Why does this term transform in this way?

• A
• MichaelJ12
In summary: It doesn't. It only implies that ##(i \gamma^\mu (\Lambda^{-1})^\nu{}_\mu \partial _\nu - m) \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) =...##.
MichaelJ12
TL;DR Summary
Trying to find out how a certain quantity transforms under Lorentz transformations.
I am trying to understand the last block of equations in the picture (after 3.31). In the first line of that block, he transforms the spinor ##\psi## which I have no problem with. What I have a problem with is the ##\gamma ^{\mu} \partial _{\mu}##. They form a Lorentz scalar, so they should not change as a whole, because ##\gamma ^{\mu} \partial _{\mu} \to \Lambda ^\mu _{\; \alpha} \gamma ^{\alpha} (\Lambda ^{-1}) ^\nu _{\; \mu} \partial _{\nu} = \gamma ^{\mu} \partial _{\mu} ##. Instead, he only transforms the derivative in this way: ##\partial _{\mu} \to (\Lambda ^{-1}) ^\nu _{\; \mu} \partial _{\nu} ##. My question is why doesn't he transform the gamma matrices as well? They have a Lorentz index too.

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Even though the (generally contravariant) index of the gammas runs from 0 to 3 in that summation, the set of 4 gammas in special relativity do not transform as the components of a 4-vector, they are constant matrices (particular forms such as chiral/Weyl, Dirac, Majorana). So no need to put an extra Lorentz transformation there.

MichaelJ12
dextercioby said:
Even though the (generally contravariant) index of the gammas runs from 0 to 3 in that summation, the set of 4 gammas in special relativity do not transform as the components of a 4-vector, they are constant matrices (particular forms such as chiral/Weyl, Dirac, Majorana). So no need to put an extra Lorentz transformation there.

Thank you, that explains it. I have a second question. How is ##\gamma ^\mu \partial _\mu## Lorentz invariant? It's transformation is

##\gamma ^{\mu} \partial _{\mu} \to\gamma ^{\mu} (\Lambda ^{-1}) ^\nu _{\; \mu} \partial _{\nu} \neq \gamma ^{\mu} \partial _{\mu} ## since the matrix ##\Lambda ^{-1}## is not a Kronecker delta in general.

MichaelJ12 said:
How is ##\gamma ^\mu \partial _\mu## Lorentz invariant?

It isn't. What is invariant is ##\gamma^\mu \partial_\mu \psi##, where ##\psi## is a spinor. The spinor transforms in just the right way to keep that expression invariant even though ##\partial_\mu## transforms as well.

vanhees71, dextercioby and MichaelJ12
PeterDonis said:
It isn't. What is invariant is ##\gamma^\mu \partial_\mu \psi##, where ##\psi## is a spinor. The spinor transforms in just the right way to keep that expression invariant even though ##\partial_\mu## transforms as well.

Thank you so much. I got confused because that page says it is a Lorentz invariant differential operator, but it is the operator acting on the spinor that remains invariant as a whole.

MichaelJ12 said:
that page says it is a Lorentz invariant differential operator

Yes, that's sloppy since it's only true when it's operating on a spinor (probably the textbook writers assumed that would be understood since that section is talking about spinors). More generally, any expression involving the gamma matrices that looks Lorentz invariant based on its indices, will only give a Lorentz invariant quantity when it's operating on a spinor (where for something like ##\gamma^\mu A_\mu##, which occurs in the QED Lagrangian, "operating on" means "multiplying to the right").

vanhees71, weirdoguy and dextercioby
PeterDonis said:
Yes, that's sloppy since it's only true when it's operating on a spinor. More generally, any expression involving the gamma matrices that looks Lorentz invariant based on its indices, will only give a Lorentz invariant quantity when it's operating on a spinor (where for something like ##\gamma^\mu A_\mu##, which occurs in the QED Lagrangian, "operating on" means "multiplying to the right").

In the third to last line at the bottom of the page, how could we change the argument of ##\psi## from ##\Lambda ^{-1} x## to ##x##?

MichaelJ12 said:
In the third to last line at the bottom of the page, how could we change the argument of ##\psi## from ##\Lambda ^{-1} x## to ##x##?

I think that's a typo since it goes back to being ##\psi(\Lambda ^{-1} x)## in the next line.

PeterDonis said:
I think that's a typo since it goes back to being ##\psi(\Lambda ^{-1} x)## in the next line.

I feel like I'm missing something important (perhaps because I am a beginner and I'm self studying this), but if we have an equation ##(i \gamma ^\mu \partial _\mu - m) \psi (x) = 0##, why does this imply that ##(i \gamma ^\mu \partial _\mu - m) \psi (\Lambda ^{-1}x) = 0##?

