Classical Field Theory - Something isn't clicking

[hawkdron496]

TL;DR Summary

I've been working through a textbook on gauge theory and can't quite put my finger on what it feels like I'm missing. I briefly run through the concepts as I understand them and then ask about specific things I'm confused about.

So, recently I've been working through "Classical Theory of Gauge Fields" by Rubakov. I've more-or-less been able to do the exercises as they've come up, but every once in a while I feel like I'm symbol pushing to get the correct answer, or ignoring certain confusions I have in favour of doing computations. I'm not entirely sure what it is that I feel like I'm missing, but I'm going to try to ask it here. Part of my confusion, I think, is that I come from a mostly math background, and physics texts use words in ways that feel slightly different from how I understand them, but seems to assume that I know what they mean by those words.

So, my understanding:

A field theory consists of a field $\phi$, which is a map from spacetime to some vector space ($\mathbb{R}$ for a scalar field, but could be any vector space). You have the lagrangian density, $\mathcal{L}$, which eats a field (and its derivatives) and gives a function $M \rightarrow \mathbb{R}$ that can be integrated over all of spacetime to give you an action.

Now, some lagrangian densities have the property that applying a transformation $T$ to the field doesn't change the action. That is, $\mathcal{L} (T\phi) = \mathcal{L} (\phi)$. However, we can extend this idea, and make $T$ depend on the spacetime point we're at. Then, we need to change any derivatives in the lagrangian density to covariant derivatives, which we can do by introducing a new field, $A$ with its own transformation properties (specifically, if $T$ is in a Lie group $G$, then $A_\mu$ is in its Lie algebra and$A_\mu \mapsto TA_\mu T^{-1} + T\partial_\mu T^{-1}$).

Then, we can construct different theories by saying (for example) "consider an SU(2) gauge theory", pick a representation of SU(2) (which is a group homomorphism from $SU(2)$ to the set of linear operators on some vector space $V$, let our field $\phi$ be a map from spacetime to $V$, and then let $A_\mu$ be an element of the induced representation of the Lie Algebra of SU(2). Then you just pick a lagrangian that's invariant under $$\phi \mapsto T(x) \phi$$ and $$A_\mu \mapsto TA_\mu T^{-1} + T\partial_\mu T^{-1}$$ and you've got yourself an SU(2) gauge theory.

So, that's great, but there are some things that I'm not entirely clear on.
1. It would be really nice to just have had a list of definitions at the start of this book. It feels like there are a lot of examples but very few definitions. Is the broad overview I gave above more-or-less correct?

2. What exactly do we mean by "An xyz field in the abc representation of G"? When I hear the term, say, "scalar field", I think we mean "A map from spacetime to $\mathbb{R}$". However, Rubakov will say things like "Construct all invariants of order up to and including four, for the scalar field in the adjoint representation of the group SU(2)." and then I'm confused. As I understand it, for a field to transform in the adjoint representation of $G$, the field must take values in the Lie algebra of $G$, and then transforms as $g \phi g^{-1}$. Since $SU(2)$'s Lie Algebra is 3D (if I recall correctly), how can $\phi$ take values in it and still be a scalar field?

I believe that these are the best questions I can come up with that get closest to the heart of what I feel I don't understand, but if more questions become clear to me I'll edit them into the post. Thank you in advance for the answers.

 

[renormalize]

Since $SU(2)$'s Lie Algebra is 3D (if I recall correctly), how can $\phi$ take values in it and still be a scalar field?

Rubakov's book is written for physicists, so in this context a "scalar" field is one that is unchanged under a spacetime Lorentz transformation; i.e., the external symmetry group. It can still carry some representation of an internal symmetry like SU(2). If the field also happens to be a scalar under a particular internal symmetry, it can be labeled, e.g., "neutral" or "uncharged" for the case of the electromagnetic U(1) gauge symmetry, or "colorless" in the case of the SU(3) gauge symmetry of chromodynamics. More broadly, a hypothetical "totally" scalar field would be described as carrying trivial representations of both the external and all internal symmetry groups. There are no such scalar fields in the standard model of elementary particle physics.

 

[hawkdron496]

So, a "scalar field" per Rubakov would be a vector (for instance) field on spacetime that at every point is unchanged by the action of a matrix in the Lorentz group? Most things that I'd consider to be Lorentz scalars (proper time, rest mass, etc...) are all still indexless objects that you get from contracting tensors (maps from spacetime to $R$). In the case of "Scalar field in the adjoint representation of $SU(2)$" are we talking about an object with multiple indicies? If the field takes values in the lie algebra of $SU(2)$ I don't understand how it could be an indexless object.

