-1
$\begingroup$

Let us consider a classical field theory with gauge fields $A_{\mu}^{a}$ and a scalar $\phi^{a}$ such that the Lagrangian is gauge-invariant under the transformation of

  1. the gauge fields $A_{\mu}^{a}$ in the adjoint representation, with dimension $D_{\bf R}$, of the gauge group $SU(N)$, and
  2. the scalar $\phi^{a}$ in the fundamental representation, with dimension $N$, of the gauge group $SU(N).$

  1. Why can we represent $\phi$ as a traceless Hermitian $N \times N$ matrix, so that $\phi = \phi^{a}T^{a}$ where the $T^a$ are the representation matrices in the fundamental representation?

  2. Why can we write down the variation of $\phi$ under a gauge transformation with gauge parameters $\theta^{a}$ as $$\delta\phi = ig[\theta^{a}T^{a},\phi]$$ and the gauge covariant derivative as $$D_{\mu}\phi = \partial_{\mu}\phi - igA_{\mu}^{a}[T^{a},\phi]?$$

$\endgroup$
  • 1
    $\begingroup$ Which reference? $\endgroup$ – Qmechanic Apr 10 '17 at 6:55
  • 1
    $\begingroup$ It seems that the post (v2) conflates fundamental, adjoint & bifundamental representations. $\endgroup$ – Qmechanic Apr 10 '17 at 7:08
2
$\begingroup$

I am not a quantum field theorist, but I think you are mixing up stuffs, so I am gonna go through some elementary concepts (some of which you might already be familiar with).

The group $\text{SU}(N)$ is a Lie-group, eg. a group, which is also a smooth manifold in the sense, it is a set of continuum cardinality elements and it can be locally parametrized by $\dim_{\mathbb{R}}\text{SU}(N)$ real parameters, in a way that two different parametrizations have smooth transition functions, and the group operations are smooth.

Such groups always admit an associated Lie-algebra, denoted as $\mathfrak{su}(N)$.

Now, $\text{SU}(N)$ is such a Lie-group, that while it exists "abstractly", it can be seen most naturally as the set of all $N\times N$ sized complex matrices, which satisfy $\Lambda^\dagger\Lambda=1$ and $\det\Lambda=1$. For such matrix Lie-groups, the Lie algebra can be seen as a set of $N\times N$ matrices too. These matrices are tangent vectors at the identity, meaning that if $\gamma:\mathbb{R}\rightarrow\text{SU}(N)$ is a smooth curve that goes through the identity at 0 ($\gamma(0)=1$), then $d\gamma/dt|_{t=0}$ is an element of $\mathfrak{su}(N)$.

We have $\gamma(t)\gamma^\dagger(t)=1$ for all $t$, so $$0=\frac{d}{dt}1= \frac{d}{dt}\gamma\gamma^\dagger|_{t=0}=\frac{d\gamma}{dt}|_{t=0}\gamma^\dagger(0)+\gamma(0)\frac{d\gamma^\dagger}{dt}|_{t=0}=\frac{d\gamma}{dt}|_{t=0}+\frac{d\gamma^\dagger}{dt}|_{t=0}=0, $$ so the elements of the Lie algebra are antihermitian matrices.

This is not the only condition though, because we also have the unit determinant condition. We can obtain (from say Jacobi's formula) that this condition also implies that $d\gamma/dt|_{t=0}$ has vanishing trace.

Conclusion: The Lie algebra $\mathfrak{su}(N)$ consists of traceless, antihermitian matrices. This set is closed under addition, scalar multiplication and also under commutators (so if $A,B\in\mathfrak{su}(N)$, then $[A,B]\in\mathfrak{su}(N)$).

Now, as convention dictates, we usually take the Lie algebra to consist of hermitian matrices instead of antihermitean ones, we can do this, because we can always convert an antihermitean matrix into a hermitean one by multiplying with $i$ (and vice versa).


Now to actually answer you question, that the scalar field $\phi^a$ is in the fundamental representation of $\text{SU}(N)$ means that the index $a$ ranges from 1 to $N$ and if we view the $\text{SU}(N)$ elements as $N\times N$ matrices, then they act on $\phi$ by $\phi^a\mapsto\Lambda^a_{\ b}\phi^b$.

Now let $A=(A^a_{\ b})$ be a matrix in $\mathfrak{su}(N)$, the Lie algebra. The adjoint representation of $\text{SU}(N)$ is when $\text{SU}(N)$ is represented on the Lie algebra itself by $A\mapsto\Lambda A\Lambda^{-1}\equiv \Lambda A\Lambda^\dagger$.

