Well Poisson's equation, \nabla^2 \phi = \rho, reduces to Laplace's equation ## \nabla^2 \phi = 0 ## at any point where the charge distribution is zero. Since the only charge in your problem iis at the origin, your potential satisfies Laplace's equation everywhere except at ##r=0##, which is...
The idea that quantum gravity implies no global symmetries is a rather widely-held belief, especially among string theorists. See for example this recent paper which made a huge splash by arguing that this is true within AdS/CFT: https://arxiv.org/abs/1810.05337. The first three citations on...
Interesting, thanks. I would guess things get hairier in cases where one doesn't have any sense of a weak-coupling, like certain CFTs. I would also guess that 2d CFTs are at least better understood since mathematicians have spent a lot of time with them, but as far as I'm aware their Hilbert...
Apologies if this is a question which merits its own thread (and if it is, I'm happy to start one), but this isn't always true, correct? One needs some further assumptions (say, a mass gap)?
Yes, viz:
\gamma^0 = i \sigma^y, \qquad \gamma^1 = \sigma^x, \qquad \gamma^2 = \sigma^x.
The gamma matrices are always even dimensional. In d space-time dimensions, the smallest representation has dimension 2^floor(d/2).
Some nice lectures I've found on the cross section of quantum field theory, quantum info, and quantum gravity:
Dan Harlow's Jerusalem lectures: https://arxiv.org/abs/1409.1231
Tom Hartman's course on "Quantum gravity and black holes": http://www.hartmanhep.net/topics2015/
I guess these are both...
This is true in the textbook perturbative string theory, but isn't in believed that the full non-perturbative theory should somehow involve a sum over spacetime topologies? I think of the first paragraph of page 8 of this Witten article: https://arxiv.org/abs/1710.01791.
It was a very esoteric and specific result, but my biggest Eureka moment as a physicist so far was my discovering the proof I published in https://arxiv.org/abs/1610.06568, the short proof given in Section II.B. I didn't believe it when I first wrote it down because it was such a simple and...
I'd forgotten about Taylor, but looking it up I see that I've actually looked at it before when I had questions about scattering, thanks! I was aiming for a higher level but this definitely looks like a good place to start when I have questions.
I'm interested in a book which treats scattering in quantum mechanics aimed at the research-level. I'm particularly interested in a text which focuses on mathematical details such as the analytic structure of the S matrix, the relation between the S matrix and various green's/two-point...
Some good references:
Tom Hartman's lecture notes: http://www.hartmanhep.net/topics2015/
Dan Harlow's lecture notes: https://arxiv.org/abs/1409.1231
The last chapter of Eduardo Fradkin's textbook Field Theories of Condensed Matter Physics concerns entanglement in QFTs and gives an...
It seems like it would be easier to turn the sums in the arguments of the exponential into products of exponentials like you did in the ##B=## case. So you get
$$
Z = \prod_i \sum_{n_i = 0,1} \exp\left( - \beta A n_i - \beta B n_i n_{i+1} \right) \approx \prod_i \sum_{n_i = 0,1} \exp\left( -...
Keep in mind we didn't even know the sign (or finiteness) of the cosmological constant for the first ~80 years of relativistic cosmology, which is to say that it takes very specific and precise experiments to even distinguish small positive/negative/zero cosmological constants from each other...
Yes, I would say this is correct.
This is a great question, and I can't guarantee that I'll answer it in complete detail. To restate your question, if we consider the set of matrices which diagonalize ##H##,
$$
M = \{ S \in SU(2) | D = S^{\dagger} H S \},
$$
you have shown that ##M \in...
Any Hermitian 2x2 matrix may be written
$$
H = a \, \mathbb{I} + b_x \sigma^x + b_y \sigma^y + b_z \sigma^z,
$$
where ##a, b_x, b_y, b_z## are real, and the ##\sigma##'s are the Pauli matrices. The relation between these four constants and the four components of ##H## is very easy to derive...