For example, if we are given that this is a fact ##(\partial _x + c) f(x) = 0 ## (where c is a constant and f is a function of x), then it does not imply ##(\partial _x + c) f(2x) = 0 ## because ##(\partial _x + c) f(2x) = 2 \frac{\partial f(2x)}{\partial (2x)} + cf(2x) ## after renaming the dummy variable x we get ##2 \frac{\partial f(x)}{\partial x} + cf(x)## which we cannot conclude is zero.

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MichaelJ12 said:
if we have an equation ##(i \gamma^\mu \partial _\mu - m) \psi (x) = 0##, why does this imply that ##(i \gamma^\mu \partial _\mu - m) \psi (\Lambda^{-1}x) = 0##?

It doesn't. It only implies that ##(i \gamma^\mu (\Lambda^{-1})^\nu{}_\mu \partial _\nu - m) \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = 0##.

dextercioby
PeterDonis said:
It doesn't. It only implies that ##(i \gamma^\mu (\Lambda^{-1})^\nu{}_\mu \partial _\nu - m) \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = 0##.

How did we conclude that the second to last equation on the page is zero?

MichaelJ12 said:
How did we conclude that the second to last equation on the page is zero?

You mean this?

$$\Lambda_{\frac{1}{2}} (i \gamma^\nu \partial _\nu - m) \psi (\Lambda^{-1}x) = 0$$

The ##\Lambda_{\frac{1}{2}}## commutes with everything inside the parentheses, so we have

$$(i \gamma^\nu \partial _\nu - m) \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = 0$$

And we have ##\Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = \psi(x)## because that's how spinors transform, so we just have the original Dirac equation back again.

PeterDonis said:
You mean this?

$$\Lambda_{\frac{1}{2}} (i \gamma^\nu \partial _\nu - m) \psi (\Lambda^{-1}x) = 0$$

The ##\Lambda_{\frac{1}{2}}## commutes with everything inside the parentheses, so we have

$$(i \gamma^\nu \partial _\nu - m) \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = 0$$

And we have ##\Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = \psi(x)## because that's how spinors transform, so we just have the original Dirac equation back again.

We know that ##\psi(x) \to \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x)##, but why can we replace the ##\to## with an ##=## and say ##\psi(x) = \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x)##?
The ##\to## represents what happens after a Lorentz transformation, right? How can we equate the field before the transformation and after the transformation?

MichaelJ12 said:
why can we replace the ##\to## with an ##=##

Because the Lorentz transformation is invertible.

PeterDonis said:
Because the Lorentz transformation is invertible.

So you mean we performed another Lorentz transformation, not just one? Why didn't we also transform the derivative?

And aren't we going around in a circle, first performing a transformation and then its inverse, which doesn't prove that the equation is invariant under all Lorentz transformations?

MichaelJ12 said:
So you mean we performed another Lorentz transformation, not just one?

No. You're making this way too hard. As you say:

MichaelJ12 said:
We know that ##\psi(x) \to \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x)##

And because the Lorentz transformation is invertible, we also know that ##\Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) \to \psi(x)##. In other words, we can freely substitute either one for the other in equations.

To put it another way, the ##\to## in the equations above is not saying "is being Lorentz transformed into". It's saying "can be replaced in equations by".

PeterDonis said:
No. You're making this way too hard. As you say:
And because the Lorentz transformation is invertible, we also know that ##\Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) \to \psi(x)##. In other words, we can freely substitute either one for the other in equations.

To put it another way, the ##\to## in the equations above is not saying "is being Lorentz transformed into". It's saying "can be replaced in equations by".

I never learned this meaning of the ##\to## symbol.

I don't feel comfortable with the meaning of ##\to## symbol. We shouldn't be able to selectively choose which part of an equation we change and which part we leave the same. The only way to think of the derivation above that makes sense to me is that the author wants to show that the Dirac equation is invariant under Lorentz transformations, so he transforms the equation (all of it).

