Get the current and the EQs of Motion of the Dirac-Lagrangian density

In summary, the conversation discusses the definition of current and its use in the context of Lagrangian mechanics. The provided equation for the current differs from the one derived by the expert summarizer, but both are valid as any constant factor can be factored out without changing the physics of the system. The expert also suggests using a different equation for computing the Noether current. Further discussions involve proving the invariance of the Lagrangian under a specific transformation and finding the equations of motion for certain variables.
  • #1
JD_PM
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Homework Statement
For a Dirac field, the transformations



$$\psi(x) \rightarrow \psi'(x) = e^{i \alpha \gamma_5} \psi(x), \ \ \ \ \ \ \ \ \psi^{\dagger} (x) \rightarrow \psi^{\dagger'}(x) = e^{-i \alpha \gamma_5} \psi(x)$$



where ##\alpha## is an arbitrary real parameter, are called chiral phase transformations.



Given the Dirac-Lagrangian density (and the definition of ##\bar{\psi}##):



$$\mathcal{L} = c \bar{\psi}\Big( i \hbar \gamma^{\mu} \partial_{\mu} - mc\Big) \psi, \ \ \ \ \ \ \ \ \ \bar{\psi} = \psi^{\dagger} \gamma^0$$



Where ##\gamma^0## is a type of Dirac matrix.



a) Show that the current associated to ##\mathcal{L}## in the limit ##m = 0## is ##j^{\mu} = \bar{\psi} \gamma^{\mu} \gamma_5 \psi##



b) Show that the equations of motion related to ##\mathcal{L}## (and ##m \neq 0##) are



$$\psi_L = \frac 1 2 (1 - \gamma_5) \psi, \ \ \ \ \ \ \psi_R = \frac 1 2 (1 + \gamma_5) \psi$$



And also show that they decouple in the limit ##m = 0##



c) After solving a) b) you have all you need to show and explain that the ##m=0## Dirac-Lagrangian density ##\mathcal{L} = i \hbar c \bar{\psi_L}\gamma^{\mu} \partial_{\mu} \psi_L## describes zero-mass fermions with negative helicity only, and zero-mass antifermions with positive helicity only. (This field is called the Weyl field and can be used to describe the neutrinos in weak interactions in the approximation of zero mass).

Source: Exercise 4.5, QFT by Mandl and Shaw.
Relevant Equations
$$j^{\mu}_{\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_a )} X_a (\phi) - F^{\mu} (\phi), \ \ \ \ \ \ \ \ \delta \mathcal{L} = \partial_{\mu} F^{\mu}; \ X_a (\phi) = \delta \phi_a$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi)} \Big) - \frac{\partial \mathcal{L}}{\partial \psi} = 0 \ \ \ \ (1)$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi^{\dagger})} \Big) - \frac{\partial \mathcal{L}}{\partial \psi^{\dagger}} = 0 \ \ \ \ (2)$$
a) Here I used the definition of current (reference: Tong notes, page 14)

$$j^{\mu}_{\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \phi_a )} X_a (\phi) - F^{\mu} (\phi), \ \ \ \ \ \ \ \ \delta \mathcal{L} = \partial_{\mu} F^{\mu}; \ X_a (\phi) = \delta \phi_a$$

In our problem we get (assuming ##\delta \mathcal{L} = 0##):

$$j^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi^{\dagger})} \delta \psi^{\dagger} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} \delta \psi$$

We know that ##\mathcal{L} = i \hbar c \bar{\psi}\gamma^{\mu} \partial_{\mu} \psi## , ##\delta \psi^{\dagger} = i \alpha \gamma_5 \psi## and ##\delta \psi^{\dagger} = -i \alpha \gamma_5 \psi^{\dagger}##

On the one hand, as ##\mathcal{L}## has no explicit dependence on ##\partial_{\mu} \psi^{\dagger}## we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi^{\dagger})} = 0$$

On the other hand we get

$$\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \psi)} = i c \hbar \bar{\psi} \gamma^{\mu}$$

Thus the current I get is

$$j^{\mu} = i c \hbar \bar{\psi} \gamma^{\mu} \delta \psi = - c \alpha \hbar \bar{\psi} \gamma^{\mu} \gamma_5 \psi$$

The provided current is

$$j^{\mu} = \bar{\psi} \gamma^{\mu} \gamma_5 \psi$$

As you can see, my current and the provided one differ; I get a constant factor ##k := -c \alpha \hbar## Is that OK?b) Here it is all about using E-L equations

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi)} \Big) - \frac{\partial \mathcal{L}}{\partial \psi} = 0 \ \ \ \ (1)$$

$$\partial_{\mu} \Big( \frac{\mathcal{L}}{\partial(\partial_{\mu} \psi^{\dagger})} \Big) - \frac{\partial \mathcal{L}}{\partial \psi^{\dagger}} = 0 \ \ \ \ (2)$$

We will get two equations of motion, as expected.

