Without electromagnetic coupling, the QM charged particle wave function is not invariant under a local gauge transformation — one with a phase that depends on space (space-time): \begin{equation} \psi \mapsto e^{-i\alpha(x)}\phi \end{equation} In order to turn on the electromagnetic coupling, we make the replacement $$\hat{\vec{p}} \mapsto \hat{\vec{p}} - \frac{ie}{c}\vec{A}$$ This means that, in the Schrodinger equation is gauge invariant because the term arising from $\vec{A}$ is cancelled through the action of $\hat{\vec{p}} = i\hbar\partial$ on the phase factor $e^{-i\alpha(x)}$ in the wave function. The math trick is clear. But physics cannot be understood from it, namely, why the requirement of local gauge invariance requires 1) the charge conservation 2) and the emergence of some kind of carrier of interaction.
122560
I am a mathematics student with a hobby interest in physics. This means that I've taken graduate courses in quantum dynamics and general relativity without the bulk of undergraduate physics courses and sheer volume of education into the physical tools and mindset that the other students who took the course had, like Noether's theorem, Lagrangian and Hamiltonian mechanics, statistical methods, and so on.  The courses themselves went well enough. My mathematical experience more or less made up for a lacking physical understanding. However, I still haven't found an elementary explanation of gauge invariance (if there is such a thing). I am aware of some examples, like how the magnetic potential is unique only up to a (time-)constant gradient. I also came across it in linearised general relativity, where there are several different perturbations to the spacetime metric that give the same observable dynamics.  However, to really understand what's going on, I like to have simpler examples. Unfortunately, I haven't been able to find any. I guess, since "gauge invariance" is such a frightening phrase, no one use that word when writing to a high school student.  So, my (very simple) question is: In many high school physics calculations, you measure or calculate time, distance, potential energy, temperature, and other quantities. These calculations very often depend only on the difference between two values, not the concrete values themselves. You are therefore free to choose a zero to your liking. Is this an example of gauge invariance in the same sense as the graduate examples above? Or are these two different concepts?
57592
In the second chapter of Peskin and Schroeder, An Introduction to Quantum Field Theory, it is said that the action is invariant if the Lagrangian density changes by a four-divergence. But if we calculate any change in Lagrangian density we observe that under the conditions of equation of motion being satisfied, it only changes by a four-divergence term.  If ${\cal L}(x) $ changes to $ {\cal L}(x) + \alpha \partial_\mu J^{\mu} (x) $ then action is invariant. But isn't this only in the case of extremization of action to obtain Euler-Lagrange equations.   Comparing this to $ \delta {\cal L}$  $$ \alpha \delta {\cal L} = \frac{\partial {\cal L}}{\partial \phi} (\alpha \delta \phi) + \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \partial_{\mu}(\alpha \delta \phi) $$  $$= \alpha \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi \right) + \alpha \left[ \frac{\partial {\cal L}}{\partial \phi} - \partial_\mu \left(\frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \right) \right] \delta \phi.    $$   Getting the second term to zero assuming application of equations of motion. Doesn't this imply that the noether's current itself is zero, rather than its derivative? That is:  $$J^{\mu} (x) =  \frac{\partial {\cal L}}{\partial \partial_{\mu}\phi} \delta \phi .$$  I add that my doubt is why changing ${\cal L}$ by a four divergence term lead to invariance of action globally when that idea itself was derived while extremizing the action which I assume is a local extremization and not a global one.
43651
I've read and heard that quantum electrodynamics is more fundamental than maxwells equations. How do you go from quantum electrodynamics to Maxwell's equations?
