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

[nrqed]

OK I was at this point

$$\psi_L = \frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1- \gamma^5) e^{i \alpha \gamma_5} \psi = \frac 1 4 e^{2 i \alpha \gamma_5} \psi + \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5} \psi - \frac 1 4 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5} \psi - \frac 1 4 \gamma^5 e^{2 i \alpha \gamma_5} \psi$$

Do you agree so far? If yes, could you please give me a hint on how could I use $\left[(1- \gamma^5), e^{i\alpha\gamma^5}\right]=0$ to show that $\psi_L\rightarrow e^{i\alpha\gamma^5}\psi_L$ ?

Thanks.

You know how $\psi$ transforms, right? Write $\psi_L$ in terms of $\psi$, apply the transformation to $\psi$ and then show that the result can be written as $e^{i\alpha\gamma^5}\psi_L$ .

 

[JD_PM]

You know how $\psi$ transforms, right? Write $\psi_L$ in terms of $\psi$, apply the transformation to $\psi$ and then show that the result can be written as $e^{i\alpha\gamma^5}\psi_L$ .

Yes, that's what I am trying. Let's show in detail what I've been doing (it is an extension of what I've done in #62)

By definition we have

$$\psi_L := \frac 1 2 (1- \gamma^5)\psi$$

Applying the given transformation $\psi \rightarrow e^{i \alpha \gamma_5} \psi$ we get

$$\psi_L = \frac 1 2 (1- \gamma^5)e^{i \alpha \gamma_5} \psi$$

We know that $\psi=\psi_L + \psi_R$; plugging it into the above equation we get

$$\psi_L = \frac 1 2 (1- \gamma^5) e^{i \alpha \gamma_5} \psi=\frac 1 2 e^{i \alpha \gamma_5} (\psi_L + \psi_R) - \frac 1 2 \gamma^5 e^{i \alpha \gamma_5} (\psi_L + \psi_R)=\frac 1 2 ( 1 - \gamma^5)e^{i \alpha \gamma_5} \psi_L + \frac 1 2 ( 1 - \gamma^5)e^{i \alpha \gamma_5}\psi_R=\frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1- \gamma^5) e^{i \alpha \gamma_5} \psi + \frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1+ \gamma^5) e^{i \alpha \gamma_5} \psi$$

Let's work out the term $\frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1- \gamma^5) e^{i \alpha \gamma_5} \psi$ explicitly

$$\frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1- \gamma^5) e^{i \alpha \gamma_5} \psi= \frac 1 4 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi + \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi-\frac 1 4 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi-\frac 1 4 \gamma^5 e^{i \alpha \gamma_5} e^{i \alpha \gamma_5}\psi=\frac 1 4 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi + \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi-\frac 1 4[\gamma^5, e^{i \alpha \gamma_5}]_+ e^{i \alpha \gamma_5} \psi$$

Now let's work out the term $\frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1+ \gamma^5) e^{i \alpha \gamma_5} \psi$ explicitly

$$\frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1+ \gamma^5) e^{i \alpha \gamma_5} \psi = \frac 1 4 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi - \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi+\frac 1 4 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi-\frac 1 4 \gamma^5 e^{i \alpha \gamma_5} e^{i \alpha \gamma_5}\psi=\frac 1 4 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi - \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi+\frac 1 4 [ e^{i \alpha \gamma_5}, \gamma^5]e^{i \alpha \gamma_5}\psi$$

Noticing that the commutation relation yields $[ e^{i \alpha \gamma_5}, \gamma^5]=0$ we get

$$\frac 1 4 ( 1 - \gamma^5)e^{i \alpha \gamma_5} (1+ \gamma^5) e^{i \alpha \gamma_5} \psi = \frac 1 4 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi - \frac 1 4 \gamma^5 e^{i \alpha \gamma_5} \gamma^5 e^{i \alpha \gamma_5}\psi$$

Putting both worked-out terms together we end up with

$$\psi_L = \frac 1 2 (1- \gamma^5) e^{i \alpha \gamma_5} \psi = \frac 1 2 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi-\frac 1 4[\gamma^5, e^{i \alpha \gamma_5}]_+ e^{i \alpha \gamma_5} \psi$$

And here's where I am stuck. How can I show that $\frac 1 2 e^{i \alpha \gamma_5}e^{i \alpha \gamma_5}\psi-\frac 1 4[\gamma^5, e^{i \alpha \gamma_5}]_+ e^{i \alpha \gamma_5} \psi=e^{i \alpha \gamma_5} \psi?$

I guess that it can be shown using the commutation relation we found out at #69 but I still do not see it :( (not due to a lack of effort; I've been over 3 hours thinking how...)

