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is there a way to get all the physics from the Lagrangian instead of using Feyman diagrams? All the Feynman diagrams are literally represented by terms in the Lagrangian. Each term in the Lagrangian has a corresponding Feynman diagram. Usually you will see terms up to quadratic powers in the fields (two vertices in the Feynman representation) or even up to ...

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In learning the Lagrangian formalism and now Feyman diagrams, when we look at Feynman diagrams we are told contruct terms at each vertex and propagator terms etc to calculate the overall amplitude and so on. But since the Lagrangian describes the physical processes, is there a way to get all the physics from the Lagrangian instead of using Feyman diagrams? ...

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The crucial point is the anticommutation of $\epsilon$ with fermion fields ($\psi,\bar\psi,\bar{c}^a,c^a$). First, we will rewrite $\mathcal{L}$ as (for simplicity, we define $B^a \equiv \xi^{-1}\partial^\mu A_\mu^a$) $$\tag{1} \mathcal{L}=-\frac{1}{4}(F^a_{\mu\nu})^2+\bar{\psi}(i\gamma^{\mu} D_{\mu}-m)\psi-\frac{\xi}{2}B^aB^a- \partial^{\mu}\bar{c}^a ... 0 Since you're trying to prove the transformation properties of these operators by acting on the vacuum, I believe there is a way to reach the conclusion of both the invariance of the vacuum and the relation$$a^\dagger_{\Lambda p}=U(\Lambda)a^\dagger_pU^{-1}(\Lambda) ,$$without assuming a priori that either of them holds. You can do this as following. ... 0 First, you need to renormalize the theory to get rid of the logarithmic divergences. Here's the completed final product, copied verbatim from my professor's lecture notes:$$ \hat{\mathcal{L}} = \frac{1}{2} ( 1 + A ) \partial_\mu \hat{\phi}_\mathrm{P} \partial^\mu \hat{\phi}_\mathrm{P} - \frac{1}{2} ( m_\mathrm{P}^2 - B ) \hat{\phi}^2_\mathrm{P} - \frac{1}{4!...

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You may be horrified or delighted to know that this question at least partially falls into the domain of interpretation. Electrons are not particles that move in classical trajectories (unless you are a devotee of Bohm). They generally don't even have a well defined position, so they don't "jump" in any physical sense. Experimentally the energy ...

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The second part of the question is about the Feynman propagators $G(p)$, and whether they are also annihilated by $(\gamma^\mu p_\mu - m)$. This is not the case; however applying this operator to $G$ does cancel the pole associated with that propagator, yielding a simple constant. So in your example, this eliminates $q_1$ and $q_2$ from the expression of the ...

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There are misconceptions in your learning experience, (you do not give your physics background). So having learned that electrons can move from one energy level to another by transmitting or taking in energy, It is not the electrons that are moving in the classical sense. It is the whole atom which has quantum mechanical solutions with energy levels ...

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The out state would be $\langle k_1^f \ldots k_n^f |= \langle 0 | a^\dagger_{k_1}\ldots a_ {k_n}^\dagger$, with $n$ number of particles in the final state. You can convince yourself (invoking an argument like in the question you reference) that, in that context, inner products between states with different numbers of particle are zero.

