# Tag Info

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Yes and no. Electrons - and all other elementary particles - may be viewed as microstates of very tiny black holes. As one considers increasingly heavy elementary particles (e.g. those in the Hagedorn spectrum of string theory), they increasingly morph into black hole microstates. When the elementary particle masses sufficiently surpass the Planck scale, ...

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In a physics of nuclear structure, by the term single particle state is typically understood an excitation, that can be attributed mostly to one proton or one neutron that jumped to a higher orbit. Contrary to collective excitation or collective state, which is an excited level, that many nucleons participate in.

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I) OP is asking about the case where the infinitesimal symmetry transformations $$\delta\phi^a~=~t^a{}_b \phi^a \tag{A}$$ are linear in the fields in the path integral$^1$ $$Z[J] ~=~\exp\left[\frac{i}{\hbar}W_c[J]\right]~=~\int \! {\cal D}\phi~\exp\left[\frac{i}{\hbar}\left( S[\phi]+J_a\phi^a \right)\right]. \tag{B}$$ Recall the Legendre transformation ...

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Hint: Eq. (1.163) represents only the quantum fluctuations in the path integral for the partition function $Z(\beta)={\rm Tr} e^{-\beta H},$ while the $\tanh$ term in eq. (1.164) comes from the classical contributions when integrating over possible boundary conditions for the classical paths, cf. eq. (1.151).

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The name of DMRG is a bit misleading: the modern view is that it does not really have much to do with the idea of real-space renormalization (although that might be a motivation for Steve White to develop it, following the success of Wilson's numerical RG). DMRG is essentially a variational method based on matrix product state(MPS) representation of the ...

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Your statement itself is not quite right. What is not conserved is the chiral current, namely the current of fermions at one of the Weyl nodes. The physics can be understood essentially in one-dimensional version of the Weyl metal: consider a 1D electron gas. There are two Fermi points, and the low-energy theory is given by two "Weyl fermions" in 1D with ...

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Using the rule (3) you can sort the four operators in the order you prefer, by swapping terms. After each swapping, you obtain a piece with four operators, plus an additional one with two operators. And so on. In your case, by swapping the second and the third term, $$a_j a_i a_k^+ a_l^+ = - a_j a_k^+ a_i a_l^+ + \delta_{ki} a_j a_l^+$$ and by swapping ...

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The statement is not true. Counter example: quantum spin ice or U(1) spin liquid. In gapless spin liquid phase, the boson (spin excitations) are emergent U(1) photons in the deconfined phase, which are gapless but not condensed.

