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-1

I think the following is a different way of saying what has been said above: How can energy be conserved AND uncertain? Remember basic vector space theory: every state can be expressed as a linear combination of other states (which form a basis). So a state which is not a state of definite energy can be expressed as a linear combination of states which ARE ...


0

I am only familiar with QM. Pauli (or maybe Dirac) wrote that there is a symmetry: energy-time is perfectly analogous to momentum-position, and one can think of energy as the momentum a thing has as it travels thru time. Einstein tells us that one man's space is another man's time, so one man's momentum IS another man's energy. How can energy be conserved ...


0

Expand the normal order: $$ :\psi^\dagger_1\psi_1\psi^\dagger_2\psi_2:=\psi^\dagger_1\psi^\dagger_2\psi_1\psi_2. $$ (creator operator before annihilation operator). Then insert unit operator $I=|0\rangle\langle0|$ between them. Meaning of this expression is that, the total amplitude include two amplitudes. One is for annihilating initial particles of ...


2

List (to be completed with more references and/or items, details of the relation to physics) there is a notion of positive energy representation (cf. Haag-Kastler axiom, "Spectrum" or "stability" condition) in which generators of translations can be choosen in the von Neumann algebra associated to the representation of the observables, but not necessarily ...


1

The first statement is correct to some extent, the second isn't. Take the case of vector gauge theories, like the ones in the Standard Model. These theories have a massless vector field, which can be described by two degrees of freedom (2 polarisations) while the classical field itself, $A_\mu$, is described by 4 components. Gauge invariance is related to ...


2

For four Majorana zero modes, if the total topological charge is $1$ there are two states $|0\rangle_{12}0\rangle_{34}$ and $|1\rangle_{12}|1\rangle_{34}$ ($i\gamma_1\gamma_2\cdot i\gamma_3\gamma_4=1$. So this system can be mapped to a qubit, with $i\gamma_1\gamma_2=\sigma_z, i\gamma_1\gamma_3=\sigma_x, i\gamma_2\gamma_3=\sigma_y$ (I did not check the signs ...


2

The state you are writing is annihilated by $c_1+ic_2$ and $c_3+ic_4$. So you need to find a state in the new basis which has the same property. If you do this, you should indeed find that this state is of the form $\vert0\rangle\vert1\rangle-i\vert1\rangle\vert0\rangle$ (or similar, depending on your conventions).


1

Electric charge is not special. Every charge is replaced by its opposite under the C conjugation. For instance, electric charge goes from positive to negative and vice versa. Color charge goes from blue to antiblue and vice versa. Mutatis mutandis with green/red color charge. Every charge is sent to its anti-charge.


1

Supposing that $p$ denotes a momentum coordinate, the function $g(p)\in L^2(\mathbb{R}^3)$ is your one particle wavefunction in the momentum representation. More precisely, the vector $\lvert 1\rangle=\Bigl(\lvert 1\rangle_0,\lvert 1\rangle_1(p_1),\dotsc,\lvert 1\rangle_n(p_1,p_2,\dotsc,p_n),\dotsc\Bigl)$ that you has written (where $\lvert ...


2

Let us make clear that the problem If proton spin emergence from quarks and gluons is mysterious, why is silver atom spin not? is a modelling problem. The spin both of the proton and the silver atom is measured and known to identify them. John's answer covers it, the energy carried by the virtual quarks and gluons within the proton are much larger ...


2

All of charge, lepton number and lepton-flavor number should be conserved in the reaction. Because the reaction is charged-current, the neutrino will be converted into a charged lepton and because it is a neutrino (and not an anti-neutrino) it must be a negatively charged lepton (to conserve lepton number). That means that the quark involved must be ...


3

The word "relativistic" means "compatible with principles of Special Relativity". This usually implies that we can no longer use the "classical" picture of universal stationary space and time. Instead we talk about 4-dimensional space-time. The word "quantum" means compatible with principles of quantum mechanics. You can look them up on wikipedia etc. But ...


0

$$\overline{\psi_L}\psi_R = \overline{\frac{1}{2}(1-\gamma^5) \psi} \frac{1}{2}(1+\gamma^5)\psi=\left(\frac{1}{2}(1-\gamma^5) \psi\right)^\dagger \gamma^0 \frac{1}{2}(1+\gamma^5)\psi = \psi^\dagger\frac{1}{2}(1-\gamma^5)\gamma^0\frac{1}{2}(1+\gamma^5)\psi$$ Since $\gamma^5\gamma^0 = -\gamma^0 \gamma^5$, it yields: $$\overline{\psi_L}\psi_R = \psi^\dagger ...


1

Your second last step is incorrect: $\bar{\psi}_L = \frac{1}{2}\psi^\dagger\gamma_0\gamma^0(1+\gamma_5)\gamma_0 = \frac{1}{2}\bar{\psi}(1-\gamma_5)$, since $\{\gamma_5,\gamma_\mu\} = 0 \implies \gamma_5\gamma_0 = -\gamma_0\gamma_5$.


