Tag Info

New answers tagged

0

The Lagrangian must be a gauge invariant and Lorentz invariant object that can be integrated over the entire spacetime $\Sigma$. So, we must first obtain an $n$-form (for $n$ the dimension of spacetime), and all that we have for that is the gauge field $A$, which itself transforms in an ugly way under gauge transformation. The only object we can build out ...


0

If we want to study the decay of an excited atomic state, the atom must be coupled to the radiation field. This can be done (and usually is) non-relativistically. The overall Hamiltonian then consists of the sum of a free atomic part, a free field part and their interaction. Now the atomic excited states are no longer bound states but turn into "resonances". ...


1

Perhaps it is now only a curiosity, but there is a classical answer to this question that can give order of magnitude estimates that are close to the answers provided by quantum mechanics in many cases. If an electron is in a circular orbit (i.e. thinking of the now obsolete Bohr theory), then if we look at it from a direction that is in the orbital plane, ...


2

Both the ground state and the excited state are eigenfunctions of the time independant Schrodinger equation. That means they are also time independant so the excited state will never decay to the ground state i.e. the transition time would be infinite. However the excited state doesn't decay to just the ground state, it decays to the ground state plus a ...


0

As far as I know, the very idea of normal-ordering implies making modifications to the quantity. For example, consider the hamiltonian of the free real field. $$ H = \int \frac {d^3 p} {(2 \pi)^3} \frac{\omega_p}{2} \left( a_p a^{\dagger}_p + a^{\dagger}_p a_p \right) = \int \frac {d^3 p} {(2 \pi)^3} \omega_p \left(a^{\dagger}_p a_p + \frac{1}{2} ...


3

If you have another model, then use it to generate a set of predictions, and let's compare those to the predictions we get in the world we see. If you think there's some ether filling the universe, then write down some properties it has, and make some predictions with those properties. Talking in vague generalities will only lead you down a rabbit hole of ...


1

As @RobinEkman mentioned, the kinetic term changes as well. This can be easily computed \begin{equation} \begin{split} {\cal L}_D = i {\bar \psi} \gamma^\mu \partial_\mu \psi &\to i {\bar \psi} e^{- i \alpha} \gamma^\mu \partial_\mu \left( e^{i \alpha} \psi \right) = i {\bar \psi} \gamma^\mu \partial_\mu \psi - \partial_\mu \alpha ( {\bar \psi} ...


1

The kinetic term $$\mathcal{L}_\text{fermion} = i\overline{\psi}\gamma^\mu \partial_\mu \psi$$ changes too, by precisely the right quantity to cancel the change in the interaction term. Thus, the total Lagrangian is invariant, and this is what matters.


2

I think this would be tricky, since any force mediator (at least from conventional thinking) must have a three-valent vertex, two of which are the charged object and one of them is the force carrier. If the force carrier is a fermion, I don't think this combination can be Lorentz invariant (spin zero combination).


0

The Weyl transformation (unlike diffeomorphisms) does not affect physical fields, only the geometry of space-time. Probably the best way to show that your equation is Weyl-invariant is to build an action which (when varied) yields the equation. In your example it would be $$ S[\Psi] = \int d^4 x \: \sqrt{-g} \: \bar{\Psi} \gamma^a e^{\mu}_a D_{\mu} \Psi. $$ ...


1

'Square root of geometry' A Dirac spinor field $\psi^\alpha(x)$ under Lorentz transformations behaves as, $$\psi^\alpha(x)\to S[\Lambda]^\alpha_\beta \psi^\beta(\Lambda^{-1}x)$$ where $\Lambda = \exp(\frac{1}{2} \Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$ and $S[\Lambda] = \exp(\frac{1}{2}\Omega_{\rho\sigma}\mathcal{M}^{\rho\sigma})$, where ...


0

Notice that in $[\partial_xO(x),O(x')]$, the partial derivative only acts on $x$, not on $x'$. So we can pull the partial derivative operator out of the bracket and get $\partial_x[O(x),O(x')]$.


