Infintesimal transformations and Noether's theorem

[CAF123]

An infinitesimal transformation of position coordinates in a d dimensional Minkowski space may be written as $$x^{'\mu} = x^{\mu} + \omega_a \frac{\delta x^{\mu}}{\delta \omega_a}$$ The corresponding change in some field defined over the space is $$\Phi '(x') = \Phi(x) + \omega_a \frac{\delta F}{\delta \omega_a},$$ where $\Phi '(x') := F(\Phi(x))$.

The $\left\{\omega_a\right\}$ are a set of infinitesimal parameters. The generator $G_a$ of a symmetry transformation is defined by the following infinitesimal transformation at a same point $$\delta_{\omega} \Phi(x) = \Phi '(x) - \Phi(x) = -i\omega_a G_a \Phi(x)$$

Could someone explain what this equation means? I have been trying to connect to what I worked on in Lie Algebras and the fact that combinations of the generators and the parameters from the identity span the Lie algebra.

Many thanks.

 

[Bill_K]

The field φ is not in general a single component scalar, rather it is a set of components φ_i. And if, for example, the coordinate transformation is a Lorentz transformation, then the φ_i will undergo a linear transformation among themselves, and this is what the operator G_a is meant to indicate.

 

[CAF123]

Hi Bill_K,

The field φ is not in general a single component scalar, rather it is a set of components φ_i.

Yes, so do you mean to say $\Phi$ may be a vector field or a scalar field over the space? So $\Phi$ is a mapping $\Phi : \mathbb{R}^d \rightarrow \mathbb{R}^{d'}$.

And if, for example, the coordinate transformation is a Lorentz transformation, then the φ_i will undergo a linear transformation among themselves, and this is what the operator G_a is meant to indicate.

I have the following definition: 'The generator G_a of an infinitesimal transformation depends on parameters $\omega_a$ which relates the changed and unchanged field at the same position.'
The left hand side of the equation is $\delta_{\omega} \Phi$ which I took to mean the change in $\Phi$ (i.e in its components) by varying the parameters without changing $x$ (since $\Phi ' (x) - \Phi (x)$ has no $x'$ dependence). But at the moment I am not seeing how this is possible. E.g take $S^1 \subset \mathbb{R}^2$. By varying the angle (or parameter of SO(2)), the position along the circle changes. Or did I misunderstand the notation?

Thanks.

 

[Bill_K]

Yes, so do you mean to say $\Phi$ may be a vector field or a scalar field over the space? (Yes)

I have the following definition: 'The generator G_a of an infinitesimal transformation depends on parameters $\omega_a$ which relates the changed and unchanged field at the same position.'

In ω_a, the subscript a is a collective index, representing different things for different transformations. In the Lorentz transformation example, it's an antisymmetric index pair μν:

x^μ → x^μ + ε λ^μ_ν x^ν

The field components φ_i will transform according to some representation b_i^j:

φ_i → φ_i + ½ ε b_i^j _μ^νλ^μ_ν φ_j

and so in this case the operator G_a represents a matrix G_μ^ν where (G_μ^ν)_i^j = ½ ε b_i^j _μ^νλ^μ_ν

 

[samalkhaiat]

Hi Bill_K,
..
The left hand side of the equation is $\delta_{\omega} \Phi$ which I took to mean the change in $\Phi$ (i.e in its components) by varying the parameters without changing $x$ (since $\Phi ' (x) - \Phi (x)$ has no $x'$ dependence). But at the moment I am not seeing how this is possible. E.g take $S^1 \subset \mathbb{R}^2$. By varying the angle (or parameter of SO(2)), the position along the circle changes. Or did I misunderstand the notation?

Thanks.

$\bar{ \Phi }( x )$ means the value of $\bar{ \Phi }( \bar{ x } )$ at a point, in the barred system, with coordinates value $\bar{ x } = x$. So, the difference $\bar{ \Phi }( x ) - \Phi ( x )$ is like the Lie derivative along the vector field, $\delta x^{ \mu } \partial_{ \mu }$,which generates the transformation.

 

[CAF123]

In ω_a, the subscript a is a collective index, representing different things for different transformations. In the Lorentz transformation example, it's an antisymmetric index pair μν:

x^μ → x^μ + ε λ^μ_ν x^ν

The presence of the ε (assumed to be <<1) is so that the shift from the original coordinate x^μ can be regarded as infinitesimal. Why is that, in a Lorentz transformation, the coordinates transform like so?

$\bar{ \Phi }( x )$ means the value of $\bar{ \Phi }( \bar{ x } )$ at a point, in the barred system, with coordinates value $\bar{ x } = x$. So, the difference $\bar{ \Phi }( x ) - \Phi ( x )$ is like the Lie derivative along the vector field, $\delta x^{ \mu } \partial_{ \mu }$,which generates the transformation.

The definition $\Phi '(x) - \Phi(x) = -iw_aG_a \Phi(x)$ may be related to the first two equations in the OP by noting that, to first order in $w_a$, $$\Phi ' (x') = \Phi(x) + w_a \frac{\delta F}{\delta w_a} (x)= \Phi(x') - w_a \frac{\delta x^{\mu}}{\delta w_a} \partial_{\mu} \Phi(x') + w_a \frac{\delta F}{\delta w_a}(x')\,\,\,\,(1)$$ I can nearly see the manipulations used to derive this.

Write $\Phi (x') = \Phi (x) + w_a \frac{\delta \Phi (x)}{\delta w_a}|_{x=x'}$, rearrange and sub in. But why is the final functional dependence of F x' at the end of (1)?

 

[samalkhaiat]

The presence of the ε (assumed to be <<1) is so that the shift from the original coordinate x^μ can be regarded as infinitesimal. Why is that, in a Lorentz transformation, the coordinates transform like so?
The definition $\Phi '(x) - \Phi(x) = -iw_aG_a \Phi(x)$ may be related to the first two equations in the OP by noting that, to first order in $w_a$, $$\Phi ' (x') = \Phi(x) + w_a \frac{\delta F}{\delta w_a} (x)= \Phi(x') - w_a \frac{\delta x^{\mu}}{\delta w_a} \partial_{\mu} \Phi(x') + w_a \frac{\delta F}{\delta w_a}(x')\,\,\,\,(1)$$ I can nearly see the manipulations used to derive this.

Write $\Phi (x') = \Phi (x) + w_a \frac{\delta \Phi (x)}{\delta w_a}|_{x=x'}$, rearrange and sub in. But why is the final functional dependence of F x' at the end of (1)?

