It just didn't happen. He all the same insists that the one-particle truncation of a quantum plain theory is perfectly consistent too causal. In particular, he repeated many times inwards his weblog post (search for the discussion "superluminal") that the relativistically modified Schrödinger's equation for 1 particle (with a foursquare root) guarantees that the moving ridge packets never spread faster than the speed of light. Oops, it's just also bad.
By these comments, Jacques says that he is ignorant most many things that I (and my instructors) considered basics of quantum plain theory since I was an undergraduate, such as:
- The special theory of relativity too quantum mechanics are consistent but their combination is constraining too has some unavoidable consequences – some basic full general properties of quantum plain theories.
- Consistent relativistic quantum mechanical theories guarantee that objects capable of emitting a particle are necessarily able to absorb them as well, too vice versa.
- For particles that are charged inwards whatever way, the existence of antiparticles becomes an unavoidable final result of relativity too quantum mechanics.
- Probabilities of processes (e.g. cross sections) that involve these antiparticles are guaranteed to hold out linked to probabilities involving the master copy particles via crossing symmetry or its generalizations.
- The duo production of particles too antiparticles becomes certainly when liberate energy \(E\gg m\) is available or when fields are squeezed at distances \(\ell \ll 1/m\) (much) shorter than the Compton wavelength.
- Only observables constructed from quantum fields may hold out attributed to regions of the Minkowski spacetime hence that they're independent from each other at spacelike separations (because they commute or anticommute).
- Wave functions that are functions of "positions of particles" unavoidably permit propagation that exceeds the speed of lite too at that spot can't hold out whatever equation that bans it. The causal propagation solely applies to quantum fields (the observables), non to moving ridge functions of particles' positions.
- Equivalently, almost all trajectories of particles that contribute to the Feynman path integral are superluminal too non-differentiable almost everywhere too this fact can't hold out avoided past times whatever relativistic version of the mathematical expressions. Causality is solely obtained past times a combination of emission too absorption, contributions from particles too antiparticles, too at the marking of quantum fields (observables).
The most well-defined disagreement is most the "relativistically corrected" Schrödinger equation\[
i\hbar\frac{\partial}{\partial t} \psi = c \sqrt{m^2c^2-\hbar^2\Delta} \psi + V(x) \psi
\] You run into that it's similar the commons one-particle equation except that the non-relativistic formula for the kinetic energy, \(E=|\vec p|^2/2m\), is replaced past times the relativistic one, \(E=\sqrt{|\vec p|^2+m^2}\), with the same Laplacian (times \(-\hbar^2\)) substituted for \(|\vec p|^2\).
Jacques believes that when you lot substitute a localized moving ridge bundle for \(\psi(x,y,z)\) at \(t=0\) too you lot hold back for fourth dimension \(t'\), it volition solely spread to the ball of radius \(t'\) away from the master copy region: it volition never propagate superluminally. Search for "superluminally" inwards his weblog post too comments. Oops, it's incorrect too embarrassingly wrong.
I recall that the simplest way to run into why he's incorrect is to realize that the equation inwards a higher house all the same has the commons non-relativistic limit. As long as you lot guarantee that \(|\vec p| \ll m\) inwards the \(c=\hbar=1\) units, the development of the moving ridge packets must hold out good approximated past times non-relativistic physics too the non-relativistic Schrödinger equation.
Consider an actual electron moving around a nucleus. In the hydrogen atom, the motion is basically non-relativistic. Consider an initial localized moving ridge bundle for the electron that has a uniform phase, is much larger than the Compton wavelength \(\hbar/mc\approx 2.4\times 10^{-12}\,{\rm m}\) (it's but \(1/m\) inwards the \(c=\hbar=1\) units) but all the same smaller than the radius of the atom. For example, the radius of the bundle is \(10^{-11}\) meters. Outside a sphere of this radius, the moving ridge part is zero.
