# Tag Info

## New answers tagged quantum-field-theory

0

Because $\bar u \gamma^0 = u^\dagger$

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The correlation function is a whole bunch of propagators joining the four points in pairs. What Srednicki writes in equation 10.6 is just one of those terms. There are others; for example, $x_1$ can be matched with $x_2$ and $x'_1$ with $x'_2$, or you can match $x_1$ with $x'_2$ and $x_2$ with $x'_1$. The structure will be almost identical. As for the ...

0

I would like to refer to the presentation of canonical quantization of scalar fields in chapter 3 of Quantum Field Theory, Srednicki (2007). To clarify this covers canonical quantization in standard Quantum Field Theory (not Quantum Gravity) which I understand is the question. Srednicki is very careful to use a Lorentz invariant differential so you can see ...

1

The issue here is actually the validity of the model. In order for a theory to have massive fermions, the mass term needs to be invariant (we assume gauge symmetry is a good symmetry here, at least up to loop corrections). Gauge symmetry forbids Majorana mass terms so the only masses you can write down are Dirac masses. However, Dirac fermions must have $g ... 1 Programs of physics can be posed in a unified mathematical formulation as follows. Consider a parameter space M and a configuration space C. The program of physics is to calculate histories x: M->C For non-relativistic mechanics we take M to be a time interval so M=R. For a single particle C=R3 its position in space. For relativistic mechanics we switch ... 1 Let me clarify the confusion regarding the Heisenberg picture and time evolution. In Schrodinger picture, states evolve with respect to time. But in the Heisenberg picture states don't evolve with respect to time. In a coordinate system$(t,x^i)$, you choose let us say the state$|\psi(t=0)\rangle$in the Schrodinger picture as your state$|\psi\rangle$in ... 1 Very important sentence can be found in section 3.1. It roughly goes as "to preserve manifest Lorentz invariance, state vectors don't change with time but instead describe whole spacetime history of the system". In usual Shrodinger quantum mechanics you have time dependent ket$| \psi (t) \rangle$and time evolution operator acting as$e^{-i(t-t_0)H} | \psi (...

0

Let we define the action as $$S[\phi] ~:=~S_2[\phi] + S_{\rm int}[\phi], \qquad S_2[\phi] ~:=~\frac{1}{2} \phi^k \Delta_{k\ell} \phi^{\ell} , \tag{1}$$ and the partition function $$Z[J] ~=~ \int {\cal D}\phi~\exp\left\{ \frac{i}{\hbar}S[\phi] +\phi^k J_k \right\}~=~ \exp\left\{ \frac{i}{\hbar}S_{\rm int}\left[\frac{\delta}{\delta J}\right] \right\} \exp\... 2 IF the number of degrees of freedom in a theory are the same in the massive and massless limit then you can approximate particles as massless at high energies. This is because position space  n -point functions in quantum field theory do not have IR divergences (i.e., they don't blow up as you probe larger and larger distances). This means that physical ... 4 At the level of representation theory, massive and massless particles are fundamentally different. As you mentioned, massless particles have 2 d.o.f. (helicities \pm h) and massive particles have 2j+1 d.o.f. This fundamental difference is quite important and leads to many structures - such as gauge invariance. It is also not possible for this reason to ... 0 I think it may be helpful for the answer to take a look at an intermediate theory. Namely, consider a scalar field, and, instead of this field being defined on \mathbb{R} or \mathbb{R}^3, using a lattice on \mathbb{Z} or \mathbb{Z}^3, or, even better, a finite lattice on on \mathbb{Z}_N or \mathbb{Z}_N^3 with periodic boundary conditions. In ... 3 "The Program of QFT" might be specified as follows. Specify a Lagrangian L for the particle content of the model in question. For instance, a scalar field of mass m would be$$L=\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi+\frac{1}{2}m\phi^2$$or a Fermion field might be$$L=i\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi-m\bar{\Psi}\Psi$$(those are four-... 1 Collisions of a state with other particles, present at finite density, influence the life-time of the state as energy can be transferred to those other particles during inelastic collisions. This change in life-time is related to a change in the width via the uncertainty relation and gives the thermal width. There is a physical effect known as "collisional ... 1 Gauge symmetry is actually not spontaneously broken in the Higgs mechanism; this is a common misconception. See What role does "spontaneously symmetry breaking" played in the "Higgs Mechanism"?. Therefore the Mermin-Wagner theorem does not apply to the Higgs mechanism, and the Higgs mechanism is possible in 1+1D. 0 There are two different ways of dealing with infinities: There is renormalization which has nothing to do with substituting some value for the cutoff. Rather, parameters of the Lagrangian are expressed in terms of measurable quantities which effectively hides the divergences. Another way of trying to get information out of these infinities is to treat a ... 0 Ok, I thnik I figured things out myself. I will follow the conventions of Wess & Bagger. If one wants to construct a free, \mathcal N=1 SUSY theory with a complex scalar and a Weyl Fermion, then the possible transformations are dictated by representation theory of the Lorentz group, dimensionalities, and the requirement that we have a free theory and ... 0 First, note that, strictly speaking, there is no such thing as spontaneous symmetry breaking in Higgs mechanism. I mean, that below and under the Higgs scale (i.e., at scale, at which non-zero Higgs VEV appears) the lagrangian can be rewritten in a gauge invariant way. How is it possible? The answer is that there are different physical states (i.e., ... 1 A part of the supersymmetry algebra is$$ \{Q_a,~{\bar Q}_{\dot b}\}~=~-2i\sigma^\mu_{a\dot b}\partial_\mu $$which is a momentum operator p_\mu~=~-i\partial_\mu. The graded Lie algebra g~=~h~+~k$$ [h,~h]~\subset~h,~[h,~k]~\subset~k,~\{k,~k\}~\subset~h, $$where the last of these contains the above anti-commutator. This model has chiral symmetry. It ... 0 Well, such form follows from the Noether theorem, which states that if action (in Minkowski space-time)$$ S = \int d^{4}xL $$of fermions \Psi is invariant under transformation$$ \tag 1 \Psi \to e^{i\alpha}\Psi , \quad \bar{\Psi} \to e^{-i\alpha}\bar{\Psi}, $$then exists the object J with one vector indice, namely$$ \tag 2 J_{\mu} = i\frac{\partial ...