Out of ##(1)## we get:

$$i \hbar c \bar{\psi} \partial_{\mu} \gamma^{\mu} - \bar{\psi} mc^2 = 0$$

Out of ##(2)## we get (note I have used the definition ##\bar{\psi} = \psi^{\dagger} \gamma^0##):

$$-i \hbar c \gamma^0 \gamma^{\mu} \partial_{\mu} \psi + c^2 \gamma^0 m \psi = 0$$

But these are not the provided solutions.

Are my solutions OK? Is there a way of going from these solutions to the provided ones?

c) As suggested, let's first get a) and b) done.
 
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  • #2
Ok, I don't know if there is more than one edition to Mandl & Shaw's book. In any case, the one I have has a slightly different statement, which I think makes much more sense than what you wrote;
First of all, you have to prove that the transformation
$$\psi \rightarrow \psi' = e^{i\alpha \gamma_5}\psi$$
leaves the Lagrangian ##\mathscr{L}=\bar{\psi}(i\not{\!\partial}-m)\psi## invariant in the limit ##m\rightarrow 0##, I would suggest you to also prove that if ##m\neq 0## then the Lagrangian is NOT invariant under such transformation.
Only then it makes sense to compute the current (I'm not aware of any use of such a current if the transformation doesn't live the action invariant).

JD_PM said:
As you can see, my current and the provided one differ; I get a constant factor ##k := -c \alpha \hbar## Is that OK?
Well, if essence, the importance of ##j^\mu## is that is conserved in the sense ##\partial_\mu j^\mu=0##, you can prove that any other current ##j'^\mu = k j^\mu## is also conserved and therefore is perfectly valid (note that even in the Lagrange level, you can always change ##\mathscr{L}\rightarrow k\mathscr{L}## without changing any physics of your system).
Also, related to this, sometimes I like to compute the Noether current using an equation given in "The Standard Model in a Nutshell" by D. Goldberg
$$j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi_a)}\frac{\partial \phi_a}{\partial \epsilon}-\frac{\partial x^\mu}{\partial \epsilon}\mathcal{L}$$
where ##\epsilon## is the parameter of the transformation (the ##\alpha## in your problem) if you do it in this way you will find that there's no ##\alpha## in ##j^\mu##.

For b) it asks to find the equations of motion FOR ##\psi_L## and ##\psi_R## defined as
$$\psi_{L}=\frac{1-\gamma_5}{2}\psi, \qquad \psi_{R}=\frac{1+\gamma_5}{2}\psi$$
and then show that for ##m\rightarrow 0## they decouple.
Try to redo parts a) and b) and then we will take a look at c)
 
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  • #3
I think we both have the second edition.

Gaussian97 said:
First of all, you have to prove that the transformation
$$\psi \rightarrow \psi' = e^{i\alpha \gamma_5}\psi$$
leaves the Lagrangian ##\mathscr{L}=\bar{\psi}(i\not{\!\partial}-m)\psi## invariant in the limit ##m\rightarrow 0##, I would suggest you to also prove that if ##m\neq 0## then the Lagrangian is NOT invariant under such transformation.

True; I omitted that due to the fact I know how to do it; it is about doing the same we discussed here: https://www.physicsforums.com/threa...f-the-complex-scalar-field-lagrangian.986077/

Thanks for the suggestion; let's make it part d). Before tacking it we better deal with the first three though.

Gaussian97 said:
Well, if essence, the importance of ##j^\mu## is that is conserved in the sense ##\partial_\mu j^\mu=0##, you can prove that any other current ##j'^\mu = k j^\mu## is also conserved and therefore is perfectly valid (note that even in the Lagrange level, you can always change ##\mathscr{L}\rightarrow k\mathscr{L}## without changing any physics of your system).
Also, related to this, sometimes I like to compute the Noether current using an equation given in "The Standard Model in a Nutshell" by D. Goldberg
$$j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi_a)}\frac{\partial \phi_a}{\partial \epsilon}-\frac{\partial x^\mu}{\partial \epsilon}\mathcal{L}$$
where ##\epsilon## is the parameter of the transformation (the ##\alpha## in your problem) if you do it in this way you will find that there's no ##\alpha## in ##j^\mu##.

