New answers tagged

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It helps to remember that invariant quantities are seen as scalars to the transformation (they have no indices in the target space). In the other hand, covariant quantities are objects that transform in a certain way. Example: Vectors in $R^{2}$, under rotation $R_{ij}$, transform covariantly since $v'_{i}=R_{ij}v_{j}$, but it's length is invariant since ...


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You can only distinguish the sublattices in this case because you've tagged them A,B. The process of inversion only exchanges identical carbon with carbon, leaving the crystal physically unchanged. If you gave me a crystal with one orientation and I then returned it to you without telling you whether or not it's been inverted, you'd have no way of knowing. ...


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Firstly, given a differentiable Lagrangian $L(q,\dot{q},t)$, we can always form the Lagrangian energy function $$\tag{1} h ~:=~\sum_ip_i \dot{q}^i-L ,\qquad p_i ~:=~\frac{\partial L }{\partial \dot{q}^i }. $$ Secondly, make the assumption that $$\tag{2} \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.} $$ ...


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Crossing symmetry tells you that you should not only exchange $$p_2\leftrightarrow -p_4$$ in the amputated matrix elements, but also replace the wavefunction polarizations $$ u^{\pm}(p_2)\rightarrow v^{\mp}(p_4) $$ (where the spin polarizations have been reversed), and finally multiply the amplitude for a factor $-1$ (Since you are crossing a fermionic ...


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For a charge distribution $\rho({\bf r'})$, the electric field at ${\bf r}$ is $${\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$$ where ${\bf R}={\bf r}-{\bf r'}$. I think the OP's claim is at positions where $\rho \ne 0$, the above integral is infinite because the integrand blows up at ${\bf R}={\bf 0}$. I think ...


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I think the first helpful fact to clarify is that there are two different kinds of topological phases: there are so-called Symmetry Protected Topological (SPT) Phases (displaying 'symmetry protected topological order') and there are (intrinsic) Topological Phases (displaying '(intrinsic) topological order'). As some quick examples: topological insulators and ...


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Your thinking is correct. At least, if the sphere is made up of small point charges, then the field will be infinite as you approach them. There is a point you are missing when you say that this contradicts Gauss' law: Gauss' law only gives you the flux of the field. To get the field of the sphere from it in the textbook-way, you have to use symmetry. You ...


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Your issue is that this interview was not transcribed by a physicist! What he said was "Gauge Symmetry" not "Gate Symmetry". Your googling should work better now, and here is one place to start: https://en.wikipedia.org/wiki/Gauge_theory


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The time reversal and chiral symmetry are special because they are antiunitary symmetries, in contrast to the other unitary symmetries like translation and rotation symmetries. Antiunitary symmetry operation involves complex conjugation of the wave function of the system, which is a non-trivial operation beyond unitary transforms. Unitary symmetries are ...


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First of all, what's the motivation for the Yang-Mills action and how should I understand the coupling constants $\theta$ and $g$? I would say motivation comes from experiments. For instance it is an experimental fact that the electric charge is conserved. The associated current is also conserved, in the sense of $$\partial_\mu J^\mu=0.$$ Therefore we ...


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You find particle hole symmetry (PHS) for example in superconductors, where you can take for the Hamiltonian the Bogoliubov-de Gennes (BdG) Hamiltonian as a mean-field approximation. This is the only experimental example I know. There are maybe other systems with particle hole symmetry that I don't know about, but I will use superconductors as an example in ...


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We do observe spontaneous symmetry restorations in nature. This is called an emergent symmetry. See e.g. this post. A system posses an emergent symmetry if it appears symmetric at large (coarse-grained) scales although the apparent symmetry is explicitly broken by the microscopic description (typically the Hamiltonian or Lagrangian). I can give two examples ...


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Going the way stated in the question's title is easy: The Euler-Lagrange condition is, inherently, a condition on the action -- the statement is that the classical path is the path for which the action takes a minimum value for the path. Since this is a statement about the value of the action, and the action is Lorentz-invariant, then this minimum value is ...