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Here's a short derivation, copied verbatim from my professor's lecture notes. $\mathcal{M}=i e^2 \epsilon_{j \nu} \epsilon^{* \nu}_j = i e^2 \eta_{\mu \nu} \epsilon^\mu_j \epsilon^{* \nu}_j = ie^2 ( \eta_{11} \epsilon^1_1 \epsilon^{* 1}_1 + \eta_{22} \epsilon^2_2 \epsilon^{* 2}_2 ) = -2i e^2$ Here, $\mathcal{M}$ is the invariant scattering amplitude; $\... 2 Yes - they are indeed just numbers. 1 You have to use the scalar product in your integral : $$\int d^3 x \frac{e^{i\vec{x} \cdot (\vec{P}_2-\vec{P}_1)}}{|\vec{x}|}=\int_{\mathbb{R}^{+}} \iint_{\Omega} e^{i|\vec{x}||\vec{P}_2-\vec{P}_1|\cos \theta} |\vec{x}| \sin \theta \,d|\vec{x}|d\theta d\varphi$$ Know it should be easier. 2 The anticomutation at spacelike separation does not cause problems because observables in volving fermions are always bilinear in the Fermi fields, for example,$\bar\psi \gamma^\mu \psi$, and these commute at spacelike separation. 3 Actually you (and Tony Zee) are correct and I am being an idiot. You do need a different$b$and$d$! Obviously (in retrospect)$\psi\ne \psi^\dagger$. The$d$and$b$anticommute $$\{b_k,d_{k'}\}=0\\ \{b^\dagger_k,d^\dagger_{k'}\}=0\\ \{b^\dagger_k, d_{k'}\}= 0\\ \{d^\dagger_k, b_{k'}\}= 0\\ \{b^\dagger_k, b_{k'}\}= (2\pi)^32 E_k \delta^3(k-k')\\ \{d^\... 1 Why must there always exist a real particle with the same mass of the virtual particle of a certain force field Because physics is not perturbation theory or mathematical objects in general. It is about observing nature, measuring accurately defined for this reason variables, and then finding mathematical models that fit the measurements and also are ... 1 The Weinberg-Witten theorem is used to deduce that the graviton is not composite using the fact that it has spin 2, not the other way around. 0 Okay so the consensus is that it is usual to define the Grassmann analogon of the effective action in a similar way as one does for scalar field theories. For example: It's done in the book from @Qmechanic or here in chapter 1.6.2. 1 The standard definition of the 1PI effective action applies to Grassmann-odd fields as well (up to sign-conventions), cf. above comment by user tbt. E.g. Ref. 1 defines in QED$$ \Gamma[A_{\rm cl},\bar{\psi}_{\rm cl},\psi_{\rm cl}]~=~W_c[J,\eta,\bar{\eta}]-\int\! d^4x (J^{\mu} A_{\mu} +\bar{\psi}_{\rm cl}\eta+ \bar{\eta}\psi_{\rm cl}).\tag{8.1.76}$$... 1 (Here I use z instead of p_0 for clarity). In your first equation, we take g(z) = \frac{e^{-iz(x_0-y_0)}}{z+E_p}. In particular, this makes the advanced propagator \frac{e^{-iE_p(x_0-y_0)}}{2E_p} equal to g(E_p). In principle, you could take any function g(z) such that g(E_p) is the advanced propagator, but this choice is particularly nice ... 1 As it has been already mentioned, radiative recombination is one possible keyword to look for, provided that one restricts one's search to atoms. Another place to look is inelastic collisions chapters, possiblly even in basic QM textbooks. E.g., Landau&Livshits have a chapter on inelastic scattering, but I am not sure, whether they discuss exactly this ... 1 Given that u^{(1)}(0) = \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} and u^{(2)}(0) =\begin{bmatrix} 0\\1\\0\\0 \end{bmatrix} form a (plane wave) basis for a rest-frame spinor, we can calculate \bar{u}(0)\gamma^i u(0) by direct matrix multiplication. However, you can also note that the \gamma^1, \gamma^2 and \gamma^3 in the Dirac basis only have off-... 0 It took me quite some time, but now I got it. The "double counting" that Seth Whitsitt mentioned is the important catchphrase. The discrete Fourier-Transformation that transforms the integration variables(see e.g. Wikipedia) corresponds to an invertible square matrix. Usually this matrix takes values from the complex numbers to the complex numbers. ... 2 I can't read Hobson's mind, but I think that when he says that the field carries information about the slits what he means is that whatever you want to calculate about the field depends on the geometry of the slits. Another way to look at it: given an interference pattern on a screen it is possible to infer what pattern of slits caused it. The field ... 0 The double slit experiment with presumed material particles, such as electrons, demonstrates that the inner reality of nature is represented by waves. In quantum field theory (QFT) the fields are regarded as most fundamental than particles, which are described as oscillations (quanta) of the fields. Note: The nodes refer to the interference pattern. 