Newest questions tagged quantum-electrodynamics - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-10-20T14:51:56Z https://physics.stackexchange.com/feeds/tag?tagnames=quantum-electrodynamics&sort=newest https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/508727 0 Why does $\zeta(3)$ appear in the expression for the anomalous magnetic moment? frauke https://physics.stackexchange.com/users/224504 2019-10-17T17:28:26Z 2019-10-17T18:06:54Z <p>The anomalous magnetic moment, to fourth order in <span class="math-container">$\alpha$</span>, contains <span class="math-container">$\zeta(3)$</span>. Is there a simple explanation for the appearance of this value of the Riemann zeta function?</p> https://physics.stackexchange.com/q/508291 1 What light "with no apparent material source" is Feynman referring to at the start of his Nobel lecture? Andrea https://physics.stackexchange.com/users/77496 2019-10-15T07:37:20Z 2019-10-15T07:37:20Z <p>Near the beginning of his 1965 Nobel lecture (end of page 2 in <a href="http://www.feynmanlectures.caltech.edu/info/other/Feynmans_Nobel_Lecture.pdf" rel="nofollow noreferrer">this pdf</a>), Feynman justifies trying to remove the degrees of freedom of the electromagnetic field since in the end the field now is always determined by some matter earlier:</p> <blockquote> <p>As a matter of fact, when we look out anywhere and see light, we can always “see” some matter as the source of the light.</p> </blockquote> <p>But right after that sentence, he mentions an exception:</p> <blockquote> <p>We don’t just see light (except recently some radio reception has been found with no apparent material source).</p> </blockquote> <p>What is this radio source he talks about?</p> <p>My first guess would be the observation of the CMB, which according to <a href="https://en.wikipedia.org/wiki/Cosmic_microwave_background#History" rel="nofollow noreferrer">wikipedia's CMB entry</a> was first observed in 1964. This same entry mentions that the interpretation was a "controversial issue", but all interpetations mentionned seem to include a matter source for the radiation, which makes me think that Feynman was maybe referring to something else.</p> https://physics.stackexchange.com/q/507930 0 What are the deciding criteria for occuring of reflection, absorption, Emission, refraction and burning? [closed] abhishek https://physics.stackexchange.com/users/231319 2019-10-13T05:29:03Z 2019-10-14T12:05:30Z <p>Why does some objects reflect, refract, absorb, emit and burn, when electromagnetic wave is thrown on them? I want to know all these things at atomic and molecular level. </p> https://physics.stackexchange.com/q/507107 0 N00b path integral formulation question Fries of Doom https://physics.stackexchange.com/users/244239 2019-10-08T15:43:39Z 2019-10-08T18:15:59Z <p>I have a question about the path integral formulation used in QED etc.</p> <p>The path integral formulation implies to me that index of refraction (and reflection or anything also calculated using this formulation) are dependent on the macro geometry of the object. In the case of a refractive block of glass, the index of refraction should change based on the size of the block, and in the case of a mirror, the actual reflected image of my face would be different a 1m^2 mirror than it would be in a 10m^2 mirror... This is non-intuitive, as the phenomena is unobserved - My face looks the same in a 1m^2 mirror as it looks in a 10m^2 mirror...</p> <p>Have I missed something? Or is it just that the coefficients in the integral have dropped off to such small numbers it makes no practical difference?</p> <p>If that is the case, would an infinitely large mirror give a different looking reflection than a 1m^2 mirror? Would it stop reflecting or get better? The integral would be including infinitely many infinitely small terms... Would it slow light down to 0?</p> https://physics.stackexchange.com/q/506754 1 Feynman’s QED: Photons travelling faster than $c$ DanHJEV https://physics.stackexchange.com/users/244082 2019-10-06T18:00:30Z 2019-10-06T19:27:48Z <p>I was reading though a Phys.SE <a href="https://physics.stackexchange.com/q/178938/244082">thread</a> on Feynman’s QED and FTL photon travel over short distances where the most popular answer explained how Feynman was not implying photons can travel faster than the conventional speed of light, and said they would be interested in a direct quote of Feynman referring to photons. I am about to start undergrad phys so I am curious as to a heuristic explanation of this as Feynman says on page 89 of QED:</p> <blockquote> <p>‘There is also an amplitude for light to go faster (or slower) than the conventional speed of light’ ... ‘It may surprise you that there is an amplitude for a photon to go at speeds faster than conventional speed,<span class="math-container">$c$</span>. The amplitudes for these possibilities are very small compared to the contribution from speed <span class="math-container">$c$</span>; in fact they cancel out when light travels over long distances.’