1

As others pointed out, the statement probably means that if you want to have a nonzero contribution from creation and annihilation operators acting on vacuum, you need to apply the creation operator first, since $\hat{a}|0\rangle = 0$, in other words, you cannot annihilate if you have nothing. In the question you link to, the situation is different, as that ...


0

That's a proverb in the beginning of Chapter I.8 (in both editions). Zee undoubtedly included it in the book because it reminded him of the fact that an annihilation operator annihilates the vacuum ket: $$\hat{a}|\Omega \rangle =0,$$ while this is not so for the creation operator $$\hat{a}^{\dagger}|\Omega \rangle \neq 0.$$ The proverb is not meant as a ...


1

No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.


0

I came here before asking myself the same question and, as I figured it out, I'd like to answer just for the record. Respecting the former answer, in the Schrödinger picture, if $\phi(x)=\sum_i\left(\alpha_i(x){U}+b_i(x){V}\right)$, the condition $[\phi(x),\phi(y)]=0$ implies $[U,V]=0$ and thus also $[\partial_i\phi(x),\phi(y)]=0$. Now, in the Heisenberg ...


0

Actually the adiabatic hypothesis is usually not required in potential scattering. In case the potential has sufficient decay in coordinate space one can show the existence of the Moller wave operators from which the scattering operator is then obtained. With some modifications this even works for the slowly decaying Coulomb potential. Thus no switching on ...


2

No, the gauge current need not be gauge invariant, since it carries a group index in non-abelian theories. You should recall that both sides of the Yang-Mills equation (and therefore the current itself) are Lie-algebra valued and therefore transform in the adjoint representation. Not even the field strength $F^a_{\mu\nu}$ is gauge invariant, but transforms ...


2

Some terms in $H_0$ can be absorbed into the definition of $\epsilon$. To understand the validity of this substitution, you must go back to chapter 7 and calculate according to eqn. (7.2)-(7.6). You will find "we can absorb the positive coefficient into $\epsilon$ to get..."


4

The four quantum field theories (QCD, QED, QFD, and EWT) unite quantum mechanics and special relativity. They are all fully understood, complete, and proven. In your quote for the standard model, there are not four distinct field theories, the electroweak has united the electromagnetic and the weak in one field theory, the electroweak theory. The ...


2

A theory with N=2 supersymmetry, where particles have two superpartners, has mirror symmetry built in. Nir Polonsky wrote some papers about an N=2 extension of the standard model (e.g.). The main problem for such a theory are the chiral Yukawa interactions between fermions and the Higgs field, which give fermions their mass in the SM. The mirror symmetry of ...


0

Fermion is left-handed spinor and anti-fermion is right-handed spinor. Firstly, you can see that $\xi = \chi ^{\dagger}$. From Sredniski, "... that hermitian conjugation swaps the two SU(2) Lie algebras...the hermitian conjugate of a field in the (2,1) representation should be a field in the (1,2) representation"... Thus $\chi$ is a right-handed fermion ...


0

Using the poisson representation of bessel function $J_n(z) = \frac{(z/2)^n}{\sqrt{\pi} \Gamma(n+1)} \int_0^{\pi} \cos{(z\cos{\theta})} \sin^{2n}{\theta} d\theta$, you can determine the constant.. In 3+1 d, for example, for a space-like distance, we can make a lorentz transformation such that $r = x-y$ is purely spatial, the amplitude is then $D(x-y) \sim ...


1

Two-dimensional surfaces are the Feynman diagrams of string theory. In quantum field theory one sums over one dimensional objects in order to calculate quantities like scattering cross sections or decay rates. This is due to the fact that in this framework, particles are represented as zero dimensional objects, i.e. their world lines are points. In ...


3

The standard treatment of identical particles starts with one-particle states and then imposes symmetry conditions. This is kind of backwards. If you look at a formula like $$|x_1,x_2\rangle = \frac{1}{\sqrt 2}(|x_1\rangle | x_2\rangle \pm |x_1\rangle| x_2\rangle)$$ you are saying that the state is a linear combination of states where particle A is at $x_1$ ...


