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

[Gaussian97]

3.

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

I still not agree with that, what is the derivative acting on? If it's acting on $\gamma^\mu$, then this is zero, because $\gamma^\mu$ is a constant matrix.

 

[JD_PM]

Oh so it should be
3.
$$\partial_\mu\frac{\partial \mathscr{L}}{\partial (\partial_\mu\psi_L)} = i\partial_{\mu}(\bar{\psi_L})\gamma^{\mu}$$
?

 

[Gaussian97]

Exact! Now, notice that you sill don't have the EoM, you have the expression for $\left[\mathscr{L}\right]_{\psi}$ , but this is not an equation, the equation of motion (i.e. Euler-Lagrange equations) are
$$\left[\mathscr{L}\right]_{\psi}=0$$
So, the equations of motion you obtain (without errors) are
$$[\mathscr{L}]_{\psi_L}=-\bar{\psi}_Rm-i\partial_{\mu}\bar{\psi}_L\gamma^{\mu}\equiv -\bar{\psi}_Rm-i\bar{\psi}_L \overleftarrow{\!\!\not{\!\partial}} =0$$
$$[\mathscr{L}]_{\psi_R}=-\bar{\psi}_Lm-i\partial_{\mu}\bar{\psi}_R\gamma^{\mu}\equiv -\bar{\psi}_Lm-i\bar{\psi}_R \overleftarrow{\!\!\not{\!\partial}} =0$$
$$[\mathscr{L}]_{\bar{\psi}_L}=i\!\!\not{\!\partial}\psi_L - \psi_R m = 0$$
$$[\mathscr{L}]_{\bar{\psi}_R}=i\!\!\not{\!\partial}\psi_R - \psi_L m = 0$$

Now, what happens when we take $m\rightarrow 0$?

 

[JD_PM]

Exact! Now, notice that you sill don't have the EoM, you have the expression for $\left[\mathscr{L}\right]_{\psi}$ , but this is not an equation, the equation of motion (i.e. Euler-Lagrange equations) are
$$\left[\mathscr{L}\right]_{\psi}=0$$
So, the equations of motion you obtain (without errors) are
$$[\mathscr{L}]_{\psi_L}=-\bar{\psi}_Rm-i\partial_{\mu}\bar{\psi}_L\gamma^{\mu}\equiv -\bar{\psi}_Rm-i\bar{\psi}_L \overleftarrow{\!\!\not{\!\partial}} =0$$
$$[\mathscr{L}]_{\psi_R}=-\bar{\psi}_Lm-i\partial_{\mu}\bar{\psi}_R\gamma^{\mu}\equiv -\bar{\psi}_Lm-i\bar{\psi}_R \overleftarrow{\!\!\not{\!\partial}} =0$$
$$[\mathscr{L}]_{\bar{\psi}_L}=i\!\!\not{\!\partial}\psi_L - \psi_R m = 0$$
$$[\mathscr{L}]_{\bar{\psi}_R}=i\!\!\not{\!\partial}\psi_R - \psi_L m = 0$$

Now, what happens when we take $m\rightarrow 0$?

What does $\overleftarrow{\!\!\not{\!\partial}}$ mean? I've never seen such notation before. Is that standard?

 

[JD_PM]

What does $\overleftarrow{\!\!\not{\!\partial}}$ mean? I've never seen such notation before. Is that standard?

I assume

$$\bar{\psi_L}\overleftarrow{\!\!\not{\!\partial}}\equiv\partial_{\mu}(\bar{\psi_L})\gamma^{\mu}$$

Now, what happens when we take $m\rightarrow 0$?

$$[\mathscr{L}]_{\psi_L}=i\bar{\psi}_L \overleftarrow{\!\!\not{\!\partial}} =0$$
$$[\mathscr{L}]_{\psi_R}=i\bar{\psi}_R \overleftarrow{\!\!\not{\!\partial}} =0$$
$$[\mathscr{L}]_{\bar{\psi}_L}=i\!\!\not{\!\partial}\psi_L = 0$$
$$[\mathscr{L}]_{\bar{\psi}_R}=i\!\!\not{\!\partial}\psi_R = 0$$

Alright so now we see that our equations of motion depend only either on $\bar{\psi}_{L,R}$ or $\ \!\!\not{\!\partial}\psi_{L,R}$ I guess that is what decoupling means. But Physically, what is this fact telling us?

I don't agree that the proof is basically the same . And I highly recommend you to go through all the steps

Alright so I think the following should be OK

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

This is what I did at #29 but you seemed unconvinced. I did not see any mistakes so far;could you please tell me where did I get wrong?

... I would suggest you to also prove that if $m\neq 0$ then the Lagrangian is NOT invariant under such transformation.

Thanks for the suggestion; let's make it part d)

Alright so I thought that it would be easy to show but apparently I was wrong.

My attempt:

Assuming $\psi \rightarrow \psi' = e^{i \alpha \gamma_5} \psi$

$$\mathscr{L} = \psi^{\dagger} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi$$

If $\psi^{\dagger} \rightarrow \psi'^{\dagger} = e^{-i \alpha \gamma_5} \psi^{\dagger}$ then the lagrangian we get is

$$\mathscr{L} = e^{-i \alpha \gamma_5} \psi^{\dagger} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi = \psi^{\dagger} \gamma^{0} (i\not{\!\partial} - m)\psi$$

Mmm but I ended up with the Lagrangian being invariant under such a transformation, which is not what you suggested I should get... where did I get wrong then?

