1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Noether current for SO(N) invariant scalar field theory

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I understand the premise of Noether's theorem, and I've read over it in as many online lectures as I can find as well as in An Introduction to Quantum Field Theory; Peskin, Schroeder but I can't seem to figure out how to actually calculate it. I feel like I'm missing a step that all of the explanations are glossing over.

    2. Relevant equations

    I want to find the Noether current [itex]J^{\mu}_{a b}[/itex] with Lagragian:

    [itex]L = \frac{1}{2} \partial_{\mu} \Phi^{a} \partial^{\mu} \Phi^{a} - \frac{1}{2} m^{2} \Phi^{a} \Phi^{a} - \frac{1}{4} \lambda ( \Phi^{a} \Phi^{a} )^{2}[/itex]

    Under the SO(N) symmetry:

    [itex]\Phi^{a} \rightarrow \Lambda^{a b} \Phi^{b}[/itex]

    Which is infinitesimally:

    [itex]\Phi^{a} \rightarrow \Phi^{a} + \epsilon^{a b} \Phi^{b}[/itex]

    And given the following relations:

    [itex]\Lambda^{T} \Lambda = I ; \Lambda = I + \epsilon ; \epsilon^{a b} = - \epsilon^{b a}[/itex]

    3. The attempt at a solution

    So I know that [itex]J^{\mu} = \frac{∂L}{∂(∂_{\mu} \Phi^{a})} Δ \Phi^{a} - F^{\mu}[/itex]


    [itex]F^{\mu} = \frac{∂L}{∂(∂_{\mu} \Phi^{a})} Δ \Phi^{a}[/itex]

    This is one of things I'm confused by, as it seems to be a meaningless statement.

    Regardless, I calculated

    [itex]\frac{∂L}{∂(∂_{\mu} \Phi^{a})} Δ \Phi^{a} = ∂^{\mu} \Phi^{a} Δ \Phi^{a}[/itex]

    And I'm fairly certain that [itex]Δ \Phi^{a} = \epsilon^{a b} \Phi^{b}[/itex]

    This would give me that:

    [itex]J^{\mu} = ∂^{\mu} \Phi^{a} \epsilon^{a b} \Phi^{b} - F^{\mu}[/itex]

    And a few of the similar examples I found seem to take F as zero here. Giving me:

    [itex]J^{\mu}_{a b} = ∂^{\mu} \Phi^{a} \epsilon^{a b} \Phi^{b}[/itex].

    Am I vastly wrong? Where am I going astray if so?

    Any help would be really appreciated, especially any help understanding the process step by step.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 25, 2012 #2


    User Avatar
    Homework Helper

    I think your definition of [itex]F^\mu[/itex] is off. I only have Peskin and Schroeder to check with at the moment, but it seems that a more appropriate definition of [itex]F^\mu[/itex] given what's in the book would be

    [tex]\Delta L = \epsilon\partial_\mu F^\mu[/tex]

    where [itex]\Delta L[/itex] is determined by applying the symmetry transformation in question to the Lagrangian as a whole.

    Other than that, you seem to be on the right track with the calculation, although take that with a grain of salt because I'm pretty tired and I'm not 100% sure I'm thinking straight at the moment.
  4. May 7, 2012 #3
    An earlier part of this question asked me to get the Euler-Lagrange equations of motion.

    I'm unsure what [itex]\frac{\partial}{\partial \phi_{\rho}} (\lambda (\phi^{a} \phi^{a}))^{2}[/itex] is though. Otherwise the equation itself seems fairly straight-forward.

    I get [itex]\partial_{\sigma} \partial^{\sigma} \phi_{\rho} + m^{2} \phi^{\rho} + \frac{1}{2}(\lambda(\phi^{a}\phi^{a}))(\frac{ \partial}{\partial \phi_{\rho}} (\lambda (\phi^{a} \phi^{a})))[/itex].

    Should the first L be the original Lagragian and the [itex]\Delta L[/itex] be the the Lagrangian with the transformation applied. That makes a lot more sense.

    I'm going to try this an post up my attempts.

    If anyone has anymore feedback it'd be much appreciated. I can't seem to find much about this subject at all.
  5. May 8, 2012 #4
    Okay, so after some digging and working at the problem I'm now down to a computational problem. I'm pretty sure this is the right method, but its getting me nowhere. Going through the whole problem from scratch:

    The Noether current [itex]J^{\mu}[/itex] is:
    [itex]J^{\mu} = \frac{\partial L}{\partial (\partial_{\mu} \phi^{a})} \alpha \Delta \phi - F^{\mu}[/itex]

    Where [itex]\Delta L = \partial_{\mu} F^{\mu}[/itex].