I don’t understand this mess.

We have group of coordinate transformation:
$$\bar{ x } = g( \omega ) \ x$$
Infinitesimally:
$$\bar{ x } = ( 1 + \omega ) \ x$$
or
$$\bar{ x }^{ \mu } = x^{ \mu } + ( \omega \cdot x )^{ \mu }$$
If the group is Lorentz, this becomes
$$\bar{ x } = x^{ \mu } + \omega^{ \mu }{}_{ \nu } \ x^{ \nu }$$

For arbitrary group of transformations, we write the above infinitesimal transformations as

$$x^{ \mu } \rightarrow \bar{ x }^{ \mu } = x^{ \mu } + \delta x^{ \mu }$$

where

$$\delta x^{ \mu } = f^{ \mu } ( x , \omega )$$

The finite components field, $\phi_{ a }( x )$, transforms by finite-dimensional (matrix) representation, $D( g )$, of the group in question:

$$\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } (g) \phi_{ b } ( x )$$

Infinitesimally, we write this (to first order in $\omega$)

$$\bar{ \phi }_{ a } ( \bar{ x } ) = \left( \delta^{ b }_{ a } + ( \omega \cdot \Sigma )_{ a }{}^{ b } \right) \phi_{ b } ( x ) \ \ (1)$$

where $\Sigma$’s are matrices satisfying the Lie algebra of whatever group you are considering.

In general, we write the above as

$$\phi_{ a } ( x ) \rightarrow \bar{ \phi }_{ a } ( \bar{ x } ) = \phi_{ a } ( x ) + \delta^{ * } \phi_{ a } ( x )$$

Eq(1) can be written as

$$\bar{ \phi }_{ a } ( x + \delta x ) = \left( \delta^{ b }_{ a } + ( \omega \cdot \Sigma )_{ a }{}^{ b } \right) \phi_{ b } ( x )$$

Expanding the left-hand-side to first order, we find

$$\bar{ \phi }_{ a } ( x ) + \delta x^{ \mu } \ \partial_{ \mu } \phi_{ a } ( x ) = \phi_{ a } ( x ) + ( \omega \cdot \Sigma )_{ a }{}^{ b } \ \phi_{ b } ( x )$$

Arrange this to the form

$$\bar{ \phi }_{ a } ( x ) - \phi ( x ) = \{ - \delta^{ b }_{ a } \ \delta x^{ \mu } \ \partial_{ \mu } + ( \omega \cdot \Sigma )_{ a }{}^{ b } \} \ \phi_{ b } ( x ) \ \ \ (2)$$

Sometimes we write this as

$$\delta \phi_{ a } ( x ) = \bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = ( \omega \cdot M )_{ a }{}^{ b } \ \phi_{ b } ( x )$$

Notice, from (1) and (2), you find

$$\delta \phi_{ a }( x ) = \delta^{ * } \phi_{ a } ( x ) - \delta x^{ \mu } \ \partial_{ \mu } \phi_{ a } ( x )$$

I hope this clears up things for you

Sam

 

[CAF123]

Thanks samalkhaiat, that was indeed helpful but I have a few questions in trying to relate what you wrote to the notation that Di Francesco uses. Please criticize:

We have group of coordinate transformation:
$$\bar{ x } = g( \omega ) \ x$$
Infinitesimally:
$$\bar{ x } = ( 1 + \omega ) \ x$$

So, $g(\omega)$ here is some function that transforms the coordinates. That expansion of $g(\omega)$ is for small $\omega$ so, infinitesimally, the new coordinates are slightly offset from the original ones.

$$\bar{ x }^{ \mu } = x^{ \mu } + ( \omega \cdot x )^{ \mu }$$

Why is it necessarily a dot product there?

The finite components field, $\phi_{ a }( x )$, transforms by finite-dimensional (matrix) representation, $D( g )$, of the group in question:

$$\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } (g) \phi_{ b } ( x )$$

To get some insight: Let $w$ be the angle parametrising the group manifold of SO(2). Then in $\mathbb{R}^2$, $\Phi$ is a 2 component field acted upon by a 2x2 rotation matrix, that matrix being D in this case. So the components of $\Phi$ transform under the representation.

Infinitesimally, we write this (to first order in $\omega$)

$$\bar{ \phi }_{ a } ( \bar{ x } ) = \left( \delta^{ b }_{ a } + ( \omega \cdot \Sigma )_{ a }{}^{ b } \right) \phi_{ b } ( x ) \ \ (1)$$

where $\Sigma$’s are matrices satisfying the Lie algebra of whatever group you are considering.

Do you mean to say the $\Sigma$'s are the generators of the Lie algebra?

$$\delta \phi_{ a } ( x ) = \bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = ( \omega \cdot M )_{ a }{}^{ b } \ \phi_{ b } ( x )$$

Is the M there the generator of the symmetry transformation? So, in the notation used in the OP, M here is G?

Finally, in your notation, what does the function F correspond to?
Many thanks.

 

[samalkhaiat]

Thanks samalkhaiat, that was indeed helpful but I have a few questions in trying to relate what you wrote to the notation that Di Francesco uses. Please criticize:

So, $g(\omega)$ here is some function that transforms the coordinates. That expansion of $g(\omega)$ is for small $\omega$ so, infinitesimally, the new coordinates are slightly offset from the original ones.

$g( \omega )$ is an element of matrix group. So it is at least $4 \times 4$ matrix with entries that are function of the group parameters $\omega$. So the 1 in the linear expansion is the unit matrix, and $\omega$ is again $4 \times 4$ matrix with entries that are linear functions of the parameters.

Why is it necessarily a dot product there?

As I just said, omega is a matrix. So it carries indices. In fact, the infinitesimal group generators are hiding in the $\omega$ matrix.

Do you mean to say the $\Sigma$'s are the generators of the Lie algebra?

Yes, they form a representation (appropriate for the field) of the Lie algebra.

Is the M there the generator of the symmetry transformation? So, in the notation used in the OP, M here is G?

Yes it is the full generator. Notice, it consists of the "internal" $\Sigma$-part and an orbital part related to $\delta x^{ \mu }\partial_{ \mu }$.

Finally, in your notation, what does the function F correspond to?
Many thanks.

It should correspond to $D(g) \phi (x)$.

