A. Neumaier said:But arbitrarily many, which makes the number of degrees of freedom infinite. An ##N##-particle system has ##3N## degrees of freedom. Already a free quantum field theory has an indeterminate number of particles. The number of degrees of freedom therefore exceeds ##3N## for any ##N##. Thus there are infinitely many degrees of freedom.
stevendaryl said:Well, I'm uncertain about the terminology, but the distinction that I'm making is between:
- An unbounded number of particles, and
- an infinite number of particles.
Well, that's not what I said. I explained in my post, what I meant by d.o.f., i.e. a state in QFT contains as much information as a state in QM. Of course, in the case of QFT, you need to embed a field algebra into the algebra of operators on the separable Hilbert space, instead of a particle algebra. However, this is not what is problematic. You can easily write down a Fock space and field operators on it and you can also write down well-defined interacting Hamiltonians in that Hilbert space. However, you won't be able to write down an interacting Hamiltonian that satisfies the Poincare algebra relations, i.e. produces a Lorentz-invariant theory. Writing down interacting theories that satisfy the Wightman axioms is of course very difficult.A. Neumaier said:This distinction is valid, but doesn't affect the fact that - in contrast to the claim of Rubi above - QFT has infinitely many degrees of freedom while QM has only finitely many. This differences is extremely important.
But this is wrong. In QM, one cannot represent the information that a system is in a superposition of arbitrarily many particles!rubi said:a state in QFT contains as much information as a state in QM.
That is not relevant to measuring how much information can be encoded into a state. Of course, different theories encode different information in the state, but the amount of information is exactly the same. I can store either Pulp Fiction or Full Metal Jacket on my hard drive, but if I choose to store Full Metal Jacket, the information about what a quarter pounder with cheese is called in France won't be represented.A. Neumaier said:But this is wrong. In QM, one cannot represent the information that a system is in a superposition of arbitrarily many particles!
But measuring information in this way is completely irrelevant. Every real number contains in your sense more information than all the physics books in the world. A classical field theory contains in your sense as much information as a quantum field theory.rubi said:That is not relevant to measuring how much information can be encoded into a state. Of course, different theories encode different information in the state, but the amount of information is exactly the same. I can store either Pulp Fiction or Full Metal Jacket on my hard drive, but if I choose to store Full Metal Jacket, the information about what a quarter pounder with cheese is called in France won't be represented.
I stated precisely what I meant by it in my post #47, so all of this is just a semantic discussion. Of course, the dimension of the Hilbert space is relevant to what quantum theories can be formulated on it. You cannot formulate QFT in ##\mathbb C^2##, but you can (in principle) formulate it in the usual ##L^2(\mathbb R^3)## space of ordinary quantum mechanics. As I explained, the fact that you are working with a field algebra instead of a particle algebra is not a priori problematic.A. Neumaier said:But measuring information in this way is completely irrelevant. Every real number contains in your sense more information than all the physics books in the world. A classical field theory contains in your sense as much information as a quantum field theory.
In any case, it is highly misleading to equate this strange measure of information with degrees of freedom, as you did in your post #47 (without being precise enough there).
No, you cannot in any sensible way since there is no natural isomorphism between a Fock space for QFT and ##L^2(\mathbb R^3)##. The structure imposed on the Hilbert space makes all the difference for its information content! Quantum mechanics does not happen in arbitrary Hilbert spaces but in Hilbert spaces with distinguished group actions!rubi said:You cannot formulate QFT in C2C2\mathbb C^2, but you can (in principle) formulate it in the usual ##L^2(\mathbb R^3)## space of ordinary quantum mechanics.
There's no state with infinitely many particles in the sense that the expectation value of the number of particles should be finite for the state to make sense. However, you are right in saying that there are states which have a non-zero overlap with any Fock state of a definite finite particle number. An example are the coherent states of the electromagnetic field, which are as close to classical em. waves as you can get. The photon-number probabilities are given by the Poisson statistics.stevendaryl said:Well, I'm uncertain about the terminology, but the distinction that I'm making is between:
The distinction is that in the case of an unbounded number of particles, if |\psi\rangle is a state with an unbounded number of particles, then there is at least one nonnegative number n and a state |\psi_n\rangle with exactly n particles such that \langle \psi|\psi_n \rangle is nonzero.