0

Let us consider what would happen if CP was actually a symmetry. Then, arguing non-perturbatively, the overlap between a $K^0$ state and a $\bar K^0$ after time $t$ would be $$\langle \bar K^0| \exp(-i H t) K^0 \rangle.$$ Let $U$ denote the operator implementing CP symmetry, such that $U^{-1}=U$. Then, since $[U,H]$ by assumption,...

1

Beautiful (in my opinion) source in which higgs mechanism nature of superconducting phenomena is discussed, is Steven Weinberg's QFT Vol. 2, sec. 21.6. Topological nature of superconucting vortices is discussed in this section. Also, there is general discussion on topological configurations in QFT, with theoretical minimum, in Chapter 23.

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I don't have enough reputation to comment but please note that Weinberg uses the (-,+,+,+) metric, which means you need a change of sign in $$e^{iP.x}$$. The field transforms as, with $x'=x+a$, $$\phi'(x')=U^{-1}(a)\phi(x')U(a)=\phi(x)$$. For (+,-,-,-) signature this is implemented by $U(a)=e^{iP.x}$, see Peskin & Schroeder, page 26 for example. In ...

1

(Note that I am only starting to study these works and I may be wrong on some points.) The paper you are quoting is indeed providing a full definition of (type II and heterotic) superstring theory (type I is missing), valid at the quantum level and for both the NS and R sectors. The definition is basically following the construction of Zwiebach (arxiv:hep-...

3

You have to keep in mind that an integral is (more or less) a sum, and therefore in a double or multiple integral, you have cross terms that have to be correctly time-ordered. The usual Riemann integral does not take in account the non-commuting nature of its integrand, so you have to make it manifest with the time-ordering symbol $\mathcal{T}$. Imagine for ...

3

Consider $$\int_0^t dt'\int_0^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right)$$ \begin{align} &= \int_0^t dt'\int_0^{t'}dt'' \mathcal{T}\left(H(t')H(t'')\right) \\ & \qquad +\int_0^t dt'\int_{t'}^{t}dt'' \mathcal{T}\left(H(t')H(t'')\right) \\&= \int_0^t dt'\int_0^{t'}dt'' H(t')H(t'') \\ & \qquad +\int_0^t dt'\int_{t'}^{t}dt''H(t'')H(t') \end{...

