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Conserved Noether current under U(1) symmetry of some Lagrangian

  1. Apr 24, 2016 #1
    1. The problem statement, all variables and given/known data

    The motion of a complex field ##\psi(x)## is governed by the Lagrangian ##\mathcal{L} = \partial_{\mu}\psi^{*}\partial^{\mu}\psi-m^{2}\psi^{*}\psi-\frac{\lambda}{2}(\psi^{*}\psi)^{2}##.

    Write down the Euler-Lagrange field equations for this system.

    Verify that the Lagrangian is invariant under the infinitesimal transformation ##\delta \psi = i \alpha \psi, \delta \psi^{*} = - i \alpha \psi^{*}##

    Derive the Noether current associated with this transformation and verify explicitly that it is conserved using the field equations satisfied by ##\psi##.

    2. Relevant equations

    3. The attempt at a solution

    ##\mathcal{L} = \partial_{\mu}\psi^{*}\partial^{\mu}\psi-m^{2}\psi^{*}\psi-\frac{\lambda}{2}(\psi^{*}\psi)^{2}##

    Here's my derivation of the Euler-Lagrange field equations:

    ##\frac{\partial \mathcal{L}}{\partial \psi} = -m^{2}\psi^{*}-\lambda (\psi^{*}\psi)\psi^{*}##

    and

    ##\frac{\partial \mathcal{L}}{\partial (\partial_{\rho}\psi)} = \frac{\partial}{\partial (\partial_{\rho}\psi)} \Big( \partial_{\mu}\psi^{*}\partial^{\mu}\psi \Big)= \frac{\partial}{\partial (\partial_{\rho}\psi)} \Big( \eta^{\mu\nu}\partial_{\mu}\psi^{*}\partial_{\nu}\psi \Big)=\eta^{\mu\nu}\frac{\partial}{\partial (\partial_{\rho}\psi)} \Big( \partial_{\mu}\psi^{*}\partial_{\nu}\psi \Big)=\eta^{\mu\nu}{\eta^{\rho}}_{\nu}\partial_{\mu}\psi^{*}={\eta^{\rho}}_{\nu}\eta^{\nu\mu}\partial_{\mu}\psi^{*}=\partial^{\rho}\psi^{*}##

    so that

    ##\frac{\partial \mathcal{L}}{\partial \psi} - \partial_{\mu}\Big(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi)}\Big)=0##

    ##-m^{2}\psi^{*}-\lambda(\psi^{*}\psi)\psi^{*}-\partial^{2}\psi^{*}=0##

    ##(\partial^{2}+m^{2}+\lambda|\psi|^{2})\psi^{*}=0##

    Furthermore,

    ##\frac{\partial \mathcal{L}}{\partial \psi^{*}} = -m^{2}\psi-\lambda (\psi^{*}\psi)\psi##

    and

    ##\frac{\partial \mathcal{L}}{\partial (\partial_{\rho}\psi^{*})} = \frac{\partial}{\partial (\partial_{\rho}\psi^{*})} \Big( \partial_{\mu}\psi^{*}\partial^{\mu}\psi \Big)={\eta^{\rho}}_{\mu}\partial^{\mu}\psi=\partial^{\rho}\psi##

    so that

    ##\frac{\partial \mathcal{L}}{\partial \psi^{*}} - \partial_{\mu}\Big(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\Big)=0##

    ##-m^{2}\psi-\lambda(\psi^{*}\psi)\psi-\partial^{2}\psi=0##

    ##(\partial^{2}+m^{2}+\lambda|\psi|^{2})\psi=0##

    Am I correct so far?
     
  2. jcsd
  3. Apr 24, 2016 #2

    stevendaryl

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    It looks right to me.
     
  4. Apr 24, 2016 #3
    Thanks!

    Next, I need to verify that the Lagrangian is invariant under the infinitesimal transformation ##\delta \psi = i \alpha \psi, \delta \psi^{*} = - i \alpha \psi^{*}##.