Will this moving ridge bundle spread superluminally? You bet. By construction, the average speed is most an social club of magnitude lower than the speed of lite which is reasonably non-relativistic. So with a 1% accuracy (squared speed), too aside from the irrelevant stage linked to the additional additive shift \(E_0=mc^2\) to the energy, the moving ridge bundle volition spread similar if it followed the non-relativistic Schrödinger equation\[
i\hbar\frac{\partial}{\partial t} \psi = -\hbar^2\frac{\Delta}{2m} \psi + V(x) \psi
\] Let's ready \(V(x)=0\). OK, how do the moving ridge packets spread according to the ordinary Schrödinger equation? Let's enquire Ron Maimon – every practiced self-didact is plenty to response such questions. Well, it's simple: the Schrödinger equation is just a diffusion (or heat) equation where the top dog parameter is imaginary. If \(m\) inwards a higher house were imaginary, \(m=i\mu\), hence the solution to the diffusion equation would be\[
\rho(x,t)\equiv \psi(x,t) = \frac{\sqrt{\mu}}{\sqrt{2\pi t}} \exp(-\mu x^2/t)
\] The width of the Gaussian bundle goes similar \(\Delta x\sim \sqrt{t/\mu}\). It's really simple.
If you lot know the graph of the foursquare root, you lot must know that the speed is initially really high. The speed \(dx/dt\) scales similar the derivative of the foursquare root of time, i.e. as \(1/\sqrt{t\mu}\). For times shorter than \(1/\mu\), the speed with which the moving ridge bundle spreads unavoidably exceeds the speed of light. It's kosher that we're looking at timescales shorter than the "Compton fourth dimension scale" of the electron. We solely assumed that the spatial size of the moving ridge bundle is longer than the Compton wavelength. Whether an analogous scaling is obeyed past times the dependence on fourth dimension depends on the equation itself too the response is clearly No. The asymmetric handling of infinite too fourth dimension inwards the equation (the foursquare root is solely used for the spatial derivatives) may hold out partly blamed for that asymmetry.
Just to hold out sure, all the scalings are the same for the value of \(\mu=-im\) that is imaginary.
If you lot don't experience certainly that our non-relativistic approximation was adequate for the question, I tin laissez passer on you lot a stronger weapon: the exact solution of the equation (Schrödinger's equation with the foursquare root). What is it? Well, it's zip else than the retarded Green's part – as taught inwards the context of the quantum Klein-Gordon field. Look e.g. at Page seven of these lectures past times Gonsalves inwards Buffalo.
The retarded part is the matrix chemical element of the development operator for the one-particle Hilbert space\[
G_{\rm ret}(x-x') = \bra{x,y,z} \exp(H(t-t')/i) \ket{x',y',z'}.
\] When the particle is initially (a delta function) at the seat \((x',y',z')\) at fourth dimension \(t'\) too you lot hold back for fourth dimension \(t-t'\) i.e. you lot evolve it past times the square-root-based Hamiltonian upward to the minute \(t'\), too you lot enquire what volition hold out the aAmplitude at the seat \((x,y,z)\), the response is zip else than the retarded Green's part of the departure betwixt the ii four-vectors.
Can the retarded Green's functions hold out analytically calculated? As long as you lot include Bessel functions with your "analytically allowed tools", the response is Yes. If nosotros ready the four-vector \(x'=0\) to zero, the retarded Green's part is simply\[
G_{\rm ret}(x) = \theta(t) \zzav{ \frac{ \delta( x^\mu x_\mu ) }{2\pi} - \frac{m}{4\pi}J_1 (mx^\mu x_\mu ) }
\] For pocket-sized too large timelike or spacelike separation, the Bessel part of the initiatory off variety used inwards the aspect asymptotically is an strange part of the declaration too behaves as (the sign is OK for positive arguments)\[
J_n(z) \sim \left\{ \begin{array}{cc} \frac{1}{n!} \zav{ \frac{z}{2} }^n&{\rm for}\,\, |z|\ll 1 \\
\sqrt{\frac{2}{\pi z}} \cos\zav{ z- \frac{(2n+1)\pi}{4} }
& {\rm for}\,\,|z|\gg 1 \end{array} \right.