1

@Ihle is right. $\gamma^5\gamma^0=-\gamma^0\gamma^5$. You don't need to insert $\gamma^0\gamma^0$.


0

As it turns out, your intuition might be in the right path. As was recently noted and published by physicists of the caliber of Arkani Hamed and Freddy Cachazo, you can avoid the trouble of doing your Feynmann integrals off-shell, by replacing the vertices with on-shell particles of complex momenta and coordinates. There is a conjecture that this replacement ...


0

Each coupling and each field normalization factor potentially gets quantum corrections due to the integration over high-momentum modes You can always consider strongly connected diagrams with appropriate external legs (e.g. two electrons for the electron normalization constant or two electrons and a photon for charge normalization). In other words, on tree ...


1

I recently stumbled upon a good comment about this in Jared Kaplan's AdS/CFT notes Any quantum field theory which has hope of having an UV-completion can be viewed as as effective theory at point in the RG flow from an UV complete theory. Field theories at the UV fixed point are conformal. Hence all 'well-defined' field theories are either CFTs or points ...


1

For your first question: the physical mass is the measured mass. If $m$ is not the measured mass, don't call it the physical mass, because it isn't. If someone does, correct them. :-) For your second question, I think we should look at the physical interpretation of all those mathematical manipulations. The hierarchy problem (as I understand it) comes from ...


0

Firstly, your are correct that in the second order phase transition the correlation time diverges, and there is a dynamical critical exponent related to it. By definition, the correlation time is defined by the time interval that two states are not correlated (similar to the definition of the correlation length), say they have quite different configurations ...


2

Energy is not a Lorentz invariant quantity, it is the zero-th component of the four-vector. Only proper orthochronous Lorentz transformations preserve the sign of the zeroth component, so if the energy is positive in one frame, a non-orthochronous Lorentz transformation would yield a frame in which the energy is negative. But we usually only allow the ...


-3

we can see negative energy solution as anti particle travelling in the -p direction in momentum space.creation operator have coefficents e^-ipx so it will create anti particle in the -p direction. here p is a four vector.four vector can be negative or positive.so we have solution in the positive p direction and solution in the - ve p direction.


0

Maybe a chemists perspective could be useful to understand why we take both energies in quantum mechanics? It's not an elegant answer but is easy to rationalise! In computational chemistry we use the variational principle to compute the orbital mixing coefficients of molecular orbitals from an atomic orbital basis set. In doing so along the way we end up ...


2

The string-net condensation is a general construction to obtain gauge fields and fermions. The chiral fermion problem refers to the fact that in the Standard Model (SM), the SU(2) gauge field only couples to the left-handed fermions but not the right-handed fermions. However in the (early version of) string-net condensation, the emergent gauge field will ...


0

You only need to use the anticommutations relations for dirac creation/annihilation operators and their linearity. Expand the definition of $P^\mu$ or $Q$ and use those relations on each term. Only one term per case will be non-zero, the one in the answers.


0

Putting the fermions on-shell or off-shell doesn't change the divergent part of counterterms. In the renormalization schemes, the counter terms are determined to cancel the divergencies. You can put $p^2=m^2$ or $p^2=-\mu^2$ or etc in the diagrams to determine the counter terms, but notice that the derived renormalization constants, $Z_1$, $Z_2$, etc, should ...


6

Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is. Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute ...


2

To understand this one shall take in quantum-mechanical approximation method namely perturbation theory into account. In perturbation theory, systems can go through intermediate virtual states which often have energies different from that of the initial and final states. This is because of time energy uncertainty principle. Consider an intermediate state ...


0

While electrons are point particles in the sense that their position eigenstates are (as far as we know) $\delta$-like. Photons can't be said to be point particles in this sense, as you cannot transform to their rest-frame (although they are featureless with respect to small scales as far as we know). The correct way to thing about electrons and photons is ...


1

The Feynman-rule for this 4-fermion-vertex is $\frac{ia}{M^2} (\sigma_{\mu\nu})_{ij} (\sigma^{\mu\nu})_{lm}$, where $i,j$ are the Spinor-indices of the neutrinos, while $l,m$ are the spinor-indices of the electrons. You see, that this vertex has no free Lorentz-indices and four free spinor-indices, as it should be. For the amplitude you find ...


3

Consider for example, simple $\lambda \phi^3$-theory with Lagrangian $$ \mathcal{L}=-\frac{1}{2}\left|\partial_\mu\phi\right|^2-\frac{m^2}{2}\phi^2+\frac{\lambda}{3!}\phi^3. $$ One can say that the $\lambda \phi^3$ term renormalizes the mass term, because the regularization and renormalization of the divergence of the one-loop diagram will lead to a ...


1

I think the confusion is due to a lack of mathematically precise definitions of what is quantum field theory? what is one trying to construct and how? etc. There are a lot of vague notions used in the physics literature: the partition function (which does not make much sense in infinite volume), the effective action,...but the bottom line is the collection ...


1

The full Green's function of an equation like the Klein-Gordon equation is the difference of the retarded and advanced Green's functions. It is only when the equation in question is an equation involving time that we often discard the advanced anti-causal part to get physically sensible solutions. In QFT, the full Green's function appears for example as the ...