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MichaelJ12 said:
Throw away this textbook or notes that you are using. To prove that the Dirac equation is Lorentz covariant, you only need the following transformation laws:
$$x^{\mu} \to \bar{x}^{\mu} = \Lambda^{\mu}{}_{\nu} \ x^{\nu} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$$$\psi (x) \to \psi^{\prime} (\bar{x}) = D(\Lambda) \psi (x) . \ \ \ \ \ (2)$$ I use $D(\Lambda)$ instead of the ugly $\Lambda_{\frac{1}{2}}$ notation. So the question of Lorentz covariance is the following: we would like to find a transformation matrix $D(\Lambda)$ such that the Dirac equation retains it’s form in the $\bar{x}$-system.
Okay, start from the Dirac equation in the x-system: $$i \gamma^{\nu} \frac{\partial}{\partial x^{\nu}} \psi (x) - m \psi (x) = 0 .$$ Multiply this from the left by $D$ and insert the identity matrix $D^{-1}D$ as follows $$i D \gamma^{\nu} \left( D^{-1}D \right) \frac{\partial}{\partial x^{\nu}} \psi (x) - m D \psi (x) = 0 . \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ Now, the chain rule and (1) give you $$\frac{\partial}{\partial x^{\nu}} =\frac{\partial \bar{x}^{\mu}}{\partial x^{\nu}} \frac{\partial}{\partial \bar{x}^{\mu}} = \Lambda^{\mu}{}_{\nu} \frac{\partial}{\partial \bar{x}^{\mu}} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$ Finally, substitute (2) and (4) in (3): $$i \left( D \gamma^{\nu} D^{-1} \Lambda^{\mu}{}_{\nu}\right) \frac{\partial}{\partial \bar{x}^{\mu}} \psi^{\prime}(\bar{x}) - m \psi^{\prime}(\bar{x}) = 0 .$$ Thus, the Dirac equation will have the same form in the $\bar{x}$-system (i.e., Lorentz covariant) if and only if the matrix $D$ satisfies $$D \gamma^{\nu} D^{-1} \Lambda^{\mu}{}_{\nu} = \gamma^{\mu} ,$$ or $$D^{-1} \gamma^{\mu}D = \Lambda^{\mu}{}_{\nu} \gamma^{\nu} .$$
Regarding the business of $\psi ( \Lambda^{-1}x )$ in the transformation law: From (1), you have $x = \Lambda^{-1} \bar{x}$. Substituting this in the transformation law (2), you get $\psi^{\prime} (\bar{x}) = D(\Lambda) \psi (\Lambda^{-1}\bar{x})$. Now you can drop the bars from the coordinates, i.e., rename your coordinates as $x$ to obtain $\psi^{\prime} (x) = D (\Lambda) \psi ( \Lambda^{-1}x)$.

dextercioby, vanhees71 and MichaelJ12
PeterDonis said:
You mean this?

$$\Lambda_{\frac{1}{2}} (i \gamma^\nu \partial _\nu - m) \psi (\Lambda^{-1}x) = 0$$

The ##\Lambda_{\frac{1}{2}}## commutes with everything inside the parentheses, so we have

$$(i \gamma^\nu \partial _\nu - m) \Lambda_{\frac{1}{2}} \psi (\Lambda^{-1}x) = 0$$
No, $\Lambda_{\frac{1}{2}}$ does not commute with $\gamma^{\nu}$.

vanhees71 and MichaelJ12
samalkhaiat said:
Throw away this textbook or notes that you are using. To prove that the Dirac equation is Lorentz covariant, you only need the following transformation laws:
$$x^{\mu} \to \bar{x}^{\mu} = \Lambda^{\mu}{}_{\nu} \ x^{\nu} , \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$$$\psi (x) \to \psi^{\prime} (\bar{x}) = D(\Lambda) \psi (x) . \ \ \ \ \ (2)$$ I use $D(\Lambda)$ instead of the ugly $\Lambda_{\frac{1}{2}}$ notation. So the question of Lorentz covariance is the following: we would like to find a transformation matrix $D(\Lambda)$ such that the Dirac equation retains it’s form in the $\bar{x}$-system.

Thank you so much! Your answer is extremely clear, simple, and obviously the correct one.

Just one little point: I want to understand why you say that ##\psi (x) \to \psi^{\prime} (\bar{x})## instead of ##\psi (x) \to \psi^{\prime} (x)##.

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MichaelJ12 said:
I want to understand why you say that ##\psi (x) \to \psi^{\prime} (\bar{x})##
I considered two coordinate systems (two observers) looking at the same point $P$ which has two different coordinate values $x$ and $\bar{x} = \Lambda x$ as seen by the two observers $(O , \psi )$ and $(O^{\prime} , \psi^{\prime})$ respectively. So, $\psi (x) \to \psi^{\prime} (\bar{x})$ simply means $\psi (P) \to \psi^{\prime} (P)$.

instead of ##\psi (x) \to \psi^{\prime} (x)##.
This is the same as the above provided you understand it to mean $$\psi (P) \to \psi^{\prime}(P) \equiv (D \psi )(P) = D \left( \psi (P)\right) .$$

dextercioby and vanhees71
PeterDonis said:
What is invariant is ##\gamma^\mu \partial_\mu \psi##, where ##\psi## is a spinor. The spinor transforms in just the right way to keep that expression invariant even though ##\partial_\mu## transforms as well.
No, $\gamma^{\mu}\partial_{\mu}\psi$ is not invariant. If it was you would not be able to add the non-invariant (spinor) term $-m \psi$ to it and the Dirac equation would make no sense. Indeed, it transforms exactly like $\psi$: $$\gamma^{\mu}\partial_{\mu} \psi \to D(\Lambda ) \ \gamma^{\nu}\partial_{\nu}\psi .$$

samalkhaiat said:
it transforms exactly like ##\psi##:

That is what I meant to convey, but you're right, "invariant" is not the right word to convey that.

