QFT: Lorentz Trans+ Field infinitesimal variation

[Mishra]

Hello,

I do not understand how to compute the infinitesimal variation of the field at fixed coordinates; under lorentz transformation . I am doing something wrong regarding the transformation of the $x$ coordinate.

I am looking for: $\Delta_a=\phi_a'(x)-\phi_a(x)$, variation appearing in Noether's theorem $j^{\mu}_{\nu}=\frac{\partial L}{\partial \partial\phi}\Delta_a - \epsilon^{\mu}_{\nu}L$

Lorentz transform:
(1): $x^{\mu}\rightarrow x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}=x^{\mu}-\omega^{\mu}_{\nu}x^{\nu}$ 4vector representation
(2): $\phi_a(x)\rightarrow\phi_a'(x')=M^{b}_a\phi_b(x')$ any field representation

From (2) we get:
$\phi_a'(x)=M^{b}_a\phi_b(\Lambda^{-1}x)$

Therefore:
$\Delta_a=\phi_a'(x)-\phi_a(x)$ becomes:
(3): $\Delta_a=M^{b}_a\phi_b(\Lambda^{-1}x)-\phi_a(x)$

Recalling the Lorentz generators:
(4): $M=exp(-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})=1-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}$

I can now plug the generator into into (3):
$\Delta_a= (1-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})^{b}_a\phi_b(\Lambda^{-1}x)-\phi_a(x)$
$\Delta_a= (1-\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})^{b}_a\phi_b(x^{\mu}+\omega^{\mu}_{\nu}x^{\nu})-\phi_a(x)$

Rearranging the terms I end up with:
$\Delta_a=\phi_a(x^{\mu}+\omega^{\mu}_{\nu}x^{\nu})-\phi_a(x) - (\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})^{b}_a\phi_b(x^{\mu}+\omega^{\mu}_{\nu}x^{\nu})$

The first two terms are a simple derivative so:
$\Delta_a=-\omega^{\mu}_{\nu}x^{\nu}\partial_{\mu}\phi_a(x) - (\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})^{b}_a\phi_b(x^{\mu}+\omega^{\mu}_{\nu}x^{\nu})$

And this is where I fail. According to pretty much every book I could find I should simply have:
$\Delta_a=-\omega^{\mu}_{\nu}x^{\nu}\partial_{\mu}\phi_a(x) - (\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})^{b}_a\phi_b(x)$

How do I get rid of this? I am probably doing something wrong at the very beginning where I define either the transformation or the infinitesimal change of the field but I don't understand why... I think I do not understand how the transformation carries the field/coordinates.

Thanks!
VM

 

[samalkhaiat]

Hello,
$\Delta_a=-\omega^{\mu}_{\nu}x^{\nu}\partial_{\mu}\phi_a(x) - (\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu})^{b}_a\phi_b(x^{\mu}+\omega^{\mu}_{\nu}x^{\nu})$
Thanks!
VM

Why didn't you expand $\varphi(x + \omega \cdot x)$ in the last term? Do that and keep only linear terms in $\omega$, because, infinitesimally $\omega^{2} \approx 0$:
$$(\omega \cdot J)\varphi(x + \omega \cdot x) = (\omega \cdot J) \ \varphi(x) + (\omega \cdot J) \ \omega \cdot x \cdot \partial \varphi (x) \approx (\omega \cdot J) \ \varphi (x)$$

 

[Mishra]

How dumb of me.

thank you very much!

 

[Mishra]

I would just have an other question. Is it correct to say that the transformation carries the field this way:
(1): $x^{\mu}\rightarrow x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}=x^{\mu}-\omega^{\mu}_{\nu}x^{\nu}$
(2): $\phi_a(x)\rightarrow\phi_a'(x')=M^{b}_a\phi_b(x')$

We then look at the variation at fixed coordinnates: $\Delta_a=\phi_a'(x)-\phi_a(x)$ by computing: $\phi_a'(x)=M^{b}_a\phi_b(\Lambda^{-1}x)$.