    Given the Lagrangian [itex]L = \frac{1}{2} \partial_{\mu} \phi^{a} \partial^{\mu} \phi^{a} - \frac{1}{2} m^{2} \phi^{a} \phi^{a} - \frac{1}{4} \lambda ( \phi^{a} \phi^{a} )^{2}[/itex]
    and the infinitesimal transformation [itex]\phi^{a} \rightarrow \phi^{a} + \epsilon^{a b} \phi^{b}[/itex] we have:

    [itex]\alpha = \epsilon^{a b}[/itex] and [itex]\Delta \phi^{a} = \phi^{b}[/itex].

    Now [itex]\alpha \Delta L = \frac{\partial L}{\partial \phi^{a}} \alpha \Delta \phi + (\frac{\partial L}{\partial (\partial_{\mu} \phi^{a})}) \partial_{\mu}(\alpha \Delta \phi)[/itex]

    So in our case we have:

    [itex]\epsilon^{ab} \partial_{\mu}(F^{\mu}) = (-m^{2}\phi^{a} - \frac{1}{2}(\lambda(\phi^{a}\phi^{a})( \frac{\partial}{\partial \phi^{a}} \lambda(\phi^{a}\phi^{a}))) \epsilon^{a b} \phi^{b} + (\partial^{\mu} \phi^{a}) (\epsilon^{ab}\partial_{\mu} \phi^{b})[/itex]

    [itex]\partial_{\mu}(F^{\mu}) = -m^{2}\phi^{a}\phi^{b} - \frac{1}{2}(\lambda(\phi^{a}\phi^{a})( \frac{\partial}{\partial \phi^{a}} \lambda(\phi^{a}\phi^{a})))\phi^{b} + \partial^{\mu} \phi^{a} \partial_{\mu} \phi^{b}[/itex]

    Here I hit a roadblock. Is there some general way [itex]\lambda[/itex] is supposed to act? Nothing is given in the question.

    How do I take a [itex]\partial_{\mu}[/itex] out of that equation? I can't see anyway to factorise it.
  6. May 8, 2012 #5


    User Avatar
    Homework Helper

    [itex]\lambda[/itex] is a constant, if that helps, so it doesn't really act in any nontrivial manner... other than that, I'll come back and take a closer look at this shortly.
  7. May 8, 2012 #6
    Thank you. I thought Lambda might be a constant, does that mean the [itex]\phi[/itex] in the brackets after it aren't saying a function [itex]\lambda[/itex] but rather that its the constant times [itex]\phi^{4}[/itex].

    That actually make sense, as these questions seem to reuse Lagrangians and some of the examples I saw had this.
  8. May 8, 2012 #7


    User Avatar
    Homework Helper

    That's right, [itex]\lambda(\phi^a\phi^a)^2 = \lambda\times(\phi^a\phi^a)\times(\phi^a\phi^a) \sim \lambda \Phi^4[/itex]. Pretty much every Lagrangian you encounter in basic QFT will be a perturbative expansion of the form [itex]\sum_n(\text{const.})(\text{fields})^n[/itex], but you don't get arbitrary functions.

    Now that I've had a chance to check out your method more closely:
    That's kind of fishy. You don't need to know what part exactly corresponds to [itex]\alpha[/itex] and what part corresponds to [itex]\Delta \phi^a[/itex], because they always occur in combination. So just leave it as
    [tex]\alpha\Delta\phi^a = \epsilon^{ab}\phi^b[/tex]
    In fact, better yet: forget about the [itex]\alpha[/itex], and let's just define
    [tex]\delta\phi^a = \epsilon^{ab}\phi^b[/tex]
    Here I'm using lowercase delta to designate the "alpha-less" notation. With this definition,
    [tex]J^\mu = \frac{\partial L}{\partial(\partial_\mu\phi^a)}\delta\phi^a - F^\mu[/tex]
    [tex]\delta L = \partial_\mu F^\mu[/tex]

