Linked Questions

2
votes
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 ...
2
votes
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 \begin{equation} \nabla_{\mu}=\partial_{\mu}+A_{\mu} \end{equation} isn't enough: One obtains \begin{equation} [\nabla_{\mu},\nabla_{\nu}]=\underbrace{[\...
2
votes
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 ...
0
votes
1answer
237 views

Scalar Field Theories

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

Gauge-covariance of the Yang-Mills field strength $F_{\mu\nu}^a$

Accordingly to Yang-Mills theories, after the introduction of a covariant derivative such that $$D_\mu = \partial_\mu - igA_\mu, \tag1$$ you can built the kinetic term for the gauge potential $A_\...
0
votes
3answers
118 views

Commutation relation of $e^{ikx}$ and $\partial_x$ in Nakahara

I'm reading through Nakahara's Geometry, Topology and Physics and I don't understand the following derivation on pg. 41: $$ \text{Now we find from the commutation relation of } \partial_x \equiv \frac{...
2
votes
1answer
330 views

Local Coordinate Expressions for Lie Derivatives

I'm currently working through the math chapters of Norbert Straumann's book on General Relativity. I have trouble understanding the coordinate expression of the Lie derivative of a basis vector. The ...
1
vote
1answer
312 views

Proof for Negele and Orland equation (2.34)

The equation (2.34) of Negele and Orland has $$\mathcal H_\text{A}(\hat{\mathbf p},\hat{\mathbf x}) = \frac{1}{2m}\left(\hat {\mathbf p} - \frac e c \mathbf A(\hat{\mathbf x})\right)^2.\tag{2.34a}$$ ...
3
votes
1answer
100 views

Schwartz QFT coupling to the photon

I am reading Schwartz's QFT textbook. In Eq. (10.104) he writes: $$ \left[i\partial_\mu-eA_\mu,i\partial_\nu-eA_\nu\right]~=~-e i[\partial_\mu A_{\nu}-\partial_\nu A_{\mu}]~=~ -e i F_{\mu \nu}. \tag{...
2
votes
1answer
108 views

Question about commutators acting on wavefunctions

Consider a commutator acting on a 1D wavefunction: $$[\frac{\hbar}{i} \frac{d}{dx},x]\psi(x)=(\frac{\hbar}{i} \frac{d}{dx}x-x\frac{\hbar}{i} \frac{d}{dx})\psi(x).$$ Now does this mean $\frac{\hbar}{...
1
vote
2answers
94 views

Commutation of position four-vector with spacetime derivatives

I am trying to understand a simple demonstration in Ashok Das' Lectures in QFT. He does the following on p. 134 $$[P_\mu,M_{\nu\lambda}]=[\partial_\mu ,x_\nu\partial_\lambda-x_\lambda\partial_\nu]=\...
3
votes
0answers
114 views

How is the derivative of an operator defined? [duplicate]

Studying quantum mechanics I have encountered several times in proofs this expression: $$ \frac{\mathrm{d}\langle\phi|\hat{H}|\psi\rangle}{\mathrm{d}t} = \biggl\langle\frac{\mathrm{d}\phi}{\mathrm{d}t}...
0
votes
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 $\frac{d}{dx}$. ...