- #36
meopemuk
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Demystifier said:No, I am saying the transformation from one observer to another must be covariant.
What does the "covariance" mean exactly, when applied to transformations of particle wavefunctions?
Demystifier said:No, I am saying the transformation from one observer to another must be covariant.
I fully agree with you that position is the most basic observable. Still, I agree with majority that the Newton-Wigner operator is not satisfying as it is not covariant. To solve the puzzle, I suggest to modify the axiom that an observable must be defined by an operator. For a concrete proposal, seemeopemuk said:I still cannot see what could go wrong with choosing the Newton-Wigner operator as a relativistic generalization of position? I think this operator provides a perfect description of position in relativistic quantum theory.
From another point of view, there must be *some* position operator in relativistic quantum theory. You cannot just say: there is no operator, so I am not going to consider the position representation. Position is the most basic observable in physics, and it remains measurable in both non-relativistic and relativistic physics. If position is an observable, then in quantum theory there must be an operator corresponding to it. It is just inevitable.
Demystifier said:Exactly! Relativistic QM does not satisfy this axiom [unitary Hamiltonian time evolution law]. This is also closely related to the fact that relativistic QM cannot be interpreted probabilistically.
No, the correct question is what does it mean when applied to transformations of particle position operator. The answer is: The position operator must be the space component of a 4-vector. Then the transformation to another observer is just a Lorentz transformation of this 4-vector, followed by taking the space part of new coordinates.meopemuk said:What does the "covariance" mean exactly, when applied to transformations of particle wavefunctions?
Yes, that is radical. But most physicists agree that you must be ready to do something radical in order to combine QM and relativity consistently.meopemuk said:This is too radical for me. I want to believe that laws of quantum mechanics remain valid in all regimes, including the relativistic one. If you say that these laws become invalid, then you need to substitute quantum mechanics with a more general (non-probabilistic?) theory. What is this theory?
meopemuk said:Apparently, particles localized in the reference frame at rest don't look like localized in the moving frame. So what? Special relativity has taught us that many things previously considered absolute are, in fact, observer-dependent. Localization is just one of these relative things. That's how I see it.
Demystifier said:I fully agree with you that position is the most basic observable. Still, I agree with majority that the Newton-Wigner operator is not satisfying as it is not covariant. To solve the puzzle, I suggest to modify the axiom that an observable must be defined by an operator. For a concrete proposal, see
http://xxx.lanl.gov/abs/0705.3542
Demystifier said:The position operator must be the space component of a 4-vector.
Hans de Vries said:A particle localized in one frame is also localized in another frame. You are
using an example where the particle is a Dirac function. So, in 4D it's a line,
(the t-axis). After a boost it will be another line, the t'-axis.
Rotations and Boost are exactly the same in moment space as they are in
configuration space, so doing a Fourier transform followed by a boost
followed by an inverse Fourier transform gives you the same result.
Now, the moving particle is of course anywhere at x at some time t, The
math in 10.1.2 doesn't consider time so that's where it might go wrong.
meopemuk said:I haven't seen a proof that (as you say) particle remains localized in all frames. If you have such a proof, it would great to compare our notes.
Regards.
Eugene.
Hans de Vries said:The easiest thing is probably to visualize this with a Minkowsky type image
like the one here:
http://www-users.york.ac.uk/~mijp1/transaction/figs/fig02_h.gif
If you confine the wave function to a point at (t,x) = (0,0) and it is allowed
to spread at any speed up to c, then it will still be confined in the future
light cone.
Any x'-axis belonging to another reference frame is rotated in the point (0,0)
over an angle between +45 and -45 degrees. This means that it will never
cut through the future (or past) light cones. It will only cut through the
particle's wave function in the point (0,0). This means any observer with a
speed v<c will see it as a point a t=t'=0.
Regards, Hans
Haelfix said:Incidentally I object to the use of the Klein Gordon equation and the Dirac equation to make some sort of argument against localization as Hegerfield (sp?) seems to argue for.
Localizing any physical particle in that framework, beyond its compton wavelength will automatically enter a regime (QFT) where pair creation becomes important and where multiple fields must be considered.
Haelfix said:Ok. Define the particle number operator in the context you are thinking off that commutes with the Newton-Wigner operator.
