Derivative of a Noether current from Dirac Equation

In summary: You can choose to play with the second term using the equations you just wrote down to see if you can get anything interesting. However, the first term will give you a non-zero contribution and that is all you need for this problem.In summary, the conversation discusses finding the expression for the divergence of the Noether current for U(1) transformations in a Lagrangian system. The equations of motion for the transformed Lagrangian are considered, but it is ultimately determined that the equations of motion for the original Lagrangian should be used to find the non-zero contribution to the divergence of the current.
  • #1
Dixanadu
254
2

Homework Statement


Hey guys,
Consider the U(1) transformations
[itex] \psi'=e^{i\alpha\gamma^{5}}\psi [/itex] and [itex] \bar{\psi}'=\bar{\psi}e^{i\alpha\gamma^{5}} [/itex] of the Lagrangian [itex]\mathcal{L}=\bar{\psi}(i\partial_{\mu}\gamma^{\mu}-m)\psi[/itex].

I am meant to find the expression for [itex]\partial_{\mu}J^{\mu}[/itex].

Homework Equations


Gamma matrices anticommute
Noether current is [itex] \delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi+\delta x \mathcal{L} [/itex]
not sure of anything else...

The Attempt at a Solution


So here's what I've done so far. Since its a U(1) transformation, the coordinates arent changing, so the Noether current is given by [itex] \delta J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)}\delta\psi [/itex]. I found that [itex]\delta\psi=i\alpha\gamma^{5}\psi[/itex], so that

[itex]\delta J^{\mu}=-\bar{\psi}\alpha\gamma^{\mu}\gamma^{5}\psi[/itex], then I drop the infinitesimal parameter to get
[itex]J^{\mu}=-\bar{\psi}\gamma^{\mu}\gamma^{5}\psi[/itex].

So the next step is to calculate the derivative of this. Doing so, I get

[itex]\partial_{\mu}J^{\mu}=-(\partial_{\mu}\bar{\psi}\gamma^{\mu}\gamma^{5}\psi+\bar{\psi}\gamma^{\mu}\gamma^{5}\partial_{\mu}\psi)[/itex]

And at this point I am stuck...im not sure if this is right and/or if I can simplify this or do something neat with it? because I think I'm meant to be using the transformed Lagrangian
[itex]\mathcal{L}'=i\bar{\psi}\partial_{\mu}\gamma^{\mu}\psi-m\bar{\psi}e^{2i\alpha\gamma^{5}}\psi[/itex]
for something but I don't really know.

Thanks guys..
 
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  • #2
What equations of motion do you have?
 
  • #3
I think its just the equations of motion from the untransformed Lagrangian, which are:
[itex](i\partial_{\mu}\gamma^{\mu}-m)\psi=0[/itex]
and
[itex]i(\partial_{\mu}\bar{\psi})\gamma^{\mu}+m\bar{\psi}=0[/itex]
 
  • #4
So I suggest using those. :)
 
  • #5
I thought of that but I'm not sure if I can use these because I'm considering the transformed Lagrangian...and the above transformations arent a symmetry unless m = 0. So how can I use these equations of motion?
 
  • #6
The fields will follow their equations of motion. Had the transformation of the fields been a symmetry of the Lagrangian, the divergence of the current would be zero. Since it is not, you will simply get a non-zero expression if you anyway chose to write down the current that would be conserved if the symmetry breaking parameter was zero.
 
  • #7
I see...so I guess I am doing it wrong? I mean I've found the equations of motion from the transformed Lagrangian
[itex](i\partial_{\mu}\gamma^{\mu}- m e^{2i\alpha\gamma^{5}})\psi=0[/itex]
and
[itex]i \partial_{\mu}\bar{\psi}\gamma^{\mu}+m\bar{\psi}e^{2i\alpha\gamma^{5}}=0[/itex]

and now I'm trying to use THESE inside my expression for [itex]\partial_{\mu}J^{\mu}[/itex]. I guess you're saying to just use the orignals I posted in post #3?
 
  • #8
I get a nonzero answer either way but I'm not sure which one is right...:(
 
  • #9
Dixanadu said:
I see...so I guess I am doing it wrong? I mean I've found the equations of motion from the transformed Lagrangian
[itex](i\partial_{\mu}\gamma^{\mu}- m e^{2i\alpha\gamma^{5}})\psi=0[/itex]
and
[itex]i \partial_{\mu}\bar{\psi}\gamma^{\mu}+m\bar{\psi}e^{2i\alpha\gamma^{5}}=0[/itex]

and now I'm trying to use THESE inside my expression for [itex]\partial_{\mu}J^{\mu}[/itex]. I guess you're saying to just use the orignals I posted in post #3?
Yes, this willbe the divergence of the current.
 

What is the Noether current in the context of the Dirac equation?

The Noether current in the context of the Dirac equation is a conserved quantity that arises from the invariance of the Dirac Lagrangian under certain transformations. It represents a symmetry of the system and can be used to derive equations of motion and conservation laws.

What is the significance of the Noether current in physics?

The Noether current is significant because it allows us to connect symmetries of a system to conservation laws. This helps us understand the fundamental principles that govern physical systems and can be applied in a wide range of fields, from classical mechanics to quantum field theory.

How is the derivative of the Noether current calculated from the Dirac equation?

The derivative of the Noether current is calculated using the Noether's theorem, which states that for every continuous symmetry of a Lagrangian, there exists a conserved quantity. In the context of the Dirac equation, the derivative of the Noether current can be obtained by taking the derivative of the Lagrangian with respect to the transformation parameters.

What are some examples of symmetries that give rise to Noether currents in the Dirac equation?

Some examples of symmetries that give rise to Noether currents in the Dirac equation include time translation symmetry, space translation symmetry, and gauge symmetry. These symmetries result in conserved quantities such as energy, momentum, and charge, respectively.

How does the Noether current of the Dirac equation relate to the conservation of energy and momentum?

The Noether current of the Dirac equation is related to the conservation of energy and momentum through the application of Noether's theorem. The conserved energy and momentum quantities are derived from the time translation and space translation symmetries, respectively, which give rise to the Noether currents.

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