 

[CAF123]

Thanks, do you have any comments about the following:
(c.f Di Francesco's book P.39) The equation that the generators of the transformations satisfy is given by: $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta w_a} \partial_{\mu} \Phi - \frac{\delta F}{\delta w_a},$$ where $\left\{w_a\right\}$ are a set of parameters for the transformation and $G_a$ is the corresponding generator. $F = F(\Phi(\mathbf{x}) )= \Phi'(\mathbf{x'})$ and $\Phi \equiv \Phi(\mathbf{x})$. (Most of this was in the OP, but I repeated it here for clarity).

He then considers some examples. For a translation, $\mathbf{x'} = \mathbf{x} + \mathbf{a}$ and $F = \Phi'(\mathbf{x+a}) = \Phi(\mathbf{x})$. I suppose the last equality there is a supposition (i.e we impose the condition that the field is invariant under translations in the coordinates). In this case, $F = \text{Id}$. I guess this warrants Francesco's statement that $\delta F/\delta w^v = 0$, but this is not general right? It is only in the case when the fields are not affected by the transformation?

Now consider a dilation. $\mathbf{x'} = \lambda \mathbf{x}$ and $F = \Phi'(\lambda \mathbf{x}) = \lambda^{-\Delta} \Phi(\mathbf{x})$, where $\Delta$ is the scaling dimension of the field. Could you explain this last equality?

We can use the first equation above to find the generator of dilations. In this case, $w_a = \lambda$ and $x'^{\mu} - x^{\mu} = \lambda x^{\mu} - x^{\mu} \Rightarrow \delta x^{\mu}/\delta \lambda = x^{\mu}.$ I know that the generator is supposed to be $D = -ix^{\mu} \partial_{\mu}$ which seems to mean that $\delta F/\delta \lambda = 0$ But how so? By chain rule, $$\frac{\delta F}{\delta \lambda^{-\Delta}} \frac{\delta \lambda^{-\Delta}}{\delta \lambda} = -\Delta \lambda^{-1-\Delta} \Phi \neq 0$$

Thanks again.

 

[samalkhaiat]

Thanks, do you have any comments about the following:
(c.f Di Francesco's book P.39) The equation that the generators of the transformations satisfy is given by: $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta w_a} \partial_{\mu} \Phi - \frac{\delta F}{\delta w_a},$$ where $\left\{w_a\right\}$ are a set of parameters for the transformation and $G_a$ is the corresponding generator. $F = F(\Phi(\mathbf{x}) )= \Phi'(\mathbf{x'})$ and $\Phi \equiv \Phi(\mathbf{x})$. (Most of this was in the OP, but I repeated it here for clarity).

That book is not the right place to learn about the transformation theory from. It is good for (and olny for) 2D conformal field theory.

He then considers some examples. For a translation, $\mathbf{x'} = \mathbf{x} + \mathbf{a}$ and $F = \Phi'(\mathbf{x+a}) = \Phi(\mathbf{x})$. I suppose the last equality there is a supposition (i.e we impose the condition that the field is invariant under translations in the coordinates). In this case, $F = \text{Id}$. I guess this warrants Francesco's statement that $\delta F/\delta w^v = 0$, but this is not general right? It is only in the case when the fields are not affected by the transformation?

All fields are invariant under translations.

Now consider a dilation. $\mathbf{x'} = \lambda \mathbf{x}$ and $F = \Phi'(\lambda \mathbf{x}) = \lambda^{-\Delta} \Phi(\mathbf{x})$, where $\Delta$ is the scaling dimension of the field. Could you explain this last equality?

See this thread
www.physicsforums.com/showthread.php?t=172461

I know that the generator is supposed to be $D = -ix^{\mu} \partial_{\mu}$ which seems to mean that $\delta F/\delta \lambda = 0$ But how so?

No it does not mean $\delta F =0$. $x^{ \mu }\partial_{ \mu }$ is the "orbital" part of the generator(see what I said in earlier posts). The full generator is $D = x \cdot \partial + \Delta$. In the link above I explained all these things.

Sam

 

[CAF123]

That book is not the right place to learn about the transformation theory from. It is good for (and olny for) 2D conformal field theory.

I am only covering a very small part of that book for a project on conformal groups over the summer. Do you have any recommendations of books for the more general treatment that you have found useful yourself?

All fields are invariant under translations.

Maybe I am visualizing this incorrectly, but if we consider a vector field in 1D like $\Phi = x \underline{e}_x$. Let x → x + a. Then $\Phi'(x') = (x+a)\underline{e}_x \neq \Phi(x) = x\underline{e}_x$ So the direction of the field is unchanged, but its magnitude has changed.

See this thread
www.physicsforums.com/showthread.php?t=172461


No it does not mean $\delta F =0$. $x^{ \mu }\partial_{ \mu }$ is the "orbital" part of the generator(see what I said in earlier posts). The full generator is $D = x \cdot \partial + \Delta$. In the link above I explained all these things.

Could you point me to the relevant posts? Most of that thread is beyond what I could understand at the moment, so it was difficult for me to find connections with what is discussed here.

Thanks.

 

[samalkhaiat]

I am only covering a very small part of that book for a project on conformal groups over the summer. Do you have any recommendations of books for the more general treatment that you have found useful yourself?

You better off using google. I think, my thread is very elementary introduction to the conformal group.

Maybe I am visualizing this incorrectly, but if we consider a vector field in 1D like $\Phi = x \underline{e}_x$. Let x → x + a. Then $\Phi'(x') = (x+a)\underline{e}_x \neq \Phi(x) = x\underline{e}_x$ So the direction of the field is unchanged, but its magnitude has changed.

Can you tell me how can the magnitude of a vector change by viewing it from translated coordinate system? Translations form an Abilean group. All its representations are (trivial) one-dimensional. So, the transformation $D_{ a }^{ b }$ matrix is simply the unit matrix $\delta^{ b }_{ a }$. So, for a vector field $V^{ \mu }( x )$, you have
$$\bar{ V }^{ \mu }( \bar{ x } ) = \delta^{ \mu }_{ \nu } V^{ \nu } ( x ) = V^{ \mu } ( x )$$

Could you point me to the relevant posts? Most of that thread is beyond what I could understand at the moment, so it was difficult for me to find connections with what is discussed here.

Thanks.

Well, the posts in that thread are connected. So, it is not easy to understand, if you just start from post #9, say. Any way, try reading posts 9,10,11 and 12. You will find that I derived the form of the scale generator (and all other generators) in at least 4 different ways.

 

[CAF123]

No it does not mean $\delta F =0$. $x^{ \mu }\partial_{ \mu }$ is the "orbital" part of the generator(see what I said in earlier posts). The full generator is $D = x \cdot \partial + \Delta$. In the link above I explained all these things.

