# Linked Questions

2answers
1k views

### For the Yang-Mills field strength defined as a commutator, why does the $A_\nu\partial_\mu - A_\mu\partial_\nu$ term vanish?

In basically every QFT book the Yang-Mills strength tensor $F_{\mu\nu}$ is defined as $$F_{\mu\nu}=[D_\mu,D_\nu]$$ where $D_\mu$ is the covariant derivative $$D_\mu=\partial_\mu-A_\mu$$ and $A_\mu$ is ...
3answers
179 views

### Is the equation $[\nabla_{\mu},\nabla_{\nu}]=F_{\mu\nu}$ correct? If yes, how does it have to be interpreted?

It seems like simply using the equation $$\nabla_{\mu}=\partial_{\mu}+A_{\mu}$$ isn't enough: One obtains [\nabla_{\mu},\nabla_{\nu}]=\underbrace{[\...
3answers
236 views

### Ordering of differential operators

If we write something like: $\partial_a X_{\mu} \partial^a X^{\mu}$ Does that mean the first derivative is only applied to the first X? ($\partial_a X_{\mu})( \partial^a X^{\mu}$) Or is the ...
1answer
237 views

### Scalar Field Theories

The Lagrangian density for a single real scalar field theory is $$\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)$$ I have often seen this written \...
2answers
389 views

2answers
94 views

1answer
64 views

### Operator $A$ only act on the neighboured state or operator but not the entire expression?

In state vector formalism $A|\psi(x)><u(x)|=(A|\psi(x)>)<u(x)|$, where $A$ only act on $|\psi(x)>$ However, in terms of wave formalism, suppose $A$ is the well known $\frac{d}{dx}$. ...