    Now, how do you compute [itex]\delta L[/itex]? You just plug the transformed versions of the fields into [itex]L[/itex], and subtract the original [itex]L[/itex] from the transformed version. You don't need to actually take a derivative and use the chain rule, as you were trying to do. (You can do that, but there are a ton of subtle ways to trip yourself up. If you're curious I can explain that in some detail.) What I would recommend is this:
    • First write down the transformations of [itex]\phi^a[/itex] and [itex]\partial_\mu\phi^a[/itex]
    • Then use those to write down the transformations of [itex]\phi^a\phi^a[/itex] and [itex]\partial_\mu\phi^a\partial^\mu\phi^a[/itex] (see the note below)
    • Finally, put those expressions into the Lagrangian to figure out the transformed version of the Lagrangian
    • Then subtract [itex]L[/itex], and what you're left with is [itex]\delta L[/itex]
    Here's the note: one thing you will need to keep in mind each pair of the form [itex]\phi^a\phi^a[/itex] acts like the multiplication of a vector with a transposed vector. Even though the two factors of [itex]\phi^a[/itex] in this term may appear to be the same field, they won't transform the same way, if the fields as a whole are to constitute an SO(N) vector representation. (This is somewhat analogous to how two functions can have the same value but different derivatives at a point) For an SO(N) transformation, one of these fields needs to transform as [itex]\phi^a \to \phi^a + \epsilon^{ab}\phi^b[/itex] and the other needs to transform as [itex]\phi^a \to \phi^a - \epsilon^{ab}\phi^b[/itex]. Remember that you can discard anything at order [itex]\epsilon^2[/itex] or higher.
  9. May 10, 2012 #8
    So [itex]\delta\phi^{a} = \epsilon^{ab}\phi^{b}[/itex] and we can get:

    [itex]\delta L = 0[/itex]

    I'm going to go through the process so that hopefully in future people will be able to find this thread to help figure out Noether currents.


    [itex]L' = \frac{1}{2}(\partial_{\mu}(\phi^{a} + \epsilon^{ab}\phi^{b}))(\partial^{\mu}(\phi^{a} + \epsilon^{ab}\phi^{b})) - \frac{m^{2}}{2}(\phi^{a} + \epsilon^{ab}\phi^{b})(\phi^{a} - \epsilon^{ab}\phi^{b}) - \frac{\lambda}{4}[(\phi^{a} + \epsilon^{ab}\phi^{b})(\phi^{a} - \epsilon^{ab}\phi^{b})]^{2}[/itex]

    [itex]= \frac{1}{2}(\partial_{\mu}\phi^{a})(\partial^{\mu}\phi^{a}) + \frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{a}\partial^{\mu}\phi^{b}) + \frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{b}\partial^{\mu}\phi^{a}) - \frac{m^{2}}{2}(\phi^{a}\phi^{a}) - \frac{\epsilon^{ab}m^{2}}{2}(\phi^{a}\phi^{b}) + \frac{\epsilon^{ab}m^{2}}{2}(\phi^{b}\phi^{a}) - \frac{\lambda}{4}[\phi^{a}\phi^{a} + \epsilon^{ab}\phi^{a}\phi^{b} - \epsilon^{ab}\phi^{b}\phi^{a}]^{2} + O(\epsilon^{2})[/itex]

    So [itex]\delta L = \frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{a}\partial^{\mu}\phi^{b}) + \frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{b}\partial^{\mu}\phi^{a})

    =\frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{a}\partial^{\mu}\phi^{b}) + \frac{\epsilon^{ba}}{2}( \partial_{\mu}\phi^{a}\partial^{\mu}\phi^{b})

    =\frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{a}\partial^{\mu}\phi^{b}) - \frac{\epsilon^{ab}}{2}( \partial_{\mu}\phi^{a}\partial^{\mu}\phi^{b})

    = 0.[/itex]

    This means we have [itex]J^{\mu}_{ab} = \frac{ \partial L}{ \partial ( \partial_{\mu} \phi^{a})} \delta \phi^{a} = (\partial^{\mu} \phi^{a}) \epsilon^{ab} \phi^{b}[/itex]

    So the next leg of my equations involves using the equations of motion to show that the Noether current is conserved (i.e.[itex]\partial_{\mu} J^{\mu}_{ab} = 0[/itex]).

    E-L: [itex]\partial_{\mu} \partial^{\mu} \phi^{a} + m^{2} \phi^{a} + \lambda (\phi^{a} \phi^{a} \phi^{a})= 0[/itex]

    and [itex]\partial_{\mu} J^{\mu}_{ab} = \epsilon^{ab}(\partial_{\mu} \partial^{\mu} \phi^{a}) \phi^{b} + \epsilon^{ab} (\partial^{\mu} \phi^{a}) (\partial_{\mu} \phi^{b})[/itex]

    I'm not sure about the next step.