- An unbounded number of particles, and
- an infinite number of particles.
With a truly infinite number of particles, there would be no overlap with any state with just finitely many particles.
Yes, but ##L^2(R^3)## doesn't have this structure but must be equipped with it in order to have it. Quantum physics is always about Hilbert spaces already equipped with the right group action.rubi said:As a mathematician, you certainly know that I can pull back any structure between isomorphic objects
But a state must only be square integrable. Therefore there are valid states (such as ##\sum_{N>0} N^{-1}|N\rangle##) where the expected number of particles is infinite. Nevertheless, upon each Born-style measurement, the number of particles would come out finite.vanhees71 said:There's no state with infinitely many particles in the sense that the expectation value of the number of particles should be finite for the state to make sense.
It would take an infinite amount of energy to create it, I believe.vanhees71 said:I'm a bit in doubt, whether it can be prepared for any type of particles.
The state ##\sum_{N\ge 0} (N+1)^{-1}|N\rangle## has the same properties but with a vacuum contribution. Thus that's not the source of the difficulties to create such a state. You can add as much vacuum as you like without changing the infinite expected particle number.vanhees71 said:This would be a state without a vacuum contribution.
An object must always be equipped with a structure in order to have it. It's just easier to write it down in some cases. I'm of course not suggesting that one should do QFT on ##L^2(\mathbb R^3)##, but that one could do it, i.e. a state vector in QM needs as much information to be fully specified as a state vector in QFT. You cannot do QFT in the Hilbert space of a spin-1/2 system. And you cannot do LQG in the Hilbert space of a QFT. The dimension of the Hilbert space is of course a measure for the information content of a quantum state.A. Neumaier said:Yes, but ##L^2(R^3)## doesn't have this structure but must be equipped with it in order to have it. Quantum physics is always about Hilbert spaces already equipped with the right group action.
Ignoring that is like saying I can do functional analysis in the set ##R_+## of positive real numbers, since I can equip any set with continuum cardinality with the structure of ##L^2(R^3)##. No mathematician in his right mind would talk like this.
No. The conventional, and the only reasonable measure of information applicable a quantum state is the entropy. What you talk about does not deserve the name information in the traditional, scientifically established sense.rubi said:The dimension of the Hilbert space is of course a measure for the information content of a quantum state.
First I don't understand what you mean by the first quoted sentence, since a "Pauli particle" (spin-1/2 particle in non-relativistic QT) is described by the usual separable Hilbert space and usually reallized as ##\mathrm{L}^2(\mathbb{R}^3,\mathbb{C}^2)## (position-spin representation or "wave mechanics").rubi said:You cannot do QFT in the Hilbert space of a spin-1/2 system. And you cannot do LQG in the Hilbert space of a QFT. The dimension of the Hilbert space is of course a measure for the information content of a quantum state.
Then please just assign whatever name you like to it, so we don't need to have this discussion about pure semantics. The dimension of the Hilbert space is a classification criterion for quantum systems.A. Neumaier said:No. The conventional, and the only reasonable measure of information applicable a quantum state is the entropy. What you talk about does not deserve the name information in the traditional, scientifically established sense.
I said "spin-1/2 system" and not "(Pauli) particle". A spin-1/2 system is conventionally understood as ##\mathcal H=\mathbb C^2## with the standard spin-1/2 representation of ##SU(2)##. People study such quantum systems without referencing the coordinates of particles. Especially in quantum information, people rarely use infinite-dimensional Hilbert spaces.vanhees71 said:First I don't understand what you mean by the first quoted sentence, since a "Pauli particle" (spin-1/2 particle in non-relativistic QT) is described by the usual separable Hilbert space and usually reallized as ##\mathrm{L}^2(\mathbb{R}^3,\mathbb{C}^2)## (position-spin representation or "wave mechanics").
Loop quantum gravity.For curiosity, what's LQG?