Alright so a) is OK then.

Gaussian97 said:
For b) it asks to find the equations of motion FOR ##\psi_L## and ##\psi_R## defined as
$$\psi_{L}=\frac{1-\gamma_5}{2}\psi, \qquad \psi_{R}=\frac{1+\gamma_5}{2}\psi$$
and then show that for ##m\rightarrow 0## they decouple.
Try to redo parts a) and b) and then we will take a look at c)

Ahhh so it is asking to use E-L equations on ##\psi_{L}## and ##\psi_{R}## instead of doing so on the Dirac-Lagrange density!

Let's redo b) then.
 
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  • #4
And just to be clear: b) is asking for the equations of motion regarding the following Lagrangians

$$\mathcal{L} = c \bar{\psi}\Big( i \hbar \gamma^{\mu} \partial_{\mu} - mc\Big) \frac{1-\gamma_5}{2}\psi, \qquad \mathcal{L} = c \bar{\psi}\Big( i \hbar \gamma^{\mu} \partial_{\mu} - mc\Big) \frac{1+\gamma_5}{2}\psi$$

Right?
 
  • #5
No, the Lagrangian is still
$$\mathscr{L}=\bar{\psi}\left(i\not{\!\partial}-m\right)\psi$$

Usually, there's some arbitrariness on what one understands by "the equation of motion of ##\phi##" some people means that you need to compute
$$[\mathscr{L}]_{\phi}\equiv\frac{\partial\mathscr{L}}{\partial \phi}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \phi)}$$
while for others is means to find an equation that involves ##\phi## and derivatives.
In any case, I think that the best you can do is to compute this four equations;
$$[\mathscr{L}]_{\psi_L}$$
$$[\mathscr{L}]_{\psi_R}$$
$$[\mathscr{L}]_{\bar{\psi}_L}$$
$$[\mathscr{L}]_{\bar{\psi}_R}$$
 
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  • #6
Gaussian97 said:
I think that the best you can do is to compute this four equations;
$$[\mathscr{L}]_{\psi_L}$$
$$[\mathscr{L}]_{\psi_R}$$
$$[\mathscr{L}]_{\bar{\psi}_L}$$
$$[\mathscr{L}]_{\bar{\psi}_R}$$

OK; out of the Lagrangian

$$\mathscr{L}=\bar{\psi}\left(i\not{\!\partial}-m\right)\psi$$

We get 4 EoM:

$$[\mathscr{L}]_{\psi_L}\equiv\frac{\partial\mathscr{L}}{\partial \psi_L}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_L)} = \frac{1-\gamma_5}{2} \bar{\psi_L} (i \not{\!\partial} - m)$$

$$[\mathscr{L}]_{\psi_R}\equiv\frac{\partial\mathscr{L}}{\partial \psi_R}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_R)} = \frac{1+\gamma_5}{2} \bar{\psi_R} (i \not{\!\partial} - m)$$

$$[\mathscr{L}]_{\bar{\psi_L}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_L}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_L})} = \frac{1-\gamma_5}{2} (i \not{\!\partial} - m) \psi_L$$

$$[\mathscr{L}]_{\bar{\psi_R}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_R}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_R})} = \frac{1+\gamma_5}{2} (i \not{\!\partial} - m) \psi_R$$

What does decoupling of the equations of motion at ##m=0## really mean though?
 
  • #7
How did you get those equations? They are not correct. Try to write first your Lagrangian out of ##\psi_L## and ##\psi_R## fields
 
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  • #8
Gaussian97 said:
Try to write first your Lagrangian out of ##\psi_L## and ##\psi_R## fields

OK I was wrong. I think the right approach is to solve for ##\psi## in the equations ##\psi_{L}=\frac{1-\gamma_5}{2}\psi, \ \ \ \psi_{R}=\frac{1+\gamma_5}{2}\psi## and plug it into the Lagrangian ##\mathscr{L}=\bar{\psi}\left(i\not{\!\partial}-m\right)\psi##

I get

$$\mathscr{L} (\psi_L, \bar{\psi_L}) = \frac{2 \bar{\psi}\left(i\not{\!\partial}-m\right) \bar{\psi_L}}{1 - \gamma_5}$$

$$\mathscr{L} (\psi_R, \bar{\psi_R}) = \frac{2 \bar{\psi}\left(i\not{\!\partial}-m\right) \bar{\psi_R}}{1 + \gamma_5}$$

Then it is all about using E-L equation on them.