Ok, I see that the book mentions that the result $D = -ix^{\mu}\partial_{\mu}$ only holds when the fields are invariant under the transformation, in which case F is trivial and indeed $\frac{\delta F}{\delta \lambda} = 0$. When you write $D = x \cdot \partial + \Delta$ is that the same usage of $\Delta$ that I am using?

From my calculation using chain rule in a previous post, I get that $$\frac{\delta F}{\delta \lambda} = -\Delta \lambda^{-\Delta -1} \Phi \Rightarrow D = x \cdot \partial - \Delta \lambda^{-\Delta -1}$$

Thanks.

 

[samalkhaiat]

Ok, I see that the book mentions that the result $D = -ix^{\mu}\partial_{\mu}$ only holds when the fields are invariant under the transformation, in which case F is trivial and indeed $\frac{\delta F}{\delta \lambda} = 0$. When you write $D = x \cdot \partial + \Delta$ is that the same usage of $\Delta$ that I am using?

From my calculation using chain rule in a previous post, I get that $$\frac{\delta F}{\delta \lambda} = -\Delta \lambda^{-\Delta -1} \Phi \Rightarrow D = x \cdot \partial - \Delta \lambda^{-\Delta -1}$$

Thanks.

Ok, I think I am repeating the same stuff again and again.

Look, under the scale transformation

$$\bar{ x }^{ \mu } = e^{ - \lambda } \ x^{ \mu } , \ \ \Rightarrow \ \ \delta x^{ \mu } = - \lambda \ x^{ \mu } ,$$

a field $\phi$ with scale dimension $\Delta$, transforms according to

$$\bar{ \phi } ( \bar{ x } ) = e^{ i \lambda \Delta } \ \phi ( x )$$
or,

$$\bar{ \phi } ( x - \lambda x ) = ( 1 + i \lambda \ \Delta ) \ \phi ( x )$$

Expand the left hand side to first order, you find

$$\bar{ \phi } ( x ) - \lambda \ x^{ \mu } \partial_{ \mu } \phi ( x ) = \phi ( x ) + i \lambda \ \Delta \ \phi ( x )$$
So, arrange this to

$$\delta \phi ( x ) = \bar{ \phi } ( x ) - \phi ( x ) = i \lambda \ ( - i x^{ \mu } \partial_{ \mu } + \Delta ) \ \phi ( x )$$

From this, you read off the generator

$$D = \Delta - i x \cdot \partial$$

Is this clear now?

 

[CAF123]

I could follow most of your steps, but there a couple I do not get and I do not see how they are equivalent to the ones presented above.

Under the scale transformation

$$\bar{ x }^{ \mu } = e^{ - \lambda } \ x^{ \mu } , \ \ \Rightarrow \ \ \delta x^{ \mu } = - \lambda \ x^{ \mu } ,$$

My book simply has $x'^{\mu} = \lambda x^{\mu}$. Why is it $x'^{\mu} = e^{-\lambda}x^{\mu}$?

a field $\phi$ with scale dimension $\Delta$, transforms according to

$$\bar{ \phi } ( \bar{ x } ) = e^{ i \lambda \Delta } \ \phi ( x )$$

Similarly, my book has $\Phi'(x') = \lambda^{-\Delta} \Phi(x)$. Why is it $\Phi'(x') = e^{i\lambda \Delta} \Phi$?

The result I got was derived using these transformation properties of the coordinates and fields from my book. Then I could use the generic equation satisfied by the generators, computing the relevant parts (i.e $\delta F/\delta \lambda$) and then obtain the generator. But the result did not match your result. If you could clarify above, it may help me see the connection.

 

[samalkhaiat]

My book simply has $x'^{\mu} = \lambda x^{\mu}$. Why is it $x'^{\mu} = e^{-\lambda}x^{\mu}$?

There is no difference. You free to scale by any factor. If you DON'T LIKE my convention, you can translate it to your book convention by putting
$$e^{ - \lambda } = \alpha$$

This translate my scaling to yours

$$\bar{ x }^{ \mu } = \alpha x^{ \mu }$$

Similarly, my book has $\Phi'(x') = \lambda^{-\Delta} \Phi(x)$. Why is it $\Phi'(x') = e^{i\lambda \Delta} \Phi$?

Similarly, take my $\Delta$ to be

$$\Delta =- i d$$

This will translate my transformation law to the one you want

$$\bar{ \phi } ( \bar{ x } ) = \left( e^{ - \lambda } \right)^{ - d } \phi ( x ) = \alpha^{ - d } \phi ( x )$$

 

[CAF123]

I see why you used the convention you did. I am trying to follow through the derivation using the convention Di Francesco adopts. So the fields transform $$\phi'(\alpha x) = \phi'(x + \alpha x - x) = \phi'(x + (\alpha - 1)x) \approx \phi'(x) + (\alpha - 1) x^{\mu} \partial_{\mu} \phi(x) = \alpha^{-i\Delta}\phi = \alpha^{-\Delta'}\phi$$ Then in the last term I could write $\alpha^{-\Delta'} = 1 + \alpha^{-\Delta'} - 1$ to obtain $$\delta \phi = (\alpha^{-\Delta} - 1) \phi - (\alpha - 1)x^{\mu} \partial_{\mu} \phi$$ but I can't quite massage this into the form to enable me to determine the corresponding generator.

Can you tell me how can the magnitude of a vector change by viewing it from translated coordinate system?

When you write it like that (i.e taking the passive viewpoint) it makes sense that a coordinate translation should not affect the magnitude of the vector. However, if we were to actively move to another point in space, then intuitively the field should change. Analogy: Weather chart - at different points the wind vector points in different direction and perhaps with a different magnitude. Or perhaps I thought about it in the wrong way.

Thanks again.

 

[samalkhaiat]

I see why you used the convention you did. I am trying to follow through the derivation using the convention Di Francesco adopts. So the fields transform $$\phi'(\alpha x) = \phi'(x + \alpha x - x) = \phi'(x + (\alpha - 1)x) \approx \phi'(x) + (\alpha - 1) x^{\mu} \partial_{\mu} \phi(x) = \alpha^{-i\Delta}\phi = \alpha^{-\Delta'}\phi$$ Then in the last term I could write $\alpha^{-\Delta'} = 1 + \alpha^{-\Delta'} - 1$ to obtain $$\delta \phi = (\alpha^{-\Delta} - 1) \phi - (\alpha - 1)x^{\mu} \partial_{\mu} \phi$$ but I can't quite massage this into the form to enable me to determine the corresponding generator.

