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February 24, 2017

Responsibility

Many years ago, when I was an assistant professor at Princeton, there was a cocktail party at Curt Callan’s house to mark the beginning of the semester. There, I found myself in the kitchen, chatting with Sacha Polyakov. I asked him what he was going to be teaching that semester, and he replied that he was very nervous because — for the first time in his life — he would be teaching an undergraduate course. After my initial surprise that he had gotten this far in life without ever having taught an undergraduate course, I asked which course it was. He said it was the advanced undergraduate Mechanics course (chaos, etc.) and we agreed that would be a fun subject to teach. We chatted some more, and then he said that, on reflection, he probably shouldn’t be quite so worried. After all, it wasn’t as if he was going to teach Quantum Field Theory, “That’s a subject I’d feel responsible for.”

This remark stuck with me, but it never seemed quite so poignant until this semester, when I find myself teaching the undergraduate particle physics course.

The textbooks (and I mean all of them) start off by “explaining” that relativistic quantum mechanics (e.g. replacing the Schrödinger equation with Klein-Gordon) make no sense (negative probabilities and all that …). And they then proceed to use it anyway (supplemented by some Feynman rules pulled out of thin air).

This drives me up the #@%^ing wall. It is precisely wrong.

There is a perfectly consistent quantum mechanical theory of free particles. The problem arises when you want to introduce interactions. In Special Relativity, there is no interaction-at-a-distance; all forces are necessarily mediated by fields. Those fields fluctuate and, when you want to study the quantum theory, you end up having to quantize them.

But the free particle is just fine. Of course it has to be: free field theory is just the theory of an (indefinite number of) free particles. So it better be true that the quantum theory of a single relativistic free particle makes sense.

So what is that theory?

  1. It has a Hilbert space, ℋ\mathcal{H}, of states. To make the action of Lorentz transformations as simple as possible, it behoves us to use a Lorentz-invariant inner product on that Hilbert space. This is most easily done in the momentum representation ⟨χ|ϕ⟩=∫d 3k→(2π) 32k→ 2+m 2χ(k→) *ϕ(k→) \langle\chi|\phi\rangle = \int \frac{d^3\vec{k}}{{(2\pi)}^3 2\sqrt{\vec{k}^2+m^2}}\, \chi(\vec{k})^* \phi(\vec{k})
  2. As usual, the time-evolution is given by a Schrödinger equation
(1)i∂ t|ψ⟩=H 0|ψ⟩i\partial_t |\psi\rangle = H_0 |\psi\rangle

where H 0=p→ 2+m 2H_0 = \sqrt{\vec{p}^2+m^2}. Now, you might object that it is hard to make sense of a pseudo-differential operator like H 0H_0. Perhaps. But it’s not any harder than making sense of U(t)=e −ip→ 2t/2mU(t)= e^{-i \vec{p}^2 t/2m}, which we routinely pretend to do in elementary quantum. In both cases, we use the fact that, in the momentum representation, the operator p→\vec{p} is represented as multiplication by k→\vec{k}.

I could go on, but let me leave the rest of the development of the theory as a series of questions.