If you agree with it, I will compute the EoM.
 
  • #9
No sorry, ##\gamma^5## is a matrix so ##\frac{1}{1\pm\gamma^5}## is not a defined quantity, also you will find the problem that ##1\pm\gamma^5## is a singular matrix so you cannot invert it, but you don't need to there's a very easy way to get rid of the ##\gamma^5## combining the expressions of ##\psi_{L,R}##.
 
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  • #10
Ahhh you mean take a linear combination of them! (i.e. add both definitions up; ##\psi = \psi_L + \psi_R##)

Thus the Lagrangian we got to work out is

$$\mathscr{L} (\psi_L, \psi_R, \bar{\psi}) = \bar{\psi}\left(i\not{\!\partial}-m\right)(\psi_L + \psi_R)$$

?

EDIT: What about ##\bar{\psi_L}## and ##\bar{\psi_R}##? Can we simply do ##\bar{\psi} = \bar{\psi_L} + \bar{\psi_R}## and get

$$\mathscr{L} (\psi_L, \psi_R, \bar{\psi_L}, \bar{\psi_R}) = (\bar{\psi_L} + \bar{\psi_R}) \left(i\not{\!\partial}-m\right)(\psi_L + \psi_R)$$

?
 
  • #11
Well, you can do it yourself if we define $$\bar{\psi}\equiv \psi^\dagger \gamma^0$$ and ##\bar{\psi}_{L,R}=\psi_{L,R}^\dagger \gamma^0## it's true that
$$\bar{\psi}=\bar{\psi}_{L}+\bar{\psi}_{R}$$
?
 
  • #12
Gaussian97 said:
Well, you can do it yourself if we define $$\bar{\psi}\equiv \psi^\dagger \gamma^0$$ and ##\bar{\psi}_{L,R}=\psi_{L,R}^\dagger \gamma^0## it's true that
$$\bar{\psi}=\bar{\psi}_{L}+\bar{\psi}_{R}$$
?

Yes, it is true (I'd say it is evident).

Thus, the Lagrangian we're looking for is

$$\mathscr{L} (\psi_L, \psi_R, \bar{\psi_L}, \bar{\psi_R}) = (\bar{\psi_L} + \bar{\psi_R}) \left(i\not{\!\partial}-m\right)(\psi_L + \psi_R)$$

Let me know if you do not agree.
 
  • #13
Yes, although you can simplify it further, notice that, if you apply distributive property you get a total of 8 terms, of those there are 4 that vanish.
Prove
$$\bar{\psi}_{L,R}\!\!\not{\!\partial}\psi_{R,L}=0, \qquad \bar{\psi}_{L,R}\psi_{L,R}=0$$
(Notice on how I write the ##L## and ##R## in the first two terms)
And then you can apply the Euler-Lagrange equations to the other 4 terms.
 
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  • #14
Gaussian97 said:
Yes, although you can simplify it further, notice that, if you apply distributive property you get a total of 8 terms, of those there are 4 that vanish.
Prove
$$\bar{\psi}_{L,R}\!\!\not{\!\partial}\psi_{R,L}=0, \qquad \bar{\psi}_{L,R}\psi_{L,R}=0$$
(Notice on how I write the ##L## and ##R## in the first two terms)
And then you can apply the Euler-Lagrange equations to the other 4 terms.

Ahhh nice! They vanish due to antisymmetry under interchange of ##\bar{\psi_L} \rightarrow \bar{\psi_R}## and ##\psi_L \rightarrow \psi_R## (as required for fermions).

I guess that ,for a rigorous proof, we need to use the ladder operators and ,in particular, the anticommutation relation

$$[a_r, a_s^{\dagger}]_+ = \delta_{rs}$$

----------------------------------------------------------

Thus we end up with the Lagrangian

$$\mathscr{L} (\psi_L, \psi_R, \bar{\psi_L}, \bar{\psi_R}) = i\bar{\psi_L} \not{\!\partial}\psi_L + i\bar{\psi_R} \not{\!\partial}\psi_R - \bar{\psi_L} m\psi_L - \bar{\psi_R} m\psi_R$$

Next I'll proceed to compute the equations of motion.
 