The dimension of $\text{SU}(N)$ is $N^2-1$, and the Lie algebra has the same dimension. Which means that the dimension of the representation space of the fundamental representation is $N$ but the dimension of the representation space of the adjoint representation is $N^2-1$, so you cannot use the same set of indices. Let $A,B,...$ take the values $1,...,N^2-1$.

The scalar field then looks like $\phi^a$ but the gauge field looks like $A^A_\mu$.

1) The scalar field cannot be written as you had written, since the indices $a$ and $A$ in general do not have the same range.

2) I don't know what is your source, but I have never ever seen a covariant derivative written like that. The point of the covariant derivative is that if you allow point-dependent gauge transformations on $\phi$, eg. you allow $\phi^a(x)\mapsto\Lambda^a_{\ b}(x)\phi^b(x)$, then the partial derivatives $\partial_\mu$ are not good differential operators on the space of $\phi$-fields, because it is not gauge-covariant.

Instead, if $x^\mu(\lambda)$ is a curve in spacetime, let's assume that the rate of change of $\phi^a$ can be split into two terms, a physical rate of change and a gauge rate of change: $$ \frac{d}{d\lambda}\phi^a=\frac{D}{d\lambda}\phi^a+\frac{\delta}{d\lambda}\phi^a, $$ and since $d\phi^a/d\lambda=\partial_\mu\phi^a\cdot dx^\mu/d\lambda$ by the chain rule, we also have $$ \frac{d}{d\lambda}\phi^a=\frac{\partial}{\partial x^\mu}\phi^a\frac{dx^\mu}{d\lambda}=\frac{dx^\mu}{d\lambda}\left(D_\mu\phi^a+\delta_\mu\phi^a\right). $$ It can be shown that the difference of two differential operators that act on real functions the same way is a linear transform, and we want the physical rate of change $D_\mu$ to act as a differential operator, $\delta_\mu$ must act as a linear transformation, so we have $\delta_\mu\phi^a=-\mathcal{A}_{\mu\ \ b}^{\ a}\phi^b$ for some linear transformation valued vector field $\mathcal{A}_{\mu\ \ b}^{\ a}$.

So we have $D_\mu\phi^a=\partial_\mu\phi^a+\mathcal{A}_{\mu\ \ b}^{\ a}\phi^b$.

But $\mathcal{A}$ must be $\mathfrak{su}(N)$-valued. Why? We say that along a curve $x^\mu(\lambda)$ the field $\phi^a$ is parallel transported, if $d\phi^a/d\lambda=-\mathcal{A}_{\mu\ \ b}^{\ a}\phi^b\frac{dx^\mu}{d\lambda}$, eg. the physical rate of change is zero.

The field $\phi^a$ at different points are incomparable, because of point-dependent gauge transformations, but let us take the curve $x^\mu(\lambda)$ to be a closed loop, and let us take a one-parameter family of such loops, $x^\mu(\lambda,\epsilon)$, where $\epsilon$ is a smooth parameter and for $\epsilon=0$, the loop is the trivial loop that stays at a point $p$ and doesn't go anywhere (so $\epsilon$ is a smallness parameter of the loop).

If we parallel transport $\phi^a$ along this loop, then the change in $\phi^a$ must be a gauge transformation, since the field did not change physically, so the difference is $\Lambda(x(\epsilon,\lambda))^a_{\ b}\phi(x)^b-\phi^a(x)$ where $\lambda$ is the endpoint parameter of the loop. The gauge transformation for small $\epsilon$s is then $\Lambda(x(\epsilon,\lambda))=1+\epsilon B(\lambda,\epsilon)$ where $B$ is an element of the Lie algebra, since it is infinitesimally close to the identity.

Now calculating the rate of change of $\phi$ gives $$ \frac{d}{d\lambda}\phi^a=\frac{d}{d\lambda}(1+\epsilon B^a_{\ b}\phi^b)=\epsilon \frac{d}{d\lambda}B^a_{\ b}\phi^b=-\mathcal{A}_{\mu\ \ b}^{\ a}\phi^b \frac{dx^\mu}{d\lambda} $$ and from this we can see that $\mathcal{A}$ must be Lie algebra valued. Then, to switch to hermitean matrices instead of antihermitean ones, we define $igA_\mu=\mathcal{A}_\mu$, and if $\{T_A\}_{A=1}^{N^2-1}$ is a basis for the Lie algebra $\mathfrak{su}(N)$, we have $$ D_\mu\phi=\partial_\mu\phi+igA_\mu^AT_A\phi. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.