\] But some other lesson of the calculation is that the Green's part is nonzero fifty-fifty for \(x^\mu x_\mu\) negative, i.e. spacelike separation – although it decreases roughly as \(\exp(-m|x|)\) over at that spot if you lot redefine the normalization past times the factor of \(2E\) inwards the momentum infinite (which is a non-local transformation inwards the seat space). See the finally displayed equation on page 2 of Gonsalves:
Relativistic Causality:Gonsalves also quotes "particle creation too annihilation" too "spin-statistics connection" as the other ii unavoidable consequences of a consistent matrimony of quantum mechanics too special relativity. He refers you lot to Chapter 2 of Peskin-Schroeder to acquire these things from a well-known source.
Quantum mechanics of a unmarried relativistic gratis signal particle is inconsistent with the regulation of relativity that signals cannot move faster than the speed of light. The probability aAmplitude for a particle of volume \(m\) to move from seat \({\bf r}_0\) to \({\bf r}\) inwards a fourth dimension interval \(t\) is\[
U(t) = \bra{{\bf r}} e^{-iHt} \ket{{\bf r}_0} =
\bra{{\bf r}} e^{-i\sqrt{{\bf p}^2+m^2}t} \ket{{\bf r}_0}\sim\\
\sim \exp(-m\sqrt{{\rm r}^2-t^2}),\quad {\rm for}\,\,{\rm spacelike}\,\, {\rm r}^2\gt t^2
\]
OK, you lot mightiness ask, what's the right modification of the moving ridge equation for 1 particle that guarantees that the moving ridge bundle never spreads luminally?
There is none. The status that the bundle never spreads superluminally would violate the dubiety principle, a key postulate of quantum mechanics.
Why is it so? I tin laissez passer on you lot a simple idea. If you lot compress the particle to a pocket-sized region, \(\Delta x \ll 1/m\), much smaller than the Compton wavelength, the dubiety regulation unavoidably says \(\Delta p \gg m\), hence the motion is ultrarelativistic. You could recall that \(\Delta p\gg m\) or \(p\gg m\) is all the same consistent with \(v\leq 1\) but the evolved moving ridge packets are unavoidably far from those that minimize the production of uncertainties too as the Bessel mathematics inwards a higher house shows, the slice inwards the spacelike part just can't precisely vanish, basically due to the non-local graphic symbol of the operators.
Similar derivations could hold out made with the assistance of the Feynman path integral. The typical trajectories contributing to the Feynman propagator are superluminal too non-differentiable almost everywhere too this fact does grip fifty-fifty inwards the calculation of the propagators inwards quantum plain theory, a relativistic theory. As I discussed inwards a weblog post inwards 2012, the superluminal or non-differentiable nature of generic paths inwards the path integral is needed for Feynman's formalism to hold out compatible with the dubiety principle. Recall that nosotros bring solved a paradox: the calculation of \(xp-px\) inwards the path integral should amount to the insertion of the classical integrand \(xp-px\) to the path integral but this classical insertion is zero. The paradox was resolved cheers to the generic paths' existence non-differentiable: the fourth dimension ordering of \(x(t)\) too \(p(t\pm \epsilon)\) mattered.
So does quantum plain theory forbid you lot from sending signals to spacelike-separated regions? And how is it achieved?
Yes, quantum plain theory perfectly prohibits whatever propagation of signals superluminally or over spacelike separations. It does hence past times using the quantum fields. Quantum fields such as \(\Phi(x,y,z,t)\) too functions of them too their derivatives are associated with spacetime points too they commute or anticommute with each other when the separation is spacelike.