1 Physics aside, quadratic action is just a Gaussian integral:$$ \int e^{-\frac{1}{2}\sum_{i,j=1}^nA_{ij}x_i x_j + \sum_{i=1}^nB_ix_i}d^nx = \int e^{-\frac{1}{2}\mathbf{x}^T\mathbf{A}\mathbf{x} + \mathbf{B}^T\mathbf{x}}d^nx \sqrt{\frac{(2\pi)^n}{\det \mathbf{A}}}e^{\frac{1}{2}\mathbf{B}^T\mathbf{A}^{-1}\mathbf{B}} $$4 Being IR or UV is not a property of the fixed point by itself but rather in reference to a trajectory of the RG. Excluding the possibility of limit cycles, a complete RG trajectory must start at a UV fixed point and end at an IR fixed point. Often the UV fixed point is the trivial or Gaussian one whereas nontrivial fixed points like WF in 3d arise as IR ... 0 So I've been watching Wolfgang Ketterle's fantastic MIT open course on AMO physics, he made a comment that, although the atom is a quantum system, a classical picture can sometimes really help your intuition, with the caveat that it also lead you astray. I think this question is highly amenable to classical analogy. The 1st problem is the reaction: 2 Z's, 4 ... 0 I have since resolved my question, after calculating the correct final result using the usual Feynman techniques as well as the spinor helicity formalism. It is indeed possible that one can obtain a particular result for the squared amplitude for one Mandelstam channel using Feynman calculus, and a different result for that same channel in the spinor ... 1 The basic answer comes from the Heisenberg uncertainty principle.$$ΔEΔt>h/4π$$In Quantum Field Theory (QFT) all uncertainty relations are due to the nonzero commutators of the relevant variables. The computation of lifetimes for the various decays in the Standard Model, a QFT model, implies that all decaying particles with a lifetime will have a width,... 2 At the end of the day, everything about RG flows, in any number of coupling-dimensions, can be understood by considering general flow equations on a manifold. A flow is defined only by a vector field which is defined everywhere on the manifold. A fixed point of such a flow is always a location where the vector field vanishes. In the special case of the RG ... 2 Just evaluate the p^0 part of the integral. For x^0>y^0, we can close the contour below and enclose the pole at p^0=E_\mathbf{p} (there's a pole at this point because p is on-shell, so p^2-m^2=0). Then we evaluate the integral using Cauchy's integral formula, (I'm playing fast and loose with the i\epsilon)$$\int \frac{d p^0}{2 \pi } \frac{ie^{... 2 Let's break down each term:$\phi^3$means that each interaction vertex has three lines attached to it 2-point means the diagram has two external lines 1-loop means that any other lines need to be contracted to form a single loop Now for these conditions, as you can verify, there are two possible diagrams [source: physicspages.com, Lancaster]: But the ... 1 The free energy per unit lenth of a chiral$c=1$Dirac fermion, or a non-chiral$c=1/2$massless Majorana is $$\beta F/L = -\int_{-\infty}^{\infty} \frac{dk}{2\pi} \ln(1+ e^{-\beta v_f \hbar |k|})\\ = - \frac{1}{\pi \beta v_f\hbar }\sum_{n=1}^\infty(-1)^{n+1}\frac 1{n^2}\\ = - \frac {\pi }{12 }\frac1 {\beta v_f\hbar } \nonumber$$ For general central ... 1 Actually, there is currently a theory which treats position as the parameter and time as the operator. It is called "space-time-symmetric formalism." In it, the Heisenberg picture and Schrodinger picture basically take the position-dependence into account just as time-dependence is taken into account via those pictures in normal quantum mechanics. ... 1 No. This argument about down-grading the position operator to a variable is misleading about the structure of quantum mechanics and field theory, which are fundamentally just different realizations of the same generic structure. In quantum mechanics (including field theory) Hilbert spaces are associated to constant-time slices of our spacetime. In a field ... 0 A photon is not an observer, and can't be because its proper time is always zero and it has no frame of reference in which it is at rest. However, Unruh radiation is black body radiation, so it does include particles with charge and mass. To get an observer from this, you can get low-probability interactions among these particles that form a Boltzmann brain. ... 0 Quantum integrable models are usually defined via the inverse scattering approach, i.e., by stating an S-matrix that fulfills a number of properties and fixing the particle spectrum. For this reason, it is possible to find a quantum model that is not the quantization of a classical one. An example for a model without classical counterpart is the scaling$Z(N)...