Do you know what infinitesimal transformation mean? Infinitesimal transformation means a small deviation $\epsilon$ from the identity transformation. So, $\alpha$ can not be a large number. So, you need to write
$$\alpha = 1 + \epsilon$$
Then you should expand
$$\alpha^{ - \Delta } = 1 - \epsilon \Delta$$

When you write it like that (i.e taking the passive viewpoint) it makes sense that a coordinate translation should not affect the magnitude of the vector. However, if we were to actively move to another point in space, then intuitively the field should change. Analogy: Weather chart - at different points the wind vector points in different direction and perhaps with a different magnitude. Or perhaps I thought about it in the wrong way.

Thanks again.

YES, YOU DID THINK WRONG. The so-called "active" and the so-called "passive" transformations are EQUIVALENT way of realizing the action of the group on space-time. So, when you translate your vector in the "active" way, you actully translate the two end points, and this clearly does not change the magnitude.

 

[CAF123]

I see how I was thinking incorrectly, thanks. Can we now touch upon the decomposition of the full symmetry generator into an 'internal' and a space/orbital part?

The full generators for translational, rotational and dilation transformations are like, respectively, $$P_v = -i\partial_{v}\,\,\,;\,\,\, L^{pv} = i(x^p \partial^{v} - x^v \partial^{p}) + S^{pv}\,\,\,;\,\,\, D = -ix^{\mu} \partial_{\mu} + \Delta,$$ where $S^{pv}$ is a matrix satisfying the Lorentz lie algebra.

I have noticed that when the fields transform trivially, this 'internal' part vanishes and we are left with only the space part in the generator. Is this correct? If so, is there a physical explanation for this? In the case of $P_v$, as I have learnt, all position-independent translations are commutative so the fields always transform trivially, so there is no 'internal' part manifest in the full generator.

The internal part seems to be the generators of the corresponding group Lie algebra, I think (They are the ∑'s in the previous posts).

 

[samalkhaiat]

I see how I was thinking incorrectly, thanks. Can we now touch upon the decomposition of the full symmetry generator into an 'internal' and a space/orbital part?

The full generators for translational, rotational and dilation transformations are like, respectively, $$P_v = -i\partial_{v}\,\,\,;\,\,\, L^{pv} = i(x^p \partial^{v} - x^v \partial^{p}) + S^{pv}\,\,\,;\,\,\, D = -ix^{\mu} \partial_{\mu} + \Delta,$$ where $S^{pv}$ is a matrix satisfying the Lorentz lie algebra.

I have noticed that when the fields transform trivially, this 'internal' part vanishes and we are left with only the space part in the generator. Is this correct? If so, is there a physical explanation for this? In the case of $P_v$, as I have learnt, all position-independent translations are commutative so the fields always transform trivially, so there is no 'internal' part manifest in the full generator.

The internal part seems to be the generators of the corresponding group Lie algebra, I think (They are the ∑'s in the previous posts).

See eq(1) and the one before it in post #7.

(field transform trivially) $\Rightarrow$ (the transformation matrix $D^{ a}_{b}$ is the identity matrix $\delta^{a}_{b}$) $\Rightarrow$ ( $S^{ \mu \nu }$ or $\Sigma^{ ab }$ is zero)

For Lorentz group, such field is called Lorentz’s scalar field, i.e. Spin-zero field or boson.

So for the Lorentz geoup the $S^{ \mu \nu }$ or $\Sigma^{ \mu \nu }$ matrix is the spin matrix of the corresponding field.

For Lorentz vector the spin matrix is
$$( \Sigma^{ \mu \nu } )^{ \rho }{}_{ \sigma } = \eta^{ \mu \rho } \delta^{ \nu }_{ \sigma } - \eta^{ \nu \rho } \delta^{ \mu }_{ \sigma }$$
Can you tell me how I got this? And I want you to work out the spin matrix for rank-2 tensor field.

For scaling (in 4D spacetime) only constant numbers have $\Delta = 0$, Lorentz scalars and vectors have $\Delta = 1$

 

[CAF123]

For Lorentz vector the spin matrix is
$$( \Sigma^{ \mu \nu } )^{ \rho }{}_{ \sigma } = \eta^{ \mu \rho } \delta^{ \nu }_{ \sigma } - \eta^{ \nu \rho } \delta^{ \mu }_{ \sigma }$$
Can you tell me how I got this? And I want you to work out the spin matrix for rank-2 tensor field.

I am not sure how you derived this - did you use the fact the spin matrices satisfy the same commutation relation as the orbital generator $L_{\mu \nu}$?

A couple of things to check: What does the notation $(\Sigma^{\mu \nu})^{a}_{\,\,b}$mean? I realize $\mu, \nu$ are labels for the entries of the matrix rep of $\Sigma$, but then what are a and b?

If we consider a spin 1/2 field, then the spin matrix is 2 dimensional so are possible reps of the generators the scaled Pauli matrices (SU(2) fundamental representation)? So the state space that this spin matrix would act on is spanned by two dimensional vectors. (2D vector space)

For a Lorentz vector, the equation you gave for $\Sigma$ is diagonal, but the Pauli matrices are not diagonal?

 

[samalkhaiat]

I am not sure how you derived this - did you use the fact the spin matrices satisfy the same commutation relation as the orbital generator $L_{\mu \nu}$?

A couple of things to check: What does the notation $(\Sigma^{\mu \nu})^{a}_{\,\,b}$mean? I realize $\mu, \nu$ are labels for the entries of the matrix rep of $\Sigma$, but then what are a and b?

We are not doing well, are we? You have in this thread every thing you need to derive the spin matrix for any field.
Do you remember the equation $\bar{\phi} ( \bar{ x } ) = D ( \omega ) \phi ( x )$?
For lorentz group, then a vector will transforms as
$$\bar{ V }^{ \mu } ( \bar{ x } ) = \Lambda^{ \mu }{}_{ \nu } ( \omega ) \ V^{ \nu } ( x )$$
Now you need the infinitesimal version of $\Lambda$. If you can not do that, you really need to do some serious reading.

If we consider a spin 1/2 field, then the spin matrix is 2 dimensional so are possible reps of the generators the scaled Pauli matrices (SU(2) fundamental representation)? So the state space that this spin matrix would act on is spanned by two dimensional vectors. (2D vector space)

Pauli's spinnors transform in the two dimensional representation of the "Lorentz" group, but Dirac's spinnor transform by the direct sum of the two fundamental Pauli's spinnors. So, it is 4-dimensional.

