32

The slave particle approach is based on the assumption of spin-charge separation in the strongly correlated electron systems (typically Mott insulators). It was proposed that the electrons can decay into spinons and chargons (holons/doublons). But to preserve the fermion statistics of the electrons, the spinon-chargon bound state must be fermionic, so the ...


25

This answer deals with a particular class of systems, where these assumptions hold: Gravity is the only important force between bodies (this would also work for any other purely attractive inverse square law). Collisions (or encounters) in the sense of processes modelled by the Fokker-Planck equation (rather than the collisionless Boltzmann equation) occur. ...


18

The full quantum analysis of the hydrogen atom is a quantum two body problem, however, one of those bodies is extremely massive compared to the other, so that this problem, as a first approximation, is analysed by solving either the first quantised (i.e. for one quantum particle in a classical environment) Schrödinger or Dirac equations for inverse square ...


14

Long wires are real macroscopic bodies, kilometers of superconducting wires are used at the LHC of CERN and the currents can be described by quantum mechanical equations. Crystals also can be described by quantum mechanical equations, and can be quite large, maybe not as large as a table. Superfluids too are in the realm of macroscopic quantum mechanics. ...


14

For researchers who study condensed matter physics (i.e. low-energy physics), it might be helpful to read following books and articles. H. Haug and A. P. Jauho: Quantum Kinetics in Transport and Optics of Semiconductors (Springer, New York, 2007). We can learn the (minimal) essence of Keldysh formalism by reading pp. 35-69 (sections 3 and 4). This article ...


14

In the context of ultracold Fermi gases, a BEC-BCS crossover means that by tuning the interaction strength (the s-wave scattering length), one goes from a BEC state to a BCS state without encountering a phase transition (thus the word "crossover"). It is also useful to know that the BEC state is a Bose-Einstein condensate of two-atom molecules, while the ...


14

None of those are really "good" answers. I will concur. Moreover, it is really more a question that ends up touching onto meta-mathematics, than physics. I am not really sure one can levy a specific reason, other than the very general reason that suitably complex equations cannot be solved exactly as a general principle. But I believe the explanation below, ...


14

The problem is not at all unique to Schrödinger's equation, it is a common feature of differential equations. The same thing happens all over the place in classical electro-magnetics. But, also to say there is no "exact" solution is misleading. It is more true to say that the solution cannot be expressed finitely in certain notations. There are theorems ...


12

There is no such thing as "Abelian anyonic commutation relations", in the sense that the "Abelian anyonic commutation relations" that you write down does not describe Abelian anyons. So the starting point of the question is not valid. Also anyons do not have a Fock space description. The standard many-body text books stress on Fock space too much, which ...


12

Is there any good definition of many body localization? Let's start with single-body (Anderson) localization. There, in a non-interacting system, a particle (e.g. an electron) becomes localized due to constructive interference with itself. This interference is induced by the presence of disorder. Turning interactions on brings us to the realm of many-body ...


12

It's just a convention to write the Dirac Hamiltonian in terms of electron and positron operators. The operator $b_p^{s\dagger}$ creates a "hole", i.e. a positron, which is the same as annihilating an electron. So we could just as well define $c_p^s = b_p^{s\dagger}$ and write the Hamiltonian in terms of $a$ and $c$ instead of $a$ and $b$. Then everything is ...


10

They are closely related: the resolvent of the eigenvalue equation of a self-adjoint operator $\hat{A}$ is the operator valued function defined as $$G_\lambda=(\hat{A}-\lambda \hat{1})^{-1}$$ We call Green's function the kernel of the resolvent [the kernel of an integral transform] which is the solution of the homogeneous differential equation $$(\hat{A} -...


10

After days of thinking, searching, discussions, and testing, I can finally answer my own question now. The answer is much more involved than I expected from such a "simple" XY model (even just for the Ising model)! All "correct solutions/spectrum" stated below are checked against results from exact diagonalization. Simply put, there's nothing wrong with ...


10

The time ordering enters as a consequence of the definition of the Hamiltonian as the generator of time translations. In the Schödinger picture: $$|\psi(t)\rangle \approx \left(1 - \frac{i}{\hbar} H(t') [t - t'] + \mathcal{O}([t-t']^2)\right) |\psi(t')\rangle,$$ where the relationship becomes exact in the limit as $t-t' \rightarrow 0^+$. It's an exercise in ...


8

Deepends on your definition of "fails". The accuracy of Born-Oppenheimer approximation is determined by the smallness of the electron/nucleus mass ratio. For hydrogen this ratio is $\approx 1/1800$. Replace the electron mass by the muon mass, which is 200 times larger, and you get $\approx 1/9$ which gives you a rough estimate for the relative accuracy in ...


