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

  • #1
1,344
33

Homework Statement



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##.

Homework Equations



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?
 

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,461
2,629

Homework Statement



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##.

Homework Equations



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?
It looks right to me.
 
  • #3
1,344
33
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
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,461
2,629
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?
Yes.
 
  • #5
1,344
33
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?
 
  • #6
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
8,461
2,629
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?
I think that's all correct.
 

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

Replies
3
Views
1K
Replies
2
Views
2K
  • Last Post
2
Replies
42
Views
2K
Replies
2
Views
2K
Replies
1
Views
601
Replies
7
Views
948
  • Last Post
Replies
16
Views
3K
Replies
1
Views
902
  • Last Post
Replies
11
Views
2K
Top