Any help is appreciated.

Thanks.

 

[Gaussian97]

My advice: Keep it simple.
Remember how easy was to prove $\gamma^5\psi_{L,R}$ and how complicated you made it? Well, to prove this is even simpler, really.
Try to play with these ideas:
What does it mean that $\left[(1\mp\gamma^5),e^{i\alpha\gamma^5}\right]=0$?
Where can you make use of that?

I think you should put aside your computations, take a new paper and try to start again the proof, without looking at your previous calculations and trying to come up with new ideas.

 

[JD_PM]

My advice: Keep it simple.

Ahhh gosh! You're absolutely right!

$$\psi_{L,R} := \frac 1 2 (1 \mp \gamma^5)e^{i\alpha \gamma_5}\psi = \frac 1 2 \Big( [(1 \mp \gamma^5), e^{i\alpha \gamma_5}] + e^{i\alpha \gamma_5}(1 \mp \gamma^5)\Big)e^{i\alpha \gamma_5}\psi=e^{i\alpha \gamma_5} \frac 1 2 (1 \mp \gamma^5) e^{i\alpha \gamma_5}\psi = e^{i\alpha \gamma_5} \psi_{L,R}$$

QED

 

[JD_PM]

OK so here we go again

So following the steps:

2. Compute $\gamma^5\psi_{L,R}$ (proved)

$$\gamma^5\psi_{L,R}=\mp \psi_{L,R}$$

3. How does $\psi_{L,R}$ transform under this transformation? (proved)

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}$$

4. Use 2. and the series expansion of $e^x$ to simplify 3.

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R}$$

Alright I see now why you suggested the following

Maybe you can try to use 2. to compute first what is the value of
$$\gamma_5^n\psi_{L}, \qquad \gamma_5^n\psi_{R}$$

We get (recalling that $n$ are all natural numbers)

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{w=0}^\infty \frac{(i \alpha)^w}{w!}\psi_{L,R} \mp \sum_{m=1}^\infty \frac{(i \alpha)^m}{m!}\psi_{L,R}$$

Where $w=0,2,4,6...$ and $m=1,3,5,7...$

$$\gamma_5^w\psi_{L,R}=\psi_{L,R},\qquad \gamma_5^m\psi_{L,R}=\mp \psi_{L,R}$$

Is this what you get at 4.?

Thanks.

 

[Gaussian97]

Ahhh gosh! You're absolutely right!

$$\psi_{L,R} := \frac 1 2 (1 \mp \gamma^5)e^{i\alpha \gamma_5}\psi = \frac 1 2 \Big( [(1 \mp \gamma^5), e^{i\alpha \gamma_5}] + e^{i\alpha \gamma_5}(1 \mp \gamma^5)\Big)e^{i\alpha \gamma_5}\psi=e^{i\alpha \gamma_5} \frac 1 2 (1 \mp \gamma^5) e^{i\alpha \gamma_5}\psi = e^{i\alpha \gamma_5} \psi_{L,R}$$

QED

I think that you have a typo and you have a $e^{i\alpha\gamma^5}$ extra, right? You mean
$$\psi_{L,R} := \frac 1 2 (1 \mp \gamma^5)e^{i\alpha \gamma_5}\psi = \frac 1 2 \Big( [(1 \mp \gamma^5), e^{i\alpha \gamma_5}] + e^{i\alpha \gamma_5}(1 \mp \gamma^5)\Big)\psi=e^{i\alpha \gamma_5} \frac 1 2 (1 \mp \gamma^5) \psi = e^{i\alpha \gamma_5} \psi_{L,R}?$$