10872
Just as background, I should say I am a mathematics grad student who is trying to learn some physics. I've been reading "The Theoretical Minimum" by Susskind and Hrabovsky and on page 134, they introduce  transformations. Here's the first example they use:  Consider a particle moving in the x,y plane under the influence of a potential, $V$, which depends only on the radius, with Lagrangian:   $L=\frac{m}{2}(\dot{x}^2+\dot{y}^2)-V(x^2+y^2)$  This is clearly invariant under rotations:  $x \rightarrow x\cos \theta + y\sin \theta$  $y \rightarrow -x\sin \theta + y\cos \theta$  All well and good. Now they say "consider what happens ... when the angle $\theta$ is replaced by an infinitesimal angle $\delta$." Already I could say "What the heck is $\delta$ really?", but I'm willing to play along with my intuition. Since $\delta$ is infinitesimal, we work to first order and say   $\cos \delta=1$ and $\sin \delta= \delta$.  Plugging this into our rotation formulas above, we obtain:  $x \rightarrow x+y\delta$  $y \rightarrow y-x\delta$  By differentiating, we see that:  $\dot{x} \rightarrow \dot{x}+\dot{y}\delta$  $\dot{y} \rightarrow \dot{y}-\dot{x}\delta$  Plugging these into the Lagrangian and ignoring terms higher than first order, we see that the Lagrangian is invariant under this transformation.  My main problem with all of this is that I don't understand what the physical nature of an infinitesimal transformation actually is. All I got from the above was that if you do this formal calculation by following rules like "only work to first order in $\delta$," then the Lagrangian is invariant. This is in contrast to the case where we have an actual transformation, like a rotation, where there is no question about what is going on physically.  I would also like to know how all of this relates to rigorous mathematics. In mathematics, I can't recall ever using infinitesimals in an argument or calculation, so it would be useful if there was some way of formulating the above in terms of limits/ derivatives/ differential forms (for example). I sense a connection to Lie Algebras, as the infinitesimal version of the rotation is $(I+A)$ where $I$ is the identity matrix and $A$ is an element of the Lie Algebra of $SO(2)$.   Here are some questions whose answers I believe may be useful to me (feel free to answer some or all):  -What is an infinitesimal quantity like $\delta$ to the physicist?   -Why do physicists argue using infinitesimals rather than "standard" calculus?  -What is the physical meaning of infinitesimal transformation? How does it relate to Lie Algebras?   -Is there a rigorous theoretical apparatus for justifying the computations shown above?  -What is meant by the Lagrangian being invariant under infinitesimal transformations?  If any of the questions seem too vague, please say so. Thanks in advance for your insights!
54636
It is well known that scattering cross-sections computed at tree level correspond to cross-sections in the classical theory. For example the tree level cross-section for electron-electron scattering in QED corresponds to scattering of classical point charges. The naive explanation for this is that the power of $\hbar$ in a term of the perturbative expansion is the number of loops in the diagram.  However, it is not clear to me how to state this correspondence in general. In the above example the classical theory regards electrons as particles and photons as a field. This seems arbitrary. Moreover, if we consider for example $\phi^4$ theory than the interaction of the $phi$-quanta is mediated by nothing except the $phi$-field itself. What is the corresponding classical theory? Does it contain both $phi$-particles and a $phi$-field?  Also, does this correspondence extend to anything besides scattering theory?  Summing up, my question is:     What is the precise statement of the correspondence between tree-level QFT and behavior of classical fields and particles?
53679
It's an old-new question (I found only one similar question with unsatisfactory (for me) answer: )  It's strange for me how all books simply pass by such an important question, and mentioning strange and mathematically unsupported reasons such as:   orbits are stationary (while as I know this is just idealization, there is no stationary orbits in reality even for Hydrogen)  electrons are actually not localized due to uncertainty principle, thus they have no acceleration (while obviously in a non-spherically symmetric orbits a kind of "charge acceleration distribution" always exist) vacuum fluctuations play a major role (according to QED).   I'm not interested in how Bohr or Schroedinger explained it, I want to see a rigorous proof with QM, QED or maybe even the standard model as whole. I would like to see how this question was closed.