</p> </blockquote> https://physics.stackexchange.com/q/505626 -2 Schwinger's calculation of $\alpha$ [duplicate] Jorge Cabezut https://physics.stackexchange.com/users/209350 2019-09-30T18:56:16Z 2019-09-30T18:56:16Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/32318/is-there-a-simple-explanation-for-schwingers-relation-g-2-frac-alpha-pi" dir="ltr">Is there a simple explanation for Schwinger&#39;s relation $g=2+\frac{\alpha}{\pi}+{\cal O}(\alpha^2)$ for the $g$-factor of the electron?</a> <span class="question-originals-answer-count"> 1 answer </span> </li> </ul> </div> <p>Could anyone shed some light as to how Schwinger first calculated the magnetic moment for the electron? </p> https://physics.stackexchange.com/q/505252 -1 Do γ gamma photons of pair production potential > 1.022MeV accelerate neutron decay? [duplicate] t8ja https://physics.stackexchange.com/users/242551 2019-09-28T17:42:05Z 2019-09-28T18:21:35Z <div class="question-status question-originals-of-duplicate"> <p>This question is an exact duplicate of:</p> <ul> <li> <a href="/questions/503480/interpretation-of-neutron-decay" dir="ltr">Interpretation of neutron decay [closed]</a> </li> </ul> </div> <p>If there are/have been studies testing this exact or the general version of this hypothesis, a reference link would be the preferred answer.</p> <p><a href="https://www.youtube.com/watch?v=ALquIdRPYkU&amp;t=4m4s" rel="nofollow noreferrer">Experimental phy. Leah Broussard describing current experimental setups, which I assume would probably be used.</a></p> https://physics.stackexchange.com/q/505052 0 What happens to photon in photoelectric effect Tushar soni https://physics.stackexchange.com/users/235367 2019-09-27T15:33:28Z 2019-09-27T19:07:21Z <p>In photoelectric effect, when the photon gives energy to the electron, then what happens to itself? Where does it go?</p> <p>For me the photon must take the place of electron after the electron escapes the atom. And now the photon acts as electron for the atom and that's the atomic spectra must be different for atoms in excited state</p> https://physics.stackexchange.com/q/504289 0 Fermat's Principle of Least Time MaiSwa https://physics.stackexchange.com/users/243089 2019-09-23T10:57:31Z 2019-09-23T15:54:37Z <p>I was going through the Feynman Lectures on optics when I came across an explanation for Fermat's Principle. I did not fully understand the explanation- can someone please break it down in a simpler way for me? </p> <blockquote> <p>Finally, we give a very crude view of what actually happens, how the whole thing really works, from what we now believe is the correct, quantum-dynamically accurate viewpoint, but of course only qualitatively described. In following the light from A to B in Fig. 26–3, we find that the light does not seem to be in the form of waves at all. Instead the rays seem to be made up of photons, and they actually produce clicks in a photon counter, if we are using one. The brightness of the light is proportional to the average number of photons that come in per second, and what we calculate is the chance that a photon gets from A to B, say by hitting the mirror. The law for that chance is the following very strange one. Take any path and find the time for that path; then make a complex number, or draw a little complex vector, <span class="math-container">$\rho e^{i\theta}$</span>, whose angle <span class="math-container">$\theta$</span> is proportional to the time. The number of turns per second is the frequency of the light. Now take another path; it has, for instance, a different time, so the vector for it is turned through a different angle—the angle being always proportional to the time. Take all the available paths and add on a little vector for each one; then the answer is that the chance of arrival of the photon is proportional to the square of the length of the final vector, from the beginning to the end!</p> <p><a href="https://i.stack.imgur.com/CRyOC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CRyOC.png" alt="enter image description here"></a></p> <p>Now let us show how this implies the principle of least time for a mirror. We consider all rays, all possible paths ADB, AEB, ACB, etc., in Fig. 26–3. The path ADB makes a certain small contribution, but the next path, AEB, takes a quite different time, so its angle θ is quite different. Let us say that point C corresponds to minimum time, where if we change the paths the times do not change. So for awhile the times do change, and then they begin to change less and less as we get near point C (Fig. 26–14). So the arrows which we have to add are coming almost exactly at the same angle for awhile near C, and then gradually the time begins to increase again, and the phases go around the other way, and so on. Eventually, we have quite a tight knot. The total probability is the distance from one end to the other, squared. Almost all of that accumulated probability occurs in the region where all the arrows are in the same direction (or in the same phase). All the contributions from the paths which have very different times as we change the path, cancel themselves out by pointing in different directions. That is why, if we hide the extreme parts of the mirror, it still reflects almost exactly the same, because all we did was to take out a piece of the diagram inside the spiral ends, and that makes only a very small change in the light. So this is the relationship between the ultimate picture of photons with a probability of arrival depending on an accumulation of arrows, and the principle of least time.