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Introducing: $a_ia_k^+=\delta_{ik}-a_k^+a_i$ Replacing 3 in 2 and solving: $\langle K|L\rangle=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+|\rangle=\delta_{ik}\langle|a_ja_l^+|\rangle-\langle|a_ja_k^+a_ia_l^+|\rangle$ $\langle K|L\rangle=\langle|a_ja_ia_k^+a_l^+|\rangle$ $$=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+\rangle$$ $$=\langle ... 1 \newcommand{\ket}[1]{\lvert #1 \rangle}Recall that the vacuum is annihilated by all annihilation operators:$$ a_i\ket{\Omega} = 0$$and that all the occupied states are created from the vacuum as$$ \ket{\chi_i} = a^\dagger_i \ket{\Omega}$$Now, if you apply an annihilation operator to a state which doesn't have the corresponding electron in it, the ... 3 No, because Haag's theorem states that there is no map between the free and interacting Hilbert spaces such that the fields and their commutation relations on one space are unitarily mapped onto the fields and their commutation relations on the other space. That is, the space of states of the interacting theory is as a representation of the commutation ... 0 Short answers: no, they are not the same; they are somewhat related. A more detailed discussion follows. Indeed, in most QFT books zero temperature is usually assumed. However, if one is interested in energy scales that are way beyond the temperature of the system, the zero-temperature approximation is a valid one. For example, the thermal energy at room ... 2 The algebraic approach gives the better idea of what the states and observables of a quantum theory are, and this holds in infinite dimensional systems as well. In the modern mathematical terminology, observables of quantum mechanics are the elements of a topological *-algebra, and states are objects of its topological dual that are positive and have norm ... 0 I met the author of that paper today. He said it's more of an empirical observation rather than a statement based on solid arguments. He asked for a counterexample, which I don't have. Please post it here once you got one. 0 An answer you might find satisfactory is that our model of spacetime is "larger" than what we observe. Yes, there is a preferred direction which gives us a (1,3) signature, but actual, real objects in spacetime must travel on timelike curves (we normalize the length of their geodesics to -1). These timelike curves only transverse part of the entire ... 0 I think a much more natural way to come to \hbar is the one used by Dirac in his principles of quantum mechanics. In it you start by stating that you want a Poisson Bracket which has the same algebraic structure as that of classical mechanics. Then, since quantum operators don't commute (unlike what happens in classical physics) you quickly find out that ... 1 Here are a couple quick and dirty ways to count these operators: Compute the conformal block expansion of the four-point function \langle \phi\phi\phi\phi\rangle. This will only contain blocks with \Delta-\ell=d-2. This is done in http://arxiv.org/abs/1009.5985, equation 64. Compute the character of the conformal group acting on operators in the ... 3 This formula is actually Euler's formula for planar graphs, and holds for all Feynman diagrams regardless of what theory we are in. The proof proceeds by induction and is easy if we first disregard the case of crossing lines: Observe that a one-loop graph has two vertices, one loop, and two internal lines, so the formula holds. Observe that a ... 2 Gauge theories become constrained Hamiltonian systems when passing from the Lagrangian L(q,\dot{q},t) to the Hamiltonian H(q,p,t) where p = \frac{\partial L}{\partial \dot{q}}. Generically, you get a constrained Hamiltonian system whenever the matrix/operator with components$$ \frac{\partial^2 L}{\partial \dot{q}^i\partial\dot{q}^j}$$is singular, ... 0 Your reasoning is essentially correct. Assuming that \alpha_\text{weak}^4 \ll \alpha_\text{EM}\alpha_\text{weak}^2, we can conclude that (a) is less probable than (b). As ACuriousMind notes, to be certain one should actually compute the full diagram, but the integrations from the loops usually give quite reasonable numbers that don't change the picture ... 7 In principle, the charged leptons may oscillate as well but it would need rather strange initial states that could be easily obtained in the ultrarelativistic limit only and the experimental arrangement would have to be very unusual, anyway. The reason why the neutrinos oscillate is that the initial state isn't an energy eigenstate. It is a superposition of ... 1 You might be thinking of the Unruh Effect. It says that observers undergoing high acceleration see an increased vacuum temperature. Mandatory Wikipedia link: http://en.wikipedia.org/wiki/Unruh_effect Remember that according to special and general relativity, observers traveling at a constant velocity (speed and direction) all have to observe the same ... 0 Regarding question 2, in Weinberg's The Quantum Theory of Fields page 295, the author defines the functional derivative of an arbitrary bosonic functional F[q,p]:$$\frac{\delta F[q,p]}{\delta q}=i[p,F[q,p]]\frac{\delta F[q,p]}{\delta p}=i[F[q,p],q]$$where q,p are a pair of operators that satisfy canonical conmutation relations. If that is indeed ... 