  1. The self-adjoint operator, x→\vec{x}, satisfies [x i,p j]=iδ j i [x^i,p_j] = i \delta^{i}_j Thus it can be written in the form x i=i(∂∂k i+f i(k→)) x^i = i\left(\frac{\partial}{\partial k_i} + f_i(\vec{k})\right) for some real function f if_i. What is f i(k→)f_i(\vec{k})?
  2. Define J 0(r→)J^0(\vec{r}) to be the probability density. That is, when the particle is in state |ϕ⟩|\phi\rangle, the probability for finding it in some Borel subset S⊂ℝ 3S\subset\mathbb{R}^3 is given by Prob(S)=∫ Sd 3r→J 0(r→) \text{Prob}(S) = \int_S d^3\vec{r} J^0(\vec{r}) Obviously, J 0(r→)J^0(\vec{r}) must take the form J 0(r→)=∫d 3k→d 3k→′(2π) 64k→ 2+m 2k→′ 2+m 2g(k→,k→′)e i(k→−k′→)⋅r→ϕ(k→)ϕ(k→′) * J^0(\vec{r}) = \int\frac{d^3\vec{k}d^3\vec{k}'}{{(2\pi)}^6 4\sqrt{\vec{k}^2+m^2}\sqrt{{\vec{k}'}^2+m^2}} g(\vec{k},\vec{k}') e^{i(\vec{k}-\vec{k'})\cdot\vec{r}}\phi(\vec{k})\phi(\vec{k}')^* Find g(k→,k→′)g(\vec{k},\vec{k}'). (Hint: you need to diagonalize the operator x→\vec{x} that you found in problem 1.)
  3. The conservation of probability says 0=∂ tJ 0+∂ iJ i 0=\partial_t J^0 + \partial_i J^i Use the Schrödinger equation (1) to find J i(r→)J^i(\vec{r}).
  4. Under Lorentz transformations, H 0H_0 and p→\vec{p} transform as the components of a 4-vector. For a boost in the zz-direction, of rapidity λ\lambda, we should have U λp→ 2+m 2U λ −1 =cosh(λ)p→ 2+m 2+sinh(λ)p 3 U λp 1U λ −1 =p 1 U λp 2U λ −1 =p 3 U λp 3U λ −1 =sinh(λ)p→ 2+m 2+cosh(λ)p 3 \begin{split} U_\lambda \sqrt{\vec{p}^2+m^2} U_\lambda^{-1} &= \cosh(\lambda) \sqrt{\vec{p}^2+m^2} + \sinh(\lambda) p_3\\ U_\lambda p_1 U_\lambda^{-1} &= p_1\\ U_\lambda p_2 U_\lambda^{-1} &= p_3\\ U_\lambda p_3 U_\lambda^{-1} &= \sinh(\lambda) \sqrt{\vec{p}^2+m^2} + \cosh(\lambda) p_3 \end{split} and we should be able to write U λ=e iλBU_\lambda = e^{i\lambda B} for some self-adjoint operator, BB. What is BB? (N.B.: by contrast the x ix^i, introduced above, do not transform in a simple way under Lorentz transformations.)

The Hilbert space of a free scalar field is now ⨁ n=0 ∞Sym nℋ\bigoplus_{n=0}^\infty \text{Sym}^n\mathcal{H}. That’s perhaps not the easiest way to get there. But it is a way …

Update:

Yike! Well, that went south pretty fast. For the first time (ever, I think) I’m closing comments on this one, and calling it a day. To summarize, for those who still care,

  1. There is a decomposition of the Hilbert space of a Free Scalar field as ℋ ϕ=⨁ n=0 ∞ℋ n \mathcal{H}_\phi = \bigoplus_{n=0}^\infty \mathcal{H}_n where ℋ n=Sym nℋ \mathcal{H}_n = \text{Sym}^n \mathcal{H} and ℋ\mathcal{H} is 1-particle Hilbert space described above (also known as the spin-00, mass-mm, irreducible unitary representation of Poincaré).
  2. The Hamiltonian of the Free Scalar field is the direct sum of the induced Hamiltonia on ℋ n\mathcal{H}_n, induced from the Hamiltonian, H=p→ 2+m 2H=\sqrt{\vec{p}^2+m^2}, on ℋ\mathcal{H}. In particular, it (along with the other Poincaré generators) is block-diagonal with respect to this decomposition.
  3. There are other interesting observables which are also block-diagonal, with respect to this decomposition (i.e., don’t change the particle number) and hence we can discuss their restriction to ℋ n\mathcal{H}_n.

Gotta keep reminding myself why I decided to foreswear blogging…

Posted by distler at February 24, 2017 6:13 PM

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29 Comments & 0 Trackbacks

Re: Responsibility

What book would you recommend for undergraduate particle physics and similar level/beginner QFT?

Posted by: Rick on February 25, 2017 12:04 AM | Permalink

Re: Responsibility

So, instead of giving the Schroedinger equation solution Psi the meaning of multi-particle thing, you insist that in the relativistic case we can have a one-particle equation (and theory) too. Is one free particle something we understand the best?

Posted by: Vladimir Kalitvianski on March 11, 2017 4:32 AM | Permalink

Re: Responsibility

Not sure what your question is, but a true statement is that the classical Klein-Gordon theory, (∂ μ∂ μ−m 2)ϕ=0 (\partial_\mu\partial^\mu-m^2)\phi = 0 when quantized, yields a Hilbert space ℋ ϕ=⨁ n=0 ∞Sym nℋ \mathcal{H}_\phi = \bigoplus_{n=0}^\infty \text{Sym}^n \mathcal{H} where ℋ\mathcal{H} is the 1-particle Hilbert space described above; the Hamiltonian H=12:∫d 3r→(π 2+(∇→ϕ) 2+m 2ϕ 2): H = \tfrac{1}{2}:\int d^3\vec{r} \left(\pi^2+{(\vec{\nabla}\phi)}^2+ m^2\phi^2\right) : is the direct sum of the single-particle Hamiltonia described above, etc.

