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

1. Apr 24, 2016

### spaghetti3451

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. Apr 24, 2016

### stevendaryl

Staff Emeritus
It looks right to me.

3. Apr 24, 2016

### spaghetti3451

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?

4. Apr 24, 2016

### stevendaryl

Staff Emeritus
Yes.

5. Apr 24, 2016

### spaghetti3451

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.