</p> </blockquote> https://physics.stackexchange.com/q/504063 0 The concept of force carrying photons Ekrem https://physics.stackexchange.com/users/241056 2019-09-21T21:02:49Z 2019-09-22T07:21:41Z <ul> <li><p>Are the force carrying photons really something physical? </p></li> <li><p>Do they really occur during electromagnetic interactions?</p></li> <li><p><strong>Does this mean that in different mediums where the speed of light differ, the interaction speed of the electric fields also differ?</strong></p></li> <li><p>Bonus: if two electrons interacted via force carrying photon, is there a possibility that the electron and force carrying photon cause a Compton scattering?</p></li> </ul> https://physics.stackexchange.com/q/503012 3 Is it a coincidence that the maximum number of protons in a stable nucleus is on the order of $\alpha_s / \alpha_{EM}$? tparker https://physics.stackexchange.com/users/92058 2019-09-15T21:27:32Z 2019-09-15T21:27:32Z <p>The equivalent to the fine-structure constant for the strong interaction is about <span class="math-container">$\alpha_s \approx 0.5$</span> at the proton rest mass energy (and it runs pretty slowly with energy, so it's on this order for any energy on the order of the parton mass). The fine structure constant is a little bigger than <span class="math-container">$1/137$</span> at this energy scale. Therefore, in an extremely rough sense, we can maybe say that the strong-interaction attraction between protons is around 70 times stronger than the electrostatic repulsion.</p> <p>The maximum number of protons in a stable nucleus (82 plus a comparable number of neutrons, in several isotopes of lead), is on the same order of magnitude. If we believe our rough story that the electrostatic repulsion is about 70 times weaker than the strong attraction at nuclear energy scales, then we could go on to argue that it therefore takes about 80 protons to build up enough electrostatic repulsion to overcome the strong attraction and destabilize the nucleus.</p> <p>Is there any truth at all to this simple story, or this just a coincidence? Nuclear stability is an incredibly complicated subject, and obviously this explanation could at best only possibly provide a very rough rule of thumb and not a precise prediction. Does it even do that? Put another way, if we imagine cranking the electromagnetic fine-structure constant down to, say, <span class="math-container">$10^{-3}$</span> while leaving the strong coupling constant unchanged, would we expect nuclei with up to about a thousand protons to become stable? Or is the upper threshold for nuclear stability determined primarily by the value of the strong coupling constant and not the electromagnetic fine-structure constant?</p> https://physics.stackexchange.com/q/502603 3 Wick Rotation and sign of the integrand in Weinberg's book Run like hell https://physics.stackexchange.com/users/121554 2019-09-13T09:08:00Z 2019-09-14T20:05:04Z <p>I'm studying from Weinberg's QFT volume 1, chapter 11. I have a problem with equation <span class="math-container">$(11.2.7)$</span>.</p> <p>Starting from eq. <span class="math-container">$(11.2.5)$</span></p> <p><span class="math-container">\begin{align} \Pi^{\rho\sigma} (q) = \frac{-ie^2}{(2\pi)^4} \int_0^1dx \int d^4p \, [p^2 + m^2 -i\epsilon + q^2x (1-x)]^{-2} \, \\ \times \, \text{Tr} \{[-i (\not\!{p} + \not\!{q} x) + m]\gamma^\rho[-i(\not\!{p} - \not\!{q}(1-x))+m]\gamma^\sigma \} \tag{11.2.5} \end{align}</span></p> <p>And we have </p> <p><span class="math-container">\begin{align} \text{Tr} \{[&amp;-i (\not\!{p} + \not\!{q} x) + m]\gamma^\rho[-i(\not\!{p} - \not\!{q}(1-x))+m]\gamma^\sigma \} \\= 4 [&amp;-(p+qx)^\rho (p-q(1-x))^\sigma + (p+qx)\cdot(p-q(1-x)) \eta^{\rho\sigma}\\ &amp;- (p+qx)^\sigma (p-q(1-x))^\rho + m^2 \eta^{\rho\sigma}] \tag{11.2.6} \end{align}</span></p> <p>Looking at the first term of <span class="math-container">$(11.2.6)$</span> the <span class="math-container">$0-0$</span> component is, for example proportional to </p> <p><span class="math-container">$$\Pi^{00} (q)\propto -i \int d^4p (-p^0 p^0) \tag{1}$$</span></p> <p>Then he performs a Wick Rotation <span class="math-container">$p^0= ip^4$</span>; <span class="math-container">$d^4p = i (d^4p)_E$</span> and obtains the following result:</p> <p><span class="math-container">\begin{align} \Pi^{\rho\sigma} (q) = \frac{4e^2}{(2\pi)^4}&amp; \int_0^1dx \int (d^4p)_E \, [p^2 + m^2 + q^2x (1-x)]^{-2} \, \\ \times \, [&amp;-(p+qx)^\rho (p-q(1-x))^\sigma + (p+qx)\cdot(p-q(1-x)) \eta^{\rho\sigma}\\ &amp;- (p+qx)^\sigma (p-q(1-x))^\rho + m^2 \eta^{\rho\sigma}] \tag{11.2.7} \end{align}</span></p> <p>And he says that <span class="math-container">$\eta$</span> can be interpreted as a euclidean metric, therefore he has that the term <span class="math-container">$4-4$</span> is</p> <p><span class="math-container">$$\Pi^{44} (q)\propto \int (d^4p)_E \, (-p^4 p^4) \tag{2}$$</span></p> <p>while I obtain that it should be </p> <p><span class="math-container">$$\Pi^{44} (q)\propto -\int (d^4p)_E \, (-p^4 p^4) \tag{3}$$</span></p> <p>Let me explain, from <span class="math-container">$(1)$</span> I have </p> <p><span class="math-container">\begin{align} \Pi^{00} (q)\propto -i \int d^4p (-p^0 p^0) \xrightarrow{\text{p^0=ip^4}} &amp;\Pi^{44} (q)\propto -i^2 \int (d^4p)_E \, (-i^2 p^4 p^4)\\ &amp;= \int (d^4p)_E \, (p^4 p^4) = -\int (d^4p)_E \, (-p^4 p^4) \end{align}</span></p> <p>Which is equal to <span class="math-container">$(3)$</span> and different from what Weinberg has, which is (2). </p> <p>Since I've seen something similar in another Padmanabhan's book too, there must be something basic that I'm missing here and it can't be a typo in the book, can you please help me and point it out?