Let's tackle c) after all this if you do not mind.

Thanks.

 

[Gaussian97]

I assume

$$\bar{\psi_L}\overleftarrow{\!\!\not{\!\partial}}\equiv\partial_{\mu}(\bar{\psi_L})\gamma^{\mu}$$

Yes, exactly

Alright so now we see that our equations of motion depend only either on $\bar{\psi}_{L,R}$ or $\ \!\!\not{\!\partial}\psi_{L,R}$ I guess that is what decoupling means. But Physically, what is this fact telling us?

Perfect, so the evolution of $\psi_L$ will not depend on $\psi_R$ and vice versa, that's exactly what it means to be decoupled. Physically $\psi_L$ and $\psi_R$ are the chiralities of a particle, so you can talk of the electron to have left and right chirality, to be decoupled (which is not the case for the electron) means that if the electron starts as a left-electron, it will remain always left. Because the electrons have mass this is not true and a left-electron gets converted into a right-electron.

Alright so I think the following should be OK

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

This is what I did at #29 but you seemed unconvinced. I did not see any mistakes so far;could you please tell me where did I get wrong?

You have an error in the second equality, something like $(1\mp\gamma^5)\psi^\dagger$ is a product of a 4x4 matrix with a 1x4 vector, this doesn't make sense. Also in the third equality, you magically correct this problem and you use $\left[\gamma^\mu,\gamma^5\right]=0$ which is not true, also I think you throw away the cross-terms very quick, I would bet that you don't really see why they cancel.
To finish the last equality is not correct, you have proved that $\gamma^0+\gamma^5\gamma^0\gamma^5=0$, but here you are using $\gamma^0-\gamma^5\gamma^0\gamma^5=0$ which is certainly not true.

Assuming $\psi \rightarrow \psi' = e^{i \alpha \gamma_5} \psi$

$$\mathscr{L} = \psi^{\dagger} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi$$

If $\psi^{\dagger} \rightarrow \psi'^{\dagger} = e^{-i \alpha \gamma_5} \psi^{\dagger}$ then the lagrangian we get is

$$\mathscr{L} = e^{-i \alpha \gamma_5} \psi^{\dagger} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi = \psi^{\dagger} \gamma^{0} (i\not{\!\partial} - m)\psi$$

Here there are few things also;
the first one may not be a mathematical error, but the way you write your sentence.
You wrote "If $\psi^{\dagger} \rightarrow \psi'^{\dagger} = e^{-i \alpha \gamma_5} \psi^{\dagger}$ then ..."
It's important to notice that you can't make any assumption on how $\psi^\dagger$ transform, the only assumption you are allowed to do is the transformation of $\psi$. So you need to prove the transformation for $\psi^\dagger$ using the transformation for $\psi$. I'm sure you did this part, I'm just telling you that then you don't have to say "If $\psi^\dagger \rightarrow$..." because this means that you are making an assumption for how $\psi^\dagger$ transform.

Now more important things, first of all, related to the previous comment, if you have deduced the transformation for $\psi^\dagger$ you did it wrong, if it was an assumption (as it's written) then it's a wrong assumption. Notice that is the same kind of error than before, $e^{-i\alpha\gamma_5}$ is a 4x4 matrix and $\psi^\dagger$ is a 1x4 vector, so the matrix product is not well-defined.
The rest of the problems are simply that, again, you are using $[\gamma^\mu,\gamma^5]=0$.The main problem in the hole thing is that I think that you are not very familiarized with Dirac notation and you forget most of the times that $\psi$ is a 4x1 vector, $\psi^\dagger$ is a 1x4 vector and that $\gamma^\mu$, $\gamma^5$ are 4x4 matrices.

 

[JD_PM]

You have an error in the second equality, something like $(1\mp\gamma^5)\psi^\dagger$ is a product of a 4x4 matrix with a 1x4 vector, this doesn't make sense. Also in the third equality, you magically correct this problem and you use $\left[\gamma^\mu,\gamma^5\right]=0$ which is not true, also I think you throw away the cross-terms very quick, I would bet that you don't really see why they cancel.
To finish the last equality is not correct, you have proved that $\gamma^0+\gamma^5\gamma^0\gamma^5=0$, but here you are using $\gamma^0-\gamma^5\gamma^0\gamma^5=0$ which is certainly not true.

If you had bet on it you'd have earned money! Let's include crossed terms as well

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

Now the issue is to prove

$$\frac 1 4 \psi^{\dagger} \gamma^0 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \gamma^5 \not{\!\partial}\psi \pm \frac 1 4 \psi^\dagger \gamma^0 \gamma^5 \not{\!\partial}\psi \mp \frac 1 4 \psi^\dagger \gamma^5 \gamma^0 \not{\!\partial}\psi=0 \ \ \ \ (1)$$

I've been trying to use the previous tricks but they do not work here. So I tried to use the following identity

$$\gamma^{\mu \dagger} \equiv \gamma^0 \gamma^{\mu} \gamma^0$$

Multiplying equation $(1)$ by $\gamma^0$ (on the right side) we get

$$\frac 1 4 \psi^{\dagger} (\gamma^0)^2 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^{5 \dagger} \not{\!\partial}\psi \pm \frac 1 4 \psi^\dagger \gamma^{5 \dagger}\not{\!\partial}\psi \mp \frac 1 4 \psi^\dagger \gamma^5 (\gamma^0)^2 \not{\!\partial}\psi=\frac 1 4 \psi^{\dagger} (\gamma^0)^2 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \not{\!\partial}\psi \pm \frac 1 4 \psi^\dagger \gamma^5\not{\!\partial}\psi \mp \frac 1 4 \psi^\dagger \gamma^5 (\gamma^0)^2 \not{\!\partial}\psi=0$$