So, of course, the single particle theory had better make sense …

Posted by: Jacques Distler on March 11, 2017 10:35 AM | Permalink | PGP Sig

Re: Responsibility

You write a theory for a free particle (relativistic or not, whatever), but you use the observable properties of this particle. So observation is included already in “one-particle” picture. Observation means interaction with this particle. In other words, a “one-particle” picture is a part of the whole picture where the interaction is already on, but weak on average, i.e., it is a multi-particle “theory”, as a matter of fact.

Posted by: Vladimir Kalitvianski on March 11, 2017 12:13 PM | Permalink

Re: Responsibility

By this approach you keep compatibility with non-relativistic physics and with the majority of intro books on QFT, but may I suggest as a counterpoint taking QFT in a Wightman-like way, as about operator-valued distributions, presented as smeared by test functions, as ϕ^ f\hat\phi_f? For the free field, we construct ϕ^ f\hat\phi_f as ϕ^ f=a f *+a f †\hat\phi_f=a_{f^*}+a_f^\dagger, with commutator [a f,a g †]=(f,g)=ℏ∫f˜ *(k)2πδ(k⋅k−m 2)θ(k 0)g˜(k)d 4k(2π) 4[a_f,a_g^\dagger]=(f,g)=\hbar\int \tilde f^*(k)2\pi\delta(k{\cdot}k-m^2)\theta(k_0)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4} and with the vacuum vector as zero eigenstate of a f|0⟩=0a_f|0\rangle=0. This is the same for any bosonic field, just the positive semi-definite form (f,g)(f,g), presented in a manifestly Poincaré invariant way, is adjusted as necessary. (f,g)(f,g) fixes the two-point VEV, which fixes the free theory; we can take the test function space to be square-integrable functions, but it’s enough to take it to be functions that are smooth in real-space and in wave-number space. This approach gives us a state over the algebra that is freely generated by the ϕ^ f\hat\phi_f, which allows us to GNS-construct a Hilbert space for the free field, without having to mention tensor products or a 1-particle Hilbert space.

This is intended to adapt the rather high mathematics of, say, Lett Math Phys (2017) 107:201–222, DOI:10.1007/s11005-016-0931-x, at an elementary to intermediate level.

FWIW, I find it helpful to think of a test function approach as a signal analysis formalism, in which ϕ^ f\hat\phi_f operates on the vacuum or other states as a stochastic modulation that is fixed by the ff. Hilbert spaces are already natural in signal analysis because of the prevalence of Fourier and other transforms. Since all QFT models ultimately are grounded in experimental datasets that are (lossily compressed digital) records of signals (CERN gives us thousands of simultaneous signals, and the compression is very lossy indeed), this gives an alternative to thinking of QFT as about particles (even mentioning the word “particle” is such a bad thing to do to students). QFT is about signals, je propose.

Posted by: Peter Morgan on March 11, 2017 12:26 PM | Permalink

Re: Responsibility

I think I disagree.

  1. GEANT is remarkable program. But I think it is neither necessary nor desirable for me to include a model of the experimentalists’ detector in my QFT calculations.
  2. I don’t think you can give a satisfactory introduction to the S-matrix, without a notion of asymptotic particle states.
  3. While it’s true that smearing against test functions gives the proper setting for undestanding QFT correlation functions (or, indeed, for understanding any distribution-valued gadget), one will never get very far in life if one is unwilling to talk about the distributions, themselves.
Posted by: Jacques Distler on March 11, 2017 6:29 PM | Permalink | PGP Sig

Re: Responsibility

All three points are well-taken, as interacting QFTs are presently construed and taught. For (2), despite its effectiveness I’ve never felt very willing to think of the S-matrix as more than a weird approximation, insofar as we do not prepare or measure anything that looks remotely like an asymptotic particle state, spread evenly over all space at t=±∞t=\pm\infty, and we cannot in fact turn on and turn off the interaction adiabatically; but once we take the S-matrix to be all we can observe, then yes. For (3), from a signal analysis perspective taking the operators to be constructed as linear functionals of the test functions seems more a first approximation than a principled choice; we have to take states to be positive semi-definite linear forms over the algebra for the sake of a probability interpretation, but we do not for this reason have to take the map from test functions into the algebra to be linear (however this veers even farther into my research POV, so there’s not enough context to discuss it properly here). For (1), I would also prefer to be able to ignore whether an experimental measurement can be verified with precision to correspond accurately to a given mathematical observable (and to ignore the details of what the widths and other finer details than just the wave-number of the prepared states and measurements might be), and for initial teaching purposes I think it may be vital to ignore such details, but I worry that it may be more important not to ignore them than whether we can compute the S-matrix to 4, 5, or 6 loops.