Or by a spacetime Lorentz transformation do we mean $\phi:M\rightarrow \mathbb{R}$ satisfies $\phi(\Lambda^{-1}(x)) = \phi(x)$ (where $\Lambda$ is a Lorentz transformation $M\rightarrow M$)

I'm not sure I fully understand the distinction between an external and internal symmetry. Is an external symmetry one that leaves the components (in a chart) of the field unchanged, and an internal symmetry one that leaves the action unchanged?

 

[renormalize]

Or by a spacetime Lorentz transformation do we mean $\phi:M\rightarrow \mathbb{R}$ satisfies $\phi(\Lambda^{-1}(x)) = \phi(x)$ (where $\Lambda$ is a Lorentz transformation $M\rightarrow M$)

Yes, that's a scalar field in physicists' parlance because of how it transforms under Lorentz transforms (the external symmetry that acts on spacetime). It carries no spacetime indices by definition. But it can still carry internal-symmetry indices. Per Rybakov pg. 12:

The 4-vector field $A_{\mu}(x)$ carries 1 spacetime index, denoted here by a Greek subscript as is often the case in physics texts. In contrast, $\phi (x)$ carries no Greek index and hence is a scalar field. Nevertheless, you can introduce a multiplet of scalar fields that together carry an internal symmetry index (often denoted by Latin indices). The complex example mentioned in the text is $\phi_1(x)+i\phi_2(x)$ which involves 2 real spacetime scalar fields that together carry a representation of the internal abelian symmetry group U(1). A nonabelian version is introduced on pg. 112:

Here, the 2 complex scalar fields are grouped as a 2D column vector that carries the fundamental representation of the internal symmetry SU(2). In contrast, the non-abelian gauge field-tensor $F_{\mu\nu}^{a}$ has 2 spacetime indices $\mu,\nu$ and the single internal index $a$ whose value ranges over $1,2,3$ and so carries the adjoint representation of SU(2). The tensor $F_{\mu\nu}^{a}$ is derived from the SU(2) vector potential $A_{\mu}^{a}$, which is actually a multiplet of 3 spacetime vectors.

 

[hawkdron496]

Here, the 2 complex scalar fields are grouped as a 2D column vector that carries the fundamental representation of the internal symmetry SU(2). In contrast, the non-abelian gauge field-tensor $F_{\mu\nu}^{a}$ has 2 spacetime indices $\mu,\nu$ and the single internal index $a$ whose value ranges over $1,2,3$ and so carries the adjoint representation of SU(2). The tensor $F_{\mu\nu}^{a}$ is derived from the SU(2) vector potential $A_{\mu}^{a}$, which is actually a multiplet of 3 spacetime vectors.

Right, okay. So how do I want to think about these things? Usually when I see something with, say, two lower indicies, I read "this is a thing that eats two tangent vectors at a point and gives me a number". In the case of something with an internal index and two spacetime indicies like $F^a_{\mu \nu}$, how should I read it? "a thing that eats two tangent vectors and (in this case) gives me an element of the SU(2) lie algebra? (or rather, $F^a_{\mu \nu} T^a = F_{\mu \nu}$), so $F^a_{\mu \nu}$ is just a scalar field for each a, but $F_{\mu \nu}$ eats two vectors and returns an element of the SU(2) lie algebra.

 

[renormalize]

Right, okay. So how do I want to think about these things? Usually when I see something with, say, two lower indicies, I read "this is a thing that eats two tangent vectors at a point and gives me a number". In the case of something with an internal index and two spacetime indicies like $F^a_{\mu \nu}$, how should I read it? "a thing that eats two tangent vectors and (in this case) gives me an element of SU(2)?"

That field tensor is not an element of the SU(2) gauge group, it is a 3-component vector that is multiplied by an adjoint element of SU(2) under a gauge transformation. But since you can always contract $F^a_{\mu \nu}(x)$ with the 3 group generators $\gamma^{a}{ }_{bc}$, you might instead call it "a thing that eats two tangent vectors and the gauge-group generators to give an element of SU(2) at each point in spacetime".

 

[renormalize]

(or rather, $F^a_{\mu \nu} T^a = F_{\mu \nu}$), so $F^a_{\mu \nu}$ is just a scalar field for each a, but $F_{\mu \nu}$ eats two vectors and returns an element of the SU(2) lie algebra.