</p> <p>EDIT: Is it possible that he's just changing the measure from Minkowskian to Euclidean but not applying the transformation to the integrand? </p> https://physics.stackexchange.com/q/502590 0 Is Lamb Shift the field correction of Darwin term? Amirali Zandie https://physics.stackexchange.com/users/240793 2019-09-13T07:24:13Z 2019-09-13T07:24:13Z <p>Darwin Term comes from the low-energy approximation of the relativistic Dirac equation for hydrogen electron by Charles Darwin and the "Zitterbewegung" effect. Though apparently there is no zitterbewegung in QED, in most hydrogen electron diagrams the Darwin Term and the Lamb Shift are two distinct corrections to the non-relativistic energy levels. Is lamb shift the quantum field correction and more exact calculation of the phenomenon? As it seems both affect orbitals with non-vanishing probability at nucleus and lamb shift will also affect the states with ℓ≠0 when higher orders of the QED corrections are considered.</p> https://physics.stackexchange.com/q/502222 0 What causes a light bulb to fluoresce? Ian https://physics.stackexchange.com/users/76310 2019-09-11T02:58:39Z 2019-09-12T20:49:52Z <p>When an incandescent light bulb or fluorescent light bulb is pumped up with energy from the power grid, presumably light is subsequently emitted via transitions from the molecular excited state to it's ground state. What is causing this emission? Is it caused by quantum fluctuating vacuum electric fields that couple the molecular eigenstates, or is the emission driven by classical electric fields due to collisions with other molecules, the electrical current from the power grid, etc...?</p> https://physics.stackexchange.com/q/501200 2 Eliminating longitudinal photons Ian https://physics.stackexchange.com/users/76310 2019-09-10T22:42:08Z 2019-09-10T23:28:45Z <p>In Feynman's paper "Mathematical Formulation of the Quantum Theory of Electromagnetic Interaction" he writes that the semiclassical Coulomb interaction can be written by eliminating longitudinal photons. I believe this was done by Fermi and Bethe in 1932, however there paper was written in German so is hard to read. Does anyone have an explanation or reference for how to eliminate longitudinal photons (in the Coulomb gauge I presume) in QED?</p> https://physics.stackexchange.com/q/501047 0 Nature of electromagnetic absorption by quantum and classical systems Ian https://physics.stackexchange.com/users/76310 2019-09-09T23:57:11Z 2019-09-10T09:33:35Z <p>Forgive me for lack of formality and possibly incorrect understanding, but hopefully someone can both help to explain the intuition and also add mathematical formalism.</p> <p>In classical electrodynamics, absorption of electromagnetic radiation by a solid, for example a collection of classical dipoles, may be thought of as a destructive interference effect. Oscillating light that is incident on a dielectric material will cause out of phase oscillation of the charges in the material, which will in turn radiate more electromagnetic fields. The incident and radiated fields conspire at lowest order to provide the exponential decay of the Beer-Lambert law. Note that this is an interference effect that relies on the <strong>addition</strong> of fields.</p> <p>Now in quantum electrodynamics, where we consider a quantum system interacting with the quantized electromagnetic field, absorption can be considered rather to be a <strong>multiplicative</strong> effect, whereby the quantum state is multiplied by an annihilation operator which removes one quantum from the electromagnetic field.</p> <p>Can someone help me rationalize the difference? I’m puzzled that quantum absorption seems to rely on “multiplicative interference” rather than “additive interference?” Thanks!</p> https://physics.stackexchange.com/q/499191 0 Mean square values of vacuum fluctuations of the radiation field yztsz https://physics.stackexchange.com/users/178814 2019-08-28T15:33:45Z 2019-08-28T15:41:37Z <p>It is well known that the mean square values of the electric and magnetic fields of vacuum fluctuations is given by(Welton, Phys.Rev.74, 1157)</p> <p><span class="math-container">$${\langle {E^2}\rangle _{Av}}={\langle {B^2}\rangle _{Av}} =\frac{{2\hbar c}}{\pi }\int_0^\infty {{k^3}} dk$$</span> What about the square value of the vector potential? Obviously, it seems that <span class="math-container">$${\langle {E^2}\rangle _{Av}} = \frac{{{\omega ^2}}}{{{c^2}}}{\langle {A^2}\rangle _{Av}}$$</span> Hence, <span class="math-container">$${\langle {A^2}\rangle _{Av}} = \frac{{\hbar c}}{{2{\pi ^2}}}\int_0^\infty {\frac{1}{k}} {d^3}k$$</span> But according to Weisskopf(Eq.25, Phys.Rev.56, 72), it is twice my result. Why？