5 votes
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A contradiction in Nonrelativistic Quantum Field Theory

There is a seeming contradiction as the integrands are in fact as you described them ($\boldsymbol{\nabla} \psi ^{*} \cdot \boldsymbol{\nabla} \psi \in \mathbb{R}$ while $\psi ^{*} \nabla ^{2} \psi \...
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  • 709
3 votes

Path Integral Quantization in Peskin and Schroeder

Perhaps P&S's notation would have been clearer if they had written $$\langle \phi_b(\cdot)|e^{-iHT}|\phi_a(\cdot)\rangle$$ $$~=~\int_{\phi(\cdot,0)=\phi_a(\cdot)}^{\phi(\cdot,T)=\phi_b(\cdot)}\,\...
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  • 170k
2 votes
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Coulomb gauge does not fix $\vec A$ uniquely. Is the solution to $\nabla^2{\vec A} = -\mu_0{\vec J}$ given in Griffiths unique?

The Helmholtz Theorem says that a vector field $\mathbf{A}$ is uniquely determined by ${\rm div}\,\mathbf{A}$, ${\rm curl}\,\mathbf{A}$, and $\mathbf{A}\rightarrow0$ as $r\rightarrow\infty$. So, when ...
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2 votes
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What is the intuitive implication behind $L' = L + \frac{df}{dt}$ not affecting equations of motion?

There is the physics SE feature of offering (in the column on the right hand side of the page) suggestions, under the heading 'Related' In this particular case the column 'Related' offers a link to a ...
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2 votes

What is the intuitive implication behind $L' = L + \frac{df}{dt}$ not affecting equations of motion?

I guess the question boils down to what the Lagrangian is in relation to the equations of motion. Yes, exactly. The clue is that the equations of motions for $q(t)$ are equivalent to the principle of ...
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2 votes

Why does Griffiths's book say that there can be no surface current since this would require an infinite electric field for an incident wave?

While Griffiths typically writes surface currents exclusively as $\mathbf{K}_f$, one can also write them as volume currents as $\mathbf{J}_f = \delta(s) \mathbf{K}_f$, where $s = 0$ corresponds to the ...
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1 vote
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Matching conditions for perfect conductors in electro*dynamics*

If we take the definition of a "perfect conductor"1 to be a medium in which $\mathbf{E}(\mathbf{r}, t) = 0$, then the magnetic field in a perfect conductor must be constant in time: $$ \frac{...
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1 vote

Is there more than one GR action (if we include boundary terms)?

The action you call $S_2$ was discovered first! (Or if not first, then at around the same time. Remember we had the EFE before the EH action). This action is known by a few names, the Einstein action, ...
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1 vote

Does this Fermion path integral have a solution?

If we consider the coherent state path integrals, they still enjoy the semigroup property (2), and we don't have to worry about functional determinants. The bosonic coherent state path integral is $$\...
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