1

First observation of a heavy flavored spin-3 particle http://home.web.cern.ch/scientists/updates/2014/07/first-observation-heavy-flavored-spin-3-particle Discovery of new subatomic particle sheds light on fundamental force of nature http://home.web.cern.ch/scientists/updates/2014/07/first-observation-heavy-flavored-spin-3-particle


0

For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)).


2

Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ...


1

At the classical level, the global gauge invariance leads via Noether's theorem to electric charge conservation, cf. e.g. this Phys.SE post. The Ward-Takahashi identity (WTI) can roughly speaking be thought of as a quantum version of this. In particular, we stress that the WTI is intimately tied to electric charge conservation. OP's observation that ...


0

I discussed this with a colleague yesterday and I think that I get it. It has to do with the normalisation of the group generators (momentum operators). Within each representation of the translation symmetry we are free to normalise the momentum operator as we like. Then it is natural to choose this normalisation such that all representations have the same ...


3

We can write the Fourier transform of $\langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle$ as $$S(p) D_{\nu\alpha}(q) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q)$$ where $S(p)$ is the full fermion propagator, $D_{\nu\alpha}(q)$ is the full photon propagator, $\Gamma^{\alpha}(p,q,p+q)$ is the proper vertex function, and an overall momentum conservation ...


4

The most simplest way is to look at arbitrary process amplitude (S-matrix), to expand it in a series of some constant, then - to use Wick theorem and, finally, to get that n-th amplitude consists of sum over multiplications of all possible numbers of propagators and normal ordering field operators. Sometimes it's convenient to use nonperturbative ...


5

We require that the number operator have the following property: $$\hat n |0\rangle = 0.$$ We know that $$\hat a |0\rangle = 0$$ and we know that $$\hat a |1\rangle = |0\rangle$$ and we know that $$\hat a^{\dagger} |0\rangle = |1\rangle. $$ Thus, it follows that $$\hat a \hat a^{\dagger} \ne \hat n$$ since $$\hat a \hat a^{\dagger} |0\rangle = ...


1

After reading through the corresponding chapters in several books I think I'm now able to give a "semi-satisfactory" answer to my own question (and to understand Lubos first comment ;) ). I write semi-satisfactory, because I hope someone with a deeper understanding of these topics will give a better answer. My explanation is still a little bit heuristic, ...


0

Oscillons and solitons are two specific types of nonlinear waves. Solitons are considered to be nonlinear partly because their phase speed depends upon amplitude and their spatial scale. I think the main differences between the two are that oscillons are solutions in granular media, they can change form when colliding with other oscillons, and oscillons can ...


0

For massless gauge bosons the only meaningful way to transform under parity is like polar vectors since they have to transform as $i\partial_\mu$ in the covariant derivative. For massive gauge boson, one in principle doesn't have this constraint, although one could assign a parity as for massless vectors assuming the mass comes from spontaneous symmetry ...


1

Hint: The equation system is a coupled set of non-linear Schrödinger equations. Here is a trick: The Lagrangian reads$^1$ $$\tag{1} L~=~ i\dot{f}f^{\ast}- i\dot{g}g^{\ast}-a|f|^2-a|g|^2+gf^{\ast}+fg^{\ast} +|f|^2|g|^2 ,$$ where we have assumed that the constant $a\in\mathbb{R}$ is real. Show that the Lagrangian (1) possesses a global $U(1)$ phase symmetry ...


2

The vacuum should have particle number $0$. In some detail: we would like $\hat{N}\ |0\rangle=0$ and $\hat{N}=a^\dagger a$ is the only ordering that does that. It follows from the usual commutation relations that $\hat{N} |n\rangle =n\ |n\rangle$ which is in sync with interpretation of $|n\rangle $ as an $n$-particle state in second-quantisation.