The right word is "covariant". A Dirac spinor field realizes the Poincare symmetry in a specific way (i.e., a representation of the Poincare group) and thus transforms under Lorentz transformations which are a subgroup of the Poincare group in a specific covariant way.

After quantizing the Dirac field you have a local unitary realization of the Poincare group, allowing to establish a valid special-relativistic QT of a spin-1/2 fermion and its anti-particle, which latter must be necessarily occur if you want a local realization of the Poincare group.

samalkhaiat said:
No, $\gamma^{\mu}\partial_{\mu}\psi$ is not invariant. If it was you would not be able to add the non-invariant (spinor) term $-m \psi$ to it and the Dirac equation would make no sense. Indeed, it transforms exactly like $\psi$: $$\gamma^{\mu}\partial_{\mu} \psi \to D(\Lambda ) \ \gamma^{\nu}\partial_{\nu}\psi .$$

Is the way you described it called the "active" transformation point of view or the "passive" point of view?

Mathematically that doesn't make much of a difference. If you have a Lorentz transformation, describing the space-time-vector transformation as
$$x'=\Lambda x, \quad \Lambda \in \mathrm{SO}(1,3)^{\uparrow}$$
Then the Dirac-spinor field transforms as
$$\psi(x) \rightarrow \psi'(x')=D(\Lambda) \psi(x)=D(\Lambda) \psi(\hat{\Lambda}^{-1} x').$$
Here ##D(\Lambda)## is a bi-spinor transformation given as the direct sum of the two two-dimensional representations of the proper orthochronous Poincare group, ##(1/2,0) \oplus (0,1/2)##.

dextercioby said:
Even though the (generally contravariant) index of the gammas runs from 0 to 3 in that summation, the set of 4 gammas in special relativity do not transform as the components of a 4-vector, they are constant matrices
Doing physics would be impossible in a world where dimensionless numbers (such as $0, 1$ and $i$, i.e., the elements of the $\gamma$’s) depend on the observer's state of motion. This the intuitive reason for not transforming the $\gamma^{\mu}$. Group theoretically, the gamma matrices play vital role in the representation theory: they project $\partial \psi$ onto the spin-$\frac{1}{2}$ Dirac representation $(\frac{1}{2} , 0) \oplus (0 , \frac{1}{2})$, i.e., they eliminate the spin-$\frac{3}{2}$ component of $\partial \psi$: $$\partial \sim ( \frac{1}{2}, \frac{1}{2}),$$$$\psi \sim (\frac{1}{2} , 0) \oplus (0 , \frac{1}{2}).$$ So, $$\partial \psi \sim ( \frac{1}{2}, \frac{1}{2}) \otimes \big\{ (\frac{1}{2} , 0) \oplus (0 , \frac{1}{2}) \big\} ,$$ or $$\partial \psi \sim \big\{ (\frac{1}{2} , 0) \oplus (0 , \frac{1}{2}) \big\}_{J = \frac{1}{2}} \oplus \big\{ (1 , \frac{1}{2}) \oplus ( \frac{1}{2} , 1) \big\}_{J = \frac{3}{2}},$$ and $$\gamma \cdot \partial \psi \sim \big\{ ( \frac{1}{2}, 0) \oplus (0 , \frac{1}{2})\big\}_{J = \frac{1}{2}} .$$

vanhees71, MichaelJ12 and dextercioby
MichaelJ12 said:
Is the way you described it called the "active" transformation point of view or the "passive" point of view?
Usually, I don’t talk about “active” vs “passive” things, because there is no meaningful distinction between the two and they often cause unnecessary confusions. But, yes I believe I used the “passive” description: That is what I meant by “two coordinate systems $O \to O^{\prime} = \Lambda O$ and one physical point $P$”. Equivalently, I could have used “one coordinate system $O$ and two physical points $P \to Q = \Lambda^{-1}P$” (i.e., the “active” thing) and obtained exactly the same result. Try it yourself, in my post, just put $\Lambda^{-1}$ instead of $\Lambda$ and use $D (\Lambda^{-1}) = D^{-1}(\Lambda)$.