    I thought i might have to use [itex]\partial_{\mu} \partial^{\mu} \phi^{a} = - m^{2} \phi^{a} - \lambda (\phi^{a} \phi^{a} \phi^{a})[/itex] but this just gives me [itex] \epsilon^{ab}(- m^{2} \phi^{a} - \lambda (\phi^{a} \phi^{a} \phi^{a})) \phi^{b} + \epsilon^{ab} (\partial^{\mu} \phi^{a}) (\partial_{\mu} \phi^{b})[/itex].

    Other than the fact that this is very reminiscent of the Lagrangian I'm not sure where to go from here.

    [itex]\epsilon^{ab} [ (\partial_{\mu} \phi^{b}) (\partial^{\mu} \phi^{a}) - m^{2} \phi^{a} \phi^{b} - \lambda ( \phi^{a} \phi^{a} \phi^{a} \phi^{b})][/itex].
  10. May 10, 2012 #9


    User Avatar
    Homework Helper

    Oh wait, I think I may have slightly misled you - I forgot about the fact that [itex]\epsilon^{ab}[/itex] is antisymmetric. That automatically takes care of the inversion of the transformation, so you don't need to make one of them [itex]\phi^a \to \phi^a - \epsilon^{ab}\phi^b[/itex]; you can just use [itex]\phi^a \to \phi^a + \epsilon^{ab}\phi^b[/itex] for both factors. There are also a couple of little sign errors in the calculation you just posted. In fact in the end, I think all those mistakes either cancel out or don't matter and you get [itex]\delta L = 0[/itex] anyway. (Here's another way to check: use the full transformation [itex]\phi^T\phi \to \phi^T\Lambda^T\Lambda\phi[/itex].)

    Also consider that in the Euler-Lagrange equation, the product [itex]\phi^a\phi^a\phi^a[/itex] doesn't really make sense. Remember that the Einstein summation convention still applies here, i.e. [itex]\phi^a\phi^a[/itex] is really [itex]\delta_{ab}\phi^a\phi^b[/itex]. So any time you see more than two of the same index, you should be a little suspicious. Keep in mind that [itex](\phi^a\phi^a)^2[/itex] is not [itex]\sum_a (\phi^a\phi^a)[/itex], it's [itex]\biggl(\sum_a\phi^a\phi^a\biggr)^2[/itex]. With that in mind, and knowing that the indices are dummy indices, what's a better way to write the last term of the E-L equation?

    Finally, don't forget that [itex]\epsilon^{ab}[/itex] is antisymmetric, so what does that mean about [itex]\epsilon^{ab}\phi^a\phi^b[/itex] if [itex]\phi^a[/itex] and [itex]\phi^b[/itex] are commuting field components?
  11. May 10, 2012 #10
    It's very late here, but I'll hazard a guess that they act like a symmetric term. So an anti-symmetric and symmetric terms go to 0.
  12. May 11, 2012 #11


    User Avatar
    Homework Helper

    I suspect you're on the right track. Try writing it out explicitly and see if that works.
  13. May 12, 2012 #12
    Okay so, I have:

    [itex]\frac{ \lambda}{4} (\phi^{a} \phi^{a})^{2} = \frac{ \lambda}{4} (\delta_{ab}\phi^{a} \phi^{b})(\delta_{ab}\phi^{a} \phi^{b})[/itex]

    [itex] \frac{\partial}{\partial \phi^{a}} (\frac{ \lambda}{4} (\delta_{ab}\phi^{a} \phi^{b})(\delta_{ab}\phi^{a} \phi^{b})) \\
    = \frac{ \lambda}{4} [\frac{\partial}{\partial \phi^{a}} (\delta_{ab}\phi^{a} \phi^{b})] (\delta_{ab}\phi^{a} \phi^{b}) + (\frac{ \lambda}{4} (\delta_{ab}\phi^{a} \phi^{b}) [\frac{\partial}{\partial \phi^{a}} (\delta_{ab}\phi^{a} \phi^{b})] \\
    = \frac{ \lambda}{4} [ \delta_{ab} \delta_{aa} \delta_{ab} \phi^{b} \phi^{a} \phi^{b} + \delta_{ab} \delta_{ab} \delta_{ab} \phi^{a} \phi^{a} \phi^{b} + \delta_{ab} \delta_{ab} \delta_{aa} \phi^{a} \phi^{b} \phi^{b} + \delta_{ab} \delta_{ab} \delta_{ab} \phi^{a} \phi^{b} \phi^{a} ] \\
    = \lambda ( \phi^{a} \phi^{b} \phi^{b} )[/itex]