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{w=0}^\infty \frac{(i \alpha)^w}{w!}\psi_{L,R} \mp \sum_{m=1}^\infty \frac{(i \alpha)^m}{m!}\psi_{L,R}$$

Where $w=0,2,4,6...$ and $m=1,3,5,7...$

$$\gamma_5^w\psi_{L,R}=\psi_{L,R},\qquad \gamma_5^m\psi_{L,R}=\mp \psi_{L,R}$$

Yes, but note that $(\mp)^w = 1$ and $(\mp)^m=\mp$, so you can join all the terms by writing
$$\gamma_5^n\psi_{L,R} = (\mp)^n\psi_{L,R}$$
You can use this inside the expansion, and then undo the expansion (resum the series), this will give you a nicer way to express the transformation.

 

[JD_PM]

I think that you have a typo and you have a $e^{i\alpha\gamma^5}$ extra, right? You mean
$$\psi_{L,R} := \frac 1 2 (1 \mp \gamma^5)e^{i\alpha \gamma_5}\psi = \frac 1 2 \Big( [(1 \mp \gamma^5), e^{i\alpha \gamma_5}] + e^{i\alpha \gamma_5}(1 \mp \gamma^5)\Big)\psi=e^{i\alpha \gamma_5} \frac 1 2 (1 \mp \gamma^5) \psi = e^{i\alpha \gamma_5} \psi_{L,R}?$$

That's right.

Yes, but note that $(\mp)^w = 1$ and $(\mp)^m=\mp$, so you can join all the terms by writing
$$\gamma_5^n\psi_{L,R} = (\mp)^n\psi_{L,R}$$
You can use this inside the expansion, and then undo the expansion (resum the series), this will give you a nicer way to express the transformation.

OK so we can write it more compactly

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\mp)^n \frac{(i \alpha)^n}{n!}\psi_{L,R}$$

Thus we can assert that

4. Use 2. and the series expansion of $e^x$ to simplify 3. (done)

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\mp)^n \frac{(i \alpha)^n}{n!}\psi_{L,R}$$

Let's go for 5.

5. Substitute in 1. and should be straightforward to see that is not invariant unless $m=0$.

Let's first show that the Lagrangian is invariant when $m=0$

The Lagrangian when $m=0$ is

$$\mathscr{L} = i\bar{\psi_L} \not{\!\partial}\psi_L + i\bar{\psi_R} \not{\!\partial}\psi_R=i\psi_L^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_L + i\psi_R^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_R$$

We analogously get the transformation for $\psi_{L,R}^{\dagger}$

$$\psi_{L,R}^{\dagger}\rightarrow e^{-i\alpha\gamma_5}\psi_{L,R}^{\dagger}=\sum_{n=0}^\infty \frac{(-i \alpha \gamma_5)^n}{n!}\psi_{L,R}^{\dagger} = \sum_{n=0}^\infty (\pm)^n \frac{(i \alpha)^n}{n!}\psi_{L,R}^{\dagger}$$

We now apply the transformations we got to the Lagrangian

$$\mathscr{L'} = i\sum_{n=0}^\infty (+)^n\frac{(i \alpha)^n}{n!}\psi_{L}^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\sum_{n=0}^\infty (-)^n \frac{(i \alpha)^n}{n!}\psi_{L} + i\sum_{n=0}^\infty (-)^n \frac{(i \alpha)^n}{n!}\psi_{R}^{\dagger}\gamma^0 \gamma^{\mu}\partial_{\mu}\sum_{n=0}^\infty (+)^n\frac{(i \alpha)^n}{n!}\psi_{R}$$

We'll all be happy if $\sum_{n=0}^\infty (+)^n\frac{(i \alpha)^n}{n!}\sum_{n=0}^\infty (-)^n \frac{(i \alpha)^n}{n!}=1$ and $\sum_{n=0}^\infty (-)^n \frac{(i \alpha)^n}{n!}\sum_{n=0}^\infty (+)^n\frac{(i \alpha)^n}{n!}=1$ which is clearly not the case.

So I guess I am missing something here...