I take the point of a good notation (we can make the scalar field about three lines, switching to the EM field adds just a different (f,g)(f,g)) to be to conceal a lot of structure, which at its best allows us to think more lucidly, however anyone expert has to understand what is concealed, how, and why. Certainly at present it is necessary for anyone learning QFT to be able to think and work in terms of the standard notation/formalism. I suppose that whether it’s worth also introducing an alternative, much more succinct notation/formalism that’s currently good only for free fields will depend on the adaptability of the students concerned.

Thank you for an interesting response.

Posted by: Peter Morgan on March 11, 2017 7:33 PM | Permalink

Re: Responsibility

Dear Jacques, I think that you overreact and it may be due to your not appreciating something.

There is a good reason why people say that one-particle relativistic quantum mechanics is no good. There is a reason why people say so. In other words, there is a refined statement of that spirit that is right.

You may truncate the Hilbert space to the one-particle subspace, indeed. But the point is that the Hilbert space of this theory isn’t equivalent to the space of wave functions that evolve locally.

To separate the one-particle Hilbert space, you need to “distinguish” the creation operators from annihilation operators in a QFT, and you can only do it non-locally, by looking at all modes of the quantum field.

In other, more physical words, the combination of relativity and QM automatically implies some phenomena characteristic of multiparticle QFT such as particles and antiparticles and their pair production. When a field is localized to the Compton wavelength or more, such things do happen. That’s why it’s not really consistent to probe the one-particle truncated theory at distances shorter than the Compton wavelength - the decoupling limit doesn’t exist.

Do you agree?

Posted by: Lubo Motl on March 12, 2017 5:01 AM | Permalink

Re: Responsibility

You may truncate the Hilbert space to the one-particle subspace, indeed. But the point is that the Hilbert space of this theory isn’t equivalent to the space of wave functions that evolve locally.

That is the statement that the Hamiltonian of the 1-particle theory, H=p→ 2+m 2H=\sqrt{\vec{p}^2+m^2}, is a pseudo-differential, rather than a differential operator.

  1. That’s true.
  2. Who cares? What’s important is that the local conservation of probability, ∂ μJ μ(r→,t)=0\partial_\mu J^\mu(\vec{r},t)=0, continues to hold. (See problem 3, above.)
Posted by: Jacques Distler on March 12, 2017 6:47 AM | Permalink | PGP Sig

Re: Responsibility

Dear Jacques, I care and everyone should care that it’s just a pseudo-differential operator because a theory for an actual classical wave that you would derive with this operator isn’t local - it generally allows (really necessitates) a superluminal propagation - so I would say that the Schr. equation with the square root involving the Laplacian isn’t even a relativistic theory, despite the fact that you built it from relativistic dispersion relations.

In QFT, one only restores locality by a combination of propagators going in both directions i.e. an interplay between particles and antiparticles.

I wrote a blog post about it on my blog. Hope you can swallow the provocative tone. In that text, I show that I disagree with you more than you are probably imagining, and I think that most HEP folks would disagree with you, too. The one-particle truncation is inconsistent even if you add any classical time-dependent potential or anything of the sort because all those things lead to particle production i.e. change of the number of particles, and render the truncation illegitimate.

Time-independent external potential is fine in your theory.

Posted by: Lubos Motl on March 12, 2017 1:07 PM | Permalink

Re: Responsibility

Dear Jacques, I care and everyone should care that it’s just a pseudo-differential operator because a theory for an actual classical wave that you would derive with this operator isn’t local …

The state of a quantum system (AKA, the wave function) is not local. Nor is it measurable. This was a point of confusion for Schrödinger in 1923, but it should not confuse anyone today.

Local observables, in this theory, evolve causally, exactly as they should. (I gave the example of the probability density, but the same remarks apply, mutatis mutandis, for any local observable.)