4

Time-ordering is needed if the Hamiltonians $H(t^{\prime})$ and $H(t^{\prime\prime})$ at different times do not commute. Example: If the Hamiltonian is $$H(t) ~=~ \left\{\begin{array}{ccl} \color{Red}{\diamondsuit}&\text{if}& t<0, \cr \clubsuit &\text{if}& t>0, \end{array} \right.$$ then the non-time-ordered product $(\int\! dt ~H(t)... 2 The evolution operator generated by a time-dependent Hamiltonian has two parameters, not just one (i.e. the initial and final time). Let's denote such evolution by$(U(t,s))_{(t,s)\in\mathbb{R}^2}$. Formally, we can write$U(t,s)=e^{-i\int_s^t H(\tau)d\tau}$(or with time-ordering if you want), but this has to be intended as follows.$(U(t,s))_{(t,s)\in\...

0

Let us consider scalar electrodynamics (Klein-Gordon-Maxwell electrodynamics) with the Lagrangian: \begin{eqnarray}\label{eq:pr6} \nonumber -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}(\psi^*_{,\mu}-ieA_\mu\psi^*)(\psi^{,\mu}+ieA^\mu\psi)-\\ -\frac{1}{2}m^2\psi^*\psi \end{eqnarray} and the equations of motion \label{eq:pr7} (\partial^\mu+...

0

I cannot find the comment you cite in the link you posted. I think the limit on the higgs mass can be easily obtained by just using the fact that $M_h^2 = \lambda v^2, \lambda < \sqrt{4\pi}$ where $\lambda$ is the Higgs self coupling. Thus one gets $M_h^2 < \sqrt{4\pi}v^2 \sim (870 GeV)^2$ The violation of the Ward Identity just requires you to add ...

2

An explicit mass term violates Gauge invariance, because left and right particles belong to different representations 2.At one loop, the lepton mass is given by $m_{1L} = M_{bare} + \Delta M_{1L}(\mu = m_{1L})$ this condition uniquely defines the bare mass. The correction $\Delta M(\mu)$ is anyway proportional to some power of the yukawa, thus is very ...

0

In any theory with chiral fermions where Left and Right fermions don't have the same charges such fermions cannot have an explicit mass therm and can acquire mass only with spontaneous symmetry breaking. In the standard Model all fermions except Right handed neutrinos are charged under the SM gauge group and all L/R counterparts live in different ...

0

When we construct bilinears with $\Gamma^a$s, the modern qft text book $\Gamma$s turn out to be useful. For a spin-$1/2$ Fermionic field $\psi$, $\bar{\psi}\psi$ transforms as a scalar, $\bar{\psi}\gamma^a\psi$ as a vector, $\bar{\psi}\gamma_5\psi$ as a pseudo scalar, $\bar{\psi}\gamma_5\gamma^a\psi$ as a pseudo vector, $\bar{\psi}\gamma^{a_1 a_2}\psi$ as a ...

1

I'd argue that the root of this is that different Gamma matrix bases give you different "good" features, and while the choice is equivalent (and amounts to a choice of basis for your Dirac spinor), which choice is "right" depends on what features you want to be obvious/trivial -- particle/antiparticle, left/right handedness, ease of performing Legendre ...

0

In 4D Minkowski space the $\Gamma^{a}$'s has standard form. As $\Gamma^{a}=\mathbf{1}_{4\times4},\gamma^{\mu},\sigma^{\mu\nu}=\frac{i}{2}[\gamma^{\mu},\gamma^{\nu}],\gamma^{5}\gamma^{\mu},\gamma^{5}$, altogether 16 matrices. Provided that $\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2\mathbf{1}_{4\times4}\eta^{\mu\nu}$.

1

I'll try to explain why there could be a critical line and not just a critical point, and hopefully that will answer your question. If you think about the Ising model, we have the standard Hamiltonian: $$-\beta H = J_1\sum_{<i,j>}s_i s_j + h\sum_{i}s_i$$ where $\sum_{<i,j>}$ is a sum over nearest neighbors. This model ...

