# Tagged Questions

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### Momentum operator of a particle in an electromagnetic field

In quantum mechanics, to all observables correspond some self-adjoint operators. In the absence of an electromagnetic field the momentum operator is clearly $\vec{P}:=\frac{\hbar}{i}\vec{\nabla}$. ...
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### Gauge symmetry for p-forms

It is well known that the Lorentz invariance of the S-matrix implies Gauge redundancy for 1-forms,'photons'. Does this argument go through to p-forms? That is does lorentz invariance of s-matrix of ...
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### The gauge covariant derivative and it's substitution

I was wondering wether it would make a difference (in general) if I were to were introduce the gauge covariant derivative $$D_\mu=\partial_\mu+ieA_\mu$$ In the Lagrangian density and then derive the ...
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### Why gauge invariance for electromagnetic fields?

What is the physical constraint that gauge invariance is a required condition for electromagnetic fields? What would happen if the electromagnetic fields were not gauge invariant?
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### Gauge theory in classical electromagnetism

I understand gauge theory as the theory of continuous transformation group which keeps Lagrangian (or dynamics) invariant. So some integral invariants could be found. In terms of classical ...
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### Local gauge invariance and fields

I have one question about local gauge invariance of the spinor and scalar theories. For the scalar complex field with lagrangian $L_{0}$ requirement of local gauge invariance leads us to the ...
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### Lorenz and Coulomb gauge-fixing conditions

Lorenz and Coulomb gauge-fixing conditions. What is physical difference between these two gauge-fixing conditions? Mathematical expression are clear but how to we choose one of these means what they ...
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### How to introduce the electromagnetic field in Quantum Field Theory?

There are many ways to introduce the electromagnrtic field in Quantum Field Theory(QFT), such as canonical quantization method which introduces the creation and annihilation oprators by treating the ...
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### The gauge-invariance of the probability current

It is simple to show that under the gauge transformation \begin{cases}\vec A\to\vec A+\nabla\chi\\ \phi\to\phi-\frac{\partial \chi}{\partial t}\\ \psi\to \psi ...
In classical electromagnetism, we are allowed to use gauge invariance through the argument that the only physical observable fields are the $E$-field and the $B$-field. So in that sense the scalar ...