2006 #1

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's probably a dumb question but what are the criteria for picking the lagrangian that leads to a certain wave equation? It is not enough to simply impose that Hamilton's principle leads to the correct wave equation so I am looking for the general principles.

    To be specific, consider Schroedinger's equation.

    The lagrangian that is given in textbooks is
    [tex] {\cal L} = {i \over 2} ( \phi^* \partial_t \phi - \phi \partial_t \phi^*) - {1 \over 2} \partial_x \phi^* \partial_x \phi + V(x) \phi \phi^* [/tex]

    and the variation wrt phi^* gives the equation for phi whereas the variation wrt phi gives the equation for phi^*.

    Fine, but one could have picked
    [tex] {\cal L} = i \phi \partial_t \phi - {1 \over 4} (\partial_x \phi)^2 + {1 \over 2} V(x) \phi^2 + {\rm terms \, in \,} \phi^*[/tex]

    instead and it seems to me that this would also lead to the correct equations (it's easy to write the correct eqs for phi^*).

    So what is wrong with this lagrangian?

    There *is* a problem because the conjugate momentum is phi (instead of being i phi^* with the usual lagrangian) and of course, the canonical quantization rule does not make sense if the conjugate momentum is equal to the field itself.

    One problem is that my lagrangian does not have the phase invariance [itex] \phi \rightarrow e^{i \alpha} \phi [/itex] so that there is no symmetry current associated to this symmetry.

    So what is the fundamental principle in determining a lagrangian? Obtaining a lagrangian which yields the correct eoms is necessary but not sufficient.

    Thanks
     
  2. jcsd
  3. Jul 10, 2006 #2

    Haelfix

    User Avatar
    Science Advisor

    "One problem is that my lagrangian does not have the phase invariance so that there is no symmetry current associated to this symmetry."

    Bingo!

    While the lagrangian you wrote down is perfectly fine classically (whatever it may describe), upon quantization it suffers from unremediable infinities.

    What makes a good lagrangian? Well the renormalization program as well as lorentz invariance fortunately leave us with only a handful of choices, all of which are known.

    If you want to relax renormalization (as is often done in modern Wilsonian treatments) then you have to work harder and look for other physical constraints (the presence of anomalies and matching conditions and so forth). Also there are strong physical theorems in say particle physics which limit any hope of going off the deep end, like the Colemann Mandula theorem.
     
  4. Jul 10, 2006 #3

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Thank you.

    What I was trying to build is a lagrangian for a quantum field theory of the Schrodinger equation. Basically, a classical lagrangian which upon quantization would produce a many-body nonrelativistics system of particles obeying the Schrodinger equation (as is often presented in some introductory QFT books).

    I was *not* preoccupied with Lorentz invariance or renormalizability.
    I know what you mean about using all those conditions as a guide in order to write lagrangians ab initio . What I am concerned here is writing lagrangians knowing the equations of motion .

    So if one strictly looks at reproducing the equations of motion, there are several lagrangian possible, in general?

    I am still confused about why the phase invariance property would matter in my example. If the Lagrangian reproduces the correct eoms, why is phase invariance an issue?

    My main problem, I think, is that I don't see how the phas invariance issue is tied with the problem with the conjugate momentum (that [itex] \pi = \phi [/itex]). I think it is connected at some deep level but I don't really understand the connection. I think it is all connected with degrees of freedom (that the Lagrangian must contain the different independent degrees of freedom interacting in a non trivial way and here they are phi and phi^*) and that somehow the phase invariance issue might be tied to that. But I feel that I am missing the "deep" idea here.

    Thanks for your input!

    Patrick
     
  5. Jul 11, 2006 #4

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I just read this thread and I'm probably making a too fast and unthoughtful comment here, but if the conjugate momentum of phi is phi itself, haven't we now a first-order differential equation instead of a second-order (and hence eliminate a lot of solutions), in other words, we have different equations of motion ?
     
  6. Jul 11, 2006 #5

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, but the equation *is* first order in time so that's ok.
    The time derivative part is [tex] \partial_t { \delta L \over \delta (\partial_t L)} [/tex]
    People always say to use [itex] L = \phi^* \partial_t \phi [/itex] (modulo some "i" factors) so that the variation wrt Phi^* gives the equation for Phi. Then the momentum conjugate to Phi is Phi^*.

    But I was wondering why I could not take [itex] L = \phi \partial_t \phi [/itex] and then do a vaiation wrt Phi itself to get directly the equation for Phi. I know that there are problems once we get to finding the conjugate momentum and then finding the hamiltonian, etc. But I was wondering, first of all, if there was any problem in getting the eoms from my lagrangian (maybe there is something I am not doing right about the way I am applying the Euler-Lagrange eqs?). And if there is *no* problem, I am wondering what are then the general principles in obtaining the correct lagrangian (if obtaining the eoms is not sufficient).

    I guess that's another related issue: is getting the correct eoms a necessary *and* sufficient condition or is it only necessary without being sufficient? In the later case, what are the other conditions?

