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4 votes
Accepted

QFT introduction: From point mechanics to the continuum

What do Peskin & Schroeder mean by $\dot{\phi}(\vec{x})$? How can we define a time derivative of a function, which depends on position only? Notation $\dot \phi$, when writing down Lagrangian/...
Ján Lalinský's user avatar
2 votes
Accepted

How to expand $(D_\mu\Phi)^\dagger(D^\mu\Phi)$ in $SU(2)$?

You appear to not appreciate your expression as a row vector dotted on a column vector (possibly sandwiching operators). I corrected your expression to $$\partial_\mu\Phi^\dagger \partial^\mu \Phi - \...
Cosmas Zachos's user avatar
2 votes

Is First-Class Constraint Generator of matter Gauge Symmetry in EM example?

From @Andrew's comment, I know how to solve it. In scalar QED, EM field would couple to current $J^{\mu}(x) = \phi^{\star}(x)\partial^{\mu}\phi(x) - \phi(x)\partial^{\mu}\phi^{\star}(x)$, and the ...
Ting-Kai Hsu's user avatar
2 votes
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Derivation of Noether Current in Condensed Matter Field Theory by Altland and Simons

A&S assumes that the action has a strict$^1$ (rather than a quasi-)symmetry, and so that the bare Noether current (1.43) without improvement terms suffice. -- $^1$ NB: It is implicitly assumed ...
Qmechanic's user avatar
  • 207k
1 vote
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How can a non-derivative interaction involve the derivative of a scalar field?

First, just a small terminology note. People normally say "derivative" instead of "4-derivative" to refer to $\partial_\mu$. People might say "4-gradient" if they want to ...
Andrew's user avatar
  • 51.2k
1 vote

How can a non-derivative interaction involve the derivative of a scalar field?

The kinetic term does not count as an interaction. Quadratic terms in the Lagrangian are present in the free theory. Terms with three or more instances of the field are interactions. Hence, in the ...
Níckolas Alves's user avatar
1 vote

How can a non-derivative interaction involve the derivative of a scalar field?

If $U(\phi)$ is not a quadratic function, then the higher-order terms (e.g. a $\phi^4$ term) can be interpreted as field self-interactions that don’t involve derivatives.
tparker's user avatar
  • 48.4k
1 vote

Is First-Class Constraint Generator of matter Gauge Symmetry in EM example?

The solution to OP's problem is to include a pertinent matter sector in the E&M Lagrangian (3) (in OP's case: a complex scalar $\phi$). This produces the source term in the first-class secondary ...
Qmechanic's user avatar
  • 207k
1 vote

When is the Lagrangian a Lorentz scalar?

An obvious, kind of dumb, answer is that the Lagrangian corresponding to a given Hamiltonian will be a Lorentz scalar if the Hamiltonian has the form, $$ \mathcal{H} = \pi^a \frac{\partial}{\partial t}...
Josh Newey's user avatar
1 vote

When is the Lagrangian a Lorentz scalar?

As far as I know, there are no good ways of stating what conditions on the Hamiltonian will cause the Lagrangian to be a Lorentz scalar other than to just say the Hamiltonian must be derived from a ...
Travis's user avatar
  • 2,167
1 vote
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What kind of object is a function in the context of gauge theory?

The terminology used in physics texts is often a bit imprecise, and thus the word "scalar" is somewhat ambiguous. Often if a function $\phi(x)$ of the coordinates is a "scalar", it ...
Bence Racskó's user avatar
1 vote
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Gauge transformation rule for $dA$, where $A$ is the gauge field

Given $$ A \to g A g^{-1} - i dg g^{-1} $$ where $A$ is the gauge field one-form $A = A_{\alpha \mu} t^\alpha dx^\mu$ and $d$ is exterior derivative, thus $$ dA\\ \to d(g A g^{-1} - i dg g^{-1}) \\...
MadMax's user avatar
  • 4,452
1 vote
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"Mass Shell" Condition on Euclidean Scalar Field

Yes, OP is right: In Euclidean (E) signature, the mass-shell condition is $-E_E^2=E^2_M={\bf p}^2+m^2$, and we're solving an elliptic boundary value problem, which is different from the hyperbolic ...
Qmechanic's user avatar
  • 207k
1 vote
Accepted

Lagrange Multiplier as chemical potential in Lagrangian Density

So I think that this is just a misunderstanding of Lagrange multipliers. But maybe I am mistaken. I find looking back at the strict mathematical definition of the Lagrange multiplier for real valued ...
hijit's user avatar
  • 99
1 vote
Accepted

Is the Dirac adjoint in the representation dual to Dirac spinor?

In order for $\overline{\psi} \psi$ to be a Lorentz scalar, we should have $$\lambda^{\dagger}\gamma^0\lambda = \gamma^0$$ or equivalently (assuming $(\gamma^0)^2 = 1$) $$\lambda^{-1} = \gamma^0\...
MadMax's user avatar
  • 4,452
1 vote
Accepted

2+1-dimensional $SU(N)$ Yang-Mills Theory

From the point of view of perturbation theory, QCD in lower dimensions is essentially identical to QCD in $d=3+1$. Namely, the beta function is negative, which means that the theory is UV complete, ...
AccidentalFourierTransform's user avatar
1 vote

2+1-dimensional $SU(N)$ Yang-Mills Theory

Two-dimensional Yang-Mills is solvable non-perturbatively. Atiyah and Bott used equivariant cohomology to study the theory. The theory is topological and in the small-coupling regime it is classically ...
Simp's user avatar
  • 21

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