    So an idea I had was that if [itex]\lambda ( \phi^{a} \phi^{b} \phi^{b} ) = \delta_{ab} \lambda ( \phi^{a} \phi^{b} \phi^{a} ) [/itex] we will get [itex]\epsilon^{ab} \delta_{ab} \lambda ( \phi^{a} \phi^{b} \phi^{a} ) \phi^{b} = \epsilon^{ab} \delta_{ab} \lambda ( \phi^{a} \phi^{b}) ( \phi^{a} \phi^{b})[/itex].

    If a [itex]\phi^{a}[/itex] and [itex]\phi^{b}[/itex] are commuting variables, [itex]\phi^{a} \phi^{b} = \phi^{b} \phi^{a}[/itex]. So it is a symmetric term. This gives us that [itex]\epsilon^{ab} \phi^{a} \phi^{b} = 0[/itex].

    So [itex]\partial_{\mu} J^{\mu}_{ab} = \epsilon^{ab} (\partial_{\mu} \phi^{b}) (\partial^{\mu} \phi^{a})[/itex].

    So I tried [itex][\partial_{\mu} \phi^{b}, \partial^{\mu} \phi^{a}] \\
    = \partial_{\mu} \phi^{b} \partial^{\mu} \phi^{a} - \partial^{\mu} \phi^{a} \partial_{\mu} \phi^{b} \\
    = \partial_{\mu} \phi^{b} \partial^{\mu} \phi^{a} - \delta_{ab} \delta_{ab} \eta^{\mu \mu} \eta^{\mu \mu} \partial_{\mu} \phi^{b} \partial^{\mu} \phi^{a} [/itex]

    So [itex](\delta_{ab})^{2} = (\eta^{\mu \mu})^{2} = 1[/itex] gives us commutativity. This implies symmetry, which once again gives us a zero term.

    Is this correct?
  14. May 12, 2012 #13


    User Avatar
    Homework Helper

    You wound up with the right answer here, but your work is a mess because you keep reusing the same variables for different indices. It makes it almost impossible to follow. Now, if you can do this for yourself without getting confused, then it's fine, but for anything you're going to be sharing with anyone else (including e.g. homework problems to be written up and handed in for a grade), you should be careful not to reuse index labels.
    [tex]\frac{\lambda}{4} (\phi^a\phi^a)^2 = \frac{\lambda}{4} (\delta_{ab}\phi^a\phi^b)(\delta_{cd}\phi^c\phi^d)[/tex]
    &= \frac{\lambda}{4}\biggl[\frac{\partial}{\partial\phi^k} (\delta_{ab}\phi^a\phi^b)\biggr] (\delta_{cd} \phi^c\phi^d) + \frac{\lambda}{4}(\delta_{ab} \phi^a\phi^b)\biggl[\frac{\partial}{\partial\phi^k} (\delta_{cd}\phi^c\phi^d)\biggr] \\
    &= \frac{\lambda}{4}\bigl[\delta_{ab}\delta_{ak}\delta_{cd}\phi^b\phi^c\phi^d + \delta_{ab}\delta_{bk}\delta_{cd}\phi^a\phi^c\phi^d + \delta_{ab}\delta_{cd}\delta_{ck}\phi^a\phi^b\phi^d + \delta_{ab}\delta_{cd}\delta_{dk}\phi^a\phi^b\phi^c\bigr] \\
    &= \frac{\lambda}{4}\bigl[\phi^k(\delta_{cd}\phi^c\phi^d) + \phi^k(\delta_{cd}\phi^c\phi^d) + (\delta_{ab}\phi^a\phi^b)\phi^k + (\delta_{ab}\phi^a\phi^b)\phi^k\bigr] \\
    &= \lambda\phi^k\phi^a\phi^a\end{align}[/tex]

    Again, you're making quite a mess out of the indices, but your procedure does basically seem to be correct.
  15. May 14, 2012 #14
    Thank you so much, I'll be much more formal and correct about my indices.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook