Problem with Commutator of Gauge Covariant Derivatives?

[tomdodd4598]

Hi there,

I have just read that the gauge field term F_μν is proportional to the commutator of covariant derivatives [D_μ,D_ν]. However, when I try to calculate this commatator, taking the symmetry group to be U(1), I get the following:

$$\left[ { D }_{ \mu },{ D }_{ \nu } \right] =\left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) \left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) -\left( { \partial }_{ \nu }-iq{ A }_{ \nu } \right) \left( { \partial }_{ \mu }-iq{ A }_{ \mu } \right) ={ \partial }_{ \mu }{ \partial }_{ \nu }-{ q }^{ 2 }{ A }_{ \mu }{ A }_{ \nu }-iq\left( { \partial }_{ \mu }{ A }_{ \mu }+{ \partial }_{ \nu }{ A }_{ \mu } \right) -{ \partial }_{ \nu }{ \partial }_{ \mu }+{ q }^{ 2 }{ A }_{ \nu }{ A }_{ \mu }+iq\left( { \partial }_{ \nu }{ A }_{ \mu }+{ \partial }_{ \mu }{ A }_{ \mu } \right) =0$$

So it seems that the commutator is zero, which doesn't seem right... where have I gone wrong?

 

[fresh_42]

If the $A_\mu$ commute, which you have used to get zero as result, then the $D_\mu$ commute as well, and vice versa, since $[D_\mu,D_\nu]=[A_\mu,A_\nu]$.

 

[vanhees71]

I'm not sure, I understand your calculation. You have to apply the operator to something. Let's take for simplicity a scalar field (Klein-Gordon field). Then
$$\mathrm{D}_{\mu} \mathrm{D}_{\nu} \phi=(\partial_{\mu} -\mathrm{i} A_{\mu}) (\partial_{\nu} -\mathrm{i} A_{\nu}) \phi = (\partial_{\mu} \partial_{\nu} -\mathrm{i} q \partial_{\mu} A_{\nu} -\mathrm{i} q A_{\nu} \partial_{\mu} - \mathrm{i} q A_{\mu} \partial_{\nu} - q^2A_{\mu} A_{\nu}) \phi.$$
From this subtract the expression with $\mu$ and $\nu$ exchanged. Finally you get
$$[\mathrm{D}_{\mu},\mathrm{D}_{\nu}]\psi=-\mathrm{i} q (\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}) \psi=-\mathrm{i} q F_{\mu \nu} \psi.$$
You can formally write
$$F_{\mu \nu}=\frac{1}{-\mathrm{i} g} [\mathrm{D}_{\mu},\mathrm{D}_{\nu}],$$
and you can generalize this to the case of non-Abelian gauge fields, where $\phi$ is some multiplet transforming under a representation of the gauge group.

In any case the point is that $F_{\mu \nu}$ (which is a "curvature" in the sense of differential geometry) transforms under the adjoint representation of the gauge group (for the Abelian case that implies that it's simply gauge invariant).

 

[tomdodd4598]

Understood - thanks! Yeh, I was being a little dumb :P

 

[DaniV]

Why $A^\mu\partial^\nu-A^\nu\partial^\mu=0?$

 

[vanhees71]

As it stands, this doesn't make sense. As I said above, a covariant derivative is an operator acting on a field, and to derive operator equations you should apply the operators to a field to see what comes out (see the calculation of the commutator in #3 as an example).

 

[Joseluis2]

6

As it stands, this doesn't make sense. As I said above, a covariant derivative is an operator acting on a field, and to derive operator equations you should apply the operators to a field to see what comes out (see the calculation of the commutator in #3 as an example).

I still don't understand what do you mean.

 

[vanhees71]

What don't you understand in the calculation in #3?

As I said the formula in #5 doesn't make sense. You have to apply the operator to a (scalar) field, and if you do so there's no reason why $(A^{\mu} \partial^{\nu}-A^{\nu} \partial^{\mu})\Phi(x)$ should be 0 for all fields $\Phi$.

 

[stevendaryl]

So it seems that the commutator is zero, which doesn't seem right... where have I gone wrong?

This is a confusing, ambiguous aspect about the meaning of an expression such as $\partial_\mu A_\nu$. In most contexts, this means "take the partial derivative of $A_\nu$". Instead, here it means "operate on [whatever] by $A_\nu$" and then operate on the result with $\partial_\mu$.

The covariant derivative $D_\mu$ is an operator that operates on a function such as $\psi$. So $\partial_\mu$ and $A_\nu$ are themselves considered operators, even though $A_\nu$ is a function, itself. The meaning of multiplication of operators is function composition: If $P$ and $Q$ are operators, then the meaning of $P\ Q$ is a third operator defined by

$P\ Q\ \psi = P\ (Q\ \psi)$

It would be clearer if people explicitly wrote $P \circ Q$ to mean the composition of $P$ and $Q$ as operators, but it's usually clear from context whether it means the composition of $P$ and $Q$ or whether it means the result of $P$ acting on $Q$.

So the meaning of $\partial_\mu A_\nu$ is that operator defined by

$\partial_\mu A_\nu \psi = \partial_\mu (A_\nu \psi) = (\partial_\mu A_\nu) \psi + A_\nu (\partial_\mu \psi)$

which can be written as:

$[(\partial_\mu A_\nu) + A_\mu \partial_\nu] \psi$

where the first term, $(\partial_\mu A_\nu)$ really does mean the result of $\partial_\mu$ acting on $A_\nu$.

So as operators, $\partial_\mu A_\nu = (\partial_\mu A_\nu) + A_\nu \partial_\mu$

I know it's confusing. But it actually comes up in ordinary quantum mechanics. People write:

$[p, x] = p\ x - x\ p = -i \hbar$

So that implies $p\ x = -i \hbar + x\ p$.

But if $p = -i \hbar \frac{\partial}{\partial x}$, then why isn't it the case that $p\ x = -i\hbar$? Why is there a $x \ p$ there on the right-hand side of the equation? Because the meaning of $p\ x$ as an operator doesn't mean $p$ acting on $x$. It means the functional composition of operators $p$ and $x$, which is defined by:

$p\ x \psi = p (x \psi)$

The expression $p (x \psi)$ can be expanded as:
$-i \hbar \frac{\partial}{\partial x} (x \psi) = -i \hbar [(\frac{\partial}{\partial x} x) \psi + x (\frac{\partial}{\partial x} \psi)]$
$=-i \hbar [ \psi + x (\frac{\partial}{\partial x} \psi)]$
$=-i \hbar \psi + x (-i \hbar \frac{\partial}{\partial x} \psi)$
$=(-i \hbar \psi+ x (p \psi)]$
$=(-i \hbar + x\ p) \psi$