Chee if we were to have the property $(\gamma^0)^2=1$, crossed terms & first and second terms would cancel out! But such a property is nowhere to be found in my book; you can only find $(\gamma^5)^2=1$. Should I use another trick instead of $\gamma^{\mu \dagger} \equiv \gamma^0 \gamma^{\mu} \gamma^0$ or I am on the right track but made a mistake?

the first one may not be a mathematical error, but the way you write your sentence.
You wrote "If $\psi^{\dagger} \rightarrow \psi'^{\dagger} = e^{-i \alpha \gamma_5} \psi^{\dagger}$ then ..."
It's important to notice that you can't make any assumption on how $\psi^\dagger$ transform, the only assumption you are allowed to do is the transformation of $\psi$. So you need to prove the transformation for $\psi^\dagger$ using the transformation for $\psi$. I'm sure you did this part, I'm just telling you that then you don't have to say "If $\psi^\dagger \rightarrow$..." because this means that you are making an assumption for how $\psi^\dagger$ transform.

Now more important things, first of all, related to the previous comment, if you have deduced the transformation for $\psi^\dagger$ you did it wrong, if it was an assumption (as it's written) then it's a wrong assumption. Notice that is the same kind of error than before, $e^{-i\alpha\gamma_5}$ is a 4x4 matrix and $\psi^\dagger$ is a 1x4 vector, so the matrix product is not well-defined.
The rest of the problems are simply that, again, you are using $[\gamma^\mu,\gamma^5]=0$.

WoW I have to say this is harder to show than I expected! Applying the given transformations to the given Lagrangian we get

$$\mathscr{L} = \psi^{\dagger} e^{-i \alpha \gamma_5} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi = i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5}\not{\!\partial}\psi - m \psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5} \psi $$

I am left to prove that

$$i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5}\not{\!\partial}\psi - m \psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5} \psi \neq i\psi^{\dagger} \gamma^{0} \not{\!\partial}\psi - m \psi^{\dagger} \gamma^0 \psi$$

But I do not know how to deal with $e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5}$ in order to prove it...

Besides it turns out that proving that the Lagrangian is invariant in the limit $m=0$ is not as straightforward as I thought due to the same issue: I am stuck in proving the following

$$i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5}\not{\!\partial}\psi = i\psi^{\dagger} \gamma^{0} \not{\!\partial}\psi$$

I am pretty sure the following holds

$$e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5} \neq \gamma^{0}$$

 

[Gaussian97]

I will hide the message inside spoiler to avoid a super large message:

Ok, first, for the $\bar{\psi}_{L,R}\not{\!\partial} \psi_{R,L}$

Spoiler: Part I

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

You have an error here, you must keep in mind that $\not{\!\partial}\equiv \gamma^\mu \partial_\mu$ is also a 4x4 matrix, and that $[\gamma^\mu,\gamma^5]\neq 0$ so the term $\not{\!\partial} \psi_{R,L}$ is:
$$\not{\!\partial} \psi_{R,L}\equiv \frac{1}{2}\!\!\not{\!\partial}\left[(1\pm \gamma^5)\psi\right]\neq \frac{1}{2}(1\pm \gamma^5)\!\!\not{\!\partial}\psi$$

Multiplying equation $(1)$ by $\gamma^0$ (on the right side) we get

$$\frac 1 4 \psi^{\dagger} (\gamma^0)^2 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^{5 \dagger} \not{\!\partial}\psi \pm \frac 1 4 \psi^\dagger \gamma^{5 \dagger}\not{\!\partial}\psi \mp \frac 1 4 \psi^\dagger \gamma^5 (\gamma^0)^2 \not{\!\partial}\psi=\frac 1 4 \psi^{\dagger} (\gamma^0)^2 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \not{\!\partial}\psi \pm \frac 1 4 \psi^\dagger \gamma^5\not{\!\partial}\psi \mp \frac 1 4 \psi^\dagger \gamma^5 (\gamma^0)^2 \not{\!\partial}\psi=0$$

Again, note that
$$\not{\!\partial}\psi \gamma^0 \neq \gamma^0\!\!\not{\!\partial}\psi$$
which throws away all this computation.

Chee if we were to have the property $(\gamma^0)^2=1$, crossed terms & first and second terms would cancel out! But such a property is nowhere to be found in my book; you can only find $(\gamma^5)^2=1$. Should I use another trick instead of $\gamma^{\mu \dagger} \equiv \gamma^0 \gamma^{\mu} \gamma^0$ or I am on the right track but made a mistake?