For a Lorentz vector, the equation you gave for $\Sigma$ is diagonal, but the Pauli matrices are not diagonal?

I did not mention any thing about the nature of the sigma matrices eccept satisfying the Lie algebra of the group in question. For the Lorentz group, $\Sigma^{ \mu \nu }$, are six matrices, $\Sigma^{ \mu \nu }= - \Sigma^{ \nu \mu }$, The extra indices {a & b} depend on the type of field, they number the row and column of the matrix. So for spacetime fields, like vectors and tensors, {a & b} become spacetime indices $\{ \mu , \nu \}$

 

[CAF123]

Do you remember the equation $\bar{\phi} ( \bar{ x } ) = D ( \omega ) \phi ( x )$?
For lorentz group, then a vector will transforms as
$$\bar{ V }^{ \mu } ( \bar{ x } ) = \Lambda^{ \mu }{}_{ \nu } ( \omega ) \ V^{ \nu } ( x )$$
Now you need the infinitesimal version of $\Lambda$.

Yes, so $$\bar{V}^{\mu} = (\delta_{\nu}^{\mu} + w_{ab}(S^{ab})^{\mu}_{\,\,\nu})V^\nu = V^{\mu} + w_{ab}(S^{ab})^{\mu}_{\,\,\nu}V^\nu$$ For the field to transform trivially, then we must have $0 = w_{ab}(S^{ab})^{\mu}_{\,\,\nu}V^\nu$ and $(S^{\mu}_{\nu})^{ab} = \eta^{a \mu} \delta^b_{\nu} - \eta^{b \mu} \delta^a_{\nu}$ does the job.

For the 2nd rank tensor field, it transforms under the Lorentz transformation like $\bar{V}^{\mu}_{\nu} = \Lambda^{\mu}_{\,\,a} \Lambda^{b}_{\,\,\nu} V^a_{\,\,b}$. Expanding each of the $\Lambda$'s to first order, ignoring the second order term in $\omega$ (because $w$ is an infinitesimal angle shift from identity so squaring it means the term becomes negligible) gives $$0 = w_{pq} (S^{pq})^b_{\,\,v}V^{\mu}_{\,\,b} + w_{pq} (S^{pq})^{\mu}_{\,\,a}V^{a}_{\,\,v}$$ or $(S^{pq})^b_{\,\,v} V^{\mu}_{\,\,b} = -(S^{pq})^{\mu}_{\,\,a}V^{a}_{\,\,v}$

How should I proceed from here?

So once they are found, these are the forms of the spin matrix for a vector and tensor field in order that the fields transform trivially under a transformation.

If the fields are not to transform trivially, then the form of the spin matrix changes, hence the spin matrix depends on the field. Correct?

 

[samalkhaiat]

Yes, so $$\bar{V}^{\mu} = (\delta_{\nu}^{\mu} + w_{ab}(S^{ab})^{\mu}_{\,\,\nu})V^\nu = V^{\mu} + w_{ab}(S^{ab})^{\mu}_{\,\,\nu}V^\nu$$ For the field to transform trivially,

Are you following me? Did you read the second line in post #21? How can a Lorentz VECTOR transforms trivially under Lorentz group? It is a VECTOR, is it not? I told you “ONLY SCALARS TRANSFORM TRIVIALLY”.

then we must have $0 = w_{ab}(S^{ab})^{\mu}_{\,\,\nu}V^\nu$ and $(S^{\mu}_{\nu})^{ab} = \eta^{a \mu} \delta^b_{\nu} - \eta^{b \mu} \delta^a_{\nu}$ does the job.

No, $\omega \cdot S \cdot V = 0$, implies $S = 0$, which is the case for scalar not vector. This is how it is done:
$$\bar{ V }^{ \mu } ( \bar{ x } ) = ( \delta^{ \mu }_{ \nu } + \omega^{ \mu }{}_{ \nu } ) V^{ \nu } ( x )$$
$$\delta^{ * } V^{ \mu } = \omega^{ \mu }{}_{ \nu } V^{ \nu } = \omega^{ \rho \sigma } \delta^{ \mu }_{ \rho } \eta_{ \nu \sigma } V^{ \nu }$$
Since omega is antisymmetric, we can write this as
$$\delta^{ * } V^{ \mu } (x) = \frac{ 1 }{ 2 } \omega^{ \rho \sigma } \ ( \delta^{ \mu }_{ \rho } \eta_{ \nu \sigma } - \delta^{ \mu }_{ \sigma } \eta_{ \nu \rho } ) \ V^{ \nu }$$
If you compare this with the general form
$$\delta^{ * } V^{ \mu } (x) = \frac{ 1 }{ 2 } \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma } )^{ \mu }{}_{ \nu } \ V^{ \nu }$$
you find the spin matrix.

For the 2nd rank tensor field, it transforms under the Lorentz transformation like $\bar{V}^{\mu}_{\nu} = \Lambda^{\mu}_{\,\,a} \Lambda^{b}_{\,\,\nu} V^a_{\,\,b}$. Expanding each of the $\Lambda$'s to first order, ignoring the second order term in $\omega$ (because $w$ is an infinitesimal angle shift from identity so squaring it means the term becomes negligible) gives $$0 = w_{pq} (S^{pq})^b_{\,\,v}V^{\mu}_{\,\,b} + w_{pq} (S^{pq})^{\mu}_{\,\,a}V^{a}_{\,\,v}$$ or $(S^{pq})^b_{\,\,v} V^{\mu}_{\,\,b} = -(S^{pq})^{\mu}_{\,\,a}V^{a}_{\,\,v}$

How should I proceed from here?

So once they are found, these are the forms of the spin matrix for a vector and tensor field in order that the fields transform trivially under a transformation.

If the fields are not to transform trivially, then the form of the spin matrix changes, hence the spin matrix depends on the field. Correct?

Never mind tensors. You should get an elementary book on Lorentz group and study how different fields transform. Usually, all books on QFT start with such introduction.
Good luck

 

[CAF123]

$$\bar{ V }^{ \mu } ( \bar{ x } ) = ( \delta^{ \mu }_{ \nu } + \omega^{ \mu }{}_{ \nu } ) V^{ \nu } ( x )$$

..

