21 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. ...
Luthien's user avatar
  • 1,644
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? ...
safesphere's user avatar
  • 12.6k
15 votes

Why can't $p^0$ change sign under a proper orthochronous Lorentz transformation?

Sometimes, diagrams are more useful than equations! Note that $p^2 = - (p^0)^2 + |\vec{p}|^2$ remains unchanged under any Lorentz transformation. Suppose that $p^2 < 0$ (say $p^2 = -1$). A plot of ...
Prahar's user avatar
  • 25.6k
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 ...
Javier's user avatar
  • 28k
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. ...
Bence Racskó's user avatar
13 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 ...
Adam's user avatar
  • 11.8k
12 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 ...
Giuseppe Rossi's user avatar
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 ...
Michael Seifert's user avatar
12 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. ...
Frederic Thomas's user avatar
11 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 ...
Nanashi No Gombe's user avatar
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 ...
Qmechanic's user avatar
  • 200k
11 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 ...
Javier's user avatar
  • 28k
10 votes
Accepted

Classical field limit of the electron quantum field

The classical limit of bosonic quantum mechanical systems with both finite and infinite degrees of freedom is pretty well understood from a mathematical standpoint (with complete rigour, and for quite ...
yuggib's user avatar
  • 11.9k
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 ...
Qmechanic's user avatar
  • 200k
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(...
knzhou's user avatar
  • 102k
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 ...
Zack's user avatar
  • 2,918
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 ...
J. Murray's user avatar
  • 68.6k
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 ...
ACuriousMind's user avatar
  • 124k
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)...
Wakabaloola's user avatar
  • 1,926
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 ...
Qmechanic's user avatar
  • 200k
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 ...
ACuriousMind's user avatar
  • 124k
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 ...
Arnold Neumaier's user avatar
8 votes
Accepted

What is the point of complex fields in classical field theory?

Two real scalar fields $\phi_1$ and $\phi_2$ satisfying an $SO(2)$ symmetry and one complex scalar field $\psi$ are equivalent. However, the latter is more convenient because the particles made by $\...
knzhou's user avatar
  • 102k
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 $\nabla^2 \phi \propto \rho^2$...
Michael Seifert's user avatar
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 ...
AccidentalFourierTransform's user avatar
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 (\...
lineage's user avatar
  • 2,668
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 ...
ACuriousMind's user avatar
  • 124k
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 ...
Cosmas Zachos's user avatar
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/...
Adam V. Steele's user avatar

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