 

[Gaussian97]

Still in 4. It is better to resum all the terms so, what is the value of the sum
$$\sum_{n=0}^{\infty}(\mp)^n\frac{(i\alpha)^n}{n!}$$

Also, I think I will never insist sufficiently in that $e^{i\alpha \gamma^5}$ is a matrix and $\psi_{L,R}$ is a vector, therefore an expression like $$e^{-i\alpha\gamma^5}\psi^\dagger_{L,R}$$ makes no sense.

 

[JD_PM]

Still in 4. It is better to resum all the terms so, what is the value of the sum
$$\sum_{n=0}^{\infty}(\mp)^n\frac{(i\alpha)^n}{n!}$$

Ahhh I assume that what you meant is this: you want to get back to exponents

$$\sum_{n=0}^{\infty}(\mp)^n\frac{(i\alpha)^n}{n!}=e^{\mp i\alpha}$$

So we end up with 4. as follows

$$\psi_{L,R}\rightarrow e^{i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\mp)^n \frac{(i \alpha)^n}{n!}\psi_{L,R}= e^{\mp i\alpha}\psi_{L,R}$$

Also, I think I will never insist sufficiently in that $e^{i\alpha \gamma^5}$ is a matrix and $\psi_{L,R}$ is a vector, therefore an expression like $$e^{-i\alpha\gamma^5}\psi^\dagger_{L,R}$$ makes no sense.

My bad, I always forget it. Actually the transformation I provided for $\psi_{L,R}^{\dagger}$ was wrong because of that. Now I think it should be OK.

Regarding 5.

5. Substitute in 1. and should be straightforward to see that is not invariant unless $m=0$.

Let's first show that the Lagrangian is invariant when $m=0$

The Lagrangian when $m=0$ is

$$\mathscr{L} = i\bar{\psi_L} \not{\!\partial}\psi_L + i\bar{\psi_R} \not{\!\partial}\psi_R=i\psi_L^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_L + i\psi_R^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_R$$

We analogously get the transformation for $\psi_{L,R}^{\dagger}$

$$\psi_{L,R}^{\dagger}\rightarrow e^{-i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(-i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\pm)^n \frac{(i \alpha)^n}{n!}\psi_{L,R} = e^{\pm i\alpha} \psi_{L,R}$$

We now apply the transformations we got to the Lagrangian (note that we can commute $e^{+ i\alpha}$ as we wish, because it is not a matrix anymore (EDIT: this is probably a wrong assumption and thus the reason why I am not getting the desired result; a hint would be appreciated if that is the case :) ).

$$\mathscr{L'} = i e^{+ i\alpha} \psi_{L} \gamma^0 \gamma^{\mu}\partial_{\mu}e^{- i\alpha}\psi_{L} + ie^{- i\alpha}\psi_{R} \gamma^0 \gamma^{\mu}\partial_{\mu}e^{+ i\alpha}\psi_{R}=i \psi_{L} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{L} + i \psi_{R} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{R}$$

Mmm... almost there!

I have to be missing something here; of course

$$i \psi_{L} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{L} + i \psi_{R} \gamma^0 \gamma^{\mu}\partial_{\mu} \psi_{R} \neq i\psi_L^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_L + i\psi_R^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_R$$

Due to the fact

$$\psi_{L,R} \neq \psi_{L,R}^{\dagger}$$

...

 

[Gaussian97]

Yes that's essentially it. You don't get it 100% correct because you did an error in

$$\psi_{L,R}^{\dagger}\rightarrow e^{-i\alpha\gamma_5}\psi_{L,R}=\sum_{n=0}^\infty \frac{(-i \alpha \gamma_5)^n}{n!}\psi_{L,R} = \sum_{n=0}^\infty (\pm)^n \frac{(i \alpha)^n}{n!}\psi_{L,R} = e^{\pm i\alpha} \psi_{L,R}$$

Of course you forget a $\dagger$ everywhere here. Also you don't need to do the expansion again. You can simply do
$$\psi_{L,R}^\dagger \rightarrow (\psi_{L,R}')^\dagger=(e^{\mp i\alpha}\psi_{L,R})^\dagger=(e^{\mp i\alpha})^*\psi_{L,R}^\dagger=e^{\pm i\alpha}\psi_{L,R}^\dagger$$
Notice that now $e^{\pm i\alpha}$ it's a number, which simplifies our lives a lot. (That's the whole point of expressing everything in terms of $\psi_{L,R}$, you could prove the transformation using $\psi$, but it's much more complicated.)
Actually, notice that when working with Dirac sponsors, the Lagrangian is always written in terms of $\bar{\psi}$, not $\psi^\dagger$, so it's better for you to compute the transformation for $\bar{\psi}_{L,R}$ and then you can work always with this fields and you don't need to modify the Lagrangian.