  • #15
No, you can prove the four equation by properties of the Dirac matrices without any quantization or antisymmetry argument.
 
  • #16
Gaussian97 said:
No, you can prove the four equation by properties of the Dirac matrices without any quantization or antisymmetry argument.

Could you please state exactly what properties of the Dirac matrices?

I'd say that the asymmetry argument should work; we are dealing with fermions here.
 
  • #17
Use the anticommutator ##\left\{\gamma^\mu, \gamma^5\right\}\equiv [\gamma^\mu,\gamma^5]_+=0## and ##\gamma^5\gamma^5=1##
 
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  • #18
The equations of motion at #6 are wrong; these should be OK:

$$[\mathscr{L}]_{\psi_L}\equiv\frac{\partial\mathscr{L}}{\partial \psi_L}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_L)} = \bar{\psi_L}(i \not{\!\partial}-m) + i\bar{\psi_R}\not{\!\partial}\psi_R - m\bar{\psi_R} \psi_R$$

$$[\mathscr{L}]_{\psi_R}\equiv\frac{\partial\mathscr{L}}{\partial \psi_R}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_R)} = \bar{\psi_R}(i \not{\!\partial}-m) + i\bar{\psi_L}\not{\!\partial}\psi_L - m\bar{\psi_L} \psi_L$$

$$[\mathscr{L}]_{\bar{\psi_L}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_L}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_L})} = (i \not{\!\partial}-m)\psi_L + i\bar{\psi_R}\not{\!\partial}\psi_R - m\bar{\psi_R} \psi_R$$

$$[\mathscr{L}]_{\bar{\psi_R}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_R}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_R})} = (i \not{\!\partial}-m)\psi_R + i\bar{\psi_L}\not{\!\partial}\psi_L - m\bar{\psi_L} \psi_L$$

If you agree, I'll proceed to discuss decoupling.
 
  • #19
No, they are still not correct, first of all, take the time to prove equations given in #13 and you will find one of your errors, also you did error when differentiating.
 
  • #20
Gaussian97 said:
No, they are still not correct, first of all, take the time to prove equations given in #13 and you will find one of your errors, also you did error when differentiating.

Alright, I got it wrong. The Lagrangian should be

$$\mathscr{L} (\psi_L, \psi_R, \bar{\psi_L}, \bar{\psi_R}) = i\bar{\psi_L} \not{\!\partial}\psi_L + i\bar{\psi_R} \not{\!\partial}\psi_R - \bar{\psi_L} m\psi_R - \bar{\psi_R} m\psi_L$$

---------------------------------------------------------------------------------------------------------------------------------------------------

Regarding the proof; I know that ##i\bar{\psi_L} \not{\!\partial}\psi_R## and ##i\bar{\psi_R} \not{\!\partial}\psi_L## cancel each other out and so they do ##\bar{\psi_L} m\psi_L## and ##\bar{\psi_R} m\psi_R##

This is how I am trying to prove the former; I am using the given definitions for ##\psi_L## and ##\psi_R## to get

$$i\bar{\psi_L} \not{\!\partial}\Big( \frac 1 2 (1+\gamma_5)\psi\Big) + i\bar{\psi_R} \not{\!\partial}\Big( \frac 1 2 (1-\gamma_5)\psi\Big) = \frac 1 2 i\bar{\psi_L} \not{\!\partial}\psi + \frac 1 2 i\bar{\psi_R} \not{\!\partial}\psi + (i\bar{\psi_L} + i\bar{\psi_R})[\gamma^{\mu}, \gamma^5]_+ \partial_{\mu} \psi =0$$

The issue I have is that I do not see how to show that ##i\bar{\psi_L} \not{\!\partial}\psi + i\bar{\psi_R} \not{\!\partial}\psi = 0##
 