See above.I also don't know what I should make of the third sentence. The information content of a quantum state, or rather our ignorance about it, is given by the (von Neumann) entropy. For a pure state it's 0 (i.e., we have full possible knowledge about the system's preparation).
A very, very weak one, since the vast majority of quantum systems of experimental interest (except for finite spin systems) have a separable infinite-dimensional Hilbert space, all of which are isomorphic if you don't have additional structure.rubi said:The dimension of the Hilbert space is a classification criterion for quantum systems.
I agree, but it seemed to me like Demystifier wasn't aware of it, which is why I explained it in my post. At first, it seems unintuitive that a Hilbert space of separable dimension is enough for field variables, because one might think that a field can have values at uncountably many space-time points. A separable Hilbert space suffices, because only smeared fields (smeared with Schwartz functions) are considered. If one wanted to include fields operators defined at sharp space-time points, one would need a Hilbert space of uncountable dimension.A. Neumaier said:A very, very weak one, since the vast majority of quantum systems of experimental interest (except for finite spin systems) have a separable infinite-dimensional Hilbert space, all of which are isomorphic if you don't have additional structure.
But one then gets into trouble with locality.rubi said:If one wanted to include fields operators defined at sharp space-time points, one would need a Hilbert space of uncountable dimension.
Possible, but why do you think so? One could require ##[\phi(x),\phi(y)]=\delta_{xy}## instead of ##[\phi(x),\phi(y)]=\delta(x-y)##.A. Neumaier said:But one then gets into trouble with locality.
Your suggestion plays havoc with everything important in QFT, since it essentially equips space-time with the discrete topology. Try to base a QFT on your suggestion and you'll see.rubi said:Possible, but why do you think so? One could require ##[\phi(x),\phi(y)]=\delta_{xy}## instead of ##[\phi(x),\phi(y)]=\delta(x-y)##.
I can see many things going wrong and I agree that one shouldn't do it. It's just an example to clarify things. I was just wondering, why you were specifically thinking about locality, because that seems pretty easy to maintain.A. Neumaier said:Your suggestion plays havoc with everything important in QFT, since it essentially equips space-time with the discrete topology. Try to base a QFT on your suggestion and you'll see.
But locality isn't conventionally defined in imaginary time ;-(vanhees71 said:Well, our lattice-QCD colleagues are pretty successful with such an approach, however only in imaginary time ;-).
The right question is this: Is it only an approximation in the limit ##a\rightarrow 0##?vanhees71 said:Well, yes. Lattice theory is an approximation as well, but tell this the lattice guys and see their reaction (which is very interesting ;-)).
I don't understand this statement. You are right, this theorem is not a rigorous theorem; it is a FAPP (For ALL Practical Purposes) theorem*. If you like, you may even call it an argument, rather then a theorem. But to criticize an argument, you must find a gap in the argument. There is no other way.rubi said:Well, apparently, this theorem is not established rigorously, so finding a gap is not necessary to criticize it.
Right. You can also view LQG as lattice quantum gravity with the continuum limit taken.vanhees71 said:Well, our lattice-QCD colleagues are pretty successful with such an approach, however only in imaginary time ;-).
Well, locality in the Lorentzian (Wightman) setting translates to the symmetry of the Schwinger functions in the Euclidean setting.A. Neumaier said:But localityisn't conventionally defined in imaginary time ;-(
A rigorous argument is an argument that doesn't have gaps. So if your argument is not rigorous, then it has gaps by definition.Demystifier said:I don't understand this statement. You are right, this theorem is not a rigorous theorem; it is a FAPP (For ALL Practical Purposes) theorem*. If you like, you may even call it an argument, rather then a theorem. But to criticize an argument, you mist find a gap in the argument. There is no other way.
Mathematical rigor is just a different name for having high standards with respect to ones arguments. This is generally a good thing. Whether Bohmian mechanics reproduces QM depends critically on a specific argument, so one ought to have high standards with respect to it. Since interpretations of QM all (claim to) make the same physical predictions, mathematical rigor is the only possible way to exclude some of them. So requiring arguments to be rigorous is the least we should expect in the interpretations business.In fact, the FAPP theorem of Bohmian mechanics could also be translated into a mathematically rigorous theorem, but in such a form it would be physically irrelevant. Yes, it would probably make some mathematical physicists happy, but still it would not be much useful for practical purposes.