</p> https://physics.stackexchange.com/q/498881 0 Quantum Electrodynamics explanation for refraction [duplicate] Saurabh Uday Shringarpure https://physics.stackexchange.com/users/29310 2019-08-26T21:53:49Z 2019-08-27T14:51:44Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/2041/how-are-classical-optics-phenomena-explained-in-qed-snells-law" dir="ltr">How are classical optics phenomena explained in QED (Snell&#39;s law)?</a> <span class="question-originals-answer-count"> 3 answers </span> </li> <li> <a href="/questions/83105/explain-reflection-laws-at-the-atomic-level" dir="ltr">Explain reflection laws at the atomic level</a> <span class="question-originals-answer-count"> 4 answers </span> </li> <li> <a href="/questions/6428/snells-law-starting-from-qed" dir="ltr">Snell&#39;s law starting from QED? [duplicate]</a> <span class="question-originals-answer-count"> 3 answers </span> </li> </ul> </div> <p>I am trying to understand the fundamentals of Quantum Electrodynamics through the simple example of refraction. Let's start with a plane wave coherent state (or a number state) impinging on a boundary along positive <span class="math-container">$x$</span> axis with momentum <span class="math-container">$\hbar k_0$</span>. The initial state is then given by <span class="math-container">$|\psi\rangle=|\alpha\rangle_{k_x=k_0,k_y=0,k_z=0}|0\rangle|0\rangle|0\rangle\dots$</span></p> <p>For the free field case, we have the Hamiltonian given simply by <span class="math-container">$\hat{H}=\hbar \omega (\hat{a}^{\dagger}\hat{a}+1/2)$</span> so in the Schrodinger picture the initial state simply gains becomes <span class="math-container">$|\psi(t)\rangle=|\alpha e^{-i\omega t}\rangle$</span> due to time evolution under this Hamiltonian. </p> <p>Now, using interaction picture for convenience, what would the Hamiltonian look like for the case when we have another medium with refractive index, say <span class="math-container">$n$</span>, which is at an angle to the <span class="math-container">$x$</span> axis, as shown below? Using this Hamiltonian how can we see that the light is bent/scattered into a different <span class="math-container">$\vec{k}$</span> mode?<a href="https://i.stack.imgur.com/dq63z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dq63z.png" alt="enter image description here"></a> </p> https://physics.stackexchange.com/q/498261 0 Can the electromagnetic field be defined for the Compton effect? Sergio Prats https://physics.stackexchange.com/users/30715 2019-08-22T22:15:34Z 2019-08-23T04:21:38Z <p>As far as I know, the Compton Effect can only be explained by using Quantum Electrodynamics, given an incoming electromagnetic wave there is a probability that a photon with different wavelength than the incoming wave is generated, this effect cannot be explained with classical electrodynamics, which only changes the wave direction but not its wavelength.</p> <p>My question is, given an incoming wave, we can get the Compton probability density for each possible generated photon, so, can we use these probabilities to figure out the classical electromagnetic field that the Compton Effect will cause?</p> https://physics.stackexchange.com/q/498054 0 Schwinger paper about charge in QED Artem Alexandrov https://physics.stackexchange.com/users/134813 2019-08-21T19:04:52Z 2019-08-21T19:21:20Z <p>I would like to find the paper by Schwinger where he discusses analytic properties of physical values in the limit <span class="math-container">$e\rightarrow 0$</span> and physical meaning of <span class="math-container">$e^2$</span> sign. I know only that this paper was written in 1952 but I cast some doubts about my results in Google Scholars.</p> <p>Also, it will be amazing if anybody recommends modern papers/reviews about analytic properties of physical values at <span class="math-container">$e=0$</span>.</p> https://physics.stackexchange.com/q/497152 1 Why are there two photons in pair production Feynman diagram? [duplicate] Ben https://physics.stackexchange.com/users/160629 2019-08-16T12:23:11Z 2019-08-16T16:38:28Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/266507/photon-pair-production-at-relativistic-speeds" dir="ltr">Photon pair production at relativistic speeds?</a> <span class="question-originals-answer-count"> 2 answers </span> </li> <li> <a href="/questions/22916/why-cant-a-single-photon-produce-an-electron-positron-pair" dir="ltr">Why can&#39;t a single photon produce an electron-positron pair?</a> <span class="question-originals-answer-count"> 7 answers </span> </li> <li> <a href="/questions/13513/spontaneous-pair-production" dir="ltr">Spontaneous pair production?</a> <span class="question-originals-answer-count"> 5 answers </span> </li> <li> <a href="/questions/453657/why-can-a-particle-decay-into-two-photons-but-not-one" dir="ltr">Why can a particle decay into two photons but not one?