If you compare this with the general form
$$\delta^{ * } V^{ \mu } (x) = \frac{ 1 }{ 2 } \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma } )^{ \mu }{}_{ \nu } \ V^{ \nu }$$

Are those $\omega$ the same though? The definition which we have (written in post #7) was $\phi'(x') = (\delta^{b}_{a} + (\omega \cdot \Sigma)_{a}^{\,\,\,b})\phi_{b}$ and so $$V'^{\mu} (x') = (\delta^{\mu}_{\nu} + (\omega \cdot \Sigma)_{\nu}^{\,\,\,\mu}) V^{\nu}(x),$$ so I was wondering what happened to the $\Sigma$ term in the top equation.

 

[samalkhaiat]

..

Are those $\omega$ the same though?

Yes. Just the indices gone up. I explained that to you.

The definition which we have (written in post #7) was $\phi'(x') = (\delta^{b}_{a} + (\omega \cdot \Sigma)_{a}^{\,\,\,b})\phi_{b}$ and so $$V'^{\mu} (x') = (\delta^{\mu}_{\nu} + (\omega \cdot \Sigma)_{\nu}^{\,\,\,\mu}) V^{\nu}(x),$$

The dot product in here means $(\omega^{ \rho \sigma } \Sigma_{ \rho \sigma })^{ \mu }{}_{ \nu } = \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma })^{ \mu }{}_{ \nu }$

so I was wondering what happened to the $\Sigma$ term in the top equation.

DIDN'T YOU ASK THIS BEFORE?

Why is it necessarily a dot product there?

AND DIDN'T I ANSWER YOU WITH THIS

As I just said, omega is a matrix. So it carries indices. In fact, the infinitesimal group generators are hiding in the ω matrix.

 

[CAF123]

As I just said, omega is a matrix. So it carries indices. In fact, the infinitesimal group generators are hiding in the $\omega$ matrix.

Yes. Just the indices gone up. I explained that to you. The dot product in here means $(\omega^{ \rho \sigma } \Sigma_{ \rho \sigma })^{ \mu }{}_{ \nu } = \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma })^{ \mu }{}_{ \nu }$

Ok so do you mean to say that $\omega^{\rho \sigma} (\Sigma_{\rho \sigma})^{\mu}_{\,\,\,\nu} \equiv {\omega}^{\mu}_{\,\,\,\nu}$?

If so, it makes sense, it just seems a peculiar notation to me.

Also, in this notation, $\left\{\rho \sigma\right\}$ define the plane of rotation. And $\left\{\mu \nu\right\}$ tells us how the action of the spin matrix transforms the components of the field? From the expression for S it seems to be the case that unless $\rho = \nu$ and $\sigma = \nu$ then the field components will transform trivially. The quantity $\omega^{\rho \sigma} (S_{\rho \sigma})^{\mu}_{\,\,\,\nu}V^{\nu}$ has summations over $\rho, \sigma$ and $\nu$ and the sum over $\nu$ has d terms. The only components of V that transform non-trivially are those lying in the $\left\{\rho \sigma\right\}$ plane. In essence, the field will always transform non-trivially but there are some components that do not transform.
Is that analysis about right?

 

[samalkhaiat]

Ok so do you mean to say that $\omega^{\rho \sigma} (\Sigma_{\rho \sigma})^{\mu}_{\,\,\,\nu} \equiv {\omega}^{\mu}_{\,\,\,\nu}$?

If so, it makes sense, it just seems a peculiar notation to me.

No, it is neither peculiar nor notation. It is a STANDARD IDENTITY. You should have realized this by now.

Also, in this notation, $\left\{\rho \sigma\right\}$ define the plane of rotation. And $\left\{\mu \nu\right\}$ tells us how the action of the spin matrix transforms the components of the field? From the expression for S it seems to be the case that unless $\rho = \nu$ and $\sigma = \nu$ then the field components will transform trivially. The quantity $\omega^{\rho \sigma} (S_{\rho \sigma})^{\mu}_{\,\,\,\nu}V^{\nu}$ has summations over $\rho, \sigma$ and $\nu$ and the sum over $\nu$ has d terms. The only components of V that transform non-trivially are those lying in the $\left\{\rho \sigma\right\}$ plane. In essence, the field will always transform non-trivially but there are some components that do not transform.
Is that analysis about right?

I am sorry, I have done enough in this thread. As I told you, you need to learn about the ABC of tensors.

 

[CAF123]

I am sorry, I have done enough in this thread. As I told you, you need to learn about the ABC of tensors.

Yes, I will be picking up some books this week. I still need to get to grips with the covariant/contra variant formulation of tensors. Anyway, do you mean to say what I wrote is incorrect?

Before you go, one last question:
Let $V^{\mu}$be a vector field defined in a Minkowski spacetime and suppose it transforms under a Lorentz transformation $V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu}$. We can write this like $V'^{\mu} = (e^{i\omega})^{\mu}_{\,\,\,\nu}V^{\nu}$ I think where $\omega$denotes a rotation in some plane spanned by indices $\left\{\rho \sigma\right\}$, say. In 2D Euclidean space time, we can write the matrix representation of $\Lambda$ as $$\begin{pmatrix} \cos \omega & \sin \omega\\-\sin \omega&\cos \omega\end{pmatrix}$$ and in Minkowski space this changes to the 'hyperbolic' rotation. Linearising the above yields $$\begin{pmatrix}1&\omega\\-\omega&1\end{pmatrix} = \text{Id} + \begin{pmatrix} 0&\omega\\-\omega&0\end{pmatrix} = \text{Id} + \omega \begin{pmatrix} 0&1\\-1&0\end{pmatrix}$$

The more general treatment gave $S$ to be $\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu}$. I am wondering how this agrees with the matrix I obtained above multiplying $\omega$. The matrix above is a rep of the generator of the SO(2) rotation group when acting on 2D vectors.

Many thanks.

 

[samalkhaiat]

...Anyway, do you mean to say what I wrote is incorrect?

No, but your remarks were utterly trivial. Of course if you do rotation in the xy-plane, the z-component of a vector will not change. There is no mystery about this.