It's true that $\left(\gamma^0\right)^2=1$, but it's not true that $\left(\gamma^5\right)^\dagger=\gamma^0\gamma^5\gamma^0$, this only works for $\mu=0,1,2,3$. I think this is not the right way to do it, although maybe you can also prove it, I don't know. Anyway, I think there's a much easier way.My advice, go extremely slow, knowing exactly what are you doing in each step, don't skip steps, and each time you commute two things think about if they commute or not.
Also, keep in mind that even if $[\gamma^\mu, \gamma^5]\neq 0$, is true that $\{\gamma^\mu, \gamma^5\}=0$.
I tell you that the cross-terms cancel each other, so maybe try to separate the problem in two parts. On one hand, consider only the cross-terms, and on the other hand consider the inner and outer terms, proof that all those terms cancel by pairs.

Then the comments about part d)

Spoiler: Part II

$$\mathscr{L} = \psi^{\dagger} e^{-i \alpha \gamma_5} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi = i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5}\not{\!\partial}\psi - m \psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5} \psi $$

Again, you are using that $\left[\not{\!\partial},e^{i\alpha\gamma^5}\right]=0$, which would be true if $\left[\gamma^\mu,\gamma^5\right]=0$, but this is not the case. So you cannot commute these matrices.

I am pretty sure the following holds

$$e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5} \neq \gamma^{0}$$

Yes, you're completely right.

Try to do these steps:
1. Write the Lagrangian in terms of $\psi_{L,R}$ (you have already done this)
2. Compute $\gamma^5\psi_{L,R}$
3. How does $\psi_{L,R}$ transform under this transformation?
4. Use 2. and the series expansion of $e^x$ to simplify 3.
5. Substitute in 1. and should be straightforward to see that is not invariant unless $m=0$.

 

[JD_PM]

I will hide the message inside spoiler to avoid a super large message:

Ok, first, for the $\bar{\psi}_{L,R}\not{\!\partial} \psi_{R,L}$

My mistake was not recalling that $\not{\!\partial}$ is a $4 \times 4$ matrix. Thus I now get

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

As suggested, let's split the proof in two:

1) Show that $\frac 1 4 \psi^{\dagger} \gamma^0 \not{\!\partial}\psi - \frac 1 4 \psi^{\dagger} \gamma^5 \gamma^0 \not{\!\partial}\gamma^5 \psi=0$

The key here is to show that (note that I assume that $\partial_{\mu}$ acts on $\psi$ and not on $\gamma^5$)

$$\gamma^0\gamma^{\mu} - \gamma^5\gamma^0\gamma^{\mu}\gamma^5=0$$

You seem to suggest we should be able to show it based on the anticommutation relation $[\gamma^{\mu}, \gamma^5]_+ =0$

I've been trying different combinations (multiplying different matrices right/left on the above equation)
but got nothing... the sign difference spoils any attempt to apply the anticommutation relation...

Could you please give me a hint on how to prove $\gamma^0\gamma^{\mu} - \gamma^5\gamma^0\gamma^{\mu}\gamma^5=0$

2) Show that $\pm \frac 1 4 \psi^\dagger \gamma^0 \not{\!\partial}\gamma^5 \psi \mp \frac 1 4 \psi^\dagger \gamma^5 \gamma^0 \not{\!\partial}\psi=0$

The key here is to show that (note that I assume that $\partial_{\mu}$ acts on $\psi$ and not on $\gamma^5$)

$$\pm\gamma^0\gamma^{\mu}\gamma^5 \mp \gamma^5\gamma^0\gamma^{\mu}=0$$

Here I have the same issue; Could you please give me a hint on how to prove $\pm\gamma^0\gamma^{\mu}\gamma^5 \mp \gamma^5\gamma^0\gamma^{\mu}=0$

Then the comments about part d)

Ahhhh I almost have this one (I am having so much fun with this one btw! )

OK let me correct matrix-commutation mistakes (note that I assume that $\partial_{\mu}$ acts on $\psi$ and not on $e^{i \alpha \gamma_5}$)

$$\mathscr{L} = \psi^{\dagger} e^{-i \alpha \gamma_5} \gamma^{0} (i\not{\!\partial} - m)e^{i \alpha \gamma_5} \psi = i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} \gamma^{\mu}e^{i \alpha \gamma_5}\partial_{\mu}\psi - m \psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} e^{i \alpha \gamma_5} \psi$$

Let me first show invariance of the Lagrangian at the limit $m=0$

We have to show that

$$i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} \gamma^{\mu}e^{i \alpha \gamma_5}\partial_{\mu}\psi=i\psi^{\dagger}\gamma^0\gamma^{\mu}\partial_{\mu}\psi$$

Applying series expansion of the exponential function up to first order we get

$$ i\psi^{\dagger} (1-i \alpha \gamma^5) \gamma^{0} \gamma^{\mu} (1+i \alpha \gamma^5) \partial_{\mu}\psi=i\psi^{\dagger}\gamma^{0} \gamma^{\mu}\partial_{\mu}\psi+i\alpha^2\psi^{\dagger}\gamma^5\gamma^0\gamma^{\mu}\gamma^5\partial_{\mu}\psi-\alpha\psi^{\dagger}\gamma^0\gamma^{\mu}\gamma^5\partial_{\mu}\psi+\alpha\psi^{\dagger}\gamma^5\gamma^0\gamma^{\mu}\partial_{\mu}\psi$$

I know that

$$-\alpha\psi^{\dagger}\gamma^0\gamma^{\mu}\gamma^5\partial_{\mu}\psi+\alpha\psi^{\dagger}\gamma^5\gamma^0\gamma^{\mu}\partial_{\mu}\psi=0$$

Due to the fact that $\gamma^0\gamma^{\mu}\gamma^5 + \gamma^5\gamma^0\gamma^{\mu}=0$. I know that this can be proven via $[\gamma^{\mu}, \gamma^5]_+=0$ but honestly, I do not know how to prove it (same issue than before; could you please give a hint on how to do so?).

