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Homework Help: Conserved current; Klein-Gordon

  1. Apr 12, 2009 #1


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

    Given the Lagrangian density of a complex relativistic scalar field
    [itex]\mathcal L=\frac{1}{2}\partial^\nu\phi^{*}\partial_\nu\phi-\frac{1}{2}m^2\phi^{*}\phi[/itex]
    where * stands for complex conjugation, compute the conserved current (using Noether's theorem).

    2. Relevant equations

    I can (should) use [itex]J^k=-\frac{\partial \mathcal L}{\partial (\partial_k \phi_I)}\phi_I[/itex], where summation over I is implied (which just means we have two terms, one for [itex]\phi[/itex] and one for [itex]\phi^{*}[/itex]).

    3. The attempt at a solution

    Well, I computed [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi)}=\partial^k\phi^{*}[/itex], and similarly [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=\partial^k\phi[/itex].

    Combining these, we simply get [itex]J^k=-\frac{\partial \mathcal L}{\partial (\partial_k \phi_I)}\phi_I=-\frac{\partial \mathcal L}{\partial (\partial_k \phi)}\phi-\frac{\partial \mathcal L}{\partial (\partial_k \phi^*)}\phi^*=-\frac{1}{2}\left(\partial^k\phi^{*}\phi+\partial^k\phi\phi^{*}\right)[/itex]

    But the correct answer should be [itex]=-\frac{1}{2}\left(\partial^k\phi^{*}\phi-\partial^k\phi\phi^{*}\right)[/itex], differing in a minus sign.

    (Just looking at the Lagrangian, [itex]\phi[/itex] and [itex]\phi^*[/itex] are symmetric, right? So the terms in the conserved current should also come in symmetric...but in the correct answer they aren't.)

    Apparently [itex]\frac{\partial \mathcal L}{\partial (\partial_k \phi^{*})}=-\partial^k\phi[/itex]?
    Last edited: Apr 12, 2009
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  3. Apr 14, 2009 #2

    George Jones

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    While [itex]\phi[/itex] and [itex]\phi^*[/itex] appear symmetrically in the Lagrangian, they don't come into play symmetrically in the application of Noether's theorem to a Klein-Gordon system.

    The Lagrangian is invariant under the continuous transformation [itex]\phi \rightarrow e^{i\alpha} \phi[/itex], which means [itex]\phi^* \rightarrow e^{-i\alpha} \phi^*[/itex].
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