  • #21
Easier, I'm not saying they cancel each other, they are exactly zero. Start with the terms like ##\bar{\psi}_{L,R}\psi_{L,R}## you have, by definition
$$\bar{\psi}_{L,R}\psi_{L,R}\equiv \psi^\dagger_{L,R}\gamma^0\psi_{L,R}=\left(\frac{1\mp \gamma^5}{2}\psi\right)^\dagger\gamma^0\left(\frac{1\mp \gamma^5}{2}\psi\right)=\psi^\dagger\left(\frac{1\mp \gamma^5}{2}\right)^\dagger\gamma^0\left(\frac{1\mp \gamma^5}{2}\right)\psi$$
Now you can work out the rest using the properties of ##\gamma^0## and ##\gamma^5##
 
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  • #22
Alright so we have

$$\psi^\dagger_{L,R}\gamma^0\psi_{L,R}=\psi^\dagger\left(\frac{1\mp \gamma^5}{2}\right)^\dagger\gamma^0\left(\frac{1\mp \gamma^5}{2}\psi\right)= \frac 1 4 \psi^{\dagger}(1 \mp \gamma^5)\gamma^0(1\mp \gamma^5)\psi=\frac 1 4 \psi^{\dagger}\gamma^0\psi + \frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \psi = 0$$

Where I have used the property ##\gamma^{5 \dagger} = \gamma^5##

But why ##\frac 1 4 \psi^{\dagger}\gamma^0\psi=0## and ##\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \psi = 0## ?
 
  • #23
Mmm... I'm not sure why on the second term of the third equality the ##\frac{1}{4}## gets converted in a ##\frac{1}{2}##. Also It's not true that ##\psi^\dagger\gamma^0\psi=0## nor ##\psi^\dagger \gamma^5\gamma^0\gamma^5\psi=0##, but rather
$$\gamma^0 + \gamma^5\gamma^0\gamma^5=0$$
To prove it use the anticommutation relation ##\left\{\gamma^0,\gamma^5\right\}=0## to factor out a ##\gamma^0##, and then use that ##\left(\gamma^5\right)^2=1##.
 
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  • #24
Gaussian97 said:
Mmm... I'm not sure why on the second term of the third equality the ##\frac{1}{4}## gets converted in a ##\frac{1}{2}##. Also It's not true that ##\psi^\dagger\gamma^0\psi=0## nor ##\psi^\dagger \gamma^5\gamma^0\gamma^5\psi=0##

OK I got it wrong; I better do not rush.

Explicitly what I get is

$$\psi^\dagger_{L,R}\gamma^0\psi_{L,R}=\frac 1 4 \psi^{\dagger}(1 \mp \gamma^5)\gamma^0(1\mp \gamma^5)\psi=\frac 1 4 \psi^{\dagger}\gamma^0\psi+\frac 1 4 \psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi+\frac 1 4 \psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi-\frac 1 4 \psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi-\frac 1 4 \psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi= \frac 1 4 \psi^{\dagger}\gamma^0\psi=0
$$

If there is no way ##\frac 1 4 \psi^{\dagger}\gamma^0\psi=0## I've had to get wrong somewhere...
 
  • #25
Now I'm really lost, how do you go from $$\frac 1 4 \psi^{\dagger}(1 \mp \gamma^5)\gamma^0(1\mp \gamma^5)\psi$$ to $$\frac{1}{4}\psi^{\dagger}\gamma^0\psi+\frac{1}{4}\psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi+\frac{1}{4}\psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi-\frac{1}{4}\psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi-\frac{1}{4}\psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi$$?
 
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  • #26
Ahhh damn! Apparently today's not my day! o_O I think I got it now though.

We have

$$\psi^\dagger_{L,R}\gamma^0\psi_{L,R}=\frac 1 4 \psi^{\dagger}(1 \mp \gamma^5)\gamma^0(1\mp \gamma^5)\psi=\frac 1 4 \psi^{\dagger} \gamma^0 \psi + \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \psi=0$$

We now prove that (which trick you have already shown at #23)

$$\frac 1 4 \psi^{\dagger} \gamma^0 \psi + \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \psi=0$$

We start by working out the anticommutation relation ##[\gamma^0, \gamma^5]=0##

$$[\gamma^0, \gamma^5] = \gamma^0 \gamma^5 + \gamma^5 \gamma^0 = 0$$

Multiplying both sides by ##\gamma^5## (to the right) we get

$$[\gamma^0, \gamma^5] = \gamma^0 (\gamma^5)^2 + \gamma^5 \gamma^0 \gamma^5= 0$$

We now use the property ##\left(\gamma^5\right)^2=1## and multiply both sides by ##\frac 1 4 \psi^{\dagger}## (left) and ##\psi## (right) to get

$$\frac 1 4 \psi^{\dagger} \gamma^0 \psi + \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \psi=0$$

QED.

