35

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 ...


26

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 ...


17

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 ...


16

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 ...


15

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, ...


15

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 ...


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

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 ...


14

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 ...


13

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 ...


13

The direct product/sum$^\dagger$ and tensor product of two vector spaces $V$ and $W$ are not the same, though they look similar on the surface. The correct terminology is given by Wikipedia; we describe composite spaces as tensor product spaces, not direct product/sum spaces. From a mathematical perspective, the difference is crucial. States in a direct ...


11

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 ...


9

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 ...


9

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 ...


8

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 ...


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

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 ...


8

The first thing to realise is that, although we often talk about BEC as a low-temperature equilibrium state, in dilute alkali gases a BEC is a highly excited, non-equilibrium, metastable state of the system. While this may sound strange at first, it becomes rather obvious after a moment's thought: the ground state of a collection of lithium atoms is not a ...


8

Direct product, tensor product, Kronecker product and outer product mean approximately the same thing$^*$ - a product without summation over the inner indices. E.g., for two vectors: $$ (\mathbf{a}\otimes\mathbf{b})_{ij} = a_i b_j $$ This is opposed to the inner product, dot product or scalar product, where $$ \mathbf{a}\cdot\mathbf{b}=\sum_i a_i b_i $$ $^*$ ...


7

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)....


7

How can a scattering process have bound states? We are familiar with bound states from our everyday experience. For example, two hydrogen atoms interact through the Coulomb force. This leads to the formation of a bound state, namely, the hydrogen molecule. The most simple model of this situation is the square-well potential. This potential has a ...


7

Because these are actually Fourier transform of the usual Green functions. Consider the Schrödinger equation : $$ \hat{\mathcal{H}}|\Psi(t)\rangle=\mathrm{i}\partial_t|\Psi(t)\rangle $$ The general solution $|\Psi(t)\rangle$ of such equation for a time-independant hamiltonian $\hat{\mathcal{H}}$ can be expressed in terms of Green function $G(x',x,t)$ : $$ \...


7

You may find the following paper useful: A Symbolic Solution of the Hubbard Model for Small Clusters, by J. Yepez. You may also want to review group theory for condensed matter physics, because your questions essentially span the basics of group and representation theory. Many texts give good overviews of the fundamentals of group theory as applied to ...


7

This question nagged me for most of Friday. It seems obvious that it is a Goldstone mode. You can translate the ICDW and the energy does not chage. However it is not clear what continuous symmetry remains since the lattice has already broken translation symmetry. To get to the bottom of the issue we should focus on the relevant Hamiltonian which is electron+...


7

The vertex function contains information about: The existence of bounded states & their spectrum Effective interactions in the theory Scattering amplitudes in the theory To illustrate these statements, let us consider the interaction $U(r_1-r_2,t_1-t_2)$ between two particles and solve the respective Bethe-Salpeter equation. First, consider 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

A projector is an observable - you can directly check that it is Hermitian $|L\rangle\langle L|^\dagger = |L\rangle \langle L|$. As to interpretation - a projector onto a single state will measure the value $1$ for definite if the system is in that state. If the system is in an orthogonal state it will measure $0$. Therefore you can think of projectors as ...


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