I have a feeling what I'm thinking off, and what you are thinking off are going to be different. Regardless I suspect this operator is going to run into issues with its commutation relations with other observables like say the hamiltonian and hence disallowed on physical grounds.
meopemuk said:It is often said in textbooks that the Klein-Gordon equation is a relativistic analog of the Schroedinger equation, and therefore it can be used for the description of time evolution. Then, as you correctly pointed out, initial conditions should include the time derivative of the wave function in addition to the wave function itself. In other word, the state of the particle at time [itex] t + \Delta t [/itex] is determined not only by its state at time [itex] t [/itex], but also by the "state derivative" at time [itex] t [/itex].
This statement is in contradiction with the fundamental equation for time evolution in quantum mechanics
[tex] |\Psi(t) \rangle = \exp(\frac{i}{\hbar}Ht) |\Psi(0) \rangle [/tex]
where H is the Hamiltonian. This equation show, in particular, that the state at time [itex] t + \Delta t [/itex] is determined by the state at time [itex] t[/itex], the Hamiltonian, and nothing else.
Demystifier said:Exactly! Relativistic QM does not satisfy this axiom. This is also closely related to the fact that relativistic QM cannot be interpreted probabilistically.
meopemuk said:To keep things simple, I will write you expression for a one-particle state localized at point [itex] \mathbf{x} [/itex]
[tex] |\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle [/tex]
Here [itex] a^{\dag}_{\mathbf{p}}[/itex] are particle creation operators and [itex] | 0 \rangle [/itex] is the vacuum vector. Of course, this state has infinite uncertainties of momentum and energy, but it is still an one-particle state. It is an eigenstate (with eigenvalue 1) of the particle number operator
[tex] N = \int d^3p a^{\dag}_{\mathbf{p}} a_{\mathbf{p}} [/tex]
No, I am Dr. Nikolic. But I do not blame you, for some reason everybody makes the same misprint in my name. There must be a reason for that, but I cannot figure it out.meopemuk said:Thanks for the reference. I enjoyed reading this well-written paper. (I have a suspicion that you are Dr. Nicolic. Is it true?)
Of course, but QFT is not the same thing as relativistic QM.jostpuur said:But isn't the time evolution in QFT defined precisly with this equation?
I agree that we agree on what you say that we agree and disagree on what you say that we disagree.meopemuk said:You can possibly argue that my departure from canons of special relativity is too radical, but I consider it less radical than your departure from laws of quantum mechanics. It is true that Einstein's special relativity cannot peacefully coexist with quantum mechanics. Something's got to give. I think we both agree about that. We disagree about ways forward.
meopemuk said:I am possibly too conservative to accept the idea that fundamental Rules of Quantum Mechanics should be abandoned.
meopemuk said:Hegerfeldt's discussion is even more general as he doesn't specify the position operator explicitly. You can find his article on the web http://www.arxiv.org/quant-ph/9809030
Bad or naive math gives plenty of opportunities to find non-causal and non-Hegerfeldt said:Could instantaneous spreading be used for the transmission of signals if it
occurred in the framework of relativistic one-particle quantum mechanics? Let
us suppose that at time t = 0 one could prepare an ensemble of strictly localized
(non-interacting) particles by laboratory means, e.g. photons in an oven. Then
one could open a window and would observe some of them at time t = " later
on the moon. Or to better proceed by repetition, suppose one could successively
prepare strictly localized individual particles in the laboratory. Preferably this
should be done with different, distinguishable, particles in order to be sure when
a detected particle was originally released. Such a signaling procedure would
have very low efficiency but still could be used for synchronization of clocks or,
for instance, for betting purposes.
meopemuk said:To keep things simple, I will write you expression for a one-particle state localized at point [itex] \mathbf{x} [/itex]
[tex] |\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle [/tex]
Here [itex] a^{\dag}_{\mathbf{p}}[/itex] are particle creation operators and [itex] | 0 \rangle [/itex] is the vacuum vector. Of course, this state has infinite uncertainties of momentum and energy, but it is still an one-particle state. It is an eigenstate (with eigenvalue 1) of the particle number operator
[tex] N = \int d^3p a^{\dag}_{\mathbf{p}} a_{\mathbf{p}} [/tex]
The problem is that the measuremeopemuk said:To keep things simple, I will write you expression for a one-particle state localized at point [itex] \mathbf{x} [/itex]
[tex] |\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle [/tex]
Demystifier said:The problem is that the measure
[tex] d^3p [/tex]
is not Lorentz invariant.
