Need some help with trace calculation in an index theorem

[Othin]

TL;DR

I am having trouble with an apparently simple step in Weinberg's index calculation for Abelian BPS vortices. Specifically, the expression $( A_1\partial_2 - A_2\partial_1)$ prevents me from finding the correct result.

I hate to create a thread for a step in a calculation, by I don't know what else to do. I'm having a lot of trouble reproducing E. Weinberg's index calculation (found here https://inspirehep.net/literature/7539) that gives the dimension of the moduli space generated by BPS solutions in the Maxwell-Higgs model. This calculation revolves around the operators
\begin{equation}
D=\begin{bmatrix}
(\partial_{1} + A_2) & (-\partial_{2} + A_1) & \phi_2 & \phi_1 \\
(\partial_{2} - A_1) & (\partial_{1}+A_2) & -\phi_1 & \phi_2\\
\phi_1 & \phi_2 & -\partial_2 & \partial_1 \\
0 & 0 & \partial_1 & \partial_2
\end{bmatrix}
\end{equation}
and its hermitian adjoint
\begin{equation}
D^{\dagger}=\begin{bmatrix}
(-\partial_{1} + A_2) & (-\partial_{2} - A_1) & \phi_1& 0 \\
(\partial_{2}+ A_1) & (-\partial_{1}+A_2) & \phi_2 & 0\\
\phi_2 & -\phi_1 & \partial_2 & -\partial_1 \\
\phi_1 & \phi_2 & -\partial_1 & -\partial_2
\end{bmatrix}
\end{equation},
where $A_j$ is an electromagnetic vector potential while $\phi_1$ and $\phi_2$ are the real and imaginary parts of a complex scalar field, and the Gauge condition $\partial_1A_1 + \partial_2A_2$ is assumed. At some point, it is stated that $D^{\dagger}D=-\nabla^2 -L_1$, $DD^{\dagger}=-\nabla^2 -L_2$, where $L_i$ are operators satisfying $\rm{Tr}(L_1-L_2)=4(\partial_1A_2 -\partial_2 A_2)$. That's what I'm having trouble with. It should be a straightforward calculation, but I get an extra factor in the result, and I don't see how it is zero. Specifically, I find $(D^{\dagger}D)_{11}=-\nabla^2 + |A|^2 - F_{12} - A_1\partial_2 + A_2\partial_1 + \phi_1^2;$ $(D^{\dagger}D)_{22}=(D^{\dagger}D)_{11} - \phi_1^2 + \phi_2^2;$ $(D^{\dagger}D)_{33}=(D^{\dagger}D)_{44}=-\nabla^2 + (\phi_1^2 + \phi_2^2)$; $(DD^{\dagger})_{11}=-\nabla^2 + |A|^2 + F_{12} + A_1\partial_2 - A_2\partial_1 + \phi_1^2 + \phi_2^2=(DD^{\dagger})_{22};$; $(DD^{\dagger})_{33}=-\nabla^2 +(\phi_1^2 + \phi_2^2)$ and, finally, $(DD^{\dagger})_{33}=-\nabla^2$. Thus, $\rm{Tr}(L_1-L_2)=4(\partial_1A_2 -\partial_2 A_2 + A_1\partial_2 - A_2\partial_1)$. At first I thought I had the signs wrong, but I always end up with this extra term, no matter how many attempts I make. So, is $\mathbf{( A_1\partial_2 - A_2\partial_1)}$ zero in the Coulomb gauge or something like that ? If not, does someone have a clue on what I messed up? It may be something silly, but I really want to understand this.

 

[Gaussian97]

Your computation (at least for $(D^\dagger D)_{11}$ which is the only one I've checked) is wrong.
In general, if you have two operators $A$ and $B$, you can define the operator $(AB)$ by requiring that $$(AB)x=A(Bx), \quad \forall x$$
Then, if your operators have the associative property you can simply write
$$(AB)x=A(Bx)=(A\cdot B) x \Longrightarrow AB = A\cdot B$$
But that's not true in general. In particular, derivatives don't have the associative property, so you can't just multiply $D$ and $D^\dagger$ as normal matrices. You need to compute $D^\dagger(D\psi)$ for an arbitrary $\psi$ and then find the corresponding operator $(D^\dagger D)$ that would give the same result.

 

[Othin]

Your computation (at least for $(D^\dagger D)_{11}$ which is the only one I've checked) is wrong.
In general, if you have two operators $A$ and $B$, you can define the operator $(AB)$ by requiring that $$(AB)x=A(Bx), \quad \forall x$$
Then, if your operators have the associative property you can simply write
$$(AB)x=A(Bx)=(A\cdot B) x \Longrightarrow AB = A\cdot B$$
But that's not true in general. In particular, derivatives don't have the associative property, so you can't just multiply $D$ and $D^\dagger$ as normal matrices. You need to compute $D^\dagger(D\psi)$ for an arbitrary $\psi$ and then find the corresponding operator $(D^\dagger D)$ that would give the same result.

Oh, that's right. Thank you very much, I now have the right result.