The second term is dropped out as it is of second order. Thus we end up with $i\psi^{\dagger}\gamma^{0} \gamma^{\mu}\partial_{\mu}\psi$, as expected.

Let's now show that the lagrangian is not invariant when $m \neq 0$

We want to show that

$$i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} \gamma^{\mu}e^{i \alpha \gamma_5}\partial_{\mu}\psi - m\psi^{\dagger}e^{-i \alpha \gamma_5}\gamma^0 e^{i \alpha \gamma_5} \psi \neq i\psi^{\dagger} \gamma^{0} \not{\!\partial}\psi - m \psi^{\dagger} \gamma^0 \psi$$

Applying series expansion of the exponential function up to first order we get (note that the term that does not have the mass factor is exactly the same that the one we got above; thus I'll not work it out again)

$$i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} \gamma^{\mu}e^{i \alpha \gamma_5}\partial_{\mu}\psi - m\psi^{\dagger}e^{-i \alpha \gamma_5}\gamma^0 e^{i \alpha \gamma_5} \psi = i\psi^{\dagger}\gamma^{0} \gamma^{\mu}\partial_{\mu}\psi - m\psi^{\dagger}(1-i \alpha \gamma_5)\gamma^0 (1+i \alpha \gamma_5) \psi= i\psi^{\dagger}\gamma^{0} \gamma^{\mu}\partial_{\mu}\psi - m\psi^{\dagger}\gamma^0\psi-m\alpha^2\psi^{\dagger}\gamma^5\gamma^0\gamma^5\psi-im\psi^{\dagger}\gamma^0\gamma^5\psi+im\alpha\psi^{\dagger}\gamma^5\gamma^0\psi$$

The key on why the lagrangian is not invariant is that

$$-im\psi^{\dagger}\gamma^0\gamma^5\psi+im\alpha\psi^{\dagger}\gamma^5\gamma^0\psi \neq 0$$

This is because $-\gamma^0\gamma^5 +\gamma^5\gamma^0 \neq 0$ due to the fact that $[\gamma^0, \gamma^5] \neq 0$

WoW I am learning a lot! Thanks Gaussian97

 

[Gaussian97]

Part I:

Well, you're almost there, note that both equations are essentially the same because if you multiply the first one by $\pm \gamma^5$ to the right, and you use $\gamma_5^2=1$ you obtain the second equations.
My hint is that you should try to use $\{\gamma^\mu,\gamma^5\}=0$ to move all the $\gamma^\mu$ to the right, and factor them out.


Part II:

Sadly, you're still not there, you almost proof what we want for infinitesimal transformations (although you still have errors because you go too fast and forget terms, or you change the signs). Anyway, you need to prove all this for finite transformations, which means that you cannot expand $e$ to the first order, you need to keep the infinitely many orders.

 

[JD_PM]

Part I:

Well, you're almost there, note that both equations are essentially the same because if you multiply the first one by $\pm \gamma^5$ to the right, and you use $\gamma_5^2=1$ you obtain the second equations.
My hint is that you should try to use $\{\gamma^\mu,\gamma^5\}=0$ to move all the $\gamma^\mu$ to the right, and factor them out.

OK so to finish part I we have to prove

$$\pm\gamma^0\gamma^{\mu}\gamma^5 \mp \gamma^5\gamma^0\gamma^{\mu}=0 \ \ \ \ (*)$$

We know that

$$\gamma^{\mu}\gamma^5=[\gamma^\mu,\gamma^5]_+ - \gamma^5\gamma^{\mu} \ \ \ \ (2*)$$

Multiplying equation $(2*)$ by $\gamma^0$ (on the left side) and plugging into $(*)$ we get

$$\pm\gamma^0[\gamma^\mu,\gamma^5]_+ \mp\gamma^0\gamma^5\gamma^{\mu}\mp \gamma^5\gamma^0\gamma^{\mu}=0 \ \ \ \ (3*)$$

$(3*)$ can be rewritten as follows

$$\pm\gamma^0[\gamma^\mu,\gamma^5]_+ \mp [\gamma^0,\gamma^5]_+ \gamma^{\mu}=0 \ \ \ \ (3*)$$

We know that $[\gamma^{\mu},\gamma^5]_+=0, \ \ \ \ [\gamma^0,\gamma^5]_+=0$

QED.

Please let me know if you agree

Part II:

Sadly, you're still not there, you almost proof what we want for infinitesimal transformations (although you still have errors because you go too fast and forget terms, or you change the signs). Anyway, you need to prove all this for finite transformations, which means that you cannot expand $e$ to the first order, you need to keep the infinitely many orders.

Let's focus first on showing that the lagrangian is not invariant when $m=0$

We have to show that

$$i\psi^{\dagger}e^{-i \alpha \gamma_5} \gamma^{0} \gamma^{\mu}e^{i \alpha \gamma_5}\partial_{\mu}\psi=i\psi^{\dagger}\gamma^0\gamma^{\mu}\partial_{\mu}\psi$$

I already know how to prove invariance of the lagrangian using first-order series expansion, as I already learned how to prove that $-\gamma^0\gamma^{\mu}\gamma^5 + \gamma^5\gamma^0\gamma^{\mu}=0$

But if we are not allowed to use the first-order series expansion, how can we proceed?