1

If a theory contains divergences, there is no need to assume it is plain wrong. It probably won't be the full theory of everything, but that's the only criticism that can be leveled at it. It can still be in perfect agreement with experiments below very-high-energy. The issue with QED is that QED is not a complete theory. In the real world, QED is just a ...


0

Finite temperature Feynman rules are simply zero temperature Feynman rules for Euclidean ($t\to i\tau$) QFT in periodic imaginary time. So instead of continuous values for the momenta, you will have a discrete spectrum for the timelike moments (such as in the infinite potential well in basic quantum mechanics). It's called the Matsubara formalism, if you ...


0

For example, when we meassure Higgs boson mass to be 125 GeV, do we think about renormalized or pole mass? Pole mass is the physical mass and independent of any renormalization scheme we use to subtract any infinite parts of the loop corrections. It is what we observe. Should the mass of the Higgs change if it is produced at higher energies? So ...


2

That is quite a general question, and I can't give you a definite answer if you don't give me some other information. For example, are the SM fermions charged under this interaction? If so, with what charges? What are exactly the gauge representations of the new fermions? How do they appear in the Lagrangian? You see, there are MANY ways of answering these ...


2

$p^2 = m^2$ is the definition (up to a minus sign) of the mass of a momentum eigenstate. He derived that the same quantity (the expectation value of $[P^2,D]$ w.r.t. $\ket{p}$) equals $0$ and $2\mathrm{i}m^2$, so $m^2 = 0$. The $s$ is the scale parameter of the scale transformation induced by $D$, and it is any real number, so, starting from a given state ...


4

Yes, the vacuum energy of a spacetime lattice with finite spacing and periodic boundary conditions within a box of finite size is finite. One would not call this "quantizing", though, rather discretizing because we are not carrying out any "quantization procedure" in the sense of going from a classical to a quantum system. In this approach, the finite size ...


0

In the standard model, there is no elementary spin 0 boson being electrically charged (but there are many charged spin 0 composite particles). However, in many extensions such as supersymmetry, there are such particles: the scalar partner of the electron, the selectron carries the same charge as the electron. The anti-selectron is the spin 0 partner of the ...


0

You can refer to chap 9 of "An introduction to quantum field theory" of Peskin & Scroeder, which includes a detailed calculation of path integral using the original physical definition of path integral. After the brutal treatment, they will show you more modern treatment using generating functional.


6

The binding energy of the electrons in a silver atom is far less than the rest energy of an electron, so there is no ambiguity about the number of electrons in a silver atom. That makes adding up the spins a straightforward business. By contrast, the combined mass of the two up and one down quarks in a proton is about 10MeV (it isn't precisely known) but ...


2

Use fig 13.2 of [2] as reference. Taking the example Qmechanic uses, the idea is that $I(\omega) = \int_{\zeta_a}^{\zeta_b}\frac{Z(\omega)}{\zeta - \xi(\omega)} d\zeta$ $I(\omega)$ needs to be analytically continued from $\omega_1 \rightarrow \omega_2$. The pole of the integrand travels from $\zeta = \xi(\omega_1) \rightarrow \zeta = \xi(\omega_2)$ in ...


1

I) Let there be given a meromorphic function $\zeta \mapsto F_{w}(\zeta)$ in the $\zeta$-plane with a single (not necessarily simple) pole at the position $\zeta=\xi(w)$, where $\xi$ is a holomorphic function, and $w\in \mathbb{C} $ is an external parameter. Ref. 1 is considering the contour integral $$\tag{A}I_{\Gamma,w} ~=~ \int_{\Gamma} \! d\zeta ...


4

I think you should not take the Peskin and Schröder quote too seriously. They are likely just using the Fourier relationship "short distance <-> high momenta" and the idea that there are propagators $\langle \phi^2 \rangle$ (which are the "fluctuation"/variance of $\phi$, see this post) associated to the Feynman lines of virtual particles, so "small ...


3

This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization. A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the ...


3

Answer to the main question: It is a well regarded fact that the terminology unified electroweak interaction is a bit of an abuse of terminology. What the term means is that both Quantum Field Theories, the Hypercharge ($U(1)_Y$) and Weak ($SU(2)_L$), are unified in a common framework, which predicts the low energy electromagnetism ($U(1)_{em}$) through the ...


0

Be careful: The spin constribution to the number current is proprotional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution ...


1

QFTs in spacetime of dimension $<4$ have their use in real applications - 3D theories to quantum surfaces, and 2D theories to quantum wires. There much of the exceptional behavior of lower-dimensional QFTs can be observed. A famous example is the fractional Hall effect with anyonic (rather than Bose or Fermi) statistics.


1

I think that the problematic part here is the notion of what demands what. For example, you state This is the standard argument by which Lorentz invariance is found to demand gauge invariance for massless particles. but I am not completely sure if I agree with this, or at least to the interpretation you are carrying with it. Taking a look at ...



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