    Therefore,

    ##\mathcal{L} = \partial_{\mu}\psi^{*}\partial^{\mu}\psi-m^{2}\psi^{*}\psi-\frac{\lambda}{2}(\psi^{*}\psi)^{2}##
    ##\implies \delta\mathcal{L} = \delta(\partial_{\mu}\psi^{*}\partial^{\mu}\psi-m^{2}\psi^{*}\psi-\frac{\lambda}{2}(\psi^{*}\psi)^{2})##
    ##\implies \delta\mathcal{L} = \partial_{\mu}(\delta\psi^{*})(\partial^{\mu}\psi)+(\partial_{\mu}\psi^{*})\partial^{\mu}(\delta\psi)-m^{2}\psi(\delta\psi^{*})-m^{2}\psi^{*}(\delta\psi)-\lambda|\psi|^{2}\psi^{*}\delta\psi-\lambda|\psi|^{2}\psi\delta\psi^{*}##
    ##\implies \delta\mathcal{L} = \partial_{\mu}(-i\alpha \psi^{*})(\partial^{\mu}\psi)+(\partial_{\mu}\psi^{*})\partial^{\mu}(i\alpha\psi)-m^{2}\psi(-i\alpha\psi^{*})-m^{2}\psi^{*}(i\alpha\psi)-\lambda|\psi|^{2}\psi^{*}(i\alpha\psi)-\lambda|\psi|^{2}\psi(-i\alpha\psi^{*})##
    ##\implies \delta\mathcal{L} = -i\alpha(\partial_{\mu}\psi^{*})(\partial^{\mu}\psi)+i\alpha(\partial_{\mu}\psi^{*})(\partial^{\mu}\psi)+i\alpha(m^{2}\psi\psi^{*})-i\alpha(m^{2}\psi^{*}\psi)-i\alpha(\lambda|\psi|^{2}\psi^{*}\psi)+i\alpha(\lambda|\psi|^{2}\psi\psi^{*})##
    ##\implies \delta\mathcal{L} = 0##, due to pairwise cancellation.

    So, the Lagrangian ##\mathcal{L} = \partial_{\mu}\psi^{*}\partial^{\mu}\psi-m^{2}\psi^{*}\psi-\frac{\lambda}{2}(\psi^{*}\psi)^{2}## is invariant under the infinitesimal transformation ##\delta \psi = i \alpha \psi, \delta \psi^{*} = - i \alpha \psi^{*}##.

    Is my working correct?
     
  5. Apr 24, 2016 #4

    stevendaryl

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    Yes.
     
  6. Apr 24, 2016 #5
    Thanks!

    Finally, I need to derive the Noether current ##j^{\mu}## associated with the infinitesimal transformation ##\delta \psi = i \alpha \psi, \delta \psi^{*} = - i \alpha \psi^{*}## and verify explicitly that the Noether current ##j^{\mu}## is conserved using the field equations ##(\partial^{2}+m^{2}+\lambda|\psi|^{2})\psi=0## and ##(\partial^{2}+m^{2}+\lambda|\psi|^{2})\psi^{*}=0## satisfied by ##\psi##.

    ##\delta\mathcal{L}=\frac{\delta\mathcal{L}}{\delta\psi}\delta\psi+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\partial_{\mu}(\delta\psi)+\frac{\delta\mathcal{L}}{\delta\psi^{*}}\delta\psi^{*}+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\partial_{\mu}(\delta\psi^{*})##
    ##\implies \delta\mathcal{L}=\frac{\delta\mathcal{L}}{\delta\psi}\delta\psi+\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi\Big)-\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\Big)\delta\psi+\frac{\delta\mathcal{L}}{\delta\psi^{*}}\delta\psi^{*}+\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\delta\psi^{*}\Big)-\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\Big)\delta\psi^{*}##
    ##\implies \delta\mathcal{L}=\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\delta\psi^{*}\Big]##
    ##\implies \partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\delta\psi^{*}\Big]=0##
    ##\implies j^{\mu} = \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi^{*})}\delta\psi^{*}##
    ##\implies j^{\mu} = (\partial^{\mu}\psi^{*})(i\alpha\psi)+(\partial^{\mu}\psi)(-i\alpha\psi^{*})##
    ##\implies j^{\mu} = i\alpha (\psi(\partial^{\mu}\psi^{*})-\psi^{*}(\partial^{\mu}\psi))##

    and

    ##\partial_{\mu}j^{\mu}=i\alpha\partial_{\mu}[\psi(\partial^{\mu}\psi^{*})-\psi^{*}(\partial^{\mu}\psi)]##
    ##\implies \partial_{\mu}j^{\mu}=i\alpha[(\partial_{\mu}\psi)(\partial^{\mu}\psi^{*})+\psi(\partial_{\mu}\partial^{\mu}\psi^{*})-(\partial_{\mu}\psi^{*})(\partial^{\mu}\psi)-\psi^{*}(\partial_{\mu}\partial^{\mu}\psi)]##
    ##\implies \partial_{\mu}j^{\mu}=i\alpha[(\partial_{\mu}\psi)(\partial^{\mu}\psi^{*})+\psi(-m^{2}-\lambda|\psi|^{2})\psi^{*}-(\partial_{\mu}\psi^{*})(\partial^{\mu}\psi)-\psi^{*}(-m^{2}-\lambda|\psi|^{2})\psi]##
    ##\implies \partial_{\mu}j^{\mu}=0##, due to pairwise cancellation.

    Are my answers correct?
     
  7. Apr 24, 2016 #6

    stevendaryl

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    I think that's all correct.
     
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