The null commutator agency that you lot may stair out them simultaneously – that the determination to stair out 1 doesn't influence the other or that the social club of the ii measurements is inconsequential. Just to hold out sure, the previous judgement doesn't say that these spacelike-separated measurements are never correlated. They may hold out correlated but correlation doesn't hateful causation. They're solely correlated if the correlation (mathematically described as entanglement within quantum mechanics) follows from the previous contact of the ii subsystems that bring evolved or moved to the spacelike-separated points.
The signal is that the outcomes themselves may hold out correlated but the human decisions – e.g. which polarization is measured on 1 photon – do non influence the statistics for the other photon itself at all. The existence of the "collapse" associated with the initiatory off mensuration doesn't alter the odds for the 2d mensuration – although if you lot know the result into which the initiatory off mensuration "collapsed", you lot must refine your predictions for the outcome of the 2d measurements because a correlation/entanglement could bring been present. OK, how does this vanishing of the spacelike-separated commutators grip with the fact that the packets spread superluminally? On page 27 of Peskin-Schroeder, you lot may run into that the "commutator Green's function" is a difference betwixt ii ordinary Green's functions too because those ii are equal inwards the spacelike region, the value just cancels inwards the spacelike region.
But again, the Fourier transform of the ordinary propagator such as \(1/(p^2-m^2+i\epsilon)\) does not vanish inwards the spacelike regions of the 4-vector \(x^\mu\). It cannot vanish because this seat infinite propagator knows most the correlation of fields at ii points of space. And the fields inwards nearby, spacelike-separated points are correlated, of course of education (very probable to hold out almost equal), peculiarly if they are closer than the Compton wavelength. You may sentiment this correlation as a final result of the escaping of high-momentum or high-energy quanta to infinity. Only low-momentum or low-energy quanta are left inwards the vacuum too its low-energy excitations – too because of the Fourier human relationship of \(x\) too \(p\), this absence of high-energy quanta agency that the quantum fields can't depend on the spatial coordinates also much.
You know, the message is that the ban on superluminal signals is compatible with quantum mechanics but the creation too annihilation of particles must hold out unavoidably allowed when you lot reconcile these ii principles, special relativity too quantum mechanics. Jacques Distler believes that relativistic causality industrial plant fifty-fifty inwards "QFT truncated to the one-particle Hilbert space" which but isn't right. He's actually misunderstanding the key argue why quantum plain theory was needed at all.
Try to calculate the expectation value of the commutator of ii fields \(F(x)\) too \(G(y)\) at ii spacelike-separated points \(x,y\). The fields \(F,G\) may hold out the Klein-Gordon \(\Phi\) itself or some bilinear constructed out of it, e.g. the factor of a electrical flow \(J^0\) that Distler talks most at some point. Imagine that you're calculating this commutator. You initiatory off expand \(F,G\) inwards damage of \(\Phi\) too its derivatives. Then you lot insert the expansions of \(\Phi\) inwards damage of the creation too annihilation operators. And you lot know the expectation values of the type \(\bra 0 \Phi(x)\Phi(y) \ket 0\). When you lot time-order \(x,y\), it's just the commons propagator inwards the seat space.
The precise calculation volition depend on the operators you lot select but a full general signal is true: There volition hold out lots of private damage that are nonzero for spacelike \(x-y\). Only if you lot amount all these damage – which volition selection creation operators from \(F\) too annihilation operators from \(G\) too vice versa etc., you lot tin attain the cancellation.
In particular, if you lot consider the operators \(F,G \sim J^0\), those volition comprise damage of the type \(a^\dagger a\) as good as \(b^\dagger b\) for a plain whose particles too antiparticles differ. Only if you lot include the correlators of from both particles too antiparticles matching betwixt the points \(x,y\), you lot may acquire a cancellation of the commutator (its expectation value).
In other words, the fact that a quantum plain is capable of both creating a particle too annihilating an antiparticle (which is the same for "real" fields) is absolutely vital for its powerfulness to commute with spacelike-separated colleagues!