72447
When Dirac found his equation for the electron $(-i\gamma^\mu\partial_\mu+m)\psi=0$ he famously discovered that it had negative energy solutions. In order to solve the problem of the stability of the ground state of the electron he invoked Pauli's exclusion principle and postulated that negative energy states well already filled by a "sea" of electrons. This allowed him to predict the positron, viewed as a hole in the sea.  This interpretation was ultimately discarded owing to its inapplicability to bosons and difficulties with explaining the invisibility of the infinite charge of the sea.  According to my understanding, the modern argument goes something like this. There is a discrete symmetry of the lagrangian called charge conjugation $\psi \rightarrow \psi^c$ which allows the negative energy solutions to be interpreted as positive energy solutions for a second mode of excitation of the electron field with opposite charge, called positrons. The decay of electrons to positrons is then suppressed by the $U(1)$ gauge symmetry of the lagrangian forcing conservation of electrical charge.  According to this interpretation What Dirac would have missed was the lagrangian formalism. Is this historically and physically correct?
101251
This is a question I've been asked several times by students and I tend to have a hard time phrasing it in terms they can understand. This is a natural question to ask and it is not usually well covered in textbooks, so I would like to know of various perspectives and explanations that I can use when teaching.  The question comes up naturally in what is usually students' second course in quantum physics / quantum mechanics. At that stage one is fairly comfortable with the concept of wavefunctions and with the Schrödinger equation, and has had some limited exposure to operators. One common case, for example, is to explain that some operators commute and that this means the corresponding observables are 'compatible' and that there exists a mutual eigenbasis; the commutation relation is usually expressed as $[A,B]=0$ but no more is said about that object.  This naturally leaves students wondering     what is, exactly, the physical significance of the object $[A,B]$ itself?   and this is not an easy question. I would like answers to address this directly, ideally at a variety of levels of abstraction and required background. Note also that I'm much more interested in the object $[A,B]$ itself than what the consequences and interpretations are when it is zero, as those are far easier and explored in much more depth in most resources.    One reason this is a hard question (and that commutators are such confusing objects for students) is that they serve a variety of purposes, with only thin connecting threads between them (at least as seen from the bottom-up perspective).   Commutation relations are usually expressed in the form $[A,B]=0$ even though, a priori, there appears to be little motivation for the introduction of such terminology. A lot of stock is placed behind the canonical commutation relation $[x,p]=i \hbar$, though it is not always clear what it means.  (In my view, the fundamental principle that this encodes is essentially de Broglie's relation $\lambda=h/p$; this is made rigorous by the  but that's quite a bit to expect a student to grasp at a first go.) From this there is a natural extension to the Heisenberg Uncertainty Principle, which in its general form includes a commutator (and an anticommutator, to make things worse). Canonically-conjugate pairs of observables are often introduced, and this is often aided by observations on commutators. (On the other hand, the energy-time and angle-angular momentum conjugacy relations cannot be expressed in terms of commutators, making things even fuzzier.) Commutators are used very frequently, for example, when studying the angular momentum algebra of quantum mechanics. It is clear they play a big role in encoding symmetries in quantum mechanics but it is hardly made clear how and why, and particularly why the combination $AB-BA$ should be important for symmetry considerations.  This becomes even more important in more rigorous treatments of quantum mechanics, where the specifics of the Hilbert space become less important and the algebra of observable operators takes centre stage. The commutator is the central operation of that algebra, but again it's not very clear why that combination should be special. An analogy is occasionally made to the Poisson brackets of hamiltonian mechanics, but this hardly helps - Poisson brackets are equally mysterious. This also ties the commutator in with time evolution, both on the classical side and via the Heisenberg equation of motion.   I can't think of any more at the moment but they are a huge number of opposing directions which can make everything very confusing, and there is rarely a uniting thread. So: what are commutators, exactly, and why are they so important?