4

If all truncated $n$-point functions vanish for $n>2$ (i.e. we are dealing with a so-called generalized free field), microcausality for vacuum expectation values and at the operator level are equivalent. The former, on its turn, follows from Lorentz invariance alone in the case of scalar (but not necessarily free) fields, as shown by Pierre-Denis MethÃ©e, ...

2

Comments to the question (v3): The time-ordered product of operators $$T[A_1(t_1)\ldots A_n(t_n)] ~:=~\sum_{\pi\in S_n} \theta \left(t_{\pi(1)} > \ldots > t_{\pi(n)} \right) (-1)^{\varepsilon_{\pi,A}} A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)}) \tag{1}$$ is graded symmetric. [Here $(-1)^{\varepsilon_{\pi,A}}$ is a sign factor in the case ...

3

Comments to the question (v4): First of all, note that a general mathematically rigorous definition of functional integrals is a well-known open problem in mathematics. One route is to try to construct the functional integral as an appropriate continuum limit of a lattice model over a discretized space-time $M$. If we for simplicity write the phase space ...

2

The diagrams you are drawing are not allowed. In the last one you have an electron going into an muon, by the emission of a photon. Try to isolate that part. If it works one way, it should also work the other way - a muon should be able to decay into an electron by the emission of a photon. This cannot happen. Muons decay to electrons in the weak interaction,...

0

The two diagrams you draw are all wrong, for the vortex is determined by Lagrangian.

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If you're only considering QED, there is only one vertex for this interaction -- electron and positron annhilate to photon, photon pair-produces muon and antimuon. Any electron, photon, muon vertex will have flavor conservation issues. If you're not JUST considering QED, there is a second diagram, identical to the first, with the photon replaced with a $... 2 For the schrodinger picture if you take a state$|\psi(t)\rangle$satisfying the schrodinger equation$i\hbar\frac{\partial }{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle$, and write a time evolution operator U such that$|\psi(t)\rangle=U(t,t_0)|\psi(t_0)\rangle$, then this gives an operator equation that$U$must satisfy namely:$$i\hbar\frac{\partial }{\... -1 There is also the confining strong interactions (quarks, gluons, color charge) which give hadrons mass independent of the Higgs (interaction, mechanism). There is also the dominant decay mode of the first detection of the Higgs boson which turned out to produce a pair of gluons, not to mention the masses of other fundamental particles / EM bosons which ... 0 That calculation has restrictions, but, one in particular should be mentioned, that master equation is supposed to be connected to this entropy, but is not necessarily, the master equation can be connected to general entropic form, and that is a fundamental idea for a more complete proff. 2 Kallen-Lehmann representation originates from the decomposition of$\hat{\phi}(x)|\Omega\rangle$into the Hamiltonian eigenstates$|\lambda_{\mathbf{p}}\rangle$. If$|\langle\Omega|\hat{\phi}(x)|\lambda_{\mathbf{p}}\rangle=0$that state doesn't give contribution into the propagator spectral representation. In the free theory$\hat{\phi}(x)|0\rangle$is ... 0 There are many ways a photon can interact with matter, but since the photon is a quantum mechanical entity, one has to define matter in the quantum regime. In dimensions commensurate to h_bar matter is composed of atoms in various combinations. One way of interacting with matter can be seen here: Atoms are modeled by electrons in orbitals around a ... 0 Consider the thing within second quantization form: the Hamiltonian contains term like$b^\dagger a_e+\text{h.c}\$, which describes the annihilation of a excitation would produce an photon, and its conjugate procedure (which you cares more): absorbing a photon and becoming excited. This kind of Hamiltonian is derived, in principally, from ...

0

Particles are represented by fluctuations in fields in quantum field theory. So if you have a photon and an electron you will have a corresponding fluctuation in each field. The two fields interact with each other and so the fluctuation in the photon field can influence the fluctuation in the electron field and disappear. If you want an analogy think of ...

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I will try in a very general sense, and then you can use your imagination, because books may not answer this for you. Every interaction takes place via some kind of force. When we push a car, we transfer our energy into the car, but the energy first coverts to force, and then goes into the car. Same way, the photon must transform into a tiny force which ...

2

It does not matter which measurement happens first, and there is no privileged frame: in both frames, after the corresponding measurements, the quantum state of one particle evolves from a sharp position, while the other particle evolves from a state with a sharp momentum. This is what "wavefunction collapse" means in that case: measuring either momentum ...

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