14
votes
If quantum fields are operator valued distributions, why aren't they always smeared?
Yes, the quantum fields must be smeared in order to become well-behaved (symmetric, densely defined) operators (in the Hilbert space of the theory). In mathematically-minded textbooks it is the ...
9
votes
What is an Intuitive example of a Gauge Symmetry?
Another example of of a Gauge Symmetry can be found in the basic $V=mgh$.
Here you can have your "ground" anywhere you want. This freedom reflects the key idea in gauge theory.
6
votes
If quantum fields are operator valued distributions, why aren't they always smeared?
The main point of quantum field theory is to study dynamics. The dynamics is given by interactions among the different fields. These interactions are point interactions. The way that these point ...
6
votes
Accepted
What is an Intuitive example of a Gauge Symmetry?
A Gauge Symmetry (not talking about large gauge transformations) refers to mathematical symmetries that are not physical but rather redundancies in our formulation.
It is a quite rich subject, but I ...
5
votes
Accepted
What does $\mathcal{N}=2$ mean?
The notation $\mathcal{N}=2$ in the context of theoretical physics, particularly in supersymmetry and supergravity, refers to a specific kind of extended supersymmetry. In simple terms, supersymmetry ...
5
votes
Accepted
Is there a standard way of calculating these transformations over creation and annihilation operators?
In the case where $Q$ is Hermitian or anti-Hermitian, it can be diagonalized using the Bogolyubov $u-v$ transform:
$$
a_x = \sum_\alpha (u_{x\alpha}c_\alpha + v_{x\alpha}c_\alpha^*), \quad
a_x^* = \...
4
votes
Accepted
If quantum fields are operator valued distributions, why aren't they always smeared?
Yes, if we are mathematically careful about the distributional nature of a quantum field, then we should indeed consistently write quantum fields only as "smeared functions" $\phi(f)$.
But ...
2
votes
Why don't we study spin-3/2 fields?
There is a theorem by Weinberg that spin-1 massless particles can only enter an interacting theory as gauge bosons and spin-2 massless particles as gravitons. This was extended by Grisaru Pendleton ...
2
votes
Stress-energy tensor spin-1 coupling
If you couple a field to the stress energy tensor in the Lagrangian, then you modify the stress energy tensor. For example, in linearized gravity, the Lagrangian contains a coupling $h_{\mu\nu} T^{\mu\...
2
votes
Accepted
Why can Principal $G$ Bundles be Trivialized when $G = SU(N)$?
A Principal G-Bundle $\pi: P\rightarrow M$ is said to be trivial if it is isomorphic to $M \times G$, which means that a global section exists.
In the case of a simply connected Lie Group G, its ...
2
votes
What is an Intuitive example of a Gauge Symmetry?
A toy model of a gauge theory is
$$ Z ~\propto~\int \! dx ~dy~ e^{iS(x)}, \tag{71.8}$$
cf. Ref. 1. The action $S(x)$ and the path integral $Z$ are invariant under a gauge transformation
$$y\quad\...
1
vote
What is an Intuitive example of a Gauge Symmetry?
Take a potential that only depends on the distance of an object’s center from a center point. Then you can consider a rotation of your object. If this transformation acts locally, so the rotation for ...
1
vote
Accepted
Gauge Boson Self-Interactions with covariant derivative
Your covariant action guess is incomplete. It is missing the gauge invariant
$$\propto \bbox[yellow]{(\partial^\mu A^\nu - \partial^\nu A^\mu) (W_\mu^+ W_\nu^- - c.c.)}\\ \propto [D_\mu,D_\nu](W_\mu^+...
1
vote
Lagrangian for quarks and pions
You have neglected to write down the chiral angle, but that's OK, one may consider the dimensional constant $1/(2f_\pi)$ as the effective angle, since its dimensions are matched by the pions, and ...
1
vote
Fermionic and bosonic degrees of freedom of a vector superfield
I think the issue is that before gauge fixing, the vector $A^\mu$ has 4 offshell degrees of freedom, not 3. That is the extra bosonic degree of freedom you are looking for, to make 8=8. When you gauge-...
1
vote
What are the equation of motion and the energy-momentum relation of a Lagrangian with a quartic potential?
The equations of motion can be found as usual using the Euler-Lagrange equation
$$\frac{\partial L}{\partial\phi} = \partial_\mu\frac{\partial L}{\partial(\partial_\mu\phi)}$$
For $L = \frac12\...
1
vote
If quantum fields are operator valued distributions, why aren't they always smeared?
I am studying these topics and I think that an answer from who is beginning to manage this field, could be helpful because often same questions arise for different people.
You should always think of ...
1
vote
How to interpret Poisson bracket of fields in terms of causality?
Poisson brackets measure the linear independence of two function over classical phase space on the basis of the dependence $m \dot x = p$ in the free case. It boils down to $[x_i,p_k] = \delta_ik$ as ...
1
vote
Problem 3.3 b) of Schwartz's Quantum Field Theory
To get from the first line to the second, you integrate by parts underneath the integral sign. This is a classic technique used in many classical field theory problems. You would have had to do it ...
1
vote
Transformation law of interacting potential in non-inertial frame?
Like knzhou said, this is a scalar field, so nothing happens when you transform to a non-inertial frame.
1
vote
Hamiltonian Field Theory in Peskin & Schroeder
The motivation is "by analogous extension" from mechanics with a finite or countable number of degrees of freedom
$$H = \sum_a \left(p_a(t) \frac{dq^a(t)}{dt} - L(t,q(t),q'(t))\right)$$
to ...
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