I wrote a blog post about it on my blog. Hope you can swallow the provocative tone.

It seems par for the course on your blog (which is why I almost never read your blog).

If you’re feeling short of things to rant about, you can write a post about how I don’t understand quantum mechanics either. Knock yourself out!

In that text, I show that I disagree with you more than you are probably imagining, and I think that most HEP folks would disagree with you, too.

I would be worried about more than the writers of undergraduate particle physics textbooks, if I thought your opinions were widely held among my theorist colleagues.

Posted by: Jacques Distler on March 12, 2017 3:06 PM | Permalink | PGP Sig

Re: Responsibility

I read your arxiv:1702.01724 with interest, however I notice that you do not include any references to what I think of as the characteristically POVM literature. I do not know that literature well, but I have on my shelf, for example, “Operational Quantum Physics”, P.Busch, M.Grabowski, and P.J.Lahti, Springer, 1995, for which you probably have institutional access), which inter alia includes a theory of unsharp measurements. Chapter IV includes a number of theorems that are applicable to your paper, the first of which is the rather general “No continuous observable admits a repeatable measurement”. Your paper may be subsumed by that book or by other papers in that literature (“may be” because I read math at that level of abstraction only quite slowly), but I suggest that you bring your paper to the attention of Paul Busch, say, who I feel certain you will find interesting. The book I reference above is very enlightening, even gorgeous, IMO.
Posted by: Peter Morgan on March 12, 2017 11:29 PM | Permalink

Re: Responsibility

Note, too, that the next exercise in the class is to couple such particles to local fields (the electromagnetic field, a scalar field which induces a “Yukawa” interaction between such particles, etc).

This eventually leads one to a full-blown QFT. But there’s a wide range of circumstances in which the particle number is conserved, and the single-particle Hilbert space remains a valid description. When that fails, it is due to the interactions, rather than to some “sickness” of the free theory itself.

Posted by: Jacques Distler on March 12, 2017 7:11 AM | Permalink | PGP Sig

Re: Responsibility

It depends on what you call “wide”. You can at most couple it to time-independent external classical potentials and put the particle’s wave equation in other stationary backgrounds.

But the scheme breaks as soon as the fields are time-dependent, as soon as you want to describe the measurement of the particle by any actual moving gadget or interaction with other quanta etc.

Your theory with the “wide” range of applications can’t justify that it’s relativistic, either. As written down, it’s not relativistic. The Laplacian - spatial derivatives - are under the square root while the time derivatives are not. This is a clear asymmetry. It gets manifested by the superluminal spreading of the wave packets, too.

Deriving this theory as a description of the 1-particle sector of a quantum field’s Hilbert space is the only plausible way to show that your theory is relativistic - the Klein-Gordon equation governing the quantum or classical field is “manifestly” local and relativistic.

Well, the relativistic-behaving inner products that you listed show that “some structures” in the theory are relativistic but yes, one wants to do something else with the particle, add interactions, at which moment you should say that there’s no “obvious” relativistic way to introduce interactions in your sqrt(m squared minus laplacian) theory.

But at the end, the statement that you dislike is much simpler. It just assumes that people would try to work with the Klein-Gordon or Dirac equation just like if it were a non-relativistic equation with some details modified. But this treatment does lead to the indefinite probability density and other sicknesses.

Posted by: Lubos Motl on March 12, 2017 2:09 PM | Permalink

Re: Responsibility

The Laplacian - spatial derivatives - are under the square root while the time derivatives are not. This is a clear asymmetry. It gets manifested by the superluminal spreading of the wave packets, too.

That is, of course, completely wrong, as Problem 4, above, will show you.

In case you are completely unaware, the theory in question was worked out by Wigner in 1939. I’ve just filled in a few details that were not his primary concern.

Posted by: Jacques Distler on March 12, 2017 3:13 PM | Permalink | PGP Sig

Re: Responsibility

If I understand the numbering of problems above, Problem 4 has nothing whatever to do with the nonlocality of the equation - it’s just some formal transformation of components of the energy-momentum in the momentum space.

The superluminality of the propagation of the packets according to the sqrt(m^2-laplace) Hamiltonian is self-evident, isn’t it? Try to start with any packet, e.g. delta3(x) which is 1 in the momentum space, and evolve it by an infinitesimal dt. You will get a new packet whose Fourier transform isn’t confined to the ball of radius dt, is it? It’s also mathematically encoded in the non-vanishing of the Green’s function in the spacelike region, see e.g. page 14 of Peskin-Schroeder or any other introduction to QFT. On page 27, you will see how the Green’s functions have to be combined to get the commutator correlators - which is where one can see that the vanishing commutators of fields at spacelike separation guarantee locality.

