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2h
answered Why does product of Moduli and Diff x Weyl Variation vanish?
3h
comment Bell State Measurement From Parametric Down Conversion Source
What do you mean exactly by "filter out". The $4$ Bell states are a basis for the $2$ qbit-state, so you may decompose $|\psi\rangle$ on this basis, with corresponding probability amplitudes, and you make a Bell measurement, so your state is projected onto one of the Bell states, with the probability = squared of the probability amplitude modulus.
3h
comment Variational derivatives of strongly connected diagrams functional in gauge theory
@AndrewMcAddams : I posted another answer. Hope it helps.
3h
answered Variational derivatives of strongly connected diagrams functional in gauge theory
4h
comment Variational derivatives of strongly connected diagrams functional in gauge theory
@AndrewMcAddams : Yes, I may provide an other answer, if you want.
5h
comment Variational derivatives of strongly connected diagrams functional in gauge theory
@AndrewMcAddams : Yes, it works with fields $A, \omega=0$. I think that one is talking about vacuum polarization, so, because it is the vacuum, $A,\omega=0$. So it seems logic, no ?
6h
comment Variational derivatives of strongly connected diagrams functional in gauge theory
@AndrewMcAddams : Sorry, my answer is wrong, as you noticed. In fact, there is a vertex ghost, anti-ghost, gluon, so certainly $\dfrac{\delta^2 \Gamma}{\delta \omega_b(y)\delta A_\mu^a(x)}$ is certainly not zero, because the derivative of this quantity relatively to $\bar \omega_c$ gives precisely this vertex . So I deleted my answer. The correct answer is necessary more subtle, I am trying to understand.Sorry again.
1d
comment kitaev-honeycomb : can't get wilson loop squared to yield +1
$w_p$ is a tensorial product (the site indices are different), not a simple product.
1d
comment Origin of phases in amplitudes in QFT
Why do you say that "Amplitudes in QFT are typically real" ? Probability amplitudes are complex numbers.
1d
answered Conservation of energy in a different frame of reference
1d
comment Polchinski equation 4.3.16
OK. I suppose that you have forgotten that the commutator $[c^I,G^g_J]$ is not zero because the anti-commutator $\{c^i,b_J\}$ is not zero. You have then extra-term which cancel with your original $G_m$ term, and other extra-term which gives you an other $-\frac{1}{4}$ factor equals to your original last term, so you get an overall $-\frac{1}{2}$ factor.
Aug
25
comment The Effect of Tortoise Coordinates
No, your potential necessarily has to be expressed in function of $r$, because there is not simple analytic function describing $r(r^*)$. See for instance, in the context of a standard Schwarzschild metrics (but applicable to your case with a function $f$), formulae $(14) \to (24)$ in this paper [be careful, for the formula $(24)$, in this paper $f = 1 - 2m/r$, so this must be changed for the AdS case]
Aug
25
comment Chern-Simons on a lattice and the framing anomaly
If one follow this paper, see formula $(6)$, the zero eigenvalues of the kernel (excluding zeroes due to translation invariance), define a set of planes (co-dimension $1$), the consequence being that the CS action is not integrable.
Aug
25
comment If $S$ is a closed achronal set in a spacetime, any timelike curve starting at a point in $I^+[S]$ and ending at a point in $I^-[S]$ interset $S$?
There is something I don't understand. If I take $S=${$x^2+y^2+z^2 \leq R^2, t=0$}, $S$ is closed, and $I^+(S)$ is disjoint from $S$. However, there are points of the spacetime manifold , for instance ($S'=$ {$x^2+y^2+z^2 > R^2, t=0$}, which do not belong to $S$, nor to $I^+(S)$, nor to $I^-(S)$, and you have timelike curves from $I^-(S)$ to $I^+(S)$ ( for instance {$x^2+y^2+z^2 = R'^2 > R^2, t$}) which do not intersect $S$. So are you speaking about a closed achronal set being a Cauchy surface, or did I miss something ?
Aug
25
comment Non translation invariant correlator in CFT
In my previous comment, I only noticed that, while $\langle\partial \tilde{\phi(z)}\partial \tilde{\phi(w)}\rangle$ is translation-invariant, $\partial \phi$ is not a primary field, and I suggest that this could be in relation with the fact $\langle\tilde{\phi(z)}\tilde{\phi(w)}\rangle$ is not translation-invariant. I have not a definitive argument about that, unfortunately.
Aug
23
comment Non translation invariant correlator in CFT
I suspect that this could be in relation with the fact that $\partial \phi$ is no more a primary field, see page $300$ (while $\partial \phi$ correlators are still translation-invariant)
Aug
23
comment Calculation of Einstein Equation
For latex, you have to put your mathematical stuff into $ \$ \$ $, like $ \$ R_ \, \{\backslash mu \backslash nu \} \$ $
Aug
23
comment Calculation of Einstein Equation
I suggest you read the basics, for instance the chapters $15.1$, $15.2$ of this paper
Aug
22
comment Question about the exclusion principle
The whole state of the $2$ electrons has to be antisymmetric. If the spin part is symmetric, then the spatial part has to be antisymmetric, that is $\psi(x_1,x_2) \sim \psi_1(x_1) \psi_2(x_2) - \psi_2(x_1) \psi_1(x_2)$.
Aug
22
comment How to tell whether photons are entangled?
You have to change the orientation of the axis of the $2$ polarization measurement apparatus, to see quantum correlations which cannot be explained by classical statistical correlations, see Bell's theorem, you may read from the paragraph CHSH inequality to the paragraph Bell inequalities are violated by quantum mechanical predictions[(a, a') and (b,b') are axis orientations]