Before you go, one last question
Let $V^{\mu}$be a vector field defined in a Minkowski spacetime and suppose it transforms under a Lorentz transformation $V'^{\mu} = \Lambda^{\mu}_{\,\,\,\nu}V^{\nu}$. We can write this like $V'^{\mu} = (e^{i\omega})^{\mu}_{\,\,\,\nu}V^{\nu}$ I think where $\omega$denotes a rotation in some plane spanned by indices $\left\{\rho \sigma\right\}$, say. In 2D Euclidean space time, we can write the matrix representation of $\Lambda$ as $$\begin{pmatrix} \cos \omega & \sin \omega\\-\sin \omega&\cos \omega\end{pmatrix}$$ and in Minkowski space this changes to the 'hyperbolic' rotation. Linearising the above yields $$\begin{pmatrix}1&\omega\\-\omega&1\end{pmatrix} = \text{Id} + \begin{pmatrix} 0&\omega\\-\omega&0\end{pmatrix} = \text{Id} + \omega \begin{pmatrix} 0&1\\-1&0\end{pmatrix}$$

The more general treatment gave $S$ to be $\delta^{\mu}_{\rho} \eta_{\sigma \nu} - \delta^{\mu}_{\sigma} \eta_{\rho \nu}$. I am wondering how this agrees with the matrix I obtained above multiplying $\omega$. The matrix above is a rep of the generator of the SO(2) rotation group when acting on 2D vectors.

Many thanks.

Again, you are asking me something which I have already answered. This is exactly what I have done in post #25, but instead of one parameter and one 2x2 generating matrix, we had 6 parameters and 6 4x4 generating matrices.
$$( I + \omega )^{ \mu }{}_{ \nu } = \delta^{ \mu }_{ \nu } + \omega^{ \mu }{ }_{ \nu } = \delta^{ \mu }_{ \nu } + \omega^{ \rho \sigma } ( \Sigma_{ \rho \sigma } )^{ \mu }{}_{ \nu }$$
$( \mu , \nu )$ give you the different matrix elements of $\Sigma$
So, for boosts, we have the 3 matrices $( \Sigma_{ 0 1 } )^{ \mu }{}_{ \nu }$, $( \Sigma_{ 0 2 } )^{ \mu }{}_{ \nu }$ and $( \Sigma_{ 0 3 } )^{ \mu }{}_{ \nu }$
And for 3 rotations, we have the generating matrices $( \Sigma_{ 1 2 } )^{ \mu }{}_{ \nu }$, $( \Sigma_{ 2 3 } )^{ \mu }{}_{ \nu }$ and $( \Sigma_{ 3 1 } )^{ \mu }{}_{ \nu }$

 

[CAF123]

Hi Sam,

The finite components field, $\phi_{ a }( x )$, transforms by finite-dimensional (matrix) representation, $D( g )$, of the group in question:

$$\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } (g) \phi_{ b } ( x )$$

Infinitesimally, we write this (to first order in $\omega$)

$$\bar{ \phi }_{ a } ( \bar{ x } ) = \left( \delta^{ b }_{ a } + ( \omega \cdot \Sigma )_{ a }{}^{ b } \right) \phi_{ b } ( x ) \ \ (1)$$

I understand the derivation you did here, however, conceptually $\Sigma$ is the spin matrix for the field $\phi$, which does not transform the coordinates so why is the LHS of the equation in the barred system?

Also here:

$$\delta \phi_{ a } ( x ) = \bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = ( \omega \cdot M )_{ a }{}^{ b } \ \phi_{ b } ( x )$$

I realize M here is the full symmetry transformation, so it is composed of a piece transforming the coordinates and the field. But the LHS of the equation does not have the coordinate x shifted, we remain at the same coordinate x throughout (i.e we do not have any instances of the barred system in this equation).

Could you provide some thoughts here?
Thanks.

 

[samalkhaiat]

Hi Sam,

I realize M here is the full symmetry transformation, so it is composed of a piece transforming the coordinates and the field. But the LHS of the equation does not have the coordinate x shifted, we remain at the same coordinate x throughout (i.e we do not have any instances of the barred system in this equation).

Could you provide some thoughts here?
Thanks.

Spacetime transformations transform the argument ( x ) of the field and mix its components. I explained all this to you in post#7.

 

[CAF123]

Spacetime transformations transform the argument ( x ) of the field and mix its components. I explained all this to you in post#7.

I understood the mathematical derivation, it is just conceptually I don't see why the equations are the way they are. E.g to be more clear, eqn (1) in post #7. The field is acted on upon by $\Sigma$ which transforms the spin indices of the field. It does nothing to the coordinates as far as I understand. But the LHS of that equation has the field and the coordinates in the barred system, where $\bar x = x + \delta x$. So it seems to be the case that the coordinates have changed. If the $\Sigma$ generators only transform the fields shouldn't we only transform to the barred system of the field?
Thanks in advance for clearing my confusion.

 

[samalkhaiat]

I understood the mathematical derivation, it is just conceptually I don't see why the equations are the way they are. E.g to be more clear, eqn (1) in post #7. The field is acted on upon by $\Sigma$ which transforms the spin indices of the field. It does nothing to the coordinates as far as I understand. But the LHS of that equation has the field and the coordinates in the barred system, where $\bar x = x + \delta x$. So it seems to be the case that the coordinates have changed. If the $\Sigma$ generators only transform the fields shouldn't we only transform to the barred system of the field?
Thanks in advance for clearing my confusion.

Observer $O$ uses system $x$ and calls his fields $\phi_{ a } ( x )$. Observer $\bar{ O }$ uses $\bar{ x }$ and $\bar{ \phi }_{ a } ( \bar{ x } )$. The very same point $P$ in space has two set of coordinates values $x$ and $\bar{x}$.
Is this new to you? If yes, then you need to learn about the use of transformations in physics. If no, then the rest is explained in post#7.

The space-time transformation group connects the two observers and their fields at $P$ as follow
$$\bar{ x }^{ \mu } = g^{ \mu }{}_{ \nu } x^{ \nu } \approx x^{ \mu } + \delta x^{ \mu } ,$$
$$\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } \phi_{ b } ( x ) \approx \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
Substitute the first equation in the second, you get
$$\bar{ \phi }_{ a } ( x + \delta x ) = \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
Now, expand the left-hand-side to first order in $\delta x$ and use the fact that $\delta x \bar{ \phi } \approx \delta x \phi$:
$$\bar{ \phi }_{ a } ( x ) + \delta x^{ \mu } \partial_{ \mu } \phi_{ a } ( x ) = \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
Arrange this to
$$\bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = - \delta x^{ \mu } \partial_{ \mu } \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
Call the left-hand-side $\delta \phi_{ a } ( x )$ and rewrite the above as
$$\delta \phi_{ a } ( x ) = - \delta x^{ \mu } \partial_{ \mu } \phi_{ a } ( x ) + \omega_{ \mu \nu } ( \Sigma^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
Now, ask yourself this: Didn’t I do all this for you in post #7? When I said “I have done every thing you need in this thread” I knew what I was talking about.

Sam