# Infintesimal transformations and Noether's theorem

1. May 27, 2014

### 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.

2. May 27, 2014

### Bill_K

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

3. May 27, 2014

### CAF123

Hi Bill_K,
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 $\Phi : \mathbb{R}^d \rightarrow \mathbb{R}^{d'}$.
I have the following definition: 'The generator Ga 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.

4. May 27, 2014

### Bill_K

In ωa, 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:

xμ → xμ + ε λμν xν

The field components φi will transform according to some representation bij:

φi → φi + ½ ε bij μνλμν φj

and so in this case the operator Ga represents a matrix Gμν where (Gμν)ij = ½ ε bij μνλμν

5. May 27, 2014

### samalkhaiat

$\bar{ \Phi }( x )$ 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.

6. May 28, 2014

### CAF123

The presence of the ε (assumed to be <<1) is so that the shift from the original coordinate xμ 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)?

7. May 28, 2014

### samalkhaiat

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

Last edited: May 28, 2014
8. May 29, 2014

### 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:
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.
Why is it necessarily a dot product there?

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.
Do you mean to say the $\Sigma$'s are the generators 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?

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

Last edited: May 29, 2014
9. May 29, 2014

### samalkhaiat

$g( \omega )$ 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.

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

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

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

10. May 30, 2014

### CAF123

(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.

11. May 30, 2014

### samalkhaiat

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.

All fields are invariant under translations.

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

Last edited: May 30, 2014
12. May 30, 2014

### CAF123

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?
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.

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.

Last edited: May 30, 2014
13. May 30, 2014

### samalkhaiat

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

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

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.

14. Jun 1, 2014

### CAF123

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.

15. Jun 1, 2014

### samalkhaiat

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?

16. Jun 1, 2014

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

17. Jun 1, 2014

### samalkhaiat

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, take my $\Delta$ to be

$$\Delta =- i d$$

This will translate my transformation law to the one you want

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

18. Jun 2, 2014

### 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.

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.

19. Jun 2, 2014

### samalkhaiat

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
$$\alpha^{ - \Delta } = 1 - \epsilon \Delta$$

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.

20. Jun 5, 2014

### 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).

21. Jun 5, 2014

### samalkhaiat

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 $\Delta = 1$

Last edited: Jun 5, 2014
22. Jun 9, 2014

### CAF123

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?

23. Jun 9, 2014

### samalkhaiat

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.

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.
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 $\{ \mu , \nu \}$

24. Jun 10, 2014

### CAF123

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?

Last edited: Jun 10, 2014
25. Jun 10, 2014

### samalkhaiat

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”.

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.

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