If you see it OK, I'll go for the equations of motion! :D
 
  • #27
The equations of motion at #6 (first shot) and at #18 (second shot) are wrong ; these should be OK:

$$[\mathscr{L}]_{\psi_L}\equiv\frac{\partial\mathscr{L}}{\partial \psi_L}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_L)}=i\bar{\psi_L} \not{\!\partial} - \bar{\psi_R}m$$

$$[\mathscr{L}]_{\psi_R}\equiv\frac{\partial\mathscr{R}}{\partial \psi_R}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_R)}=i\bar{\psi_R} \not{\!\partial} - \bar{\psi_L}m$$

$$[\mathscr{L}]_{\bar{\psi_L}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_L}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_L})}=i\not{\!\partial}\psi_L - \psi_R m$$

$$[\mathscr{L}]_{\bar{\psi_L}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_L}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_L})}=i\not{\!\partial}\psi_R - \psi_L m$$
 
  • #28
Well, first of all, you still need to prove ##\bar{\psi}_{L,R}\gamma^\mu\psi_{R,L}=0## 😅 😅 .
Then, of the four equations, the last two seems right to me, but you should be careful with the first two, what is the derivative acting on?
 
  • #29
Alright so let's prove ##\bar{\psi_{L,R}}\not{\!\partial} \psi_{R,L}=0##

$$\bar{\psi_{L,R}}\not{\!\partial} \psi_{R,L}=\psi^\dagger_{L,R}\gamma^0 \not{\!\partial} \psi_{R,L}=\frac 1 4 \psi^{\dagger}(1 \mp \gamma^5)\gamma^0\not{\!\partial}(1\pm \gamma^5)\psi=-\frac 1 4 \psi^{\dagger} \gamma^0 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \not{\!\partial}\psi=0$$

I'd say we can prove ##\psi^{\dagger} \gamma^0 \not{\!\partial}\psi + \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \not{\!\partial}\psi=0## by using the exact same previous idea:

$$\gamma^0 + \gamma^5\gamma^0\gamma^5=0$$

So the proof is basically the same that the one at #26; the only difference is that we deal here with ##\not{\!\partial}\psi## instead of ##\psi##
 
  • #30
Gaussian97 said:
you should be careful with the first two

OK you seem to be suggesting I better apply ##\not{\!\partial} := \gamma^{\mu}\partial_{\mu}##. In that case, the first two EoM look as follows$$[\mathscr{L}]_{\psi_L}\equiv\frac{\partial\mathscr{L}}{\partial \psi_L}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_L)}=i\bar{\psi_L} \gamma^{\mu} - \bar{\psi_R}m$$

$$[\mathscr{L}]_{\psi_R}\equiv\frac{\partial\mathscr{L}}{\partial \psi_R}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_R)}=i\bar{\psi_R} \gamma^{\mu} - \bar{\psi_L}m$$

Mmm but something tells me this is wrong: we should not have free indices in our equations of motion and I've got ##\gamma^{\mu}## on both...
 
  • #31
JD_PM said:
Alright so let's prove ##\bar{\psi_{L,R}}\not{\!\partial} \psi_{R,L}=0##

$$\bar{\psi_{L,R}}\not{\!\partial} \psi_{R,L}=\psi^\dagger_{L,R}\gamma^0 \not{\!\partial} \psi_{R,L}=\frac 1 4 \psi^{\dagger}(1 \mp \gamma^5)\gamma^0\not{\!\partial}(1\pm \gamma^5)\psi=-\frac 1 4 \psi^{\dagger} \gamma^0 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \not{\!\partial}\psi=0$$

I'd say we can prove ##\psi^{\dagger} \gamma^0 \not{\!\partial}\psi + \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \not{\!\partial}\psi=0## by using the exact same previous idea:

$$\gamma^0 + \gamma^5\gamma^0\gamma^5=0$$

So the proof is basically the same that the one at #26; the only difference is that we deal here with ##\not{\!\partial}\psi## instead of ##\psi##
I think you should go slower in you computations, you create - signs from nothing and you're commuting matrices that don't commute. What you did is not correct, but it's true that the final expression will vanish because the property ##\gamma^0 +\gamma^5\gamma^0\gamma^5##, which of course you have already proved and you don't need to prove it again.