I don’t know whether Wigner wrote something wrong in the 1930s but that question is about the history of physics, not current physics itself.

Posted by: Lubos Motl on March 12, 2017 3:56 PM | Permalink

Re: Responsibility

The relevant question is how the probability density, J 0(r→,t)J^0(\vec{r},t) evolves. Start with some initial “nice” probability density, and evolve it according to the equations above. See what you get …

Posted by: Jacques Distler on March 12, 2017 4:00 PM | Permalink | PGP Sig

Re: Responsibility

The continuity equation holds for a diffusion equation, for heat conduction equation, for Schrödinger equation, etc., so what? Superluminal spreading of the initially localized solution takes place too.

Posted by: Vladimir Kalitvianski on March 15, 2017 1:05 PM | Permalink

Re: Responsibility

The spreading is not superluminal. Work it out for yourself (as Lubos is clearly too lazy to do so for you).

Posted by: Jacques Distler on March 15, 2017 1:08 PM | Permalink | PGP Sig

Re: Responsibility

OK, I will work it out for myself. But let me note that usually the continuity equation is derived from the Schroedinger equation. At first glance I do not see how one can obtain the term div(j) from your H_0. Do you postulate the continuity equation? In that case one may try to find the space components of the four-current J, as you propose. Otherwise J is uncertain.

Posted by: Vladimir Kalitvianski on March 16, 2017 1:57 AM | Permalink

Re: Responsibility

I asked: “Is one free particle something we understand the best?” and I think it is something we understand the worst. Because, in Classical Mechanics (Physics, not Mathematics) a single particle is an approximation with the corresponding inequalities. I will mention two ones: an extended body size is much smaller than the other sizes of the problem, so we content ourself with a point-like approximation (description). And the second one is that the extended body is permanently observable, say, due to emitting light. In other words, the body in question changes, not stays the same, and the light implies multi-photon emission, i.e., an inclusive picture. That is what I meant while saying about factually multi-particle picture of one-particle description. In my opinion, only this understanding may help us describe correctly interactions, which are already switched on in a one-particle description.

Posted by: Vladimir Kalitvianski on March 13, 2017 1:45 AM | Permalink

Re: Responsibility

When we write an equations for a “free” physical particle, we mean that the interaction has much smaller contribution than the kinetic energy, i.e., it is an (our) approximation of always physical = interacting particle.

Posted by: Vladimir Kalitvianski on March 15, 2017 3:19 AM | Permalink

Re: Responsibility

Dear Distler,

Already single particle does not make sense since even the electron cannot actually have a charge unless it interacts. That is what Dirac equation(e does not appear in it) says AND QFT.

Posted by: Adel Sadeq on March 14, 2017 11:45 AM | Permalink

Re: Responsibility

The function f i(k→)f_i(\vec{k}) can be any function since the commutator does not impose any restriction to it. Let us see: i⋅[∂∂k i+f i(k→)]p j−i⋅p j[∂∂k i+f i(k→)]=i⋅δ j ii\cdot\left[\frac{\partial}{\partial k_i}+f_i(\vec{k})\right]p_j - i\cdot p_j \left[\frac{\partial}{\partial k_i}+f_i(\vec{k})\right]=i\cdot\delta^i_j The imaginary unit ii drops out. The operator ∂∂k ip j\frac{\partial}{\partial k_i}p_j reads p j∂∂k i+δ j ip_j \frac{\partial}{\partial k_i}+\delta^i_j The latter term cancels with the right hand side. And the remaining left hand side is an identity like A−A=0A-A=0. Probably I missed something important. Please put me on the right track, Jacques.
Posted by: Vladimir Kalitvianski on March 16, 2017 8:03 AM | Permalink

Re: Responsibility

For the sake of coherence with x ix^i, we should have written the function ff with the upper index (superscript), like this: f i(k→)f^i(\vec{k}), but it changes nothing.

Posted by: Vladimir Kalitvianski on March 16, 2017 8:21 AM | Permalink

Re: Responsibility

As I got stuck at f i(k→)f^i(\vec{k}) and the continuity equation is postulated rather than derived from (1), I am not sure that any other solution to (1) but a plane wave makes sense and spreads out. Maybe it is collapsing?