7

Since you define e.g. in the bosonic case $c_j^\dagger: H_N^S \rightarrow H_{N+1}^S,\quad c_j^\dagger | \ldots n_j \ldots \rangle := \sqrt{n_j+1} |\ldots n_j+1 \ldots \rangle$ $c_j: H_N^S \rightarrow H_{N-1}^S,\quad c_j | \ldots n_j \ldots \rangle := \sqrt{n_j} |\ldots n_j-1 \ldots \rangle$ it makes more sence to use $a^\dagger a$ which will give you ...


1

It seems OP's main question concerns the systematics of gauge-fixing. We interpret/reformulate OP's questions as essentially the following. The original gauge-invariant action $S_0$ is unsuitable for quantization, so we add a non-gauge invariant gauge-fixing term to the action. Obviously we cannot add any non-gauge invariant term to the action. ...


0

99.9% of the mass of a hadron or a meson comes from confinement in QCD. Confinement is a special feature of QCD due to its non abelian symmetry which leads to a negative beta function. It is confinement that also leads to a breaking of the chiral symmetry at about 200 MeV or the radius of a hadron (about 1 femto meter).


2

I wonder if this may be of some help. The optical theorem relates the total cross-section of a scattering with the forward scattering amplitude. For example for a $2\to2$ scattering you get that: $\left\langle n\left|S\right|m\right\rangle \equiv\delta_{mn}+i\left(2\pi\right)^{4}\delta^{4}\left(p_{m}^{\mu}-p_{n}^{\mu}\right)\left\langle ...


1

I don't know if that is the case, but you may have been misguided by the formalism. For convenience, the language of second quantization is often utilized in non-relativistic many body physics. So you may have stumbled upon creation and annihilation operators in that context. Nevertheless, even if written in second quantization, the Hamiltonians of many ...


0

If I understand correctly your question, you are essentially asking why is it, that we always see a potential of this form: Rather than a, say $V=-\phi^2 +\lambda \phi^4$ graph. Where in the first case the coordinate can roll-down only one direction, whereas in the other case it can roll down the other direction as well. It doesn't matter which you ...


1

In a weakly coupled fluid the viscosity scales as $\eta\sim 1/g^a$, where $a$ is a positive power and $g$ is the coupling. In QCD, up to logs, $a=4$. This means that in perturbation theory $\eta$ is parametrically large, and strong coupling is required to describe low viscosity fluids. There are many counter-intuitive aspects of viscosity. For a recent ...


4

It's Stokes's theorem. Consider a field $F = dA + A \wedge A$ such that $A$ is pure gauge at infinity, that is, $\lim_{x\to\infty} A(x) = \omega\, d \omega^{-1}$ for some $\omega : S^3 \to SU(2) \sim S^3$ where $\omega$ is a function on the 3-sphere because the limit can depend on the direction out to infinity. In differential forms the first expression is ...


1

First, the second equation starting with $S^a\propto\dots $ should probably say $S^b\propto\dots$. Now, the first two equations for the operators $S^a$ and $S^b$ which are the relevant parts of $S=S^a+S^b$ have the positive plus sign – the additional factors that are omitted don't differ by any extra sign because there is a well-defined factor (and sign) ...


1

For the massless case, one needs to show that $W^\mu = \lambda P^\mu$. Equation (10.53) provides a basis for an arbitrary four-vector and then expands $W^\mu$ in that basis. Imposing the two conditions $W\cdot P=0$ and $W \cdot W=0$ completes the proof by showing that all other "components" in that basis vanish.


0

The two-particle interaction in the Hubbard model is non-zero only when these two particles occupy the same site, ie. get as close as it gets for the given band structure. This is similar to balls with hard cores - they only interact if the cores overlap.


4

There is but one truth about time evolution in quantum mechanics: The Hamiltonian is the generator of time translations. In the Heisenberg picture, this means we directly borrow the quantized1 version of the evolution of observables on phase space, i.e. $$ \frac{\mathrm{d}}{\mathrm{d}t}A(\vec x,t) = \mathrm{i}[H,A(\vec x, t)]$$ holds for all ...



Top 50 recent answers are included