Show Noether Currents Given Dirac Eq & Conjugate

[latentcorpse]

Given the Dirac equation and its conjugate, how do I show that $j^\mu = \bar{\psi} \gamma^\mu \psi$ is a Noether current?

Is there even a standard formula for Noether currents or does it vary depending on each individual case? And if it does vary, how do I go about figuring how to apply it to each particular example?

Thanks!

 

[fzero]

The Noether current is always associated with a symmetry. Given the symmetry operation and Lagrangian, you can always compute the Noether current. The general formula for Noether currents from scalar field theory can be adapted to fermions.

Given the Noether current, it's a little bit trickier to invert the relationship, but you can usually do it on a case by case basis.

 

[latentcorpse]

The Noether current is always associated with a symmetry. Given the symmetry operation and Lagrangian, you can always compute the Noether current. The general formula for Noether currents from scalar field theory can be adapted to fermions.

Given the Noether current, it's a little bit trickier to invert the relationship, but you can usually do it on a case by case basis.

Sorry. I still don't understand what I'm supposed to do? You talk about inverting the relationship - am I expected to find the symmetry, and if so, how?

Thanks.

 

[The_Duck]

There is a formula to get the Noether current corresponding to a symmetry of a given Lagrangian. You can find a derivation of this formula for classical field theories embedded in http://www.physics.utoronto.ca/~luke/PHY2403/References.html" - start at page 40. I think it is for a scalar theory but you can figure out how to adapt it to a field with multiple components.

To apply this I think you will need to first figure out or guess what symmetry the current you are given corresponds to.

 

[latentcorpse]

There is a formula to get the Noether current corresponding to a symmetry of a given Lagrangian. You can find a derivation of this formula for classical field theories embedded in http://www.physics.utoronto.ca/~luke/PHY2403/References.html" - start at page 40. I think it is for a scalar theory but you can figure out how to adapt it to a field with multiple components.

To apply this I think you will need to first figure out or guess what symmetry the current you are given corresponds to.

"guesswork" doesn't seem very mathematically precise though?

Also, what is $\Pi$ in the notation in your notes?

Thanks.

 

[The_Duck]

"guesswork" doesn't seem very mathematically precise though?

Also, what is $\Pi$ in the notation in your notes?

Thanks.

Rereading a little I see the notes do indeed seem to be treating the case of a field $\phi_a$ with several components indexed by 'a'.

$\Pi_a^\mu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}$ is the "canonically conjugate momentum" to the field component $\phi_a$:

The solution process I am imagining here is: write down the classical Lagrangian which gives an equation of motion that is the Dirac equation (you can look this up). Find the symmetries of this Lagrangian and plug them into the formula for Noether currents, until you find the one that gives the desired current.

I only learned this stuff very recently, so perhaps you are supposed to be doing something else here. Have you done classical field theory from Lagrangians? You can read about it in those notes if not, if that's what you're supposed to be doing.

 

[latentcorpse]

Rereading a little I see the notes do indeed seem to be treating the case of a field $\phi_a$ with several components indexed by 'a'.

$\Pi_a^\mu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_a)}$ is the "canonically conjugate momentum" to the field component $\phi_a$:

The solution process I am imagining here is: write down the classical Lagrangian which gives an equation of motion that is the Dirac equation (you can look this up). Find the symmetries of this Lagrangian and plug them into the formula for Noether currents, until you find the one that gives the desired current.

I only learned this stuff very recently, so perhaps you are supposed to be doing something else here. Have you done classical field theory from Lagrangians? You can read about it in those notes if not, if that's what you're supposed to be doing.

okay so the Lagrangian for the Dirac field is

$L= \bar{\psi(x)} \left( i \gamma^\mu \partial_\mu - m \right) \psi(x)$

Now in a set of notes that we have, I have read that the Dirac Lagrangian is invariant under rotating the phase of the spinor, $\psi \rightarrow e^{-i \alpha} \psi$, and that it is this symmetry that gives rise to this particular Noether current.

So I guess the next step is to use this symmetry in the formula to show this is the Noether current.

On page 43 of your notes, the formula for the Noether current is:

$j^\mu = \Pi^\mu_a D \phi_a - F^\mu$

Is this the general form that we can adapt to any situation?

If so, I can sub $\psi$ for $\phi_a$ to get

$j^\mu = \frac{ \partial L}{ \partial ( \partial_\mu \psi) } D \psi - F^\mu$

Is this ok so far?