I am asking this because I see a lot of these in various lecture notes:
$x^{\mu}\rightarrow x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}=x^{\mu}-\omega^{\mu}_{\nu}x^{\nu}$
$\phi_a(x)\rightarrow\phi_a'(x')=M^{b}_a\phi_b(x)$ (unprimed $x$ here)

Is this just a common way of writting it (after the expansion you were mentionning) ?

 

[haushofer]

I'd say your own expression equals these textbook expressions. If you replace the unprimed x by (Lamda-1)x in your last expression, and also at the LHS, you see the equality.

 

[vanhees71]

I would just have an other question. Is it correct to say that the transformation carries the field this way:
(1): $x^{\mu}\rightarrow x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}=x^{\mu}-\omega^{\mu}_{\nu}x^{\nu}$
(2): $\phi_a(x)\rightarrow\phi_a'(x')=M^{b}_a\phi_b(x')$

We then look at the variation at fixed coordinnates: $\Delta_a=\phi_a'(x)-\phi_a(x)$ by computing: $\phi_a'(x)=M^{b}_a\phi_b(\Lambda^{-1}x)$.

I am asking this because I see a lot of these in various lecture notes:
$x^{\mu}\rightarrow x'^{\mu}=\Lambda^{\mu}_{\nu}x^{\nu}=x^{\mu}-\omega^{\mu}_{\nu}x^{\nu}$
$\phi_a(x)\rightarrow\phi_a'(x')=M^{b}_a\phi_b(x)$ (unprimed $x$ here)

Is this just a common way of writting it (after the expansion you were mentionning) ?

The latter is the correct definition of a tensor or spinor field (depending on the meaning of the latin indices and the matrix in your equation). This is what's called a local representation of the (proper orthochronous) Lorentz group.

 

[Mishra]

The latter is the correct definition of a tensor or spinor field (depending on the meaning of the latin indices and the matrix in your equation). This is what's called a local representation of the (proper orthochronous) Lorentz group.

This is what I do not understand. When you apply the transformation to $\phi(x)$ why would you only end up with $M\phi(x)$ ?
It seems to me that the coordinates change too and I should get $M\phi(x')$ (where M is a non trivial representation acting on the field).

edit: Also, this would not be consistant with the computations in my original post no ?

 

[samalkhaiat]

This is what I do not understand. When you apply the transformation to $\phi(x)$ why would you only end up with $M\phi(x)$ ?
It seems to me that the coordinates change too and I should get $M\phi(x')$ (where M is a non trivial representation acting on the field).

edit: Also, this would not be consistant with the computations in my original post no ?

Think of $x$ and $\bar{x} = \Lambda x$ as two coordinate values of the same point $P$ of the space-time manifold. So, $\varphi_{a}(x)$ and $\bar{\varphi}_{a}(\bar{x})$ are the values of the field at $P$ as seen in the unbarred and barred systems respectively. The relation between the two is determined by the nature of the index $a$, i.e., the dimension of the representation space to which the field $\varphi$ belongs:
$$\bar{\varphi}_{a} (\bar{x}) = M_{ab}(\Lambda) \ \varphi_{b}(x) , \ \ \ \ \ \ \ \ (1)$$
where $M(\Lambda)$ is the appropriate matrix representation of the Lorentz group:
$$M(\Lambda_{1}) M(\Lambda_{2}) = M(\Lambda_{1}\Lambda_{2}) .$$
You can also rewrite (1) as
$$\bar{\varphi}_{a} (\bar{x}) = M_{ab}(\Lambda) \ \varphi_{b}(\Lambda^{-1}\bar{x}) ,$$
and rename the coordinates as $x$
$$\bar{\varphi}_{a} (x) = M_{ab}(\Lambda) \ \varphi_{b}(\Lambda^{-1}x) .$$
Have a look at the PDH below.

 

[vanhees71]

This is what I do not understand. When you apply the transformation to $\phi(x)$ why would you only end up with $M\phi(x)$ ?
It seems to me that the coordinates change too and I should get $M\phi(x')$ (where M is a non trivial representation acting on the field).

edit: Also, this would not be consistant with the computations in my original post no ?

No, the correct transformation is
$$\phi'(x')=\hat{M} \phi(x)=\hat{M} \phi(\Lambda^{-1} x').$$
This makes much sense, because the field describes a certain physical situation, and the transformation describes how its components change between two (inertial) observers. Each of them uses his space-time coordinates and his components of the field.