 

[JD_PM]

Alright now let's prove that the Lagrangian is not invariant when $m \neq 0$

$$\mathscr{L} = i\psi_L^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_L + i\psi_R^{\dagger} \gamma^0 \gamma^{\mu}\partial_{\mu}\psi_R - m\psi_L^{\dagger} \gamma^0 \psi_R - m\psi_R^{\dagger} \gamma^0 \psi_L $$

Applying the transformations

$$\mathscr{L'} = i e^{+ i\alpha}\psi_{L}^\dagger \gamma^0 \gamma^{\mu}\partial_{\mu} e^{- i\alpha}\psi_{L} + i e^{- i\alpha}\psi_{R}^\dagger\gamma^0 \gamma^{\mu}\partial_{\mu}e^{+ i\alpha}\psi_R - m e^{+ i\alpha}\psi_{L}^\dagger \gamma^0 e^{+ i\alpha}\psi_R - me^{- i\alpha}\psi_{R}^\dagger \gamma^0 e^{- i\alpha}\psi_{L}=i e^{+ i\alpha}\psi_{L}^\dagger \gamma^0 \gamma^{\mu}\partial_{\mu} e^{- i\alpha}\psi_{L} + i e^{- i\alpha}\psi_{R}^\dagger\gamma^0 \gamma^{\mu}\partial_{\mu}e^{+ i\alpha}\psi_R-m e^{2i\alpha} \psi_{L}^\dagger \gamma^0 \psi_R-m e^{-2i\alpha} \psi_{R}^\dagger \gamma^0 \psi_L \neq \mathscr{L}$$

QED

 

[JD_PM]

The exercise is almost done. Let's tackle c)

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).

But what are we supposed to explain and/or prove here besides what we've discussed so far?

 

[Gaussian97]

Well, the definition of helicity is defined (in Mandl & Shaw) in 4.47, but this a much more useful for us is the equation A.42. There you see that the helicity operator (for $m=0$) is $\gamma^5$. Therefore you have to prove that $\psi_L$ have negative helicity (is an eigenstate of $\gamma^5$ with eigenvalue -1) and that the antiparticle $\bar{\psi}_L$ is an eigenstate of positive helicity.

 

[JD_PM]

Well, the definition of helicity is defined (in Mandl & Shaw) in 4.47, but this a much more useful for us is the equation A.42. There you see that the helicity operator (for $m=0$) is $\gamma^5$. Therefore you have to prove that $\psi_L$ have negative helicity (is an eigenstate of $\gamma^5$ with eigenvalue -1) and that the antiparticle $\bar{\psi}_L$ is an eigenstate of positive helicity.

Mmm I've been reading information regarding equation A.42 (page 457). Let me post it.

We define the operators related to the Dirac equation as follows:

$$\Pi^{\pm}( \mathbf p ) = \frac 1 2 (1 \pm \sigma_{\mathbf p})$$

Where $\sigma_{\mathbf p}$ is the $4 \times 4$ spin matrix.

For a zero-mass Dirac particle, we can express the helicity projection operators in terms of the $\gamma^5$ matrix. For $m=0$, we have $p_0=|\mathbf p|$ and the spinors $u_r(\mathbf p)$ and $v_r(\mathbf p)$ satisfy the following equations ($r=1,2$)

$$\gamma^0|\mathbf p|u_r(\mathbf p)=-\gamma^k p_k u_r(\mathbf p)=\gamma^k p^k u_r(\mathbf p), \ \ \ \ \gamma^0|\mathbf p|v_r(\mathbf p)=-\gamma^k p_k v_r(\mathbf p)=\gamma^k p^k v_r(\mathbf p)$$