 

[Gaussian97]

Ok, now you have proved Part I. For Part II, I have already given you clues in #43 that you haven't explored yet.

 

[JD_PM]

OK let's follow the steps you propose

Try to do these steps:
1. Write the Lagrangian in terms of $\psi_{L,R}$ (you have already done this)
2. Compute $\gamma^5\psi_{L,R}$
3. How does $\psi_{L,R}$ transform under this transformation?
4. Use 2. and the series expansion of $e^x$ to simplify 3.
5. Substitute in 1. and should be straightforward to see that is not invariant unless $m=0$.

1. Write the Lagrangian in terms of $\psi_{L,R}$ (you have already done this)

Just to be clear: You meant in terms of $\psi_{L,R}$ and $\bar{\psi_{L,R}}$, right?

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

 

[Gaussian97]

Yes, but we know how to simplify this, right? (look at #13).

 

[JD_PM]

Alright.

1. Write the Lagrangian in terms of $\psi_{L,R}$ (you have already done this)

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

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

It is not really clear to me what you mean here, so I will show the Lagrangian explicitly:

$$\mathscr{L} (\psi_L, \psi_R, \bar{\psi_L}, \bar{\psi_R}) = i\bar{\psi_L} \not{\!\partial}\psi_L + i\bar{\psi_R} \not{\!\partial}\psi_R - \bar{\psi_L} m\psi_R - \bar{\psi_R} m\psi_L=\frac 1 4 i \psi^{\dagger}(1-\gamma^5)\gamma^0\gamma^{\mu}\partial_{\mu}(1-\gamma^5)\psi+\frac 1 4 i \psi^{\dagger}(1+\gamma^5)\gamma^0\gamma^{\mu}\partial_{\mu}(1+\gamma^5)\psi-\frac m 4 \psi^{\dagger}(1-\gamma^5)\gamma^0(1+\gamma^5)\psi-\frac m 4 \psi^{\dagger}(1+\gamma^5)\gamma^0(1-\gamma^5)\psi$$

Did you mean I have to expand out the above equation and then use the given transformations?

 

[Gaussian97]

For 1 perfect, but then in 2. you are going backwards, you already know the expression of $\mathscr{L}(\psi, \bar{\psi})$. What I tell you is much more simple, just compute $\gamma^5 \psi_L$ and $\gamma^5\psi_R$

 

[JD_PM]

NOTE: I do not explicitly show computations on $\bar{\psi_{L,R}}$; it is analogous.

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

$$\gamma^5 \psi_{L,R} =\frac 1 2 \gamma^5(1\mp \gamma^5)\psi=\frac 1 2 \gamma^5 \psi \mp \frac 1 2 \psi$$3. How does $\psi_{L,R}$ transform under this transformation?

Applying the transformation to $\psi_{L,R}$ we get

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

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

Well, we get

$$\frac 1 2 (1+ i \alpha \gamma_5 + ... ) \psi \mp \frac 1 2 \gamma^5 (1+ i \alpha \gamma_5 + ... ) \psi$$

But how can I use 2. to simplify it even further?

 

[Gaussian97]

We are working with $\psi_{L,R}$ so it's always a good idea to write all the expressions with $\psi_{L,R}$, this actually simplifies 2. and 3. a lot.
Then for 3, you need to use all the terms, so the dots are only hidden you relevant information, write the expansion explicitly to all orders.

 

[JD_PM]

We are working with $\psi_{L,R}$ so it's always a good idea to write all the expressions with $\psi_{L,R}$, this actually simplifies 2. and 3. a lot.
Then for 3, you need to use all the terms, so the dots are only hidden you relevant information, write the expansion explicitly to all orders.

If what I did at #52 is not what you meant regarding 2. 3., then I have no clue... could you please give an explicit hint?

Then you say I have to include all terms of the series expansion... there are infinitely many though! I must be missing your point...

Bottom line is this: I think I need explicit hints regarding 2. 3. and 4.

Thanks

 

[Gaussian97]

What you did in 2 and 3 is OK, but you have the result involving $\psi$, while we are working with $\psi_{L,R}$, so try to write the result in terms of $\psi_{L,R}$.
For 4. you should know that
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
here I'm writing the infinite terms.

 

[JD_PM]

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

$$\gamma^5 \psi_{L,R} =\frac 1 2 \gamma^5(1\mp \gamma^5)\psi=\frac 1 2 \gamma^5 \psi \mp \frac 1 2 \psi = \frac 1 2 \gamma^5 (\psi_L + \psi_R) \mp \frac 1 2 (\psi_L + \psi_R)=\frac 1 2 (\gamma^5 \mp 1) \psi_L + \frac 1 2 (\gamma^5 \mp 1) \psi_R$$

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

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

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

Note: Here I assumed that $e^{i \alpha \gamma_5}$ and $\psi_{L,R}$ commute.

$$\frac 1 2 ( 1 \mp \gamma^5)e^{i \alpha \gamma_5} \psi_L + \frac 1 2 ( 1 \mp \gamma^5)e^{i \alpha \gamma_5}\psi_R=\frac 1 2 ( 1 \mp \gamma^5)\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!} \psi_L + \frac 1 2 ( 1 \mp \gamma^5)\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}\psi_R=\frac 1 2 ( 1 \mp \gamma^5) \psi_L \sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!} + \frac 1 2 ( 1 \mp \gamma^5)\psi_R \sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}=\gamma^5 \psi_{L,R}\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!} $$

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

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

It does not look straightforward to me that it is not invariant unless $m=0$; I guess there has to be something wrong at 4. ...