121309
The  (disclaimer: I don't know this wikipedia entry) is usually mentioned as the reason why gravitons may not be composite particles. I do understand the proof of the theorem, but not the previous conclusion.   The theorem states that in an interacting and Poincare invariant quantum field theory, there are not massless, spin-2 particles unless there exists a gauge symmetry — which makes the energy-momentum tensor non-covariant (actually covariant up to a gauge transformation) under Lorentz transformations in the Fock space. So the immediate conclusion of the theorem is that the existence of a massless, spin-2 particle (like a graviton) requires linearized diffeomorphisms.  My question is: why do linearized diffeomorphisms imply that gravitons are elementary particles? Or, more in general, why  the particle corresponding to a gauge field must be elementary (I know that a gauge symmetry must be exact, but why this implies that the corresponding particle must be elementary?).
25446
In talking about Higgs mechanism, the first part is always some introduction to the concept of spontaneously  symmetry breaking (SSB), some people saying that Higgs mechanism is the results of SSB of local gauge symmetry, some people says that we can formulate Higgs mechanism in a gauge invariant way, some people also says that we need only a non-zero vaccum expectation value... I am confused about this different or maybe same point of views.  In this post:  , the most highly voted answer, I still can't feel how SSB worked in Higgs mechanism. It seem that the validity of last part, the appearance of a mass term for $A$, is guaranteed if we have a non-zero equilibrium value $\phi_0$ to expand around. I do not see that the requirement that the phase of the field $\phi$ need to be fixed at some particular value to generate mass term. Thus it seems to me it is not true that SSB is really indispensable for Higgs mechanism.  To put it simply:  The spontanously breaking of what is attributed to Higgs mechanism?   local gauge symmetry global symmetry, since breaking of a "gauge symmetry" should not have any effect on physics. In higgs mechanism, the really broken symmetry is a global one. Mathematically, it is similar in looking as fixing a gauge, but one should not think it as a spontanously breakdown of local gauge symmetry. other.   Is SSB really indispensable for Higgs mechanism?    yes, Higgs mechanism is relied on the SSB of some symmetry (above question), the other approches of description eventually has spontanously broke some symmetry. No, the SSB is just one way to describe Higgs mechanism (or even not a complete way), what is really need is the non-zero vaccum expectation value, for example in the linked post the requirement for the mass term to occur is to have some non-zero expectation value of $\phi$ to expand around, we do not need the phase of the field to be fixed, thus the symmetry is not broken. Other.   some reference materials:    States that SSB of local gauge symmetry is impossible.  in the abstract states that:       gauge symmetries merely reflect a redundancy in the state description   and therefore the spontaneous breaking can not be an essential   ingredient. Indeed, as already shown by Higgs and Kibble, the   mechanism can be explained in terms of gauge invariant variables,   without invoking spontaneous symmetry breaking     In the introduction it says:      In particular, we emphasize that global U(1) phase rotation symmetry,   and not gauge symmetry, is spontaneously violated, and show that the   BCS wave function is, contrary to claims in the literature, fully   gauge invariant
102265
The electromagnet spectrum looks to be continuous, yet energy is discrete. Shouldn't one see evidence of the electron's "quantum leap"? What's going on?     Moderator, please size my photo. Thank you.
