# Tag Info

0

I) The first part of OP's construction is directly related to the covariant Hamiltonian formalism for a real scalar field with Lagrangian density $${\cal L} ~=~ \frac{1}{2}\partial_{\alpha} \phi ~\partial^{\alpha} \phi -{\cal V}(\phi), \tag{CW4}$$ see e.g. Ref. [CW] and this Phys.SE post. Here we use the $(+,-,-,-)$ Minkowski signature convention. OP's ...

0

Purely mathematically, a function $f(x)$ (that can be differentiated enough times and so on) can be Taylor expanded around a point $a$ as $$f(x) = f(a) + (x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)+..$$ Now if we're describing a physical system with $f$ and the point $a$ is an equilibrium of the system, $f'(a)=0$. Then we have \begin{...

0

I have talked with people more knowledgeable than I on SUSY, and they agree that something isn't right here. I'm posting my correction to the above in case someone has this problem in the future. This is missing from the errata. We start with (2.60) $$\mathcal{L}=-\frac{1}{2}W^{jk}\psi^{\alpha}_{j}\psi_{\alpha k} + W^{j}F_{j} +h.c.$$ We have the following ...

1

No, many other couplings are possible. For example, in the very simple Lagrangian $$L = \frac{1}{2} mv^2 - mgh$$ we have coupled the particle to the gravitational field $\phi = gh$. This is already in relativistically invariant form, since both $m$ and $\phi$ are scalars. (Of course the real story for coupling to gravity is more complicated, but this works ...

1

This might help you out a bit: In what sense is a quantum field an infinite set of harmonic oscillators? From my understanding, most people think it provides a useful way to conceptualize uncoupled quantum fields physically. It doesn't, however, work for coupled quantum fields. The main problem with this seems to be that infinite harmonic oscillators give ...

0

Sort of, except that you can't generally decompose a rank-2 tensor into a product of rank-1 tensors. Let $\Lambda^{\mu}{}_{\nu}$ be an arbitrary Lorentz transformation. As you probably saw in Peskin, this transformation acts on vectors as $$x^{\mu} \mapsto \Lambda^{\mu}{}_{\nu} x^{\nu}.$$ We can extend this principle to a tensor with an arbitrary number ...

2

Okay, let's give it a try. $SU(2)$ sector of Standard Model Lagrangian is rather involved, so we will take a look at something simpler. Neutron-proton interaction comes to my mind. In low energy limit it is mediated by a massive scalar particle — a pion. We will be very qualitative about this, in reality there are a lot of details. Lagrangian will look ...

0

Because in an isotropic space x is equivalent to -x, this is just Noether theorem. If you want to see how parity violation is explained in the standard model look here

0

Electric and magnetic fields are not relativistically invariant. What you measure will depend on the frame of reference you are in. In your example, the moving charge in frame A will be responsible for both an electric field and a magnetic field. In frame B where the charge is stationary, then an observer would only see an electric field. Exactly the ...

0

Well I'll give it a try. I guess you have to be careful that after transforming you get two different quantized momenta. You're then summing/integrating over $q',q$. Leaving out the normalization constants: \begin{align} H[\tilde{\phi}] &= \int dx \left[k_1 \left( \partial_x \int dq \tilde{\phi}(q)e^{-iqx}\right)\left(\partial_x \int dq' \tilde{\phi}(q'...

0

No you can't do that. The unit vector $\hat{a}_{\rho}$ does not mean: $\rho=1$, $\phi=0$, $z=0$. $\hat{a}_{\rho}$ actually depends on $\phi$ as it's $\left( cos\phi,sin\phi,0 \right)$.

0

Yes but with lots of subtleties to be described in (2). The main questions to be settled are: whether we count just the fields (configuration space) or fields and their derivatives (phase space) whether we count the fermions along with bosons, or separately (I will count them separately), and whether we double the number for them because the Standard Model ...

Top 50 recent answers are included