</a> <span class="question-originals-answer-count"> 2 answers </span> </li> </ul> </div> <p>Given</p> <p><a href="https://i.stack.imgur.com/fYmoQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fYmoQ.png" alt="https://www.researchgate.net/publication/323384550_ELECTROWEAK_INTERACTION_AND_ITS_TESTS_Bachelor_Thesis"></a></p> <p>I wonder why are there two photons entering in a) pair production? Isn't it one photon resulting in <span class="math-container">$e^+e^-$</span> ?</p> https://physics.stackexchange.com/q/496663 6 Are photons emitted by a magnet? Luke https://physics.stackexchange.com/users/106906 2019-08-13T21:30:41Z 2019-08-14T18:14:33Z <p>If you put a photon detector near a magnet (with the magnetic field static in time), is there some probability that the photon detector will detect a photon?</p> <p>Does QFT not predict that a photon could be detected? If we have a uniform static classical <span class="math-container">$\vec{B}$</span> field in the <span class="math-container">$z$</span> direction, e.g. <span class="math-container">$A_{classical}^{\mu}=(0,-By,0,0)$</span> and we couple electrons to this classical field by adding the term <span class="math-container">$H_{interaction}=\bar{\psi}\gamma_{\mu}\psi A^{\mu}_{classical}$</span> to the QED Lagrangian, then there are then Feynman diagrams which seem to indicate that the classical field can produce real photons. (e.g. the Feynman diagram where the classical source emits a photon which fluctuates into an <span class="math-container">$e^{+}e^{-}$</span> bubble with a real photon emitted off of the bubble). </p> https://physics.stackexchange.com/q/496334 7 How can we deduce that a hydrogen atom is stable in relativistic QED? Chiral Anomaly https://physics.stackexchange.com/users/206691 2019-08-12T04:17:24Z 2019-08-12T06:04:52Z <p>Consider relativistic quantum electrodynamics (QED) with three quantum fields: the electromagnetic field <span class="math-container">$A_\mu$</span>, one fermion field <span class="math-container">$\psi$</span> for electrons/positrons, and one fermion field <span class="math-container">$\psi'$</span> for protons/antiprotons. The protons/antiprotons are treated as elementary in this model, just like electron/positrons but with opposite charge and larger mass. The lagrangian is <span class="math-container">\begin{align*} L &amp;\sim \overline\psi\gamma^\mu(i\partial_\mu-A_\mu)\psi +m\overline\psi\psi \\ &amp;+ \overline\psi'\gamma^\mu(i\partial_\mu-A_\mu)\psi' +m'\overline\psi'\psi' \\ &amp;-\frac{1}{4q^2}F_{\mu\nu}F^{\mu\nu} \end{align*}</span> where <span class="math-container">$q$</span> is the elementary charge. In this model, the electron/positron electric charge observable <span class="math-container">$Q$</span> and the proton/antiproton electric charge observable <span class="math-container">$Q'$</span> are separately conserved. The justification for this claim is given below. The lowest-energy state in the <span class="math-container">$(Q,Q')=(-q,+q)$</span> subspace is <strong>presumably</strong> the state with a single hydrogen atom having zero total momentum. If that presumption is true, it would immediately prove that the hydrogen atom is stable (in this model), because there is no state with lower energy into which it could decay without violating the <span class="math-container">$Q$</span> or <span class="math-container">$Q'$</span> conservation laws.</p> <p><strong>Question:</strong> Is that presumption true? Is the lowest-energy state in the <span class="math-container">$(Q,Q')=(-q,+q)$</span> subspace really a zero-momentum hydrogen atom? </p> <p>I'm sure it is true, but my confidence is based on intuition from non-relativistic models, along with some reputable but hand-waving effective-field-theory arguments about how those non-relativistic models are related to relativistic QED. It's the hand-waving part that bugs me. Such arguments have stood the test of time, and I've often used them myself without flinching, but mathematically it's a weak link. I'm looking for a more <em>mathematical</em> argument for the stability of hydrogen that uses relativistic QED itself, specifically using the model described above.</p> <hr> <p>Comments:</p> <ul> <li><p>There are other questions on Physics SE about the stability of hydrogen, but the ones I found all either use a non-relativistic approximation (sometimes with relativistic perturbations) or address peripheral issues using perturbation theory. I'm asking a non-perturbative question and seeking a non-perturbative answer.</p></li> <li><p>Relativistic QED can be constructed non-perturbatively with no mathematical ambiguity if we replace continuous space with a finite lattice. (We can make the lattice so fine and so large that it might as well be continuous and infinite.) So I think the question is well-posed, except that I haven't defined what I mean by "hydrogen atom" ... </p></li> <li><p>By "hydrogen atom," I mean a state in which the correlation function between the charge densities <span class="math-container">$j_0(x)$</span> and <span class="math-container">$j_0'(y)$</span> falls off exponentially with increasing spacelike distance <span class="math-container">$|x-y|$</span>, but falls off more slowly (if at all) when <span class="math-container">$|x-y|$</span> is less than a characteristic scale that can be identified as the size of the atom. The charge densities <span class="math-container">$j_0$</span> and <span class="math-container">$j'_0$</span> correspond to the charges <span class="math-container">$$Q=\int d^3x\ j_0(x) \hskip2cm Q'=\int d^3x\ j'_0(x).