Multiplying $\gamma^k p^k u_r(\mathbf p), \ \gamma^k p^k v_r(\mathbf p)$ by $\gamma^5 \gamma^0$ on the left and using the identity $\sigma^{i j} = - \gamma^0 \gamma^5 \gamma^k$ we get

$$\gamma^5 u_r(\mathbf p)=\sigma_{\mathbf p}u_r(\mathbf p), \ \gamma^5 v_r(\mathbf p)=\sigma_{\mathbf p}v_r(\mathbf p) \ \ \ \ (A.42)$$

And the previously shown helicity projection operators become, for $m=0$

$$\Pi^{\pm}( \mathbf p ) = \frac 1 2 (1 \pm \gamma^5) \ \ \ \ (A.43)$$

Therefore you have to prove that $\psi_L$ have negative helicity (is an eigenstate of $\gamma^5$ with eigenvalue -1) and that the antiparticle $\bar{\psi}_L$ is an eigenstate of positive helicity.

Do you mean I have to prove (A.43) out of (A.42)?

 

[Gaussian97]

Well, what you have to prove, as I said, is that $\psi_L$ and $\bar{\psi}_L$ are eigenstates of the helicity operator (so $\gamma^5$).
Of course, it's best if you understand why $\gamma^5$ is the helicity operator when $m=0$ and, as in a lot of cases, a good exercise can be to see why $\gamma^5$ is NOT the helicity operator for massive particles. (So yes, you should prove all of the equations of section A.6)

 

[JD_PM]

Well, what you have to prove, as I said, is that $\psi_L$ and $\bar{\psi}_L$ are eigenstates of the helicity operator (so $\gamma^5$).

Honestly, after reading section A.6, I still have no idea how could I do so.

Could you please give me a hint? I know I have to focus on (A.42) and (A.43) but I do not know how to move on...

 

[Gaussian97]

Well, just to have it clear, this exercise has two parts:
Part 1: Proof (or understand) why $\gamma^5$ is the helicity operator. This is a purely theoretical mathematical demonstration that has nothing to do with our exercise, is completely general and is what it's done in section A.6, if you want to discuss some of this theory, we can do that.

Part 2: Proof that our Lagrangian describes a particle of negative helicity. This is what we need to do for this example, which, from what we know is simply to prove that $\psi_L$ is an eigenstate of $\gamma^5$. This has nothing to do with section A.6 and you will find no help there.

 

[JD_PM]

Well, just to have it clear, this exercise has two parts:
Part 1: Proof (or understand) why $\gamma^5$ is the helicity operator. This is a purely theoretical mathematical demonstration that has nothing to do with our exercise, is completely general and is what it's done in section A.6, if you want to discuss some of this theory, we can do that.

Part 2: Proof that our Lagrangian describes a particle of negative helicity. This is what we need to do for this example, which, from what we know is simply to prove that $\psi_L$ is an eigenstate of $\gamma^5$. This has nothing to do with section A.6 and you will find no help there.

Thanks for the clarification.

I think we can assume by now part 1 and decide if we tackle it later and now go for part 2, which is what part c) is asking for. If you agree, could you please give me a hint on how I could approach Part 2?

Thank you.

 

[Gaussian97]

Well, you simply have to prove $\gamma^5 \psi_L = -\psi_L$ and $\gamma^5\bar{\psi}_L^T=+\bar{\psi}_L^T$

 

[JD_PM]

Let's prove $\gamma^5 \psi_L = -\psi_L$ (this was already proven at #62)

$$\gamma^5 \psi_L = \frac{1}{4}(\gamma^5 - 1)(1-\gamma^5)\psi = -\frac{1}{4}(1-\gamma^5)(1-\gamma^5)\psi = -\frac 1 4 \psi - \frac 1 4 \psi + \frac 1 4\gamma^5 \psi + \frac 1 4 \gamma^5 \psi = -\frac 1 2 (1-\gamma^5) \psi = -\psi_L$$