I'm now replying to your edited post #5. No, $F^a_{\mu \nu}$ is not "just a scalar field for each a", rather it is a rank-2, antisymmetric spacetime tensor field for each $a$ because it carries the two indices $\mu,\nu$.

 

[hawkdron496]

I'm now replying to your edited post #5. No, $F^a_{\mu \nu}$ is not "just a scalar field for each a", rather it is a rank-2, antisymmetric spacetime tensor field for each $a$ because it carries the two indices $\mu,\nu$.

I'm sorry, I think I'm confused again. I was under the impression that by definition if we have a rank 2 antisymmetric tensor $F$, and local coordinates $(x^1, ... x^n)$ in an open set $U$, that $F_{\mu \nu} = F(\partial_\mu, \partial_\nu)$ and is a locally defined function $U \rightarrow \mathbb{R}$ on the manifold (I shouldn't have said scalar field, that's not correct).

Then, for a given $a, \mu, \nu$, $F^a_{\mu \nu}$ is a locally defined scalar function, for a given $\mu, \nu$ $F^a_{\mu \nu} t_a$ is an element of the lie algebra, and the object $F$ eats two tangent vectors at a point (say $\partial_\mu, \partial_\nu$) and returns an element of the SU(2) lie algebra ($F_{\mu \nu}^a t_a$).

Is that not correct?

 

[renormalize]

...for a given $a, \mu, \nu$, $F^a_{\mu \nu}$ is a locally defined scalar function...

This statement is the only thing I object to. I would write instead: "for a given $a, \mu, \nu$, $F^a_{\mu \nu}$ is a locally defined single function". From a physicist's perspective, the term scalar is reserved for a single function that happens to transform trivially (is unchanged) under any spacetime Lorentz transformation. In contrast, each component of the tensor $F^a_{\mu \nu}$ is a single function which transforms non-trivially (is changed) under a general Lorentz transformation (as well as under a general SU(2) gauge transformation). In other words, for spacetime transformations, scalar $\Rightarrow$ single-function but single-function $\nRightarrow$ scalar.

 

[haushofer]

Yes, that's a scalar field in physicists' parlance because of how it transforms under Lorentz transforms (the external symmetry that acts on spacetime).

I haven't read the whole topic, so maybe I'm missing something, but isn't there a prime missing on the scalar field in that definition? The definition there seems to say that the field is just constant over spacetime.

Sorry if I confuse things.

 

[hawkdron496]

This statement is the only thing I object to. I would write instead: "for a given $a, \mu, \nu$, $F^a_{\mu \nu}$ is a locally defined single function". From a physicist's perspective, the term scalar is reserved for a single function that happens to transform trivially (is unchanged) under any spacetime Lorentz transformation. In contrast, each component of the tensor $F^a_{\mu \nu}$ is a single function which transforms non-trivially (is changed) under a general Lorentz transformation (as well as under a general SU(2) gauge transformation). In other words, for spacetime transformations, scalar $\Rightarrow$ single-function but single-function $\nRightarrow$ scalar.

Right, okay. So I need to be more careful about the use of the word "scalar". Thank you for clarifying.

I think I may still be missing something, though. Rubakov defines a current:

#\varphi# transforms under the fundamental representation of $SU(n)$, which, if I'm understanding things correctly, means that it's a scalar under Lorentz transformations but has one "internal" index, so if we were, say, considering $SU(2)$:

$$\varphi = \begin{pmatrix} \varphi^1\\\varphi^2 \end{pmatrix}$$

and for $\omega$ in $SU(n)$ transforms as $\varphi \mapsto \omega \varphi$, $D_\mu \varphi \mapsto \omega D_\mu \varphi$.


Then, he writes:

My understanding is that this means that the $SU(n)$ algebra valued 1-form $j_\mu = j_\mu^a T^a$ should transform as:

$$j_\mu \mapsto j_\mu^a \omega T^a \omega^\dagger. \tag 1$$


First of all, my understanding is that, since $\varphi \in \mathbb{C}^n$ each point and $T^a$ is in the fundamental representation of the Lia algebra at each point, then just by multiplying out the matricies, $[\varphi^\dagger T^a D_\mu \varphi - (D_\mu \varphi)^\dagger \ T^a \varphi]$ should be a real number at each point (for a fixed $\mu$). That is, $j^a_\mu$ is a purely imaginary number at each point (for fixed $a$ and $\mu$).