Of course, mathematicians also study, how big ##N## must be in order to have the probability of deviating from the ##N\rightarrow\infty## limit to be sufficiently small.Demystifier said:To explain what I mean by a FAPP theorem, let me give an example: the law of large numbers in probability theory. In the limit ##N\rightarrow\infty## it is a rigorous theorem. But as such, it is pretty useless. It is only useful for a big but finite ##N##, sometimes as small as ##N=1000##. For finite ##N##, the law of large numbers is only a FAPP theorem.
Well that seems to be Arnold's point also. Standard quantum mechanics makes correct predictions about the hydrogen atom even without including extra baggage and ending up with a complicated model (of which we don't even rigorously know whether it works, even if we include all the extra baggage).vanhees71 said:Well, my only quibble is, what's "practical" about Bohmian mechanics.
It adds to quantum theory a host of unobservable (and hence unverifiable and unfalsifiable) degrees of freedom to give its believers the illusion that particle have infinitely accurate positions at all times. This comes at the cost of lots of other counterintuitive features:vanhees71 said:So what's physics wise the merit of Bohmian mechanics compared to minimally interpreted quantum theory?
Which does not mean that one does not need to criticize the specific gaps in order to criticize the final conclusion of the argument.rubi said:So if your argument is not rigorous, then it has gaps by definition.
To paraphrase Feynman, Bohmian mechanics is like sex. Sure, it may have practical applicationsvanhees71 said:Well, my only quibble is, what's "practical" about Bohmian mechanics. I've never been able to make sense of the claimed trajectories, which cannot be verified empirically. So what's physics wise the merit of Bohmian mechanics compared to minimally interpreted quantum theory? At best it's a nice academic mathematical exercise to calculate the unobservable trajectories, right?
Demystifier said:The question for everybody: What do you do when you see a magician trick?
For a poor spectator the entry fee is worth the show only under option 1.Demystifier said:for a poor spectator there are only a few options:
1) Watch and enjoy the show. (The analog of shut up and calculate for QM.)
So you never try to figure out how did he did it?A. Neumaier said:For a poor spectator the entry fee is worth the show only under option 1.
I am not a poor spactator. I have the thermal interpretation. It is enough to show that particles have an approximate position whenever the state is such that the position can be observed. No hidden variables are needed for that.Demystifier said:So you never try to figure out how did he did it?
Fair enough. And is there an analog of thermal interpretation for the rabbit-from-the-hat phenomenon?A. Neumaier said:I am not a poor spactator. I have the thermal interpretation.
Demystifier said:The question for everybody: What do you do when you see a magician trick?
Yes:Demystifier said:Fair enough. And is there an analog of thermal interpretation for the rabbit-from-the-hat phenomenon?
Does your attempted explanation involve some hidden variables, like those in 5) above?Spinnor said:Accept and enjoy that the magician did a great trick and then try and figure out how he did it.
I still don't understand how that helps to explain the rabbit-from-the-hat phenomenon.A. Neumaier said:Yes:
6) Study physics and extrapolate from what the physicists really do when they compare experiments with theory.
This is how I discovered the thermal interpretation.
Does the rabbit have a definite position before it is pulled out of the hat ? Or is that interpretation dependent ?Spinnor said:Accept and enjoy that the magician did a great trick and then try and figure out how he did it.
Without physics, how can you explain the appearance of the rabbit? Physics tells you that mass is conserved (and much more). So you know that the rabbit must have been there before.Demystifier said:I still don't understand how that helps to explain the rabbit-from-the-hat phenomenon.
That still doesn't explain the trick.A. Neumaier said:Without physics, how can you explain the appearance of the rabbit? Physics tells you that mass is conserved. So you know that the rabbit must have been there before.
It also tells you that brains are not very reliable detectors and may fail when their attention is somewhere else.Demystifier said:That still doesn't explain the trick.
