22
votes
Accepted
How does canonical quantization work with Grassmann variables?
When you are dealing with Grassmann numbers you have a "left derivative" and a "right" derivative. A left derivative removes the variable from the left, a right derivative removes it from the right.
...
16
votes
Are all fields in the universe we know of quantum fields?
Currently all fundamental fields are quantum, except for gravity. For this reason Quantum Gravity is a hot area of research, but the full Quantum Gravity theory has not been developed yet. Why not?
...
15
votes
Accepted
When is Schwartz's method for "integrating out" a field valid?
I find that Schwartz is using the laziest way to discuss the integration of a field. What is going on can be made much more transparent rather easily.
To simplify the discussion, I will assume that ...
14
votes
Does the Newtonian gravitational field have momentum analogous to the Poynting vector?
No, in Newtonian mechanics there is no gravitational momentum. Physically, the reason is that the gravitational field doesn't propagate: it responds instantaneously to changes in the matter ...
13
votes
Invariance of Action vs. Lagrangian in Noether's theorem?
No, they are not the same. To see why, even in classical mechanics, suppose we have symmetry transformation $q \rightarrow q + \epsilon K$ that leaves the Lagrangian invariant. This means that we must ...
13
votes
Accepted
Can Einstein-Hilbert action be derived from symmetry considerations?
Well, one can reason as follows:
One wants a diffeomorphism-invariant action, which must be of the form $$ S=\int d^4x\sqrt{-g}L, $$ where $L$ is a scalar in terms of transformation properties.
...
13
votes
Why do we need to make a tensor for the electromagnetic field?
The main reason why one needs a (2,0)-tensor is the fact that neither the electrical field $\mathbf{E}$ (3d-vector) nor the magnetic field $\mathbf{B}$ (3d-vector) can be fitted into a 4-vector, i.e. ...
13
votes
Gross asymmetry in Maxwell Equations
At the level of the vacuum equations, you can certainly find some form $C$ such that $\star F = dC$ (up to the usual statements about the topology of the underlying space.) You would then have $d(\...
12
votes
Accepted
Counting Degrees of Freedom in Field Theories
You really should split your question. I will answer the part where you do not understand how counting of degrees of freedom work.
Basically we count the number of propagating (physical) degrees of ...
12
votes
Klein gordon field and positive/negative energy solutions
You say you're doing classical field theory, but the terminology comes from QM: these terms are only positive and negative energy if you interpret the field $\phi$ as a wavefunction, as people did ...
12
votes
Accepted
What is "gradient energy" in classical field theory?
The best intuition is from the example of a "loaded string": a set of $N$ masses that are connected by a massless elastic string. If we imagine the limit as $N \to \infty$ but with the ...
11
votes
Accepted
What is the actual form of Noether current in field theory?
Eq. (5) is (up to factors of the infinitesimal parameter $\varepsilon$) the standard expression for the full Noether current. Here:
$\delta x^{\mu}$ is the so-called horizontal component of the ...
10
votes
Why is action a functional of $q$ only?
The notation $q$ in the functional $S[q]$ stands for the whole parametrized curve/path $q:[t_i,t_f]\to \mathbb{R}$, not just a single position. In particular, the parametrized path already carries all ...
10
votes
Accepted
Are fixed points of RG evolution really scale-invariant?
No, dimensionful couplings do not have to be all set to zero at an RG fixed point. An RG fixed point is one where all of the beta functions vanish, and beta functions generally have the form
$$\beta(...
10
votes
Accepted
Hamiltonian Field Theory in Peskin & Schroeder
When you transform from the Lagrangian to the Hamiltonian picture, you necessarily must choose a particular foliation of spacetime -- that is, you must single out a particular time direction, and ...
10
votes
Accepted
Why do we need to make a tensor for the electromagnetic field?
I think it's important to emphasize a point which was never emphasized to me when I was taking my courses. When a theory such as electromagnetism can be formulated in (roughly) equivalent ways at ...
9
votes
Accepted
Why is a nonzero VEV for a spinor field said to break Lorentz invariance?
The $v$ you write is itself a spinor, not a scalar. A non-zero spinor is obviously not invariant under Lorentz transformations, so a non-zero spinorial VEV breaks Lorentz invariance of the 1-point ...
9
votes
Accepted
What is a gauge theory?
A gauge symmetry is simply a symmetry transformation of the action that depends non-trivially on spacetime. You can ask for all physical theories whether such transformations exist. For the case of ...
9
votes
Accepted
When can we handle a quantum field like a classical field?
A clean way to make the concept of a classical field precise it to phrase things in terms of a quantum effective action: given a generating function of connected and renormalised Green functions, $W(J)...
9
votes
Accepted
Uniqueness of the definition of Noether current
In Noether's first theorem, the continuity equation$^1$
$$ d_{\mu} J^{\mu}~\approx~0 \tag{*}$$
is an on-shell equation, i.e. it holds if the EOMs [= Euler-Lagrange (EL) equations] are satisfied. It ...
9
votes
How can a gauge field have physical effects if it only reflects a redundancy in our mathematical description of physical reality?
You just need to phrase both your points more carefully. That changing the gauge has no physical effect does not mean the gauge field does not have any physical effect (for one, since not all possible ...
8
votes
Motivation for covariant phase space
Given a dynamical system $\dot x=F(x)$ with continuously differentiable $F$, there is a canonical bijection between initial conditions at a fixed time and solution trajectories.
If the dynamical ...
8
votes
Accepted
Scalar Field Theory for Gravity
If we linearize the equations of motion about a $\phi = 0$ background (and assume the usual sort of kinetic term), we will find a non-relativistic limit of something like $\nabla^2 \phi \propto \rho^2$...
8
votes
Transformation of $d^4x$ under translation disregarded?
Perhaps the most clear way to see what's going on is to compare the action
$$
S_1[\phi]=\int_{\mathbb R^d} \phi(x)^2\mathrm dx\tag1
$$
to the action
$$
S_2[\phi]=\int_{\mathbb R^d} x^2\phi(x)^2\mathrm ...
8
votes
Accepted
Is the Four-gradient of a Scalar Field a Four-Vector?
The $4$-gradient is a $4$- vector.
Formally, when $x^\mu\to x'^\mu=\Lambda^\mu{}_\nu x^\nu$
$$
\begin{align*}
\partial'_\mu
&=\frac{\partial}{\partial x'^\mu}\\
&=\frac{\partial}{\partial (\...
8
votes
Accepted
Gauge symmetry of massive vector field
Symmetries of the action must be considered without use of the equations of motion. An on-shell symmetry is a vacuous notion - if you use the equations of motion, as you do when using eq. (5) to ...
8
votes
Accepted
Schroedinger equation for wave functional (QFT)
Recall, from Hatfield's textbook (QFT of point particles and strings) & Jackiw's review that the functional equation you are emulating is just that, a functional equation the extension of an ...
7
votes
Accepted
Charged particle as observed from an inertial and a non-inertial frame of reference
First, I'll note that unless the non-inertial frame has a changing acceleration, there is some doubt as to whether it radiates at all.
https://en.wikipedia.org/wiki/...
7
votes
Accepted
Relation between Noether's charge and the generator of a $U(1)$ symmetry
Your confusion resides in the notation. To make it clear I'll use a hat over the operators. The Noether symmetry is
$$
\hat\phi'\equiv\mathrm e^{i\theta\hat Q}\hat\phi\mathrm e^{-i\theta\hat Q}\equiv \...
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