This insight may hold out formulated inwards yet some other equivalent way. You just can't build a localized – relativistically causally well-behaved – plain operator at a given signal that would only comprise damage of a given creation-annihilation schematic type, e.g. solely \(a^\dagger a\) but no \(b^\dagger b\), solely \(a^\dagger\) but no \(b\), too hence on. Any operator that has a well-defined "number of particles of each type that it creates or annihilates" is unavoidably "non-local" too can't precisely commute with its spacelike-separated counterparts!
If you lot wanted to written report the truncation of the quantum plain theory to a one-particle Hilbert infinite where the publish of particles is \(N=1\), too the publish of antiparticles (and all other particle species) is zero, hence all "first-quantized" operators on your Hilbert infinite represent to some combination of operators of the \(a_k^\dagger a_m\) form. You annihilate 1 particle too create 1 particle. But no such combination of operators may hold out strictly confined to a part hence that it would commute with itself at spacelike-separation.
Students who bring carefully done some basic calculations inwards quantum plain theory know this fact from many "happy cancellations" that weren't obvious for some time. For example, consider the quantized electromagnetic field. Write the total liberate energy as\[
H = \int d^3 x\,\frac{1}{2}\zav{B^2+ E^2},
\] i.e. the integral of the electrical too magnetic liberate energy density. Substitute \(\vec A\) too its derivatives for \(\vec B,\vec E\), too write \(A\) too its derivatives inwards damage of creation too annihilation operators for photons. So you lot volition acquire damage of the shape \(a^\dagger a\), \(aa\), too \(a^\dagger a^\dagger\). At the end, the total Hamiltonian solely contains the damage of the \(a^\dagger a\) "mixed" type but this simplified shape is solely obtained 1 time you lot integrate over \(\int d^3 x\) which makes the damage \(a a\) too \(a^\dagger a^\dagger\) vanish because of their oscillating dependence on \(x\). If you lot solely write the liberate energy density itself, it volition unavoidably comprise the operators of the type \(aa\) too \(a^\dagger a^\dagger\) – annihilating or creating ii photons – too. And the damage of all these forms are as of import for the quantum plain to hold out well-behaved, peculiarly for the vanishing of its commutators at spacelike separations.
The broader lesson is that of import principles of physics are ultimately reconcilable but the reconciliation is oftentimes non-trivial too implies insights, principles, too processes that didn't seem to unavoidably follow from the principles separately. So the combination of relativity too quantum mechanics implies the basic phenomena of quantum plain theory – antiparticles, duo production, the inseparability of creation too annihilation, spin-statistics relations, too a few other things.
In the same way, maybe a to a greater extent than extreme one, the unification of quantum mechanics too full general relativity is possible but whatever consistent theory obeying both principles has to honour some qualitative features nosotros know from quantum gravity – as exemplified past times string theory, in all probability the solely possible precise Definition of a consistent theory of quantum gravity. In particular, dark holes must send a finite entropy, hold out practically indistinguishable from heavy particle species, too such heavy particle species must exist. The processes around dark holes too those involving unproblematic particles are unavoidably linked past times some UV-IR relationships too string theory's modular invariance is the most explicit known trial (or toy model?) of such relationships.
In combination, the known of import principles of physics are far to a greater extent than constraining than the principles are separately too they imply that the "kind of a theory nosotros need" or fifty-fifty "the precise theory" is basically unique. This strictness is ultimately practiced news. If it didn't exist, nosotros would hold out drowning inwards the infinite plain of possibilities. Because of the "bonus" strictness resulting from the combination of of import principles of physics, nosotros know that a theory combining quantum mechanics too special relativity must piece of occupation similar quantum plain theory too a theory that also respects gravity as inwards full general relativity has to hold out string/M-theory.
BTW I recommend you lot Feynman's Reason for Antiparticles (PDF, 75-minute video), a Dirac lecture past times RPF. Go to Page 8/60 or read the finally pages to run into that Feynman ended upward with the same conclusions. If you lot solely permit positive liberate energy one-particle states, you lot can't always confine the moving ridge functions within the lite cone. He makes this declaration explicitly many times.
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