JD_PM said:
OK you seem to be suggesting I better apply ##\not{\!\partial} := \gamma^{\mu}\partial_{\mu}##. In that case, the first two EoM look as follows$$[\mathscr{L}]_{\psi_L}\equiv\frac{\partial\mathscr{L}}{\partial \psi_L}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_L)}=i\bar{\psi_L} \gamma^{\mu} - \bar{\psi_R}m$$

$$[\mathscr{L}]_{\psi_R}\equiv\frac{\partial\mathscr{L}}{\partial \psi_R}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_R)}=i\bar{\psi_R} \gamma^{\mu} - \bar{\psi_L}m$$

Mmm but something tells me this is wrong: we should not have free indices in our equations of motion and I've got ##\gamma^{\mu}## on both...

Again, you go too fast and you have forget to apply the derivative.
 
Last edited:
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  • #32
Gaussian97 said:
I think you should go slower in you computations

True. But we agree the proof is basically the same so let's jump into the EoM.

Gaussian97 said:
Again, you go too fast and you have forget to apply the derivative.

Alright. We know that the Lagrangian is

$$\mathscr{L} = i\bar{\psi_L} \gamma^{\mu}\partial_{\mu} \psi_L + i\bar{\psi_R} \gamma^{\mu}\partial_{\mu} \psi_R - \bar{\psi_L} m\psi_R - \bar{\psi_R} m\psi_L$$

Based on it we should get the above equations of motion; I've checked it and I do not see my mistakes.

You said that 'I forgot to apply the derivative'. Could you please be more specific on where I got wrong?
 
  • #33
I don't agree that the proof is basically the same :smile:. And I highly recommend you to go through all the steps

For the EoM you simply need to do it step by step:
1. What is ##\frac{\partial \mathscr{L}}{\partial \psi_L}##?
2. What is ##\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)}##
3. Take the derivative of 2. to compute ##\partial_\mu\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)}##
4. Substract 1. and 3. to get ##\left[\mathscr{L}\right]_{\psi_L}##
 
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  • #34
Gaussian97 said:
I don't agree that the proof is basically the same :smile:. And I highly recommend you to go through all the steps

I agree it is important but let's postpone it if you do not mind; let's tackle the EoM

Gaussian97 said:
For the EoM you simply need to do it step by step:
1. What is ##\frac{\partial \mathscr{L}}{\partial \psi_L}##?
2. What is ##\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)}##
3. Take the derivative of 2. to compute ##\partial_\mu\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)}##
4. Substract 1. and 3. to get ##\left[\mathscr{L}\right]_{\psi_L}##

Yes, I missed the derivative! Thanks for your patience.

1.

$$\frac{\partial \mathscr{L}}{\partial \psi_L}=-\bar{\psi_R}m$$2.

$$\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)}=i\bar{\psi_L}\gamma^{\mu}$$

3.

$$\partial_\mu\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)} = i\bar{\psi_L}\partial_{\mu}\gamma^{\mu}$$

4. Thus the equation of motion is

$$\left[\mathscr{L}\right]_{\psi_L}=-\bar{\psi_R}m-i\bar{\psi_L}\partial_{\mu}\gamma^{\mu}$$
 
  • #35
The equations of motion at #6 (first shot), #18(second shot) and #27(third shot) are wrong; these should be OK:

$$[\mathscr{L}]_{\psi_L}\equiv\frac{\partial\mathscr{L}}{\partial \psi_L}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_L)} = -\bar{\psi_R}m-i\bar{\psi_L}\partial_{\mu}\gamma^{\mu}$$

$$[\mathscr{L}]_{\psi_R}\equiv\frac{\partial\mathscr{L}}{\partial \psi_R}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \psi_R)} = -\bar{\psi_L}m-i\bar{\psi_R}\partial_{\mu}\gamma^{\mu}$$

$$[\mathscr{L}]_{\bar{\psi_L}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_L}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_L})}=i\not{\!\partial}\psi_L - \psi_R m$$

$$[\mathscr{L}]_{\bar{\psi_L}}\equiv\frac{\partial\mathscr{L}}{\partial \bar{\psi_L}}-\partial_\mu \frac{\partial \mathscr{L}}{\partial(\partial_\mu \bar{\psi_L})}=i\not{\!\partial}\psi_R - \psi_L m$$

If you agree I'll go on by discussing decoupling.
 

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