 

[Gaussian97]

You can still simplify 2. What are the values of
$$\frac{1}{2}(\gamma^5 - 1)\psi_L, \qquad \frac{1}{2}(\gamma^5 - 1)\psi_R, \qquad \frac{1}{2}(\gamma^5 + 1)\psi_L, \qquad \frac{1}{2}(\gamma^5 + 1)\psi_R$$
?

I think that maybe it will be easier for you to work with $\psi_L$ and $\psi_R$ separately, so the equations read:
$$\gamma^5 \psi_L = \frac{1}{2}(\gamma^5 - 1)\psi_L + \frac{1}{2}(\gamma^5 - 1)\psi_R$$
$$\gamma^5 \psi_R = \frac{1}{2}(\gamma^5 + 1)\psi_L + \frac{1}{2}(\gamma^5 + 1)\psi_R$$

 

[JD_PM]

You can still simplify 2. What are the values of
$$\frac{1}{2}(\gamma^5 - 1)\psi_L, \qquad \frac{1}{2}(\gamma^5 - 1)\psi_R, \qquad \frac{1}{2}(\gamma^5 + 1)\psi_L, \qquad \frac{1}{2}(\gamma^5 + 1)\psi_R$$
?

I think that maybe it will be easier for you to work with $\psi_L$ and $\psi_R$ separately, so the equations read:
$$\gamma^5 \psi_L = \frac{1}{2}(\gamma^5 - 1)\psi_L + \frac{1}{2}(\gamma^5 - 1)\psi_R$$
$$\gamma^5 \psi_R = \frac{1}{2}(\gamma^5 + 1)\psi_L + \frac{1}{2}(\gamma^5 + 1)\psi_R$$

I guess you want me to expand it out

$$\gamma^5 \psi_{L,R} =\frac 1 2 \gamma^5(1\mp \gamma^5)\psi=\frac 1 2 (\gamma^5 \mp 1) \psi_L + \frac 1 2 (\gamma^5 \mp 1) \psi_R=\frac 1 2 \gamma^5 \psi_L \mp \frac 1 2 \psi_L + \frac 1 2 \gamma^5 \psi_R \mp \frac 1 2 \psi_R$$

?

By the way: Do you agree on what I did at 4.? (at #56)

 

[Gaussian97]

No, what I want you to prove is
$$\gamma^5\psi_{L,R}=\mp \psi_{L,R}$$

 

[JD_PM]

No, what I want you to prove is
$$\gamma^5\psi_{L,R}=\mp \psi_{L,R}$$

Alright. Could we postpone such a proof (I've been trying it for over 2 hours and got nothing...)? I'd like to prove the non-invariance of the Lagrangian (unless $m=0$) first.

So following the steps:

2. Compute $\gamma^5\psi_{L,R}$ (I assume it for now)

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

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

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

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

Note: Here I assumed that $e^{i \alpha \gamma_5}$ and $\psi_{L,R}$ commute.

$$\frac 1 2 ( 1 \mp \gamma^5)e^{i \alpha \gamma_5} \psi_L + \frac 1 2 ( 1 \mp \gamma^5)e^{i \alpha \gamma_5}\psi_R=\gamma^5 \psi_{L,R}\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!} = \mp \psi_{L,R}\sum_{n=0}^\infty \frac{(i \alpha \gamma_5)^n}{n!}$$

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

But I still do not see how can we plug the above results into the below Lagrangian to check invariance...

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

If you think it would be more useful to first prove 2. please let me know but please tell me what's the trick because I am completely stuck.

Thanks.

 

[Gaussian97]

Well, we can leave 2 for later if you want, although is really very easy once you realize. I give you more hints, let's define the matrices
$$P_L=\frac{1-\gamma^5}{2}, \qquad P_R=\frac{1+\gamma^5}{2}$$
Then, obviously $\psi_L=P_L \psi$ and $\psi_R=P_R\psi$ compute
$$P_L\cdot P_R = \frac{1}{4}(1-\gamma^5)(1+\gamma^5), \qquad P_R\cdot P_L = \frac{1}{4}(1+\gamma^5)(1-\gamma^5)$$
Then compute $P_R\psi_L$ and $P_L\psi_R$, can you see it why
$$\gamma^5 \psi_{L,R}=\mp \psi_{L,R}$$
?

3 is OK, but again, you can simplify it a lot, try to compute the commutator
$$\left[(1\pm \gamma^5), e^{i\alpha \gamma^5}\right]$$
And proof that
$$\psi_L\rightarrow e^{i\alpha \gamma^5}\psi_L, \qquad \psi_R\rightarrow e^{i\alpha \gamma^5}\psi_R$$

4 is not correct, notice that an expression like $\psi_L\gamma^5$ makes no sense because $\psi_L$ is a row vector while $\gamma^5$ is a matrix, so makes no sense to talk about the commutator of $\psi_L$ and a matrix like $e^{i\alpha \gamma^5}$, so no they don't commute. Maybe you can try to use 2. to compute first what is the value of
$$\gamma_5^n\psi_{L}, \qquad \gamma_5^n\psi_{R}$$

 

[JD_PM]

No, what I want you to prove is
$$\gamma^5\psi_{L,R}=\mp \psi_{L,R}$$

OK thanks to your hints at #61 I got it!