$$</span> I'm not quite happy with this definition of "hydrogen atom," but hopefully the intent is clear.</p></li> <li><p>I claimed that <span class="math-container">$Q$</span> and <span class="math-container">$Q'$</span> are separately conserved in this model. Here's the basis for that claim. The model has two independent global <span class="math-container">$U(1)$</span> symmetries, namely <span class="math-container">$$\psi(x)\to\exp(i\theta)\psi(x) \hskip2cm \psi'(x)\to\exp(i\theta')\psi'(x)$$</span> The transformation parameters <span class="math-container">$\theta$</span> and <span class="math-container">$\theta'$</span> are independent of each other because the model doesn't have any terms in which these two fields are multiplied together. They are coupled only via the electromagnetic field, which isn't affected by these global transformations. Using Noether's theorem, we get two separately conserved currents associated with these two independent symmetries. We can also verify these conservation laws directly, by using the Heisenberg equations of motion to calculate <span class="math-container">$\partial_\mu j^\mu$</span> and <span class="math-container">$\partial_\mu (j')^\mu$</span>, with <span class="math-container">$$j^\mu\sim q\overline\psi\gamma^\mu\psi \hskip2cm (j')^\mu\sim q\overline\psi'\gamma^\mu\psi'.$$</span></p></li> <li><p>Of course, <span class="math-container">$Q$</span> and <span class="math-container">$Q'$</span> are not separately conserved in the real world (a neutron can decay into a proton, electron, and neutrino), but that's because there is more to the real world than QED. I'm asking this question in the context of QED. A similar comment applies to the possibility of proton instability, which may occur in the real world but not in the model described above.</p></li> </ul> https://physics.stackexchange.com/q/496027 0 Need some literature recommendation - Quasi-Free electron scattering (Nuclear Physics) Matej Bajec https://physics.stackexchange.com/users/238733 2019-08-09T17:48:55Z 2019-08-09T18:19:17Z <p>I'm not sure if this is the right place for this kind of questions, but here it goes. I'm a <em>last year undegrad</em> and I'm currently working on a project with my professor. I just finished a smaller project, as an introduction, regarding elastic electron scattering on a 12C nucleus. Just so I'm familiar with some of the terminology. I began working on my actual assignment (longitudinal and transverse 12C nucleus response from e- scattering) earlier this week and I started to experience some problems. </p> <p>Most of my info comes from articles, but I just can't understand most of the theory behind their result interpretations. <strong>I'd like to ask, if anyone knows some literature (undergrad-grad level) for inelastic/quasi-free electron scattering from nucleus.</strong> I've done courses on special relativity, QM, nuclear physics, and have some basic scattering information from my intro. More specifically, I'd like to know more about virtual photon polarization and electric and magnetic form factors (I assume QED and relativistic QM is needed for understanding these concepts?)</p> <p>As help, perhaps, Feynman's book The theory of fundamental processes was a bit too advanced for me, while this article <a href="https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.80.189" rel="nofollow noreferrer">https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.80.189</a> seems very insightful, but I'd preferably liked some additional (more 'basic'?) info to build my intuition from.</p> https://physics.stackexchange.com/q/495800 2 How Gupta-Bleuler condition implies $(a_p^3-a_p^0)| \phi \rangle=0$? eigenvalue https://physics.stackexchange.com/users/150163 2019-08-08T05:40:31Z 2019-08-08T07:23:05Z <p>Gupta-Bleuler condition is <span class="math-container">$$\partial^\mu A_\mu^+ | \phi \rangle=0\tag{6.54}$$</span> where <span class="math-container">$$A_\mu^+= \int\frac{d^3\mathbf p}{(2\pi)^3 \sqrt{2|\mathbf p|}} \sum_{\lambda=0}^3 \epsilon^\lambda_\mu a_p^\lambda e^{-ip\cdot x}.$$</span></p> <p>David Tong's <a href="http://www.damtp.cam.ac.uk/user/tong/qft.html" rel="nofollow noreferrer">QFT lecture note</a> claims that this implies <span class="math-container">$$(a_p^3-a_p^0)| \phi \rangle=0.\tag{6.56}$$</span> I can't see how to obtain that. When I apply the Gupta-Bleuler condition and substitute the above equation, I obtain <span class="math-container">$$\int\frac{d^3\mathbf p}{(2\pi)^3 \sqrt{2|\mathbf p|}} \sum_{\lambda=0}^3 (-ip^\mu)\epsilon^\lambda_\mu a_p^\lambda e^{-ip\cdot x}| \phi \rangle=0$$</span> Since <span class="math-container">$\epsilon^1, \epsilon^2$</span> are transverse photon and they are orthogonal to <span class="math-container">$p$</span>, only <span class="math-container">$\lambda=0$</span> and <span class="math-container">$\lambda=3$</span> term contributes. Then a sufficient condition for Gupta-Bleuler condition is that <span class="math-container">$$\{(p\cdot \epsilon^0)a_p^0-(p\cdot\epsilon^3)a_p^3\}|\phi\rangle=0$$</span> but this is different from the lecture note!