Is $\gamma^5\bar{\psi}_L^T=+\bar{\psi}_L^T$ meant to be equivalent to $\gamma^5\psi_R=+\psi_R$? If yes, we also proved it at #62

$$\gamma^5 \psi_R = \frac 1 2 (\gamma^5 + 1) \psi_L + \frac 1 2 (\gamma^5 + 1) \psi_R = \frac{1}{4}(\gamma^5 + 1)(1-\gamma^5)\psi + \frac{1}{4}(\gamma^5 + 1)(\gamma^5 + 1)\psi = \frac 1 4 \psi + \frac 1 4 \psi + \frac 1 4\gamma^5 \psi + \frac 1 4 \gamma^5 \psi = \frac 1 2 (1+\gamma^5) \psi = \psi_R$$

 

[Gaussian97]

Is $\gamma^5\bar{\psi}_L^T=+\bar{\psi}_L^T$ meant to be equivalent to $\gamma^5\psi_R=+\psi_R$? If yes, we also proved it at #62

Well, you have to prove this and you are done

 

[JD_PM]

Well, you have to prove this and you are done

We have

$$\gamma^5\bar{\psi}_L^T=+\bar{\psi}_L^T$$

And we want to end up with

$$\gamma^5\psi_R=+\psi_R $$

We recall that

$$\psi_{L,R} := \frac{1 \mp \gamma^5}{2} \psi$$

$$\bar{\psi} := \psi^{\dagger}\gamma^0$$

$$\psi^{\dagger} := (\psi^*)^T$$

The LHS can be written as follows

$$\gamma^5\bar{\psi}_L^T=\gamma^5(\psi^{\dagger}_L\gamma^0)^T=\gamma^5 \Big( (\psi_L^*)^T\gamma^0\Big)^T= \gamma^5 \psi_L^* \gamma^0= \frac 1 2 \gamma^5 (1+\gamma^5)\psi \gamma^0=\gamma^5 \psi_R \gamma^0$$

Mmm but

$$\gamma^5 \psi_R \gamma^0 \neq \bar{\psi}_L^T$$

So one of the following assumptions I made has to be wrong:

1) $(\gamma^0)^T= \gamma^0$

2) $(\gamma^5)^* = -\gamma^5$

3) $(\psi)^*=\psi$

 

[Gaussian97]

The same problem as always, in the third equality, $\psi_L \gamma^0$ makes no sense.
Also assumptions 2) and 3) are not true.

 

[JD_PM]

Alright so the issue is that this equality doesn't hold

$$\gamma^5 \Big( (\psi_L^*)^T\gamma^0\Big)^T= \gamma^5 \psi_L^* \gamma^0$$

This is because the $4 \times 1$ matrix $\psi$ (contained in $\psi_L^*$) cannot be multiplied by $\gamma^0$ (a $4 \times 4$ matrix).

So my mistake had to be made before reaching that equality. Do you agree? I still do not see it though...

 

[Gaussian97]

No, the LHS is correct, the problem is precisely this equality, I could tell you why is wrong, but I think would be more instructive that you try to show all the hidden steps and then write explicitly what properties are you using in each step.

 

[JD_PM]

I think would be more instructive that you try to show all the hidden steps and then write explicitly what properties are you using in each step.

I agree. You are doing right in not telling me what the solution is straight away.

Let's start over. This time I am going to use the fact that $\bar{\psi}=\bar{\psi_L}+\bar{\psi_R}$

$\gamma^5\bar{\psi}_L^T=\gamma^5\Big(\bar{\psi}-\bar{\psi}_R \Big)^T=\gamma^5\Big(\psi^{\dagger} \gamma^0-\psi^{\dagger}_R \gamma^0 \Big)^T=\gamma^5\Big(\psi^{\dagger}\gamma^0-\frac 1 2 \psi^{\dagger} \gamma^0-\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0\Big)^T=\gamma^5\Big(\frac 1 2\psi^{\dagger}\gamma^0-\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0\Big)^T$

Do you agree? At point I think everything makes sense: before taking the transpose, inside the parenthesis we have $1\times 4$ matrices; once we take the transpose we get a $4 \times 1$ matrix. Thus the product $\gamma^5(...)$ is OK.

Mmm now the issue is to show that

$$\Big(\frac 1 2\psi^{\dagger}\gamma^0-\frac 1 2 \psi^{\dagger} \gamma^5 \gamma^0\Big)^T=\psi_R$$

Could you please give me a hint on how to do so?