 

[The_Duck]

Yeah, so now you will need to calculate that expression for the phase-rotation symmetry.

 

[latentcorpse]

Yeah, so now you will need to calculate that expression for the phase-rotation symmetry.

well in your notes it says $D \psi = \frac{ \partial \psi}{ \partial \lambda} |_{ \lambda = 0 }$ but I don't understand how to evaluate this...

actually, i have an idea...
if we have a transformation $\phi_a(x) = \phi_a(x,0) \rightarrow \phi_a(x, \lambda)$
then$D \phi_a = \frac{\partial \phi_a}{\partial \lambda} |_{\lambda=0}$

and so in our case $\psi(x) \rightarrow e^{-i \alpha} \psi(x) = \psi(x, \alpha)$
and so $D \psi = \frac{\partial \psi}{\partial \alpha} |_{\alpha=0}=-i \psi$

and then since $\frac{\partial L}{\partial ( \partial_u \psi)} = i \bar{\psi} \gamma^\mu$ we find

$j^\mu = i \bar{\psi} \gamma^\mu ( -i \psi) = \bar{\psi} \gamma^\mu \psi$ as required.

The only slight "hiccup" with this is that $D \phi_a$ would have to be defined as $D \phi_a = \frac{\partial \phi_a(x, \lambda)}{\partial \lambda}|_{\lambda=0}$ whereas in your notes it is just written $D \phi_a = \frac{\partial \phi_a}{\partial \lambda} |_{\lambda=0}$. Is this alright?

And also, how do we justify setting $F^\mu=0$

 

[The_Duck]

\The only slight "hiccup" with this is that $D \phi_a$ would have to be defined as $D \phi_a = \frac{\partial \phi_a(x, \lambda)}{\partial \lambda}|_{\lambda=0}$ whereas in your notes it is just written $D \phi_a = \frac{\partial \phi_a}{\partial \lambda} |_{\lambda=0}$. Is this alright?

Yes, the arguments of the phi are just left unwritten in that equation.

And also, how do we justify setting $F^\mu=0$

Here $F^\mu$ is defined by

$\frac{\partial \mathcal{L}}{\partial \lambda}|_{\lambda = 0} = \partial_\mu F^\mu$

so you can figure out what F must be based on how the Lagrangian changes under your symmetry. This is how you can tell if you have a symmetry and thus a conserved current: if you can write the left hand side of the above in the form of the right hand side, then the notes show how you can use this information to construct a conserved current. In this problem, F is very simple because the Lagrangian is completely unchanged by the symmetry transformation, so the left hand side is zero.

Have you worked through the derivation given in those notes? I think if you do you'll see why this recipe for getting a conserved current works.

 

[latentcorpse]

Yes, the arguments of the phi are just left unwritten in that equation.



Here $F^\mu$ is defined by

$\frac{\partial \mathcal{L}}{\partial \lambda}|_{\lambda = 0} = \partial_\mu F^\mu$

so you can figure out what F must be based on how the Lagrangian changes under your symmetry. This is how you can tell if you have a symmetry and thus a conserved current: if you can write the left hand side of the above in the form of the right hand side, then the notes show how you can use this information to construct a conserved current. In this problem, F is very simple because the Lagrangian is completely unchanged by the symmetry transformation, so the left hand side is zero.

Have you worked through the derivation given in those notes? I think if you do you'll see why this recipe for getting a conserved current works.

Is this formula for $F^\mu$ true for all examples?

Also, surely if the LHS is 0, $F^\mu$ doesn't ahve to be 0, it just has to be constant, no?

 

[The_Duck]

Is this formula for $F^\mu$ true for all examples?

If F satisfies that equation then the other formula that uses F gives a conserved current. Work through the derivation to see why F is defined the way it is.

Also, surely if the LHS is 0, $F^\mu$ doesn't ahve to be 0, it just has to be constant, no?

Indeed. And if you add a constant 4-vector to a conserved current you still get a conserved current.

 

[latentcorpse]

If F satisfies that equation then the other formula that uses F gives a conserved current. Work through the derivation to see why F is defined the way it is.



Indeed. And if you add a constant 4-vector to a conserved current you still get a conserved current.

so our answer for the Noether current is defined up to a constant and then what - do we argue that we might as well just set this constant equal to zero?

 

[dextercioby]

The boundary conditions at infinity would <kill> any constant added to the 4-current.