# Tag Info

Accepted

Time ordering $T$ (like any other operator ordering, such as, normal ordering $: ~:$, radial ordering ${\cal R}$, etc.) is technically speaking a linear map from symbols to operators, not a linear map ...
• 172k

### What are the adequate Hilbert spaces for Schrödinger, Schrödinger–Pauli, Dirac equations, and QFT?

Ordinary QM has essentially a fixed Hilbert space of $L^2(\mathbb{R}^n)\otimes S$, where $n$ is the number of spatial dimensions and $S$ some representation of the rotation group. This is due to the ...
• 108k
Accepted

### What's the importance of all four fundamental forces being "curvature"?

When we study non-gravitational fundamental interactions, we distinguish internal symmetries associated with only such interactions from the external symmetries of spacetime. For all fundamental ...
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### Euler-Maclaurin formula for Casimir Effect

Trick 1: Rewrite the integral so it has the bounds you want. \int_0^\infty \nu d \nu = \int_0^1\nu d \nu + \int_1^\infty \nu d \nu = \frac{1}{2} + \int_1^\infty \nu d \nu \end{...
• 35.3k
Accepted

### Is it just a mnemonic to call $\phi (x)|0\rangle$ a particle at position $x$?

$\phi(x)|0\rangle$ is not the state of a particle (I stress that $\phi(x)|0\rangle$ is a one-particle state since I am referring to a free field) with position $x$ (when the temporal component of $x$ ...
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### Euler-Maclaurin formula for Casimir Effect

I want to add something to @Andrews answer. They gave a good answer which is technically correct, but I think there should be at least some things added to Trick 3. In the present answer there are ...
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### Weinberg, Effective Field Theories

Unitarity says that the sum of the probabilities for all possible out states (for a given in state) must be 1. Since the probability is the square of an amplitude, if any individual amplitude has a ...
• 35.3k

### BRST Symmetry and Single Particle States

It isn't obvious from that transformation alone. Remember that in P&S, the forward and backward polarization vectors are defined as: \epsilon ^{\pm}_{\mu} = \frac{1}{\sqrt{2} |\...
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Accepted

### BRST Symmetry and Single Particle States

I will post an answer because I have understood things in a certain way and I would like to share it. In this way, if it is wrong, I will get to know why it is wrong (if someone is kind enough to ...
• 1,587
Accepted

• 231

### Source of spontaneous symmetry breaking

What you need for spontaneous symmetry breaking - and not just in QFT - is an unstable equilibrium, i.e. a potential maximum. The term $-\mu^2 \phi^2$ is the most simple expression for such a ...
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### Muon pair production

You can answer this question by drawing the diagrams for the Mandelstam variables. Below, from Wikipedia, with time increasing to the right: In the s-channel, the intermediate trajectory is timelike, ...
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$$\mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}_I(t') \right] =...$$ Is that just a definition? Seems like it must be, yes. Or perhaps better thought of as ...
• 8,436
Accepted

There is a useful class of dummy integration variable trick, which applies also later when calculating Feynman diagrams. You can make the following observation: $$\frac{1}{x} = \int_0^{+\infty}dt e^{-... • 2,355 3 votes ### Why 't Hooft says: field configuration in Euclidean space that have the vacuum (or a gauge transformation thereof) at the boundary By `vacuum' he means a vanishing gauge field A and by gauge transformation thereof he means some arbitrary A which is related to the vanishing configuration by a non-Abelian gauge transformation. ... • 7,788 3 votes Accepted ### Transverse and longitudinal component of a photon propagator For example, the paragraph below (12.56)  D^{\mu v}(q)=D(q)(g^{\mu v}-\frac{q^\mu q^{v}}{q^2})+\frac{-i}{q^2} \frac{q^\mu q^v}{q^2}says that the first term here corresponding to the transverse ... • 8,436 3 votes ### Relationship between Wilson's RG and the Callan-Symanzik Equation's normalization scale Clearly, Wilson's approach and perturbative RG based on dimensional regularization are NOT the same thing. In Wilson's approach, you use a UV cut off and your flow equations explicitly depend on it. ... • 1,155 2 votes ### Why implementing \partial_\mu A^\mu=0 as an operator equation is not valid? If you write out the functional dependencies in the equation you are referring to, you might see this more clearly. Note that$$ [\partial_i A^i(0,\mathbf{x}), A^\nu(0,\mathbf{y})] = \partial_{x,i} [A^...
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