8 votes

Why do we "normal-order" instead of just subtracting off vacuum energy?

A normal ordered operatpr has finite matrix elements between any pair of states. Subtracting a constant only gurantees you a finite number for the diagonal matrix element between one specific state ...
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  • 41.8k
5 votes
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Quantum Fields living in finite dimensional non-unitary irreducible representations of the Lorentz group

The finite-dimensional representation is on the target space of the classical fields, not on the space of fields itself. Of course, the "target space" of an operator-valued distribution is a ...
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  • 107k
4 votes

Questions on RG flows, CFTs, and UV and IR theories

The theory with characteristic length $R_\text{c}=0$ is the UV CFT, since it is probing things at the smallest possible length scale, i.e. at the largest possible energy. In other words, its ...
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3 votes

$\Gamma^{1PI}$ contribution to the effective action $\Gamma$

The effective action $\Gamma[\phi_{\rm cl}]$ is the generating functional of 1PI diagrams, cf. e.g. this Phys.SE post. The effective action $\Gamma[\phi_{\rm cl}] = S[\phi_{\rm cl}] +{\cal O}(\hbar)...
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2 votes
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Correct way to expand the generating functional

The two series for $Z_0[J]$ are equivalent. This generating functional is defined by : \begin{align} Z_0[J] &= \mathcal{N} \exp\Biggl\{-\frac{1}{2} \int dz \int dw\ J(z) \Delta_F(z - w) J(w)\Biggr\...
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  • 4,026
2 votes
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On the Ward Identity in QED

Using translation operators the position space amplitude can be written $$\mathcal{M}^\mu(x)\equiv \langle f|j^{\mu}(x)|i\rangle=\langle f|j^{\mu}(0)|i\rangle e^{i(\sum p_f-p_i)x}$$ After Fourier ...
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  • 7,628
2 votes
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Why is the Vacuum state got by a limit to imaginary time?

Given a (non-relativistic) propagator $K_t(A,B)$ giving the 'conditional amplitude' to go from state $B$ to state $A$ in time $t$, it is known that... It is helpful to "look inside" this ...
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  • 7,623
2 votes
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Non-perturbative approach to high-energy physics

In the current day calculations, Lattice-QCD usually refers to calculations in Euclidean time, after wick rotation. The reason for this is that it transforms the usual complex measure (that has lots ...
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  • 5,872
1 vote

Klein-Gordon solution's Fourier image

First some general comments. The Dirac delta distribution you find merely constrains the values of $p$ that span the space of solutions to the Klein-Gordon equation. When you try to solve the KG ...
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1 vote
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Why is it that the Quantised Dirac field spin has non-half integral eigenvalues?

Don't forget that the $\Sigma_i$ here are non-relativistic notions! When you do $p\approx 0$, i.e. the slow approximation, then indeed the number there is always $\approx\frac{1}{2}$, meaning ...
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  • 107k
1 vote

Is randomness in the collapse of the wavefunction in quantum mechanics simply a manifestation of chaos, and not inherently random at all?

My two cents of the euro, you ask: Is randomness in the collapse of the wavefunction in quantum mechanics simply a manifestation of chaos, and not inherently random at all? Do an experiment with a ...
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  • 222k
1 vote

Is randomness in the collapse of the wavefunction in quantum mechanics simply a manifestation of chaos, and not inherently random at all?

No. Even if one can reproduce exactly the initial state and the measurements are ideal (thus even if the initial conditions are exactly identical) the outcomes can be randomly different. Thus, even ...
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1 vote

Operator inversion to get propagator

We have: \begin{equation} \left(p_\mu p_\nu-p^2 g_{\mu\nu}+\frac{n_\mu n_\nu}{\xi}\right)p^\nu p^\rho =\frac{n^\sigma p_\sigma}{\xi}n_\mu p^\rho. \end{equation} \begin{equation} \left(p_\mu p_\nu-p^2 ...
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1 vote

How to determine matching coefficients in a Effective Field Theory?

I don't see how we can determine both coefficients with only the above information. We can't. We need one observable for each EFT quantity that we want to match to the UV. The physical mass $m_\text{...
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1 vote

Feynman rule for trace over fermion loop

traces are not up to the spinor indices though. This is incorrect. Maximum trace path is equivalent to the Dirac Spinors only, hence the fermion propagators being correct is only up to the trace where ...
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