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Of course an anomalous global symmetry destroys the associated Ward identity, but...we don't care so much about that. The Ward identity of global symmetries is not needed for consistency of the theory. However, a broken local Ward identity completely destroys the associated gauge theory, in particular since the decoupling of the unphysical degrees of freedom ...


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When we say that a lattice has a particular symmetry we mean that the lattice is mapped onto itself by the symmetry. So if I have a (2d) material which has inversion symmetry in the bulk and which has an atom at a point $(x,y)$ then inversion symmetry tells me that there is another, identical atom at $(-x, -y)$. At the surface, however, this is no longer ...


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The precise statement of "self-adjoint operators generate continuous unitary symmetries" is Stone's theorem. It guarantees that there is a bijection between self-adjoint operators $O$ on a Hilbert space and unitary strongly continuous one-parameter groups $U(t)$ that is given by $O\mapsto \mathrm{e}^{\mathrm{i}tO}$. The definition of the exponential for an ...


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Naively, suppose you have a field F in your theory. If you look for the mass term of the field F in the Lagrangian it appears like -L $\supset$ $m^2$ $\bar{F}F$, which is invariant under the charge operator C, i.e., -$L^c$ $\supset$ $m^2$ $F^\dagger$$C^\dagger$$\gamma^0$$C$$F$ = $m^2$ $F^\dagger$$\gamma^0$$F$ =$m^2$ $\bar{F}F$ = -L, which means that the same ...


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If $\rho$ is a generic physical quantity, e.g. mass density in this case, then spherical symmetry is represented in the form of $\rho = \rho(\lvert \vec r\rvert)$ and not $\rho = \rho(\vec r)$ with $\vec r$ being the position vector of the point at which the quantity $\rho$ is being measured.It is assumed that the center of mass of the distribution coincides ...


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I think this should be an alternative answer from first principles. Assume the Hamiltonian is one dimensional, and $V(x)$ is an even function. Given a general solution $\psi (x)$, we also have that $\psi (-x)$ is a solution. If the eigenspace of the Hamiltonian is one dimensional, then we must have $$ \psi (x) = \alpha \psi ( -x) $$ where $|\alpha | = 1$. ...


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Translational symmetry in the sense of the standard formulation of Noether theorems means that the Lagrangian is invariant under the action of the group of spatial translations. This is not the case in your example because $U$ does not admit such invariance. However there is another, more physical, version of the idea of translational invariance for a ...


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I) Let the Lagrangian be $$\tag{1} L~=~\frac{m}{2}v^2-U(x), \qquad v~:=~\dot{x}.$$ Let the force $$\tag{2} F~=~-U'(x) $$ be a constant. II) Infinitesimal translations $$\tag{3} \delta x~=~\varepsilon $$ is a quasi-symmetry $$\tag{4} \delta L ~=~\varepsilon \frac{df}{dt}, \qquad f~:=~Ft $$ of the Lagrangian (1). Here $\varepsilon$ is an ...


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Noether's theorem tells us that a conserved quantity is related to a symmetry of the action, where the action $S$ is given by: $$ S = \int L dt $$ where $L$ is the Lagrangian given by: $$ L = T - V $$ Since the potential $V$ is a function of position the Lagrangian and hence the action is not symmetric under displacements in space.


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The Lagrangian is \begin{equation} L = \frac{1}{2} \dot{x}^2 - ax-b. \end{equation} Introducing spatial translation $x \rightarrow x+\Delta$ for constant $\Delta$ we see that \begin{equation} L \rightarrow L' = \frac{1}{2} \dot{x}^2 - ax - a\Delta -b. \end{equation} Therefore the action changes as \begin{equation} \delta S = \int{dxdt \; (L'-L)} = ...


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Lets start from the beginning. I will drop $r$ (to prove my lazyness). By definition $p_x \Psi = -i \sum_i c_i \frac{x}{a}(w(R_i+\hat{x})-w(R_i)) $ $p_y \Psi = -i \sum_i c_i \frac{y}{a}(w(R_i+\hat{y})-w(R_i)) $ Ignore the $\hbar$. I keep $x$ and $y$ to make the expression general. Now, $\Psi(\sigma_y p_x - \sigma_x p_y) \Psi\\ = \sum_{ij} c_i^\dagger ...



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