Posted by: Vladimir Kalitvianski on March 17, 2017 12:10 PM | Permalink

Re: Responsibility

Jaques,

The crucial idea required to wed quantum mechanics and relativity is the idea of antiparticles, which arise naturally when you try to understand the positive and negative energies that arise out of E^2 - p^2 = m^2.

If you just define a hamiltonian with the positive square root, it is a violation of relativity and will lead to superluminal signals.

I suggest you read Feynman’s “the reason for antiparticles”, which explains this nicely.

A quote:
0
“If a function f(t) can be fourier decomposed into positive frequencies only f(t) = int_0^inifinity exp(-iwt) F(w) dw, then f cannot be zero for any finite range of t unless it is zero everywhere”

So as long as all energies are only postive, you canont confine it within the lightcone. You absolutely must have antiparticles.

Posted by: dx on March 18, 2017 8:34 AM | Permalink

Re: Responsibility

The theorem about Fourier Transforms is correct (and, of course, has nothing to do with Relativity). Your leap from that to a conclusion about superluminal signals is incorrect.

I suggest that you (and the others who seem to think that parroting half-understood truisms is a demonstration of your erudition) shut up and calculate the spreading of (say, Gaussian) wave packet, J 0(r→,0)J^0(\vec{r},0), using the above setup.

If you find a superluminal answer (alas, you won’t), that’ll be something worthwhile that you can bring to the discussion.

Posted by: Jacques Distler on March 18, 2017 10:37 AM | Permalink | PGP Sig

The answer

It appears (from the handful of inquiries I’ve gotten since this was published — I just got another one yesterday) that the set of people both willing and able to compute the probability density (that I left as an exercise) is ∅\emptyset.

So let me just record the answer.

(1)J 0(r→,t) =|∫d 3k→(2π) 32ω ke −i(ω kt−k→⋅r→)ϕ(k→)| 2 J→(r→,t) =∫d 3k→d 3k→′(2π) 62ω kω k′k→+k→′ω k+ω k′e −i((ω k−ω k′)t−(k→−k→′)⋅r→)ϕ(k→′)¯ϕ(k→)\begin{aligned} J^0(\vec{r},t)&= \left\vert {\int{\frac{d^3\vec{k}}{(2\pi)^3\sqrt{2\omega_k}} e^{-i(\omega_k t-\vec{k}\cdot\vec{r})} \phi(\vec{k})}} \right\vert^2\\ \vec{J}(\vec{r},t)&= \int\frac{d^3\vec{k}d^3\vec{k}'}{(2\pi)^6 2\sqrt{\omega_k}\sqrt{\omega_{k'}}} \frac{\vec{k}+\vec{k}'}{\omega_k+\omega_{k'}}e^{-i\bigl((\omega_k-\omega_{k'})t-(\vec{k}-\vec{k}')\cdot\vec{r}\bigr)}\overline{\phi(\vec{k}')}\phi(\vec{k}) \end{aligned}

where I’ve abbreviated ω k=k→ 2+m 2\omega_k=\sqrt{\vec{k}^2+m^2}. The probability density, J 0(r→,t)J^0(\vec{r},t), is manifestly positive and normalized ( 1=⟨ϕ|ϕ⟩=∫d 3k→(2π) 32ω k|ϕ(k→)| 2 1=\langle\phi|\phi\rangle= \int\frac{d^3\vec{k}}{(2\pi)^3 2\omega_k} |\phi(\vec{k})|^2 implies ∫d 3r→J 0(r→,t)=1\int d^3\vec{r} J^0(\vec{r},t)=1) and obeys the conservation equation

(2)∂ 0J 0+∇→⋅J→=0 \partial_0 J^0 +\vec{\nabla}\cdot\vec{J}=0

Manifestly, from (1), the probability density spreads no faster than the speed of light.

What you can’t do (which many seem to want to do) is demand that J 0(r→,0)J^0(\vec{r},0) vanish outside of some compact region in ℝ 3\mathbb{R}^3. That’s just not compatible with the normalizability of |ϕ⟩|\phi\rangle and finiteness of the energy: ⟨ϕ|ϕ⟩=1\langle\phi|\phi\rangle=1, ⟨ϕ|H|ϕ⟩<∞\langle\phi|H|\phi\rangle\lt\infty.

Posted by: Jacques Distler on January 20, 2024 11:02 AM | Permalink | PGP Sig