</p> https://physics.stackexchange.com/q/495093 0 Could channeling effects explain cold fusion? [closed] Cristian Dumitrescu https://physics.stackexchange.com/users/31339 2019-08-03T21:08:44Z 2019-08-04T22:17:33Z <p>Anomalous QED effects related to <a href="https://doi.org/10.1016/0370-1573(85)90021-3" rel="nofollow noreferrer">channeling effects</a> in  crystal lattices, could explain an enhanced fusion cross section, related to a wide class of fusion reactions.</p> <p>Following <a href="https://www.infinite-energy.com/iemagazine/issue1/colfusthe.html" rel="nofollow noreferrer">Schwinger</a> , I consider the reaction:</p> <p>proton + deuteron   = Helium-3 + lattice energy.</p> <p>Could channelling effects in the Palladium lattice explain the alleged positive results of the original Pons-Fleischmann experiment? If not, because the energy range of the quantum systems involved is low (in the original Pons-Fleischmann experiments), could high energy protons and deuterons channeled through the Palladium lattice lead to enhanced fusion rates?</p> https://physics.stackexchange.com/q/495064 0 Gamma radiation incident on material generating $e^+/e^-$ pair and Schwinger Limit Riddick State https://physics.stackexchange.com/users/237846 2019-08-03T17:47:41Z 2019-08-03T18:06:32Z <p>Does a gamma photon incident upon a material creating an electron-positron pair plus kinetic energy transfer momentum to an EM field and if so does the intensity approach the Schwinger Limit during the short duration prior to the pair generation? If true, doesn't this validate the predictions that pairs are created by and EM field near the Schwinger limit?</p> https://physics.stackexchange.com/q/493611 1 Why does light come as quanta of the harmonic oscillator? user2757771 https://physics.stackexchange.com/users/199017 2019-07-25T17:57:00Z 2019-07-25T21:25:50Z <p>I've recently been learning the basics of Quantum optics and it seems to be a fundamental concept that light is best described in the framework of the Quantum Harmonic Oscillator.</p> <p>This lead to a relation for the Hamiltonian which is not clear to me <span class="math-container">$$\int \frac{\varepsilon_{0}}{2}\left(\varepsilon \hat{E}^{2}+\frac{c^{2}}{\mu} \hat{B}^{2}\right) \mathrm{d} x=\sum_{k} \hbar \omega_{k}\left(\hat{a}_{k}^{\dagger} \hat{a}_{k}+\frac{1}{2}\right)$$</span></p> <p>Why must every particle be treated as an identical H.O., is this just a good model of is there more mathematical significance that I'm missing? </p> https://physics.stackexchange.com/q/493137 0 Why are 2 gamma photons created? [duplicate] John Hon https://physics.stackexchange.com/users/115337 2019-07-23T08:40:55Z 2019-07-23T09:13:11Z <div class="question-status question-originals-of-duplicate"> <p>This question already has an answer here:</p> <ul> <li> <a href="/questions/216886/positron-electron-annihilation-can-more-than-two-photons-be-created" dir="ltr">Positron-electron annihilation - can more than two photons be created?</a> <span class="question-originals-answer-count"> 2 answers </span> </li> </ul> </div> <p>When a positron and an electron come together they annihilate and produce 2 gamma photons <span class="math-container">$$e^+ +e^- \rightarrow 2\gamma$$</span></p> <p>I can understand that they must be produced in pairs to conserve momentum. However, that still leaves all other even numbers. Why 2 photons, why not 4,6, 8, 10... photons?</p> <p>Note it is NOT a duplicate of <a href="https://physics.stackexchange.com/questions/301658/why-are-two-photons-created-in-annihilation">Why are two photons created in annihilation?</a> because that question only asks how can mass-less particles have momentum.</p> https://physics.stackexchange.com/q/493076 5 Does the concept of the photon as a particle exist in QFT in the path integral formalism? Does the concept of a particle exist? doetoe https://physics.stackexchange.com/users/25794 2019-07-22T21:23:44Z 2019-07-23T15:14:08Z <p>In the second quantization approach to quantum field theory, how I understand it, the field is decomposed in components of definite momentum which are treated as non-interacting harmonic oscillators, which are then quantized. </p> <p>The spectrum of a quantum harmonic oscillator consists essentially of the non-negative integers, and it is natural to interpret the state corresponding to the eigenvalue <span class="math-container">$n$</span> to be the state that contains <span class="math-container">$n$</span> particles of the corresponding momentum. (Please let me know where you think that my understanding of second quantization is mistaken, if you think it is.)</p> <p>In other formalisms, especially in the path integral formalism, of course we still have a photon field, fermion fields, etc, but the nicely countable discrete excitations are not explicitly present, for as far as I can tell.</p> <p>Could we say that the concept of a particle in QFT as a discrete and countable quantity, in other words one that somewhat resembles a particle in our everyday experience, is something specific to second quantization (more generally: the formalism we're working in), rather than a concept inherent in QFT?</p>