However, when I actually try to do the calculation, I get (noting that for $\omega \in SU(n)$, we have $\omega [\varphi^\dagger T^a D_\mu \varphi - (D_\mu \varphi)^\dagger \ T^a \varphi] = [\varphi^\dagger T^a D_\mu \varphi - (D_\mu \varphi)^\dagger \ T^a \varphi] \omega$):

$$\begin{align}
j_\mu &\mapsto -i[\varphi^\dagger \omega^\dagger T^a \omega D_\mu \varphi - (D_\mu \varphi)^\dagger \omega^\dagger T^a \omega \varphi] T^a
\end{align}$$

and it's very non-obvious to me that this can be transformed into (1). I've tried defining a new orthonormal basis for the $SU(n)$ Lie algebra $k^a = \omega^\dagger T^a \omega$, which gives:
$$\begin{align}
&= -i[\varphi^\dagger k^a D_\mu \varphi - (D_\mu \varphi)^\dagger k^a \varphi] \omega k^a \omega^\dagger\\
&= Ad(\omega) \left(-i[\varphi^\dagger k^a D_\mu \varphi - (D_\mu \varphi)^\dagger k^a \varphi] k^a\right)
\end{align}$$

so if I could show that
$$-i[\varphi^\dagger k^a D_\mu \varphi - (D_\mu \varphi)^\dagger k^a \varphi] k^a = j_\mu^a T^a$$
I'd be done, but it's not clear to me why this would be true (and in fact I think it'd be false).

Am I misunderstanding something still? Or am I just missing a step in the calculation?

 

[renormalize]

$[\varphi^\dagger T^a D_\mu \varphi - (D_\mu \varphi)^\dagger \ T^a \varphi]$ should be a real number at each point (for a fixed $\mu$). That is, $j^a_\mu$ is a purely imaginary number at each point (for fixed $a$ and $\mu$).

Let me comment on this particular statement. First, starting from $j_{\nu}^{a}=-i\left[\varphi^{\dagger}T^{a}D_{\nu}\varphi-\left(D_{\nu}\varphi\right)^{\dagger}T^{a}\varphi\right]$, hermitian-conjugate to obtain:
\begin{align}
\left(j_{\nu}^{a}\right)^{\dagger} & = +i\left[\left(D_{\nu}\varphi\right)^{\dagger}\left(T^{a}\right)^{\dagger}\varphi-\varphi^{\dagger}\left(T^{a}\right)^{\dagger}D_{\nu}\varphi\right] \nonumber \\
& = -i\left[\varphi^{\dagger}\left(T^{a}\right)^{\dagger}D_{\nu}\varphi-\left(D_{\nu}\varphi\right)^{\dagger}\left(T^{a}\right)^{\dagger}\varphi\right] \nonumber\\
& = j_{\nu}^{a}\nonumber
\end{align} since the group generators are hermitian, $\left(T^{a}\right)^{\dagger}=T^{a}$, as shown for the cases of SU(2) and SU(3) on pgs. 45-46 of Rybakov. Hence, I claim that each component of $j^a_\mu$ is not "a purely imaginary number at each point", rather it is a real number at each point (for fixed $a$ and $\mu$).
Are you with me to this point?

 

[hawkdron496]

Let me comment on this particular statement. First, starting from jνa=−i[φ†TaDνφ−(Dνφ)†Taφ], hermitian-conjugate to obtain:
\begin{align}
\left(j_{\nu}^{a}\right)^{\dagger} & = +i\left[\left(D_{\nu}\varphi\right)^{\dagger}\left(T^{a}\right)^{\dagger}\varphi-\varphi^{\dagger}\left(T^{a}\right)^{\dagger}D_{\nu}\varphi\right] \nonumber \
& = -i\left[\varphi^{\dagger}\left(T^{a}\right)^{\dagger}D_{\nu}\varphi-\left(D_{\nu}\varphi\right)^{\dagger}\left(T^{a}\right)^{\dagger}\varphi\right] \nonumber\
& = j_{\nu}^{a}\nonumber
\end{align} since the group generators are hermitian, (Ta)†=Ta, as shown for the cases of SU(2) and SU(3) on pgs. 45-46 of Rybakov. Hence, I claim that each component of jμa is not "a purely imaginary number at each point", rather it is a real number at each point (for fixed a and μ).
Are you with me to this point?

Yep, I follow. In retrospect that should have been clear given that we're subtracting a number from its conjugate and then multiplying by $i$.