68412
I am trying to prove the Lorentz invariance of the (left-handed) Weyl Lagrangian: $$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi$$  A Lorentz transformation is realized as $\psi\to M\psi$, where $M=\exp[-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}]$, and $S$ are the generators of the representation, i.e., $S^{0i}=-\frac{i}{2}\sigma^i$ and $S^{ij}=\frac{1}{2}\epsilon^{ijk}\sigma^k$. Therefore, the Lagrangian transforms as $$\mathcal L=i\psi^\dagger\bar\sigma^\mu\partial_\mu\psi\to i\psi^\dagger M^\dagger\bar\sigma^\mu M\partial_\mu \psi$$ so that if it actually is invariant, we must have $M^\dagger\bar\sigma^\mu M=\bar\sigma^\mu$ (and, in particular, $M^\dagger M=\mathbb I$). I think this is a nice relation, which expresses the invariance of the symbol $\bar\sigma^\mu$.  Now the thing is, I think this relation is not true. Take, for example, a boost in the $z$-direction with rapidity $\omega$. This means $$M=\exp[-i\omega S^{03}]=\exp\left[\frac{\omega}{2}\begin{pmatrix}-1&amp; 0\\0&amp;1\end{pmatrix}\right]=\begin{pmatrix}e^{-\omega/2}&amp;0\\0&amp;e^{\omega/2}\end{pmatrix}$$  In this case, it is easly seen that $M^\dagger M\neq\mathbb I$. My question is: what went wrong in my argument?
72578
Is the Schrodinger equation covariant under Galilean transformations? I am only asking this question so that I can write an answer myself with the content found here:  and here:  I learned about these pages in a comment to  by Ron Maimon. I think Ron Maimon is the original writer of this content. This is creative commons, so it's ok to copy it here. It is not on any textbook on non-relativistic quantum mechanics that I know of, and I thought it would be more accessible (if to no one else, at least for myself) and safe here. I hope this type of question is not in disagreement with site policy.
10946
Are there any massless (zero invariant mass) particles carrying electric charge?  If not, why not? Do we expect to see any or are they a theoretical impossibility?
71288
Suppose there is an electromagnetic wave moving forward in the $\mathbf{\hat{k}}$ direction.  Its electric/magnetic field components are given by: $$\mathbf{E} = E_0 \sin(kz - \omega t) \mathbf{\hat{i}}$$ $$\mathbf{B} = B_0 \sin(kz - \omega t) \mathbf{\hat{j}}$$ If a particle of charge $q$ was lying on the wave's trajectory, the Lorentz force law says that the force is given $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. However, is an electromagnetic wave a combination of both E and B fields, requiring both fields to be plugged into the equation, or does the force on the electron only depend on one of the fields, and is  an EM wave only either an electric field or a magnetic field at one instant? Edit: changed x to z in expression for EM wave.
124191
is a question about the canonical momentum that I had asked some days ago, but I still have one point that I am not understand.  Considering a particle moves in a magnetic field with charge $q$ and mass $m$, its hamiltonian is $$H=\frac{\vec{P}^2}{2m}=\frac{(\vec{p}+q\vec{A})^2}{2m}$$ where $\vec{p}$ is the momentum of the particle, $\vec{A}$ is the vector potential of the magnetic field and $\vec{P}$ is the canonical momentum of the particle.  I think, because of the expression of the hamiltonian, the canonical momentum $\vec{P}$ is a conserved quantity.  But by the answer in , it seems that the canonical momentum is not conserved even in a simple example that a particle moves in a homogeneous magnetic field.  I am confused about this question. Is the canonical momentum conserved when a particle moves in magnetic field?
91861
For one dimensional quantum mechanics $$[\hat{x},\hat{p}]=i\hbar. $$  Does this fix univocally the form of the $\hat{p}$ operator? My bet is no because $\hat{p}$ actually depends if we are on coordinate or momentum representation, but I don't know if that statement constitutes a proof. Moreover if we choose $\hat{x}\psi=x\psi$ is the answer of the following question yes?  For the second one   $$(\hat{x}\hat{p}-\hat{p}\hat{x})\psi=x\hat{p}\psi-\hat{p}x\psi=i\hbar\psi, $$  but I don't see how can I say that $\hat{p}$ must be proportional to $\frac{\partial}{\partial x}$. I don't know if trying to see that $\hat{p}$ must satisfy the Leibniz rule and thus it should be proportional to the $x$ derivative could help. Or using the fact that $\hat{x}$ and $\hat{p}$ must be hermitian  Any hint will be appreciated.