samalkhaiat said:
Doing physics would be impossible in a world where dimensionless numbers (such as $0, 1$ and $i$, i.e., the elements of the $\gamma$’s) depend on the observer's state of motion. This the intuitive reason for not transforming the $\gamma^{\mu}$. Group theoretically, the gamma matrices play vital role in the representation theory: they project $\partial \psi$ onto the spin-$\frac{1}{2}$ Dirac representation $(\frac{1}{2} , 0) \oplus (0 , \frac{1}{2})$, i.e., they eliminate the spin-$\frac{3}{2}$ component of $\partial \psi$: $$\partial \sim ( \frac{1}{2}, \frac{1}{2}),$$$$\psi \sim (\frac{1}{2} , 0) \oplus (0 , \frac{1}{2}).$$ So, $$\partial \psi \sim ( \frac{1}{2}, \frac{1}{2}) \otimes \big\{ (\frac{1}{2} , 0) \oplus (0 , \frac{1}{2}) \big\} ,$$ or $$\partial \psi \sim \big\{ (\frac{1}{2} , 0) \oplus (0 , \frac{1}{2}) \big\}_{J = \frac{1}{2}} \oplus \big\{ (1 , \frac{1}{2}) \oplus ( \frac{1}{2} , 1) \big\}_{J = \frac{3}{2}},$$ and $$\gamma \cdot \partial \psi \sim \big\{ ( \frac{1}{2}, 0) \oplus (0 , \frac{1}{2})\big\}_{J = \frac{1}{2}} .$$

How did you go from the third line to the fourth line?
Also, where did you learn this beautiful and enlightening kind of stuff? Could you recommend a good book?

A very good book about all the group-theoretical aspects is

R. U. Sexl, H. K. Urbantke, Relativity, Groups, Particles,
Springer, Wien (2001).

vanhees71 said:
A very good book about all the group-theoretical aspects is

R. U. Sexl, H. K. Urbantke, Relativity, Groups, Particles,
Springer, Wien (2001).

Thank you very much. I love those kinds of books. Have you read other books that belong to the same category as Relativity, Groups, Particles, meaning that they are meant to impart understanding? I only want books about "advanced" physics or mathematics like quantum field theory, general relativity, differential geometry, topology, and any other branches of pure mathematics.

MichaelJ12 said:
How did you go from the third line to the fourth line?
$$(A , B) \otimes (A^{\prime} , B^{\prime}) = (A + A^{\prime} , B + B^{\prime}) \oplus (A + A^{\prime} - 1 , B + B^{\prime}) \oplus \cdots \oplus (|A - A^{\prime}| , B + B^{\prime}) \oplus (A + A^{\prime} , B + B^{\prime} - 1) \oplus \cdots \oplus (|A - A^{\prime}| , |B - B^{\prime}|) .$$
Could you recommend a good book?
Almost all books on group representations in general and the representation theory of the Lorentz group $\mbox{SL}(2 , \mathbb{C})$ in particular teach you these stuff. The latter can be found in all textbooks on Supersymmetry.

samalkhaiat said:
$$(A , B) \otimes (A^{\prime} , B^{\prime}) = (A + A^{\prime} , B + B^{\prime}) \oplus (A + A^{\prime} - 1 , B + B^{\prime}) \oplus \cdots \oplus (|A - A^{\prime}| , B + B^{\prime}) \oplus (A + A^{\prime} , B + B^{\prime} - 1) \oplus \cdots \oplus (|A - A^{\prime}| , |B - B^{\prime}|) .$$

Almost all books on group representations in general and the representation theory of the Lorentz group $\mbox{SL}(2 , \mathbb{C})$ in particular teach you these stuff. The latter can be found in all textbooks on Supersymmetry.

I meant where did you learn the physical meaning and significance of the gamma matrices projecting out certain spins.

MichaelJ12 said:
I meant where did you learn the physical meaning and significance of the gamma matrices projecting out certain spins.
I have already said that you can learn about this stuff from all textbooks on supersymmetry. I just know this stuff because it is my business to know it.

MichaelJ12 and weirdoguy
MichaelJ12 said:
Thank you very much. I love those kinds of books. Have you read other books that belong to the same category as Relativity, Groups, Particles, meaning that they are meant to impart understanding? I only want books about "advanced" physics or mathematics like quantum field theory, general relativity, differential geometry, topology, and any other branches of pure mathematics.
Then all textbooks by Weinberg are right for you :-)).

MichaelJ12

1. Why does this term change when it is transformed?

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