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

We note that the second term vanishes due to $(\gamma^5 - 1)(\gamma^5 + 1)=(\gamma^5)^2-1=1-1=0$

Thus we end up with

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

Analogously we get

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

QED.

3 is OK, but again, you can simplify it a lot, try to compute the commutator
$$\left[(1\pm \gamma^5), e^{i\alpha \gamma^5}\right]$$
And proof that
$$\psi_L\rightarrow e^{i\alpha \gamma^5}\psi_L, \qquad \psi_R\rightarrow e^{i\alpha \gamma^5}\psi_R$$

This is what I have got far; let's prove first that $\psi_L\rightarrow e^{i\alpha \gamma^5}\psi_L$

Note that I am going to apply the given transformation $\psi\rightarrow e^{i\alpha \gamma_5}\psi$

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

The second term vanishes, so we get

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

And here's where I am stuck in. I've been trying to use the commutator you provided

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

But I still do not see it.

Am I on the right track?

Should I keep trying to solve for one of the commutator factors and get something like $\psi_L = ... \left[(1 - \gamma^5), e^{i\alpha \gamma_5}\right] ...$ ??

Thanks.

 

[Gaussian97]

Perfect, note that was as simple as
$$\gamma^5\psi_{L,R}=\gamma^5\frac{1\mp\gamma^5}{2}\psi = \frac{\gamma^5\mp \gamma_5^2}{2}\psi = \frac{\gamma^5\mp 1}{2}\psi = \mp \frac{\mp\gamma^5+1}{2}\psi=\mp \psi_{L,R}$$.

For 3., for a general operator $A$ and a general analytic function $f(x)$, do you know what is the value of the commutator
$$[A, f(A)]$$
?

 

[JD_PM]

Perfect, note that was as simple as
$$\gamma^5\psi_{L,R}=\gamma^5\frac{1\mp\gamma^5}{2}\psi = \frac{\gamma^5\mp \gamma_5^2}{2}\psi = \frac{\gamma^5\mp 1}{2}\psi = \mp \frac{\mp\gamma^5+1}{2}\psi=\mp \psi_{L,R}$$.

Beautiful!

For 3., for a general operator $A$ and a general analytic function $f(x)$, do you know what is the value of the commutator
$$[A, f(A)]$$
?

I'd say

$$[A, f(A)] := A f(A) - f(A) A$$

Right?

 

[nrqed]

Beautiful!
I'd say

$$[A, f(A)] := A f(A) - f(A) A$$

Right?

What is $[A, A^n ]$ for $n \geq 0$? Now, assuming that $f(A)$ can be Taylor expanded, what can we say about $[A,f(A)]$? (if $f(A)$ can't be represented as a Taylor expansion, it is a bit more tricky).

 

[JD_PM]

For 3., for a general operator $A$ and a general analytic function $f(x)$, do you know what is the value of the commutator
$$[A, f(A)]$$

?

What is $[A, A^n ]$ for $n \geq 0$? Now, assuming that $f(A)$ can be Taylor expanded, what can we say about $[A,f(A)]$? (if $f(A)$ can't be represented as a Taylor expansion, it is a bit more tricky).

After some googling I found that given two operators $x_1$ and $x_2$ with a constant commutation relation $c$, such that $[x_1, x_2]=c$, it can be shown that

$$[x_1, f(x_2)]=cf'(x_2)$$

So in answer to you question (I guess you really meant $[A,f(x)]$ instead of $[A,f(A)]$); we get

$$[A,f(x)]=kf'(x), \ \ \ \ \ \ [A,x]=k$$

If you really meant $[A,f(A)]$ we get

$$[A,f(A)]=g f'(A)=0, \ \ \ \ \ \ [A,A]=g=0$$

But I still do not see how is $[A,f(x)]$ going to help me in order to move forward in 3. (work shown at #62)...

Source: see page 4 of attached PDF.

 

[Gaussian97]

Nop, I mean $[A,f(A)]$, don't you see why this can help you to compute the commutator we are interested in?
$$\left[(1\mp\gamma^5),e^{i\alpha\gamma^5}\right]$$
Note that $e^x$ is an analytic function.

 

[JD_PM]

Nop, I mean $[A,f(A)]$, don't you see why this can help you to compute the commutator we are interested in?
$$\left[(1\mp\gamma^5),e^{i\alpha\gamma^5}\right]$$
Note that $e^x$ is an analytic function.

Ahhh OK; applying it to our case I get

$$\left[(1\mp\gamma^5),e^{i\alpha\gamma^5}\right]= k (i\alpha e^{i\alpha\gamma_5})$$

But now my question is: isn't $k=0$?

By definition I get that

$$k:=[(1\mp\gamma^5),\gamma^5]=0$$

...

 

[Gaussian97]

Yes, so you have it
$$\left[(1\mp \gamma^5), e^{i\alpha\gamma^5}\right]=0$$
Does this help you to prove that
$$\psi_L\rightarrow e^{i\alpha\gamma^5}\psi_L, \qquad \psi_R\rightarrow e^{i\alpha\gamma^5}\psi_R$$

 

[JD_PM]

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.