    (Aside: In the book by Hatfield he seems to be using symmetries and current conservation as a fundamental guiding pricniple in determining the Lagrangian. Of course, the question is then: what if one has a system an done does not notice a certain symmetry!? )

    Regards

    PAtrick
     
  7. Jul 11, 2006 #6

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You should listen more carefully to what I say. When I say that my comments are probably too fast and unthoughtful, you should take that seriously :blushing: :biggrin:

    Hum. I get your point now. :shy:
     
  8. Jul 12, 2006 #7

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This seems obvious: the equations of motion specify the extrema of the lagrangian, and don't say anything of what happens outside of them. It is as if you were to give a list of maxima of a function (list = equations of motion) and then try to find the function that has these maxima (function = lagrangian). There are of course zillions of functions that will have the same maxima, so when taking the extrema of the function, you'll find back your list of maxima.

    However, I'm not sure how much of this "almost total liberty" remains of the lagrangian when certain smoothness and other conditions are imposed. But a priori, I'd think that there are miriads of different lagrangians that can reproduce the same set of extremal curves (= solutions to the equations of motion).

    In that case, the transition from EOM --> Feynman path integral is ambiguous, because in the Feynman path integral, you NEED the values of the lagrangian in the non-solutions (they contribute), while these values are not specified by the EOM, and could in principle be anything.
     
  9. Jul 12, 2006 #8
    In my first attempt at resolving this difficulty, I convinced myself that the Euler-Lagrange equations are not always true.

    ... And then almost immediately after posting, I realized that that claim was rubbish. :yuck:

    It was strange. When I used the E-L equations on the proposed Lagrangian, I would get the Schrödinger equation (modulo a sign in front of the ∂x2φ term). But when I performed the variation δS with the actual expression for L under the integrals I would not! In that case, I would find a term of the form "iφδ(∂tφ)", which could be rewritten as "iφ∂t(δφ)" and then integrated by parts to give
    "-i(∂tφ)(δφ)" ... resulting in a cancellation whereby "∂tφ" drops completely out of the picture.

    Now I see my error in applying the E-L equations. I was writing

    L/∂φ = V(x)φ ,

    when really I should have been writing

    L/∂φ = V(x)φ + i∂tφ .


    ... What's wrong with L is that it doesn't lead to the correct equations.
     
  10. Jul 12, 2006 #9

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Uh, yes, I guess you're right. I took it for granted that things came out correctly, but you're right.

    The term [tex]\phi \partial_t \phi [/tex] will always, in the [tex] \frac{\delta L}{\delta \phi}[/tex] term, give rise to [tex] \partial_t \phi [/tex], and in the term [tex]\partial_t \frac{\delta L}{\delta \partial_t \phi}[/tex], it will give [tex] \partial_t \phi[/tex], so the subtraction of both in the E-L equation will make the time derivative drop out. No coefficient in front of the term in L can change this.
    However, in the original lagrangian with the conjugate phi and the phi, there is a SIGN CHANGE between the two terms containing the partial derivatives to t, so that the above game can be played USING THE TWO DIFFERENT TERMS, and hence the two different signs.

    I mean, the derivative to phi works on the second term, while the derivative to \partial_t phi works on the first term. It is the sign difference which avoids the dropping out.
    If both contributions are, however, generated by the same term, there cannot be such a sign change.
     
  11. Jul 12, 2006 #10

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see what you are saying. But since all this can eb applied between spacetime points which are infinitesimally close to one another, it was not clear to me that one could really write different Lagrangians yielding the correct eoms at all points (beside the obvious liberty of adding different constants). I would be interested in seeing an example of two different lagrangians (with different dependence on the degrees of freedom) yielding the same eoms. I will give it some thought.
    Indeed. So a possible non-uniqueness of the lagrangian would be an acute problem here. Food for thoughts...
     
  12. Jul 12, 2006 #11

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Of course!
    Sorry for my dumb question:frown: :frown: :cry:

    That was really dumb of me.

    Ok, that makes sense now.
    Thanks Eye in the Sky for correcting my stupid mistake and thanks Vanesch for further comments.

    This is all interesting because for eqs linear in the time derivative, the momentum conjugate to the field is the complex conjugate to the field (this is the case for Dirac equation too) whereas for eqs with second order time derivative, the conjugate momentum is the time derivative of the field.

    But that brings up an obvious question: it's not possible to have a field equation first order in time for a real field? Let's say I just remove the i in Schrodinger's equation and treat this as a classical wave equqation for a real field?!

    There is something that I am missing. It obviously has something to do with the number of degrees of freedom. With a complex field, one has to specify both Phi and Phi^* at t=0. And then one evolves in time. But I don;t see the problem with specifying a real field at t=0 and then evolving in time. :uhh: :uhh:

    Thanks again for correcting my stupid mistake!

    Patrick
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: What is wrong with this lagrangian
  1. What goes wrong here? (Replies: 6)

Loading...