If you replace it with a Lorentz invariant measure
[tex] d^3p/2E [/tex]
then the state is no longer localized, but "semilocalized" as Haelfix said.
jostpuur said:This is probably not what you are talking about, but I'll continue with "very simple QFT questions". What would
[tex]
\int\frac{d^3p}{(2\pi\hbar)^3}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}} e^{-i\boldsymbol{p}\cdot\boldsymbol{x}/\hbar} a^{\dagger}_{\boldsymbol{p}} |0\rangle
[/tex]
be?
Demystifier said:No, I am Dr. Nikolic. But I do not blame you, for some reason everybody makes the same misprint in my name. There must be a reason for that, but I cannot figure it out.
Demystifier said:Of course, but QFT is not the same thing as relativistic QM.
Fra said:What is your interpretation of probability? Frequentists? Bayesian? Do you just stick to the Kolmogorovs axioms of the formalism? What I see clearly is that while there isn't anything obviously wrong with the axioms themselves from the point of view of mathematics. The problem is howto make a satisfactory connection to reality, which is supposedly the business here at least for me. I consider this to be one of the prime issues. I can't accept that way these parts are rushed over as if it's obvious that while the particle position isn't deterministic, it's probability is? It's very typical of physics, to grab a nice mathematical formalism and adapt it. Are everything you need to define the probability space itself, observables? It seems not. This bothers me. And I won't sleep until it's fixed.
Fra said:I find it easier to see what I think is wrong now when I've had a break from physics. I try to forget what it's "supposed" to be like, and instead try to think from scratch in a critical manner. I found that to be very hard to do when you are in the middle of something, a course for example, because they you are kind of fighting against yourself and it makes everything 10 times harder becuase you are trying to "learn something" and the critical to it at the same time. This is what I strongly disliked during the courses I took. I tried to be critical, which annoyed the teachers because it was "counterproductive" for the classwork. So clearly, you were encouraged to accept the ideas, trying to be critical only made it harder from everyone by arguing that what is taught doesn't quite make sense.
/Fredrik
Haelfix said:Yes but this still begs the question, this one-particle state is still only physically realized on the order of the Compton wavelength, beyond that extra degrees of freedom arise.
Haelfix said:Moreover this is also in the strict confines of the free particle, turn on interactions (the physical regime) and you can't even define this.
Demystifier said:The problem is that the measure
[tex] d^3p [/tex]
is not Lorentz invariant.
If you replace it with a Lorentz invariant measure
[tex] d^3p/2E [/tex]
then the state is no longer localized, but "semilocalized" as Haelfix said.
jostpuur said:Except that I think you should have the square root there like in my previous post. With the square root the norm [itex]\langle\psi|\psi\rangle[/itex] becomes Lorentz's invariant. At least if we have [itex][a_p,a^{\dagger}_{p'}]=(2\pi\hbar)^3\delta^3(p-p')[/itex]. This is the convention P&S use. Do other sources put [itex]2E_p[/itex] in front of the delta function?
Or maybe not. I don't have the book right here, and instead just looked at my notes, but now I started doubting if I have them correctly... Srednicki seems to indeed put that energy factor in front of the delta function. Blaa...
meopemuk said:I described my interpretation of probability in section 2.3 of http://www.arxiv.org/physics/0504062 . Probabilities are normally assigned to experimental "propositions", i.e., statements that can have two values: either "true" or "false". We prepare N identical copies of the same system and perform N measurements of the proposition. Then we count the number of instances (M) in which the value of the proposition was found "true". Then we take the limit [itex] N \to \infty [/itex] and say that the probability of the proposition to be true is equal to the ratio [itex] M/N [/itex] in this limit. So, in principle, probabilities can be measured to an arbitrary precision.
Fra said:Thanks Meopemuk for your comments.
I have saved your large 600+ page paper and I hope to find in it your core ideas some day. I haven't gotten around to it in detail yet.
/Fredrik