In the quantized version the transformation law reads
$$\hat{U} \hat{\phi}(x') \hat{U}^{-1}=\hat{M} \hat{\phi}(\Lambda^{-1} x').$$

 

[Mishra]

The more I think about it the more I realize I am missing something big here.

To me it would seem that the field and the coordinates are tranformed. if point P has coordinates $x'=\Lambda x$ in the two different frames (related by the transformation $\Lambda$). Then the field $\phi(x)$ seen at $x$, is transformed to the field $\phi'(x')$ seen at $x'$.
The "old field" in the old coordinates is transformed into the new field in the new coordinates.

What am I missing here ?

edit:
Just in case, I also found a mistake in my original post. I Somehow found the correct $\phi'(x)-\phi(x)=M\phi(\Lambda^{-1}x)-\phi(x)$ while having equation (2) wrong.

 

[vanhees71]

I thinks, you don't miss anything there. Why do you think, you did?

 

[Mishra]

Well, in such case I should have:

Old field at x = new field at x'
ie: $\phi(x)=M\phi(x')$

 

[vanhees71]

Why that? Take the most simple example of a scalar field first, say the temperature field of the air in your room. It's defined by the temperature values at any point in your room at a given time, which defines a scalar field $T(x)$. Now suppose Alice is moving with some constant velocity relative to your room. She'll measure the temperature too, and according to the rules, how temperature (nowadays!) is defined, she does this with a thermometer momentarily at rest with the air in your room. Then she'll get the same values at any given point, but she's using her own space-time coordinates. So she will get the field
$$T'(x')=T(x).$$
Now you have
$$x' =\Lambda x,$$
with a Lorentz matrix $\Lambda$. Then the above equation reads
$$T'(x')=T(x)=T(\Lambda^{-1} x').$$
Of course, now you can write any variable for the space-time positions, e.g., $x'=y$:
$$T'(y)=T(\Lambda^{-1} y).$$
Of course, now you can also set $y=x$, and you have
$$T'(x)=T(\Lambda^{-1} x).$$
That's a bit confusing in the beginning (at least it confused me a lot when I started to learn about Noether's theorem in field theory), but it's just an issue mathematical notation. Note that in the latter form we have only one set of space-time coordinates on each side, while in the defining equation, we have two sets of space-time coordinates, yours (we then called $x$) and Alice's (we then called $x'$).

The latter form is important, because it explains why the definition of the unitary transformations of the Poincare group in the quantized theory look as they do, i.e., for a scalar field and a boost
$$\hat{\phi}'(x)=\hat{U}(\Lambda) \hat{\phi}(x) \hat{U}^{-1}(\Lambda) = \hat{U}(\Lambda) \hat{\phi}(x) \hat{U}^{\dagger}(\Lambda) = \hat{\phi}(\Lambda^{-1} x).$$

 

[samalkhaiat]

Well, in such case I should have:

Old field at x = new field at x'
ie: $\phi(x)=M\phi(x')$

No, the functional change in the field is given by $\bar{\varphi} = M \varphi$. In group theory language, this means that $\varphi$ and $\bar{\varphi}$ belong to the same representation space (same multiplet). Now, evaluate both side at the point $P$, i.e., $\bar{\varphi} (P)= M \varphi (P)$. If, you are sitting in the $\bar{S}$-system, you take $\bar{x}$ to be the coordinates of the point $P$, while, your friend at the $S$-system uses $x$, so $\bar{\varphi} (\bar{x})= M \varphi (x)$. The transformation matrix $M^{a}{}_{b}$ depends on the nature of the field $\varphi_{a}$ ,i.e., the representation space to which the field belong. For example, a scalar field transforms in the trivial (one dimensional) representation, so $M^{a}{}_{b} = \delta^{a}_{b}$, implying that $\bar{\varphi}_{a}(\bar{x}) = \varphi_{a}(x)$.

 

[Mishra]

I got it now!
You transform the old field you get a new one, in a new frame. This was so simple...

Thanks a lot!

 

[haushofer]

To make it less simple you could now consider passive and active interpretations, and in the context of general relativity. Ho ho ho, merry christmas :P