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Fields are "fundamental entities" whether there are particles or not: Particles are just excitations of fields. In their vacuum state quantum fields fluctuates according to Heisenberg uncertainty principle for time and energy. So the vacuum state of the gravitational field would be like seafoam (like Jean-Pierre Luminet's analogy). By definition of ...

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It's sort of both. Spin directly corresponds to rotations, but through them also does define the behaviour of boosts. I can only remember a loose verbal description rather than the rigorous maths, but I feel that might be what you really want anyway. The relevant thing to look up if you can be bothered is "Wigner's classification" of the ...

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Yes. You need a field $\lambda(x)$ in @Alexandro Nikolaenko's answer above. One lambda for each site or point $x$. $$\int d[\lambda(x)]e^{\int dx \lambda(x)(b^\dagger(x) b(x)+...)}$$

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The fact that in your two equations there's $|0_M \rangle$ comes from what you want to measure. The first one is trivial to interpret: It is the number of particles in the Minkowski vacuum for the Minkowski observer. But the second one is a bit tricky: It is the number of particles in the Minkowski vacuum from the point of view of the Rindler observer. You ...

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yes, but in quantum theory the word 'static' here means that there is no evolution over time (except a global phase), it does not mean that a sequence of observations will all yield the same result yes; 'indivisible' in the sense that it is continuous and you can't separate one part from another so as to leave a hole where there is no quantum field yes, ...

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It seems that the main culprit behind OP's question is the mathematical notation $u[f]$ where $u$ is a distribution, and $f$ is a test function. It is not a product of $u$ and $f$. In particular, the LHS of OP's eq. (2) is not just a product of ${\rm PV}\left( \frac{1}{x} \right)$ and $f(x)$. For starters it also involves an implicitly written integration ...

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Simple scattering theory would predict exactly the sort of superposition you are describing. Your factor of $1/2$ would essentially be the modulus square of some scattering amplitude (possibly integrated over a range of angles). An electron in a metal, however, is part of a (typically strongly interacting) many-body system, and so the fact you have removed a ...

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Electrons are indivisible Within the context of condensed matter physics, electrons are indivisible particles, i.e. their particle number operator has integer eigenvalues. Therefore one cannot meaningfully speak about half of an electron. Probabilities can be less than 1 However, the setup described can be thought of as a scattering problem: a photon is ...

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The electron is an elementary particle described by a quantum mechanical wave function, normalize to 1, because the wavefunction is connected with the probability of finding the electron at (x,y,z,t), and probabilities have to be normalized to 1 by construction. All of our data validate the statement that the electron is elementary,and is always whole.

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Start from $$\langle 0|e^{-iHT} = \sum_n \langle 0|e^{-iHT} |n\rangle \langle n|= \sum_n \langle 0|e^{-iE_nT} |n\rangle \langle n|$$ from which $$\lim_{T\rightarrow \infty(1-i\varepsilon)}\langle 0|e^{-iHT}= e^{-iE_\Omega T} \langle 0 |\Omega \rangle \langle \Omega|$$ and \begin{align} \langle \Omega| =&\, \lim_{T\rightarrow \infty(1-i\varepsilon)}\...

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Below I give a general sketch of how the arguments go. For the details, see the short book written by Zamolodchikov and Zamolodchikov "Conformal Field Theory and Critical Phenomena in Two-Dimensional Systems." They are only interested in the two-dimensional case because that's where the C-Theorem can be proved, but the part of the argument I ...

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It isn’t necessary that the GUT gauge bosons obtain masses that are the same as the GUT breaking scale. If these are different there will be threshold effects that are calculable functions of the ratio of these scales and group theoretic factors related to the reps the gauge bosons transform in.

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Actually, notwithstanding the no-go result, there is a position vector for photons; but it is singular in much the same sense that spherical coordinates are singular. The issue can be best addressed by looking at the Wigner classification - but within the framework of symplectic geometry, rather than Hilbert spaces. The real meaning and import of the no-go ...

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