8

One of my favorite non-trivial, exact many-body ground states is the solution of a very specific spin-1 magnetic insulator in 1D, with a hamiltonian $$H_{AKLT}=\sum_{\langle ij\rangle}\vec{S}_i \cdot \vec{S}_j + \frac{1}{3}(\vec{S}_i \cdot \vec{S}_j)^2$$ It turns out that you can construct the ground state by looking at the spin-1 operators as a projection ...


8

Heisenberg's Hamiltonian, if I understood what you're referring to, is a Hamiltonian operator for a system whose components are a family of spins. What is relevant here is that $H$ is a sum of terms of the kind $\mathbf{s}_i\cdot\mathbf{s}_j$. $SU(2)$ is a group of transformations which in this case represents rotations (it is the covering of the $SO(3)$ ...


8

Although these equations are correct, it is not quite correct to treat them as eigenvalue equations. In quantum theory, both bosonic and fermionic operators act on a Hilbert space, thus have ordinary numerical matrix elements and eigenvalues. One of the possible realizations of the Hilbert space in the case of bosonic operators is as a Hilbert space of ...


8

The Plummer model has a potential of the form $$ \Phi(r)=-\frac{1}{\sqrt{r^2+1}} $$ (obviously ignoring all constants). Equating the above with the kinetic energy, you get $$ \frac12v_e^2+\Phi(r)=0\to v_e(r)=\sqrt{2}\left(r+1\right)^{-1/4} $$ This velocity is the maximum velocity you can have at a radius $r$, so we must have that $0\leq v\leq v_{e}$. In ...


8

Don't be fooled by the zero inside the ket. That is just a label. For example in Scalar QFT, the vacuum state of interacting theories is usually denoted as $|\Omega\rangle$ rather than $|0\rangle$. So no, the vacuum state does not represent the null vector of the Hilbert space. Rather, the vacuum state is defined to be the state with the lowest possible ...


8

Even Newtonian gravity is extremely complicated if the number of moving bodies exceeds 2. If you add nonlinearity of general relativity and existence of gravitational radiation, then generally, a system of several black holes could be investigated only by numerically solving full system of Einstein equations: a very computationally intensive system of ...


7

Those degrees of freedom of a quantum system that are described by a pure partial state must be very well shielded from unwanted interactions with the environment, otherwise they will be decoherered to a mixed state in a moment. This shielding can be done for a few degrees of freedom (like a superconducting current) but not for position and momentum of ...


7

There are a number of schemes that have been developed over the years to describe the dynamics of the reduced density matrix. The problem that you encounter is that you have to make a choice between computability versus generality. Let me make an overview of some of the approaches you can find. We typically think of system consisting of a some particle in ...


7

They are equivalent. But one might say that that Hubbard-Stratanovich transformation are more systematic, as it might be easier to figure out how to go beyond mean-field. It might also be easier to combine different kinds of channels (for instance, you have selected a particle-hole channel in your example, whereas in the case of superconductivity, one would ...


7

This is an important question that is asked by many people entering the field of density functional theory. I think that it should be answered with a high degree of detail and thus I would like to add a few aspects to the answer of supermarche. As mentioned the Hohenberg-Kohn theorem states that (up to a constant energy shift) the external potential of the ...


7

In simple terms, the reason that free electron theory works is that the (non-interacting) electrons in a metal form a fermi-surface. If in the absence of interactions the fermi surface is the ground state of the system. If you add coulomb interaction this will enable electrons to interact. However, due to the Pauli exclusion principle (and conservation of ...


7

The question of whether an analytic solution exists for a given dynamical system or PDE goes under the name "integrability". This is a notoriously slippery concept, with precious few useful and rigorous ideas. However, there is one extremely important result (very relevant to the present question) in this area: The KAM Theorem. One of the consequences ...


6

This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that do not yet have commentary. J. Berges, Introduction to Nonequilibrium ...


6

First, the direct sum $S=V\oplus U$, of two vector spaces $V$ and $U$ is just the vector space constituted of "sums of vectors" $s=v+u:=(v,u)$ of each original vector space. The multiplication by scalar is viewed as obbeying the distributive property: $\alpha s=\alpha(v+u)=(\alpha v,\alpha u)$. The sum of two vectors of $S$ is just $s_1+s_2=(v_1+v_2,u_1+u_2)....


6

According to this lecture from the University of Edinburgh, numerical simulations of N-body systems suggest a half-mass relaxation time: $$ t_\text{rh} = 0.138\frac{N^{1/2}r_\text{h}^{3/2}}{m^{1/2}G^{1/2}\ln(\gamma N)} $$ where $r_\text{h}$ is the radius that initially contains half the mass of the system, $G$ is the gravitational constant, $m$ is the ...


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