0 A Weyl transformation is a conformal transformation. They just call it a Weyl transformation because its within General Relativity. This mean that your attempt to understand the statement (i.e. questions "1." and "2.") is invalid and has little to do with the relationship between the Weyl symmetry and conformal scalar symmetry. The statement you are ... 4 Comment to the question (v4): Classically, the Lagrangian for a fermion system reads$$ L ~=~ \int\! d^3x~ i\psi^{\dagger}\dot{\psi}-H.\tag{A}$$The Legendre transformation from the Lagrangian to the Hamiltonian formalism is tricky for at least three reasons: The traditional Dirac-Bergmann analysis leads to constraints. See e.g. my Phys.SE answers here ... 0 As with a lot of my questions, I need a lot more background even to pose the question properly, but I think this link Cooper Pairs and the excerpts below have pointed me in the right direction. In particular they have motivated me to do more basic background on Fermi-Dirac statistics before posting similiar questions. I have included two excerpts from the ... 0 In asymptotic region (high energy s\rightarrow \infty and small momentum transfer t), the eikonal approximation means we drop out any diagram that has connections between internal lines; or it is corresponding to taking infinite ladder and cross-ladder Feynman's diagrams (including tree-level diagrams) in calculation of scattering amplitude and ... 0 I never did the quantitative calculation. But possibly it can be understand in this way. Note that a free electron cannot absorb a photon. Therefore, it is not so surprising that a loosely bound electron has less possibility to be kicked out by a high energy photon. 2 Ref. 1 writes the correct formula$$ U(t,t^{\prime})~=~e^{iH_0(t-t_0)} e^{-iH(t-t^{\prime})}e^{-iH_0(t^{\prime}-t_0)} , \qquad t~\geq~ t^{\prime},\tag{4.25}$$which satisfies$$ U(t_1,t_2)U(t_2,t_3)~=~U(t_1,t_3) , \qquad t_1~\geq~ t_2~\geq~ t_3.\tag{4.26}$$Here t_0 is an arbitrary but fixed fiducial initial instant where operators and states in the ... 3 All quarks have baryon number 1/3, so that the nucleons can be built up. Baryon number is a conserved quantity in the standard model . In models where the proton can decay, it is not conserved, but no proton decays have been detected up to now. One has to realize that the quantum numbers S,C,B,T are attributes for all quarks and have a value, even if that ... 2 They do. It's the third component of isospin, which came about before Murray Gell-Mann's quark model. $$I_3 = \frac{(N_u-N_\bar{u})-(N_d-N_\bar{d})}{2}$$ 0 First, you took the second order Taylor's expansions of e^{-itH} and e^{-itH_0}, so, if your following calculations were right, H and H_0 would commute with each other just to second order of time t, not for all order of t (or just approximatively commute). Second, for 4 operators A,B,H,K:$$ AHB=AKB \rightarrow A(H-K)B=0. $$From this ... 1 The fact that$$ U_I(t_1,t_2)U_I(t_2,t_3) = U_I(t_1,t_3)\tag{1}$$in the interaction picture does not rely on H_0 and H_\text{int} commuting, but can be derived without that assumption from the Tomonaga-Schwinger equation$$ \mathrm{i}\partial_t U_I(t,t_0) = H_I(t)U_I(t,t_0)$$with H_I(t):=\mathrm{e}^{\mathrm{i}H_0 ... 1 It does vanish and it should, because in a non-broken symmetry phase the current operator cannot create a single particle state. You should have summed over all intermediate states, including the vacuum and all multi particle states, whereas you seem to have only inserted a single particle state. 2 Using elementary graph theory identities one can show that the number of loops in a connected diagram is related to the number of external lines and the number of vertices of type i each of which has n_i lines attached to it, is related by$$ \sum \left(\frac{n_i}{2}-1\right) V_i -\tfrac{1}{2}E +1= L $$So you can see that for a fixed process (fixed ... 0 Let's call \Delta^+(x-y;m^2) the 2-point function for a free scalar field \varphi of mass m: denoting the ground state as \Psi_0$$ \Delta^+(x-y;m^2)\equiv \langle\varphi(x)\varphi(y)\rangle \equiv \left(\Psi_0, \varphi(x)\varphi(y) \Psi_0 \right)= \int\frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_m(\mathbf{k})}e^{-ik\cdot(x-y)} $$where implicitly ... 1 I did not get my copy of Srednicki out but from what you have written... Srednicki is referencing the method of steepest descent. Although these notes look to be better than wikipedia. Another page that is directly applicable to the quantum field theory case is here. In short, exponential integrals may be estimated by the saddle points of the integrand. ... 0 Let's check that parity is violated by the weak interaction lagrangian:$$\mathcal{L}(x) = \bar{\psi}(x) \gamma^\mu \frac{(1-\gamma^5)}{2} \psi(x) W_\mu(x) Saying that parity is violated means that the transformed lagrangian $\mathcal{L}'(x)$ is not equal to the old lagrangian resulting from new coordinates $\mathcal{L}(x')$ where $x'^0 = x^0$ and ...

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